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TOPIC | Revision Videos | Mind Maps |
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General Organic Chemistry | Revise | Mindmap |

Acidic and Basic Strength | Revise | Mindmap |

Hydrocarbons | Revise | Mindmap |

Optical Isomerism | Revise | Mindmap |

Haloalkane & Haloarenes | Revise | Mindmap |

Alcohol, Phenol and Ethers | Revise | Mindmap |

Carbonyl Compounds | Revise | Mindmap |

Aldol, Cannizzaro and Haloform Reaction | Revise | Mindmap |

Name Reaction | Revise | Mindmap |

Oxidation Reaction | Revise | Mindmap |

Reduction Reaction | Revise | Mindmap |

Carboxylic Acid | Revise | Mindmap |

Nitrogen Containing Compounds | Revise | Mindmap |

Bio-molecules | Revise | Mindmap |

Polymers | Revise | Mindmap |

Chemistry in Everyday Life | Revise | Mindmap |

Atomic Structure | Revise | Mindmap |

Chemical Bonding | Revise | Mindmap |

Periodic Table | Revise | Mindmap |

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**Very Short Answer (1 Mark)**

**Q. Give the IUPAC names of the following compounds :**

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**Sol.**2-bromo- 4-methyl pentan- 3-one

**Q. Give the IUPAC name of the alkane having the lowest molecular mass that contain a quaternary**carbon.

**Q. Write the structural formula of (i) sec-butyl and (ii) isobutyl groups.**

**Q. Lassaigne’s test is not shown by diazonium salts, though they contain nitrogen. Why ?**

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**Sol.**Diazonium salts $\left(C_{6} H_{5} N_{2}^{+} X^{-}\right)$ readily lose $N_{2}$ on heating before reacting with fused sodium metal. Therefore, these do not give positive Lassaigne’s test for nitrogen.

**Q. Suggest a suitable technique of separating naphthalene from kerosene present in a mixture.**

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**Sol.**By differential extraction.

**Q. How will you separate a mixture of**

*o*-nitro-phenol and*p*-nitrophenol ?**Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...**

**Sol.**A mixture of

*o-*nitrophenol and

*p-*nitrophenol can be separated by steam distillation.

*o*-nitrophenol being less volatile distils over along with water while

*p*-nitrophenol being non-volatile remains in the flask.

**Q. Will $C C l_{4}$ give white precipitate of $A g C l$ on heating with silver nitrate? Give reason for your answer.**

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**Sol.**The precipitate of $A g C l$ will not be formed because $C C l_{4}$ is covalent compound and does not ionise to give $C l^{-}$ ions to react with $A g N O_{3}$. $C C l_{4}+A g N O_{3} \longrightarrow$ No reaction

**Q. Which is expected to be more stable, $\mathrm{O}_{2} N \mathrm{CH}_{2} \mathrm{CHO}$ or $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{O}^{-}$ and why?**

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**Sol.**$\mathrm{O}_{2} \mathrm{N} \mathrm{CH}_{2} \mathrm{CHO}$ is expected to be more stable because each atom has complete its octet and has no charge.

**Q. How many $\sigma$ and $\pi$ bonds are present in each of the following molecules?**(i) $\quad \mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{C} \equiv \mathrm{CH}$ (ii) $\quad \mathrm{HC} \equiv \mathrm{C}-\mathrm{CH}_{2}-\mathrm{CH}=\mathrm{CH}_{2}$

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**Sol.**(i) $\quad \sigma_{C-C}=4, \sigma_{C-H}=6, \pi_{C=C}=3$ (ii) $\quad \sigma_{\mathrm{C}-\mathrm{C}}=4, \sigma_{\mathrm{C}-\mathrm{H}}=6, \pi_{\mathrm{C}=\mathrm{C}}=3$

**Q. Which of the following pairs of structures do not constitute resonance structures :**

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**Sol.**Pair of structures shown in (i) and (iv) do not constitute resonance structures.

**Q. Classify the following molecules/ions as nucleophiles or electrophiles: $H \ddot{S}^{-}, B F_{3}, C H_{3} C H_{2} O^{-}$**

**$\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}, \mathrm{Cl}^{+}, \mathrm{CH}_{3} \mathrm{C}=\mathrm{O}, \mathrm{H}_{2} \mathrm{N}:, \mathrm{NO}_{2}^{+-}$**

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**Sol.**Electrophiles : $B F_{3} . C l^{+}, C H_{3}^{+} C=O, N O_{2}^{+}$ Nucleophiles : $H S, C H_{3} C H_{2} O^{-},\left(C H_{3}\right)_{3} \ddot{N}, \dot{H}_{2} \bar{N}$:

**In $s p^{3}, s p^{2}$ and sp hybrid orbitals which hybrid orbitals show the electronegativity?**

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**Sol.**$s p^{3}$ hybrid orbitals.

**Q. What type of hybridization is shown by second carbon atom.**

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**Sol.**

*sp*hybridization.

**Q. Explain why $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}^{+}$ is more stable than $C H_{3}^{+} C H_{2}$ and $^{+} C H_{3}$ is the least stable cation.**

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**Sol.**the nodal plane of the vacant 2

*p*orbitals and hence cannot overlap with it. Thus lacks hyperconjugative stability.

**Q. State bond length and strength of $s p, s p^{2}$ and $s p^{3}$ hybrid**

**orbitals.**

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**Sol.**The sp hybrid orbital contains more $s$ character and hence it is closer to its nucleus and forms shorter and stronger bonds than the $s p^{3}$ hybrid orbital. The $s p^{2}$ hybrid orbital is intermediate between $s p$ and hybrid orbitals, hence the length and enthalpy of the bonds it forms is also intermediate between them.

**Q. On the basis of type of hybridization, predict the shape of the following molecules :**(i) $\quad H_{2} C=O \quad$ (ii) $C H_{3} F \quad$ (iii) $\mathrm{HC} \equiv \mathrm{N}$

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**Sol.**(i) $s p^{2}$ hybridised carbon and therefore, shape is trigonal planar. (ii) $s p^{3}$ hybridised carbon and shape is tetrahedral. (iii) $s p$ hybridised carbon and shape is linear.

**Short Answer (2 or 3 Marks)**

**Q. Write IUPAC nomenclature of the following :**

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**Sol.**(i) 2-ethyl 3-methyl pentan-1-ol (ii) 1-chloro propan-2-one (iii) Hexa 1, 3-diene-5-yne (iv) 2, 4, 6 – tri bromo phenol

**Q. Give IUPAC name for the following :**

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**Sol.**(i) 2 methyl propan-1-ol (ii) 4-ethyl-2-methyl aniline

**Q. Give IUPAC name of the following organic molecules :**

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**Sol.**(i) 5-methoxy-2-nitrophenol (ii) 4-ethyl-2-methyl anisol (iii) 2-chloro-4-methoxy benzoic acid (iv) 2-ethoxy-4-methyl aniline

**Q. Write the IUPAC names of :**

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**Sol.**(i) Trans-2-hexene (ii) Cis-2-hexene (iii) (E)-3-methyl-2-pentene (iv) (Z)-3-methyl-2-pentene (v) Trans-1, 4-dimethyl cyclo hexane.

**Q. Give the IUPAC name of the following compounds :**

**Q. Which of the following represents the correct IUPAC name for the compounds concerned :**

**(i) 2, 2–dimethylpentane or 2–dimethylpentane.**

**(ii) 2, 3–dimethyl pentane or 3, 4–dimethyl pentane.**

**(iii) 2, 4, 7–trimethyloctane or 2, 5, 7–trimethyloctane,**

**(iv) 2 –chloro–4–methylpentane or 4–chloro 2–methylpentane,**

**(v) But–3–yn-1-ol or But–4–ol–1–yne.**

**[NCERT]**

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**Sol.**(i) 2, 2-dimethyl pentane (ii) 2, 3-dimethyl pentane (iii) 2, 4, 7-trimethyl octane (iv) 2-chloro-4-methyl pentane (v) But-3-yne-1-ol

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**Sol.**(i) 1-ethoxy propan-2-ol (ii) Ethyl benzoate (iii) 2-phenyl propanal (iv) 6-chloro-3-methyl-hexan-2-one

**Q. Draw formulae for the first five members of each homologous series beginning with the following compounds.**

**(i) $\mathrm{HCOOH},$ (ii) $\mathrm{CH}_{3} \mathrm{COCH}_{3},\left(\text { iii) } \mathrm{CH}_{2}=\mathrm{CH}_{2}\right.$**

**Q. Identify the functional groups in the following compounds:**

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**Sol.**(iii)

*N*-Substituted amide and acid chloride

**Q. For the following bond cleavages, use curved-arrows to show the electron flow and classify each as homolysis or heterolysis. Identify reactive intermediate products as free radical, carbocation and carbanion.**

**[NCERT]**

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**Sol.**(i) Homolysis, Free radicals are formed.

**Q. Explain the order of stability of primary, secondary and tertiary carbonium ions.**

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**Sol.**$3^{\circ}>2^{\circ}>1^{\circ}$ is order of stability of carbocation. 3 ” carbocation are most stable due to maximum positive inductive effect of three alkyl groups. $2^{\circ}$ carbocation is morestable than $1^{\circ}$ carbocation due togreater inductive effect of two alkyl groups as compared to $1^{\circ} C$ carbocation in which there is one alkyl group.

**Q. 0.378**

*g*of an organic acid gave on combustion 0.264*g*of carbon dioxide and 0.162 g of water vapours. Calculate the percentage of*C*and*H*.**[NCERT]**

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**Sol.**Mass of organic compound $=0.378 \mathrm{g}$ Mass of $\mathrm{CO}_{2}$ formed $=0.264 \mathrm{g}$ Mass of $\mathrm{H}_{2} \mathrm{O}$ formed $=0.162 \mathrm{g}$ (i) Percentage of carbon $44 \mathrm{g}$ of contains carbon $0.264 \mathrm{g}$ of contains carbon $=12 \mathrm{g}$ $=\frac{12}{44} \times 0.264=0.072 \mathrm{g}$ Percentage of carbon $=\frac{0.072}{0.378} \times 100=19.04 \%$ $18 g$ of contains hydrogen $=2 g$ $0.162 g$ of contains hydrogen $=\frac{2}{18} \times 0.162=0.018 \mathrm{g}$ Percentage of hydrogen $=\frac{0.018}{0.378} \times 100=4.76 \%$

**(ii) Percentage of hydrogen**

**Q. During nitrogen estimation of an organic compound by Kjeldahl’s method, the ammonia evolved by $0.5 g$ of the**compound neutralised $10 \mathrm{mL}$ of $1 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4} .$ Calculate the percentage of nitrogen in the compound.

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**Sol.**$1 \mathrm{M}$ of $10 \mathrm{mL} \mathrm{H}_{2} \mathrm{SO}_{4} \equiv 1 \mathrm{M}$ of $20 \mathrm{mL}$ of $\mathrm{NH}_{3}$ $1000 \mathrm{mL}$ of $1 \mathrm{M}$ ammonia contains $=14 \mathrm{g}$ nitrogen $20 \mathrm{mL}$ of $1 \mathrm{M}$ ammonia contain $=\frac{14 \times 20}{1000} \mathrm{g}$ nitrogen $\therefore$ Percentage of nitrogen $=\frac{14 \times 20}{1000 \times 0.5} \times 100=56.0 \%$

**Q. (i) In sulphur estimation, 0.157 g of organic compound gave $0.4813 \mathrm{g}$ of $\mathrm{BaSO}_{4} .$ What is the percentage of sulphur in organic compound?**

**(ii) $0.092 g$ of organic compound on heating in carius tube and subsequence ignition gave $0.111 g$ of $M g_{2} P_{2} O_{7}$. Calculate the percentage of phosphorus in organic compound.**

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**Sol.**(i) Mass of $B a S O_{4}=0.4813 \mathrm{g}$ Mass of organic compound $=0.157 \mathrm{g}$ $\% \mathrm{S}=\frac{32 \times W_{B a S O_{4}} \times 100}{233 \times W_{\mathrm{Substance}}}=\frac{32 \times 0.4813 \times 100}{233 \times 0.157}=42.10$ (ii) Mass of organic compound $=0.092 \mathrm{g}$ Mass of $\mathrm{Mg}_{2} \mathrm{P}_{2} \mathrm{O}_{7}=0.111 \mathrm{g}$ $\mathrm{g}_{\mathrm{bof}} \mathrm{P}=\frac{62 \times \mathrm{W}_{\mathrm{Mg}_{2} \mathrm{P}_{2} \mathrm{O}_{7}}}{222 \times \mathrm{W}_{\mathrm{Substance}}} \times 10 \mathrm{C}=\frac{62 \times 0.111 \times 100}{222 \times 0.092}=33690024

**Q. (i) A mixture contains benzoic acid and nitrobenzene. How can this mixture be separated into its constituents by the technique of extraction using an appropriate chemical reagent?**

**(ii) Write the formula of iron (**

*III*) hexa cyanoferrate (*II*).**[NCERT]**

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**Sol.**(i) The mixture is shaken with a dilute solution of $\mathrm{NaHCO}_{3}$ and extracted with ether or chloroform when nitrobenzene goes into the organic layer. Distillation of the solvent gives nitrobenzene. The filtrate is acidified with dil.

*HCl*when benzoic acid gets precipitated. The solution is cooled and benzoic acid is obtained by filtration (ii) $\quad F e_{4}\left[F e(C N)_{6}\right]$

**Q. Give hybridization state of each carbon in following compounds :**

**$\mathrm{CH}_{2}=\mathrm{C}=\mathrm{O}, \mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}_{2},\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO}, \mathrm{CH}_{2}=\mathrm{CHCN},$**

**$\mathrm{CH}_{3} \mathrm{CH}_{2}^{-}, \mathrm{CH}_{3} \mathrm{CH}_{2}^{+}, \mathrm{CH}_{3} \mathrm{CH}_{2}$**

**Q. (i) Giving justification categorise the following species as nucleophile or electrophile :**

**$\begin{array}{ll}{\text { (a) } B F_{3}} & {\text { (b) } C_{2} H_{5} O^{-}}\end{array}$**

**(ii) Write the IUPAC names of the following compounds :**

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**Sol.**(i) $\quad$ (a) $B F_{3}$ is electrophile because it is electron deficient, i.e. octet of boron is not complete. (b) $C_{2} H_{5} O^{-}$ is nucleophile because it is negatively charged.

**Q. Indicate the $\sigma$ and $\pi$ -bonds in the following molecules:**

**(i) $\quad C_{6} H_{6}$**

**(ii) $\quad C_{6} H_{12}$**

**(iii) $C H_{2} C l_{2}$**

**(iv) $\quad \mathrm{CH}_{3} \mathrm{NO}_{2}$**

**(v) $\quad H C O N H C H_{3}$**

**[NCERT]**

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**Sol.**(i) $\quad C_{6} H_{6}$ is benzene and structural formulae is:

**Q. Draw the complete structures of bromomethane, bromoethane. 2–bromopropane and tert–butylbromide. Arrange them in order of decreasing steric hindrance.**

**[NCERT]**

**Q. Draw the resonating structures for the following compounds. Show the electron shift using curved arrow notation.**(i) $\quad C_{6} H_{5} O H$ (ii) $\quad C_{6} H_{5} N O_{2}$ (iii) $C_{6} H_{5} C^{\oplus} H_{2}$ (iv) $\quad \mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCHO}$ (v) $\quad \mathrm{CH}_{3} \mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{2}^{\oplus}$ (vi) $\quad C_{6} H_{5} \mathrm{CHO}$ (vii) $C H_{2}=C H O C H_{3}$

**[NCERT]**

**Q. Explain why alkyl groups act as electron donors when attached to a $\pi-$ system.**

**[NCERT]**

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**Sol.**Alkyl group has $s p^{3}$ hybridization whereas $\pi$ -bond atom is $s p^{2}$ hybridized which is more electronegative therefore alkyl group acts as electron donors.

**Q. Classify the reagents shown in bold in the following equation as nucleophiles or electrophiles. Use curved-arrow notation to show the electron movement.**

**(i) $\mathrm{CH}_{3} \mathrm{COOH}+\mathrm{HO}^{-} \longrightarrow \mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{H}_{2} \mathrm{O}$**

**(ii) $\mathrm{CH}_{3} \mathrm{COCH}_{3}+^{-} \mathrm{NC} \longrightarrow \mathrm{CH}_{3} \mathrm{C}(\mathrm{CN}) \mathrm{OHCH}_{3}$**

**(üii) $\mathrm{C}_{6} \mathrm{H}_{6}+\mathrm{CH}_{3} \mathrm{C}^{+} \mathrm{O} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COCH}_{3}$**

**[NCERT]**

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**Sol.**(i) $\quad O H^{-}$ is nucleophile

**Q. Classify the following reactions in one of the reaction type:**

**(i) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br}+\mathrm{HS}^{-} \longrightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{SH}+\mathrm{Br}^{-}$**

**(ii) $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{CH}_{2}+\mathrm{HCl} \longrightarrow\left(\mathrm{CH}_{3}\right)_{2} \mathrm{ClC}-\mathrm{CH}_{3}$**

**(iii) $\left.\mathrm{(CH}_{3}\right)_{2} \mathrm{CH}-\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{HBr} \rightarrow\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CBr}_{2} \mathrm{CH}_{3}$**

**(iv) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br}+\mathrm{HO} \longrightarrow \mathrm{CH}_{2}=\mathrm{CH}_{2}$**

**[NCERT]**

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**Sol.**(i) Nucleophilic Substitution reaction (ii) Addition reaction (iii) Nucleophilic Substitution reaction (iv) Elimination reaction.

**Q. Explain with the help of examples that Geometrical isomerism is different from conformation.**

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**Sol.**Conformation are obtained by rotation around $\sigma$ -bond,

*e.g.,*eclipsed and staggered conformation of ethane.

**Q. What is the relationship between the members of following pairs of structures ? Are they identical, structural or geometrical isomers, or resonance contributors ?**

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**Sol.**(i) Structural Isomers. (ii) Geometrical Isomers. (iii) Resonating structures. (iv) Geometrical Isomers.

**Long Answer (5 Mark)**

**Q. (i) Given the IUPAC name for the amine.**(ii) Discuss the hybridization of carbon atoms in allene $\left(C_{3} H_{4}\right)$ and show the $\pi$ – orbital overlaps.

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**Sol.**(i) $3, N, N$ -Trimethylpentan- $-3$ -amine. (ii) $\quad$ The structure of allene $\left(C_{3} H_{4}\right)$ is The carbon atoms 1 and 3 are $s p^{2}$-hybridized since each one of them is joined by a double bond. In contrast, carbon atom 2 is -hybridized since it has two double bonds. Thus, the two $\pi-$bonds in allene like in acetylene are perpendicular to each other as shown below : Whereas $H_{c}$ and $H_{d}$ lie in the plane of the paper while $H_{a}$ and $H_{b}$ lie in a plane perpendicular to the plane of the paper.

**Q. Give IUPAC names of the following compounds :**

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**Sol.**In all the questions given above the ring phenyl group. In case the phenyl group is further substituted, the carbon atoms of the ring are separately numbered starting from the carbon atom of the ring directly attached to the parent chain in such a way that substituent on ring gets lowest possible position. According the IUPAC name are : (i) 2-(4-chlorophenyl) propanoic acid (iv) If the compound contains more than one similar complex radicals/substituents, then the numerical prefixes such as

*di, tri, tetra*, etc. are replaced by bis,tris, tetrakis, etc. respectively. Hence, the IUPAC name is : 1, 1, 1-Trichloro-2, 2-bis-(chlorophenyl) ethane.

**Q. (i) A mixture contains two components**

*A*and*B*. The solubilities of*A*and*B*in water near its boiling point are 10 grams per 100*ml*and 2*g*per 100*ml*respectively. How will you separate*A*and*B*from this mixture?**(ii) Explain why an organic liquid vapourizes at a temperature below its boiling point in steam distillation?**

**(iii) Which technique can be used to separate naphthalene from kerosene oil present in its mixture.**

**[NCERT]**

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**Sol.**(i)

*A*and

*B*from the mixture can be separated by using

*Fractional crystallization.*When the saturated hot solution of this mixture is allowed to cool, the less soluble component

*B*crystallizes out first leaving the more soluble component

*A*in the mother liquor. (ii) A liquid boils when its vapour pressure becomes equal to the atmospheric pressure. During steam distillation, the mixture boils when sum of the vapour pressure of water and organic liquid becomes equal to the atmospheric pressure. Since the vapour pressure of water is appreciably higher than that of organic liquid, therefore, the organic liquid will vapourize at a temperature much lower than its normal boiling point. (iii) Since naphthalene and kerosene have large difference in their boiling points. Hence they can be separated by

*simple distillation.*

**Q. (i) The $R_{f}$ value of**

*A*and*B*in a mixture determined by*TLC*in a solvent mixture are 0.65 and 0.42 respectively. If the mixture is separated by column, chromatography using the same solvent mixture as a mobile phase, which of the two components,*A*or*B*, will elute first ? Explain.**(ii) A mixture contains 71 percent calcium sulphate and 29 percent camphor. Name a suitable technique of separation of the components of this mixture.**

**(iii) Without using column chromatography, how will you separate a mixture of camphor and benzoic acid?**

**[NCERT]**

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**Sol.**(i) Since the $R_{f}$ value of

*A*is 0.65, therefore, it is less strongly adsorbed as compared to compound

*B*with value of 0.42. Therefore, on extraction of the column, A will elute first. (ii)

*Sublimation*process can be used for the separation of camphor and calcium sulphate. Camphor sublimes where as is non-volatile. (iii) Sublimation cannot be used since both camphor and benzoic acid sublime on heating. Therefore, a chemical method using solution is used when benzoic acid dissolves leaving camphor behind. The filtrate is cooled and then acidified with dil.

*HCl*to get benzoic acid.

**Q. What are reactive intermediates ? How are they generated by bond fission?**

**[NCERT]**

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**Sol.**Generally, the organic reactions complete through the involvement of certain chemical species which are short lived ($10^{-6}$ seconds to a few seconds) and highly reactive. The short lived and highly reactive species are called reactive intermediates such as carbocations, carboanions, free radicals, carbenes and nitrenes. Heterolytic fission of covalent bond results in carbocations and carboanions. Free radicals are formed by homolytic fission. Carbenes are neutral carbon species in which carbon atom is bonded to two monovalent atoms or groups and also contains one non-bonding pair of electrons. They are produced in photolysis (irradiation with

*UV*light or thermolysis or pyrolysis (action of heat) on diazoalkanes or ketones.)

**Q. What is the relationship between the members of following pairs of structures ? Are they identical, structural or geometrical isomers, or resonance contributors ?**

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**Sol.**(i) Structural Isomers. (ii) Geometrical Isomers. (iii) Resonating structures. (iv) Geometrical Isomers.

**Q. What are the different types of structural isomerism ? Describe them with suitable examples.**

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**Sol.**Structural isomerism is of following types :

**(i) Chain isomerism :**“When two or more compounds have same molecular formula but different carbon skeleton are said to have chain isomerism.” For example, $\mathrm{C}_{5} \mathrm{H}_{12}$ shows the following three isomers:

**(ii) Position Isomerism :**‘When two or more compounds have same molecular formula but different position of substituent atoms or groups on the carbon skeleton, are said to have position isomerism.’ For example, $C_{3} H_{8} O$ shows the following two position isomers of alcohol :

**(iii) Functional group isomerism :**‘When two or more compounds have same molecular formula but different functional groups, are said to have functional group isomerism.’ For example, $C_{4} H_{10} O$ represents the following two compounds.

**Q. Why do alkenes and cycloalkanes show cis-trans isomerism?**

**Or**

**What is the origin of geometrical isomerism in alkenes?**

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**Sol.**Presence of carbon-carbon double bond, in a molecule is the cause of geometrical isomerism. This is because when a carbon-carbon double bond is present, the molecule cannot rotate freely about the double bond. It is called

**hindered**or

**restricted rotation**of the molecule about the double bond. During the formation of double bond, if two similar atoms or groups are present on the same side of the plane of the double bond it is called a

*cis*isomer and if present on the opposite sides, it is called a

*trans*isomer. Because of restriction in free rotation one form cannot change to the other. Therefore presence of

*C = C*bond is the origin of geometrical isomerism . This restriction in rotation may be present in rigid of cyclic molecules. Therefore cyclic compounds also shows geometrical isomerism.

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**Sol.**Group I elements are highly reactive and with water (moisture in the atmosphere) form strong alkalies, so they are called alkali metals.

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**Sol.**Washing soda is sodium carbonate $\left(N a_{2} C O_{3}\right)$.

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**Sol.**They are highly reactive, therefore, they occur in combined state and do not occur in free state.

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**Sol.**The important minerals of lithium are:

(i) Lipidolite $(L i . N a . K)_{2} A l_{2}\left(S i O_{3}\right)_{3} \cdot F(O H)$ (ii) Spodumene $L i A I S i_{2} O_{3}$ (iii) Amblygonite $\operatorname{LiAl}\left(P O_{4}\right) F$

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**Sol.**At large scale, sodium hydroxide is prepared by castner kellner cell.

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**Sol.**Because solubility of $\mathrm{KHCO}_{3}$ is fairly large as compared to $\mathrm{NaHCO}_{3}$.

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**Sol.**Plaster of paris is calcium sulphate hemihydrate : $\mathrm{CaSO}_{4} \cdot \frac{1}{2} \mathrm{H}_{2} \mathrm{O}$ or $\left(\mathrm{CaSO}_{4}\right)_{2} \cdot \mathrm{H}_{2} \mathrm{O}$

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**Sol.**Due to weak metallic bonds and large atomic size, their density is low.

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**Sol.**Alkali metals have bigger atomic size, therefore they have lower first I.E. than group 2 elements.

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**Sol.**Because they have a strong tendency to lose outer most electron.

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**Sol.**$K I$ In the chemical bond is ionic in character. On the other hand due to small size of lithium ion and its high polarising power the bond in is predominently covalent in character. Hence LiI is more soluble than in ethanol.

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**Sol.**Both $L i^{+}$ and $H^{-}$ have small size and their combination has high lattice energy. Therefore LiH is stable as compared with $\mathrm{NaH}$

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**Sol.**$B e C l_{2}$ is covalent, therefore soluble in organic salvents.

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**Sol.**Lithium floats on water without any apparent reaction with it.

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**Sol.**The element is beryllium, its oxide $B e O$ is soluble in excess of $\mathrm{NaOH}$ $B e O+2 N a O H \longrightarrow N a_{2} B e O_{2}+H_{2} O$ Its dipositive ion has electronic configuration $\left(B e^{2+}=1 s^{2}\right)$

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**Sol.**Because $B e C l_{2}$ is covalent compound.

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**Sol.**The charge over radius ratio, i.e., polarizing power is similar, that is the cause of diagonal relationship. [esquestion]. Name the alkali metals which forms superoxide when heated in excess of air.

*K*,

*Mg*,

*Ca*&

*Al*form amphoteric oxide ?

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**Sol.**form amphoteric oxide, i.e., acids as well as basic in nature.

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**Sol.**Sodium metal removes moisture from benzene by reacting with water. However, ethanol cannot be dried by using sodium because it reacts with sodium. $2 \mathrm{Na}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{ONa}+\mathrm{H}_{2}$

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**Sol.**$\mathrm{CaCl}_{2}$ reduces melting point of $N a C l$ and increases electrical conductivity.

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**Sol.**$\mathrm{CaCO}_{3}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2} \longrightarrow \mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}$

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**Sol.**We get amorphous sodium carbonate becouse it loses water molecules.

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**Sol.**$M g_{3} N_{2}$ is a salt of a strong base, $M g(O H)_{2}$ and a weak acid $\left(N H_{3}\right)$ and hence gets hydrolysed to give $N H_{3} .$ In contrast, $M g C l_{2}$ is a salt of a strong base, and a strong acid, and hence does not undergo hydrolysis to give

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**Sol.**Caesium has the lowest while lithium has the highest ionization enthalpy among all the alkali metals. Hence, caesium can lose electron very easily while lithium cannot.

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**Sol.**Strontium and Barium nitrates are used in pyrotechnics for giving red and green flames.

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**Sol.**The elements in which the last electron enters the -orbital of their outermost energy level are called -block elements. It consists of Group 1 and Group 2 elements. Their electronic configuration is

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**Sol.**(i) $\quad N a$(ii) $\mathrm{Ca}$ (iii) $M g \quad$ (iv) $\quad A l$

**[NCERT]**

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**Sol.**$\mathrm{CaO}=50$ to $60 \% \quad \mathrm{SiO}_{2}=20$ to $25 \%, \quad \mathrm{Al}_{2} \mathrm{O}_{3}=5$ to $10 \%$ $M g O=2$ to $3 \%, \quad F e_{2} O_{3}=1$ to $2 \%, \quad S O_{2}=1 \quad$ to $2 \% \quad$ is The ratio of $S i O_{2}$ (silica) to alumina $\left(A l_{2} O_{3}\right)$ should be between 2.5 and 4.0 and the ratio of lime $(\mathrm{CaO})$ to total oxides of silicon $\mathrm{SiO}_{2}, \mathrm{Al}_{2} \mathrm{O}_{3}$ and $\mathrm{Fe}_{2} \mathrm{O}_{3}$ should be as close to 2 as possible.

*Mg*from sea water.

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**Sol.**Like alkali metals, alkaline earth metals are also highly electropositive and strong reducing agents. Same difficulties, we face in the extraction of these metals. Therefore, these metals are extracted by electrolysis of their fused metal halides. [esquestion]. Starting with sodium chloride how would you proceed to prepare (state the steps only) : (i) sodium metal (ii) sodium hydroxide (iii) sodium peroxide (iv) sodium carbonate #tag#

**[NCERT]**

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**Sol.**(i) They do not show variable oxidation states. (ii) They are soft metals having low melting and boiling point. (iii) They are highly electropositive and most reactive metals.

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**Sol.**$L i F$ is an ionic compound containing small ions and hence has very high lattice enthalpy. Enthalpy of hydration in this case is not sufficient to compensate for high lattice enthalpy. Hence, is insoluble in water. has partial covalent character due to polarization of chloride ion by ion. Thus, has partial covalent character and partial ionic character and hence is soluble in water as well as less polar solvents such as acetone.

**[NCERT]**

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**Sol.**When an alkali metal is dissolved in liquid ammonia it produces a blue coloured conducting solution due to formation of ammoniated cation and ammoniated electron as given below : When the concentration is above the colour of solution is copper-bronze. This colour change is because the concentrated solution contains clusters of metal ions and hence possess metallic lustre.

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**Sol.**[esquestion]. Alkali metals have low ionisation enthalpies.Why is it so?

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**Sol.**$K^{\prime}$ has lowest ionisation energy due to larger atomic size among these elements. The force of attraction between valence electron and nucleus is less, therefore it can loose electron easily.

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**Sol.**The solvated electron, $e\left(N H_{3}\right)_{x}$ or ammoniated electron is responsible for blue colour of alkali metal solution in It absorbs light from visible regions and radiates complimentary colour: $2 N a(s)+2 N H_{3}(l) \longrightarrow 2 N a N H_{2}(s)+H_{2}(g)+e\left(N H_{3}\right)_{x}$

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**Sol.**(i) They form unipositive ions. (ii) Their second ionisation energy is very high. (iii) They have weak metallic bonds due to larger size and only one valence electron.

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**Sol.**(i) $L i^{+}$ ions are smallest in size therefore most hydrated, that is why they have lowest mobility in aqueous solution. $C s^{+}$ ions are largest in size, least hydrated, therefore have highest mobility. Size of hydrated cation decreases, therefore, mobility of ions increases down the group. (ii) $L i$ is smallest in size and best reducing agent, therefore, it forms nitride with $N_{2}$ $6 L i+N_{2} \longrightarrow 2 L i_{3} N$ (Lithium nitride) (iii) It is due to less difference in their standard reduction potentials which is resultant of sublimation energy, ionisation energy and hydration energy. (iv) is least ionic as compared to other fluorides of alkali metals. It has high lattice energy, therefore, it is least soluble. [esquestion]. Why do alkali metals impart characteristic colours to the flame of a bunsen burner ? What is the colour imparted to the flame by each of the following metals ? Lithium, Sodium and Potassium.

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**Sol.**Duralium and Magnaliam are alloys of $\mathrm{Al}, \mathrm{Mg}$ is lighter in density than $A l$ therefore, it makes the alloys lighter. These alloys are used in automobile engines and aeroplanes.

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**Sol.**$-B e(O H)_{2}<M g(O H)_{2}<C a(O H)_{2}<S r(O H)_{2}<B a(O H)_{2}$ Solubility of hydroxide goes on increasing down the group because hydration energy dominates over lattice energy. $B e S O_{4}>M g S O_{4}>C a S O_{4}>S r S O_{4}>B a S O_{4}$ Solubility of sulphate goes on decreasing down the group because lattice energy dominates over hydration energy.

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**Sol.**Properties of beryllium different than other elements of the group (i) Beryllium is harder than other elements. (ii) Melting and boiling points are higher than that of other elements. (iii) It does not react with water but other elements do. (iv) It does not react with acids to form hydrogen. (v) It forms covalent compounds but others form ionic compounds. (vi) Beryllium oxide is amphoteric but other oxides are basic.

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**Sol.**(i) Atomic size decrease from group 1 to group 2 due to increase in effective nuclear charge. (ii) First ionisation enthalpy increases from group 1 to group 2 due to decrease in atomic size. (iii) Density increases from group 1 to 2. (iv) Melting points increase from group 1 to 2.

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**Sol.**(i) Group 1 metals form oxides of strong basic nature (except $L i_{2} O$. Group 2 metal form oxide of less basic nature. (ii) Carbonates : Carbonates of Group 1st are soluble and stable, solubility in case of Group 2nd decreases down the group. (iii) Polarizing power of cations increases. (iv) Reactivity and reducing power decreases.

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**Sol.**(i) On heating sodium bicarbonate it forms sodium carbonate \[ 2 \mathrm{NaHCO}_{3} \stackrel{\Delta}{\longrightarrow} \mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{O}+\mathrm{CO}_{2}\] (ii) When sodium amalgam, $N a / H g,$ is formed the vigourosity of reaction of sodium with water decreases. \[ 2 \mathrm{Na} / \mathrm{Hg}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{NaOH}+\mathrm{H}_{2} \] (iii) Sodium reacts with ammonia to form amide. \[2 \mathrm{Na}+2 \mathrm{NH}_{3} \stackrel{\Delta}{\longrightarrow} 2 \mathrm{NaNH}_{2}+\mathrm{H}_{2} \]

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**Sol.**(i) An aqueous solution of sodium carbonate has a large concentration of hydroxyl ions making it alkaline in nature. (ii) Sodium is a very strong reducing agent therefore it cannot be extracted by the reduction of its ore (chemical method). Thus the best way to prepare sodium is by carrying electrolysis of its molten salts containing impurities of calcium chloride.

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**Sol.**(i) $L i^{+}$ ions is smaller in size. It is stabilized more by smaller anion, oxide ion $\left(O^{2-}\right)$ as compared to larger anion, peroxide ion $\left(O_{2}^{2-}\right)$. (ii) $\quad K^{+}, R b^{+}, C s^{+},$ are large cations. A large cation is more stabilized by large anions. since superoxide ion, $O_{2}^{-}$ is quite large, $K, R b$ and $C s$ form superoxides in preference to oxides and peroxides. (iii) In aqueous solution sodium carbonate undergoes hydrolysis forming sodium hydroxide. $\mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{NaHCO}_{3}+\mathrm{NaOH}$ [esquestion]. Write balanced equations or reactions between :

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**Sol.**$L i_{2} C O_{3}$ is thermally unstable because it is covalent. It decomposes to form $L i_{2} O$ and $C O_{2}$ $L i_{2} C O_{3} \stackrel{\Delta}{\longrightarrow} L i_{2} O+C O_{2}$ [esquestion]. Why are ionic hydrides of only alkali metals and alkaline earth metals are known ? Give two examples.

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**Sol.**It is because $K F$ is more ionic than $N a F$

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**Sol.**s-block elements are highly reactive, therefore they never occur in free state rather occur in combined state in the form of halides, carbonates, sulphates. They are generally prepared by electrolysis of their molten salts

.

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**Sol.**mg occurs in the form of $M g C l_{2}$ in sea water from which it can be extracted. Sea water containing $\mathrm{MgCl}_{2}$ is concentrated under the sun and is treated with $\mathrm{Ca}(\mathrm{OH})_{2} \cdot \mathrm{Mg}(\mathrm{OH})_{2}$ is thus precipitated, filtered and heated to give the oxide. The oxide is treated with $\mathrm{C}$ and $\mathrm{Cl}_{2}$ to get $\mathrm{MgCl}_{2}$. $\mathrm{MgO}+\mathrm{C}+\mathrm{Cl}_{2} \stackrel{\text { heat }}{\longrightarrow} \mathrm{MgCl}_{2}+\mathrm{CO}$ $\mathrm{MgCl}_{2}$ is fused with $\mathrm{NaCl}$ and $\mathrm{CaCl}_{2}$ at $970-1023 \mathrm{K}$ and molten mixture is electrolysed. Magnesium is liberated at cathode and Chlorine is evolved at the anode. At Cathode : $\mathrm{Mg}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Mg}$ At Anode : $2 \mathrm{Cl}^{-} \longrightarrow \mathrm{Cl}_{2}+2 \mathrm{e}^{-}$ A steam of coal gas is blown through the cell to prevent oxidation of Mg metal.

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**Sol.**Sea water contains $M g C l_{2}$ which is concentrated under the sun and is treated with calcium hydroxide, $\mathrm{Ca}(\mathrm{OH})_{2}$ , magnesium hydroxide is thus precipitated, filtered and heated to give oxide. The oxide is treated with carbon and $C l_{2}$ to get $M g C l_{2}$. $M g O+C+C l_{2} \longrightarrow M g C l_{2}+C O$ is mixed with sodium chloride so as to reduce its melting point and increase its electrical conductivity. Molten mixture is electrolysed using steel cathode and carbon anode. A steam of coal gas is blown through the cell to prevent oxidation. of metal. At cathode Mg’t $+2 \mathrm{e}^{-} \longrightarrow \mathrm{Mg}$ At anode $\quad 2 \mathrm{Cl}^{-}-2 \mathrm{e}^{-} \longrightarrow \mathrm{Cl}_{2}$ mg obtained in liquid state is further distilled to give pure mg When burns in air, it forms magnesium oxide and magnesium nitride. $3 M g+N_{2} \longrightarrow M g_{3} N_{2} ; 2 \mathrm{Mg}+\mathrm{O}_{2} \longrightarrow 2 \mathrm{MgO}$ [esquestion]. (i) Draw a neat and labelled diagram of Castner-Kellner cell for the manufacture of caustic soda. (ii) Give chemical equations of the reaction of caustic soda with (a) ammonium chloride, and (b) carbon dioxide.

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**Sol.**(i) $2 N a+2 H_{2} O \longrightarrow 2 N a O H+H_{2}$ ; sodium hydroxide is formed with evolution of $H_{2}(g)$ The hydrogen gas catches fire due to highly exothermic process. (ii) $\quad 2 \mathrm{Na}+\mathrm{O}_{2} \longrightarrow \mathrm{Na}_{2} \mathrm{O}_{2}$ sodium peroxide is formed. (iii) Sodium hydroxide and hydrogen peroxide are formed. $\mathrm{Na}_{2} \mathrm{O}_{2}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow 2 \mathrm{NaOH}+\mathrm{H}_{2} \mathrm{O}_{2}$

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**Sol.**The lattice enthalpies of hydroxides and carbonates of magnesium and calcium are very high due to the presence of divalent cations. The enthalpy of hydration cannot compensate for the energy required to break the lattice in these compounds. Hence, they are sparingly soluble in water. On the otherhand, the lattice enthalpies of hydroxides and carbonates of sodium and potassium are low due to the presence of monovalent cations. The enthalpy of hydration in this case is sufficient to break the lattice in these compounds. Hence, hydroxides and carbonates of sodium and potassium are easily soluble in water.[esquestion]. What happens when : (i) magnesium is burnt in air (ii) quicklime is heated with silica (iii) chlorine reacts with slaked lime (iv) calcium nitrate is heated #tag# [NCERT]

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**Sol.**(i) Be forms amphoteric oxide whereas others form basic acids. (ii) $B e C l_{2}$ is covalent, others form ionic halides. (iii) does not react even with hot water where as others react easily with water. has smallest atomic size, highest ionisation energy, high polarising power which makes it anomalous in this group.[esquestion]. Beryllium exhibits some similarities with aluminium. Point out three such properties. [NCERT]

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**Sol.**(i) Nitrates of alkali metals are more stable than that of alkaline earth metals. All nitrates are soluble in water. (ii) Carbonates of alkali metals are more stable and more soluble in water than that of alkaline earth metals. (iii) Sulphates of alkali metals are more stable and more soluble in water than that of alkaline earth metal.[esquestion]. Discuss the anomalous behaviour of Lithium. Give its diagonal relationship with Magnesium.

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**Sol.**(i) $\mathrm{Na}_{2} \mathrm{CO}_{3}$ is alkaline due to its hydrolysis, it forms $O H^{-}$ more than $H^{+}$ because $H_{2} C O_{3}$ is weak acid, therefore, is alkaline. (ii) Alkali metals are strong reducing agents and highly reactive with water, therefore, they are prepared by electrolysis of their fused Chlorides. (iii) Sodium is more useful than potassium because sodium is less reactive and found in more abundance than K

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**Sol.**(i) Sodium carbonate does not decompose whereas calcium carbonate decomposes on heating. $\quad \mathrm{CaCO}_{3} \stackrel{\Delta}{\longrightarrow} \mathrm{CaO}+\mathrm{CO}_{2}$ (ii) $\mathrm{MgCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O} \stackrel{\Delta}{\longrightarrow} \mathrm{MgO}+2 \mathrm{HCl}+5 \mathrm{H}_{2} \mathrm{O}$ $\mathrm{CaCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{CaCl}_{2}+6 \mathrm{H}_{2} \mathrm{O}$ (iii) $2 \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2} \stackrel{\Delta}{\longrightarrow} 2 \mathrm{CaO}+4 \mathrm{NO}_{2}+\mathrm{O}_{2}$ $2 \mathrm{NaNO}_{3} \stackrel{\Delta}{\longrightarrow} 2 \mathrm{NaNO}_{2}+\mathrm{O}_{2} |$

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**Sol.**Be predominantly forms covalent compounds due to smaller size, higher ionisation energy and high polarising power. $B e O, B e C l_{2}$ and $B e F_{2}$ are covalent and get hydrolysed by water. BeO is least soluble in water due to covalent character. $B e O+H_{2} O \longrightarrow B e(O H)_{2}$ $B e C l_{2}+2 H_{2} O \longrightarrow B e(O H)_{2}+2 H C l$ $B e F_{2}+2 H_{2} O \longrightarrow B e(O H)_{2}+2 H F$ They are less soluble in water but more soluble in organic solvents, which shows they are covalent in nature.

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**Sol.**(i) Oxides of group 1 elements are more basic than that of group 2. (ii) Solubility and thermal stability of carbonates of group 1 is higher than that of group 2. (iii) Polarizing power of cations of group 2 is higher than that of group 1. (iv) Reactivity and reducing power of group 1 elements is higher than corresponding group 2 elements.

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**Sol.**(i)

*Lime stone*: (a) It is used in manufacture of glass and cement. (b) It is used as flux in extraction of iron. (ii)

*Cement*: (a) It is used as building material. (b) It is used in concrete and reinforced concrete, in plastering and in construction of bridges, dams and buildings. (iii)

*Plaster of Paris*: (a) It is used for manufacture of chalk. (b) It is used for plastering fractured bones. (c) It is used for making casts and moulds. (d) It is also used in dentistry, in ornaments work and for taking casts of statues.

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**Sol.**(i) Group-I consists of lithium, sodium, potassium, rubidium, caesium and francium. (ii) Elements have electronic configuration $n s^{1}$ (iii) Elements have 1 electron in the outermost s-orbital and have a strong tendency to lose this electron, so : (a) These are highly electropositive metals. (b) Never found in free state due to their high reactivity. (c) They form $M^{+}$ ion. (iv)

*Atomic radii*: Atomic radii of alkali metals are largest in their respective periods. (v)

*Density*: Their densities are quite low. Lithium is the lightest known metal. (vi)

*Oxidation state*: The alkali metal atoms show only +1 oxidation state. (vii)

*Reducing agents*: Due to very low value of ionisation energy, alkali metals are strong reducing agents and reducing character increase $N a$ to $C s$ but $L$ but is the strongest reducing agent. (viii) When alkali metals are heated in air, Lithium forms normal oxide $\left(L i_{2} O\right)$ sodium forms peroxide $\left(N a_{2} O_{2}\right)$potassium, rubidium and caesium form superoxides with peroxides. (ix) The alkali metals form hydrides of type reacting with hydrogen at about $673 K \quad \text { (Li forms hydride at } 1073 K)$ forms hydride at Ionic character of hydrides increases from to

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**Sol.**Sodium carbonate $\mathrm{Na}_{2} \mathrm{CO}_{3} \cdot \mathrm{H}_{2} \mathrm{O}$ or washing soda is manufactured by solvay-process.

*Principle of process*: Carbondioxide gas is passed through brine solution (about 28% $N a C l$ saturated with ammonia, sodium carbonate is formed. Plant used for the manufacture of washing soda.

*Process*: It completes in the following steps : (i)

*Saturation of brine with ammonia*: Lime stone (Calcium carbonate is strongly heated to form carbon dioxide. Ammonia and carbondioxide mixture is passed through a tower in which saturated brine is poured down. Ammoniated brine is filtered to remove impurities of calcium and magnesium carbonate. (ii) The milky solution is removed and filtered with the help of a vacuum pump. (iii)

*Ammonia recovery tower*: The filtrate is step 2 is mixed with calcium hydroxide and heated with steam. Ammonia obtained is recycled with carbondioxides. Potassium carbonate cannot be prepared by Solvay process as the solubility of $\mathrm{KHCO}_{3}$ is fairly large as compared to $\mathrm{NaHCO}_{3}$

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**Sol.**The raw materials used in this process are sodium chloride and lime stone Calcium chloride is the by-product in this process.

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**Sol.**

*Industrial uses of caustic soda.*(i) It is used for manufacture of soap. (ii) It is used in paper industry. (iii) It is used in textile industries. Sodium hydroxide can be prepared by electrolysis of saturted solution of brine $(N a C l)$ (iii) [esquestion]. (i) How is plaster of Paris prepared ? Describe its chief property due to which it is widely used. (ii) How would you explain ? (a) $\quad$ BeO is insoluble but $B e S O_{4}$ in soluble in water. (b) $\quad B a O$ is soluble but $B a S O_{4}$ is insoluble in water. (c) $\quad$ Lil is more soluble than $K I$. (d) $\quad \mathrm{NaHCO}_{3}$ is known in solid state but $\mathrm{Ca}\left(\mathrm{HCO}_{3}\right)_{2}$ is not isolated in solid state. \ #tag# [NCERT]

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**Sol.**(i)

*Beryllium :*It is invariably divalent. Its oxide is soluble in and its dipositive ion has a noble gas core. (ii)

*Quick lime :*It is calcium oxide $(C a O)$ It is produced by heating $\mathrm{CaCO}_{3}$

*Lime water :*The clear aqueous solution of Calcium hydroxide in water is called lime water. It is formed when $\mathrm{Ca}(\mathrm{OH})_{2}$ is dissolved in excess of $H_{2} O$

*Slaked Lime :*Calcium hydroxide solid is known as Slaked Lime. It is produced when water is added to $\mathrm{CaO}$

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**Sol.**(i)

*Action of heat on calcium carbonate*: When calcium carbonate is heated, a colourless gas carbon dioxide is given out and a white residue of calcium oxide is left behind. (ii)

*Action of heat on magnesium chloride hexahydrate*: It loses water and hydrogen chloride and yields a residue of magnesium oxide. (iii)

*Action of heat on gypsum*: It forms a hemi-hydrate, called plaster of paris. (iv)

*Action of heat on magnesium sulphate heptahydrate*: It loses water of crystallization and forms anhydrous salt.

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**Sol.**(i) (a) Caustic soda is used in prepartion of pure fats and oils. It is also used for preparation of rayon (artificial silk). (b) Sodium carbonate is used for manufacture of glass. It is used in softening of hard water. (c) Quick lime is used for white washing. It is used for manufacturing of glass and cement.

[Polymeric structure of $\left.B e C l_{2}\right]$ in solid state

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**Sol.**$\mathrm{N}_{2}<\mathrm{SO}_{2}<\mathrm{ClF}_{3}<\mathrm{K}_{2} \mathrm{O}<\mathrm{LiF}$

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**Sol.**$\mathrm{C}-\mathrm{H}<\mathrm{Br}-\mathrm{H}<\mathrm{Li}-\mathrm{Cl}<\mathrm{Na}-\mathrm{I}<\mathrm{K}-\mathrm{F}$

**[NCERT]**

**[NCERT]**

**[NCERT]**

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**Sol.**It is because extent of overlapping is more in $\sigma$ -bond than $\pi$ bond.

**[NCERT]**

*Z*-axis as the inter nuclear axis, which out of the following will form a sigma bond :

**(i) $1 \mathrm{s}$ and $1 \mathrm{s}$**

**(ii) $1 \mathrm{s}$ and $2 \mathrm{p}_{\mathrm{z}}$**

**(iii) $2 p_{x}$ and $2 p_{x}$**

**(iv) $2 p_{x}$ and $2 p_{y}**

**(\mathrm{v}) 2 p_{y}$ and $2 p_{y}$**

**[NCERT]**

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**Sol.**(i) and (ii)

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**Sol.**$s p^{3} d$ hybridisation.

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**Sol.**BeH_ $_{2}$ molecule is linear due to presence of sp-hybridization in $B e .$ Therefore, dipole moments due to the polar $B e-H$ bond get cancel out. Therefore, molecule has zero dipole moment.

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**Sol.**$-O C S$ will have higher dipole moment because bond moment of $O=$ $-C$ and that of $C=S$ do not cancel each other

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**Sol.**No, these cannot be taken as canonical forms because the position of atoms have changed.

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**Sol.**Molecular orbital configuration of $B e_{2}$ is $\left(\sigma_{1 s}\right)^{2},\left(\sigma^{*}_{1 s}\right)^{2}$ $\left(\sigma_{2 s}\right)^{2},\left(\sigma_{2 s}\right)^{2} .$ Bond order of \[ B e_{2}=\frac{1}{2}\left(N_{b}-N_{a}\right)=\frac{1}{2}(4-4)=0 \]

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**Sol.**$C_{2}^{-2}(B . O=3)$ Diamagnetic is more stable than $\quad O_{2}^{-}\left(B . O=1 \frac{1}{2}\right)$ paramagnetic

*bi*-pyramidal arrangement in preference to an axial position.

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**Sol.**In $S F_{4}$ molecule, lone pair of electron occupy an equatorial position because $l . p .-b . p .$ repulsions are less in that case. As a result, energy is less which leads to more stability.

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**Sol.**Hydrogen bond is attractive force which binds H-atom of one molecule with electronegative atom of another molecule. It is stronger than vander waals force.

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**Sol.**In a non-polar molecule, there may be instantaneous dipole created by specific position of electrons. This instantaneous dipole can induce a dipole in a nearby non-polar molecules.

*H*-bond.

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**Sol.**H-bond is force of attraction between hydrogen and most electronegative element like

*F, O*and

*N.*It is a weak bond.

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**Sol.**(i) High electronegativity of atom bonded to hydrogen. (ii) Small size of electronegative atom.

*X……H-X*).

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**Sol.**

*S < Cl < N < O < F.*.

**$S \quad$ and $\quad S^{2-} ; \quad P$ and $P^{3-} ; \quad N a$ and $N a^{+} ; \quad A l$ and $A l^{3+}$; $H$ and $H^{-}$**

**[NCERT]**

**[NCERT]**

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**Sol.**In $B C l_{3},$ Octet rule is not obeyed because boron contains only six electrons around it.

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**Sol.**

**Nitrite :**The Lewis structure of nitrite ion is

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**Sol.**In $A I C I_{3}$ the hybrid state of aluminium is $s p^{2}$ but in $A I C I_{4}^{-}$ the hybrid state of aluminium is $s p^{3} .$ The $A l C l_{4}^{-}$ adopts tetrahedral arrangement. Thus hybrid state of $A l$ atom changes from $s p^{2}$ to $s p^{3}$

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**Sol.**All the hybrid orbitals have same shape, $i . e .,$ one lobe larger than the other, however, their sizes are in the order $s p<s p^{2}<s p^{3}$

*B*and

*N*atoms as a result of the following reaction : $B F_{3}+N H_{3} \longrightarrow F_{3} B \leftarrow N H_{3} ?$

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**Sol.**In $B F_{3}, B$ is $s p^{2}$ hybridised and in $N H_{3}, N$ is $s p^{3}$ hybridised. After the reaction, hybridisation of $B$ changes to $s p^{3}$ but that of $N$ remains unchanged. Hybridisation of Boron changes to involve a vacant orbital to accommodate electron pair coming from $N H_{3} .$

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**Sol.**In $C H_{4},$ central carbon atom is surrounded by four bond pairs of electrons. These bond pairs arrange themselves in tetrahedral manner with maximum possible bond angle of $109^{\circ} 28^{\prime} .$ In square planar geometry, the bond angles will be $90^{\circ}$ with larger bond pair-bond pair repulsions. Therefore, square planar geometry is not possible.

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**Sol.**(i) 5 valence electron in nitrogen. 3 bond pairs and one lone pair of electron. Therefore it’s shape is pyramidal. since the Bond pairBond pair repulsion <bond pair-lone pair, repulsion. Therefore, it has pyramidal or distorted tetrahedral shape. (ii) $C H_{4}$ is tetrahedral because it has four bonded pair of electrons.

**$H_{2} \mathrm{O}, P C l_{3}, N H_{3}, N F_{3}$**

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**Sol.**Out of the given structure $1,2$ and 3 are important for describing the bonding in $C O_{2}$ molecule. These structure are isovalent (same number of bonds) and thus have higher contribution. Structure 4 has unsaturated carbon with only 4 electrons and thus has negligible stability.

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**Sol.**(i) Bonding molecular orbital has a single lobe and no node whereas antibonding molecular orbital has two lobes which are separated by a node. (ii) Bonding molecular orbital has less energy whereas antibonding has higher energy than atomic orbitals.

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**Sol.**When $N_{2}$ change to $N_{2}^{+}$ the electron is removed from bonding molecular orbital while when $O_{2}$ changes to $O_{2}^{+},$ the electron is removed from antibonding molecular orbital.

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**Sol.**Bond order cannot be used for quantitative comparison of the strengths of chemical bonds. It can give us only approximate idea about the strength of the bond. Greater the bond order, greater is the strength of the bond. $N_{2}(3), O_{2}(2), O_{2}^{+}(2.5) . O_{2}^{-}(1.5)$

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**Sol.**$N H_{3}$ molecule has one lone pair while $H_{2} O$ has two lone pairs of electrons. Due to the presence of lone pairs the geometries of and are distorted. Due to the presence of stronger $l p-b p$ repulsion than $b p-b p$ repulsion the bond angle in is reduced from normal tetrahedral bond angle $\left(109.5^{\circ}\right)$ to $107^{\circ} .$ In case of, two $l p$ of electrons force the $O-H$ bonds more closely than $N-H$ bonds in. So bond angle decreases to a larger extent i.e. to $104.5^{\circ}$

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**Sol.**$\mathrm{CO}_{2}$ has zero dipole moment. This means that molecules is linear so that the two $C=O$ bond moments get cancelled giving zero resultant dipole moment. However, molecule has resultant dipole moment showing that it cannot be linear. The two $O-H$ bonds are arranged in angular shape and the bond moment of two $O-H$ bonds give resultant dipole moment.

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**Sol.**According to Lewis concept, bond order is number of bonds present between two atoms of a molecule or ion. Bond order in $N \quad N\left(N_{2}\right)$ is 3 Bond order in $\mathrm{O} \quad \mathrm{O}\left(\mathrm{O}_{2}\right)$ is 2 Bond order in $C \quad O(C O)$ is 3

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**Sol.**$\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}$ is trimethyl amine. It has a pyramidal geometry as shown in structure I. $\left[\left(C H_{3}\right)_{3} \mathrm{Si}\right]_{3} N$ is planar as shown is structure II. Thus, the two species are not isostructural. In $\left(C H_{3}\right)_{3} N . N$ eatom assumes $s p^{3}$ hybrid state whereas in , the $N$ atom assumes Inybrid state.

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**Sol.**(i) $\quad B e C l_{2}$ has zero dipole moment because it has linear shape both dipole are equal and opposite such that net dipole moment is zero. (ii) In resonating structure, the position of atom is same. The bond lengths are equal due to resonance. Larger the number of canonical structures, greater will be the stability of molecule / ion. (iii) $\quad H_{2} O$ is liquid because water molecule are associated with inter molecular $H$ -bonding where as $H_{2} S$ is not associated with inter molecular $H$ -bonding that is why it is a gas.

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**Sol.**But since, the population of antibonding orbitals is more in case of and thus, is slightly less stable than.

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**[NCERT]**

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**Sol.**The properties of elements are periodic function of their atomic number.

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**Sol.**Because they have similar valence shell electronic configuration.

*Z*= 14 ?

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**Sol.**Period – 3, Group – 14.

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**Sol.**In the modern periodic table, each period begins with the filling with a new shell, therefore, the period indicates the value of principal quantum number. Thus option (iii) is correct.

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**Sol.**The roots for $1,2$ and 0 are $u n, b i,$ and nil respectively. $\therefore$ Name of element: Unbinilium Symbol : Ubn.

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**Sol.**(i) Lawrencium (

*Z*=103) and Berkelium (

*Z*=97). (ii) Seaborgium (

*Z*=106)

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**Sol.**The electronegativity of nitrogen will not be 3.0 in all its compounds. It depends upon other atom attached to it.

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**Sol.**$\mathrm{Be}^{-2}, \mathrm{Mg}^{-2}, \mathrm{Na}^{+}, \mathrm{Cl}^{-}, \mathrm{s}^{-2}, \mathrm{Br}^{-}$

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**Sol.**Isotopes are atoms of the same element which have same atomic number but different mass number. Therefore, they have same number of electrons and nuclear charge (protons). Thus, they will have same first ionization enthalpies.

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**Sol.**Isoelectronic species are those which have same number of electrons. (i) $F^{-}\left(10 e^{-}\right): O^{2-}$ (ii) $A r\left(18 e^{-}\right): C l^{-}$ (iii) $M g^{2+}\left(10 e^{-}\right): N a^{+}$ (iv) $R b^{+}\left(36 e^{-}\right): K r$

**[NCERT]**

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**Sol.**– No, the oxidation state of $A l$ is $+3$ and covalency is 6

**[NCERT]**

**(NCERT)**

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**Sol.**The basic theme of organisation of elements in the periodic table is to simplify and systematize the study of the properties of all the elements and millions of their compounds. On the basis of similarities in chemical properties, the various elements has now been divided into different groups. This has made the study simple because the properties of elements are now studied in form of groups rather than individually.

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**Sol.**In the third period the filling up of only 3

*s*– and 3

*p*– orbitals occur. Therefore, in this period there are only two

*s*– and six

*p*-block elements. Since third period starts with and ends at , therefore, elements with and are

*s*– block elements. The next six elements with to 18 are

*p*– block elements and belong to groups 13, 14, 15, 16, 17 and 18. Therefore, the elements which will lie in seventeenth group will have

*Z*= 114 ? So, it belongs to $7^{\text {th }}$ period and $14^{\text {th }}$ group.

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**Sol.**The electronic configuration of the element with $Z=114$ would be $[R n\} 5 f^{14} 6 d^{10} 7 s^{2} 7 p^{2} .$ since it has $n=7$ for the valence shell, it belong to the $7^{\text {th }}$ period. It receives the last electron in $p$ -orbital. therefore, it belongs to $p$ -block. The group number will be $10+4$ (No. of electrons in $n s \text { and } n p \text { orbitals })=14$ So, it belongs to $7^{\text {th }}$ period and $14^{\text {th }}$ group.

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**Sol.**The fifth period beings with the filling of 5

*s*orbital and continues till the filling of sixth energy level (6

*s*) starts. The sub-shell which follow 5

*s*are .Thus, the elements which involve filling of 5

*s*, 4

*d*and 5

*p*sub-shell are accommodated in the fifth period. These sub-shells have nine orbitals that can accept 18 electrons in all. Hence, there are 18 elements in the fifth period.

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**Sol.**In the sixth period the subshell filled are and . There is a total of 16 orbitals which can accommodate 32 electrons. Hence there are 32 elements in the sixth period.

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**Sol.**(i) All these ions have same number (10) of electrons. Therefore, there are also called isoelectronic species. (ii) Since the number of electrons are same, the ionic size decrease with increase in nuclear charge. Therefore, the ions can be arranged in increasing order of ionic radii as $A l^{3+}<M g^{2+}<N a^{+}<F^{-}<O^{2-}<N^{3-}$

*Z*) (iii) Nuclear mass (iv) Number of core electrons.

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**Sol.**Nuclear mass does not affect the valence shell because nucleus consists of protons and neutrons. Where as protons

*i.e.*, nuclear charge affects the valence shell but neutrons do not. Thus option (iii) is wrong.

*p*-block has six columns, because a maximum of 6 electrons can occupy all the orbitals in a

*p*-subshell. (ii) The

*d*-block has 8 columns, because a maximum of 8 electrons can occupy all the orbitals in a

*d*-subshell.

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**Sol.**Statement (ii) is incorrect while other statements are correct. The correct statement (ii) is : the

*d*-block has 10 columns, Statement (ii) is incorrect while other statements are correct. The correct statement (ii) is : the

*d*-block has 10 columns,

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**Sol.**Mendeleev Periodic Law states that the physical and chemical properties of the elements are a periodic function of their atomic weights while Modern Periodic Law states that physical and chemical properties of elements are a periodic function of their atomic number. Thus, the basic difference in approach between Mendeleev’s Periodic Law and Modern Periodic Law is the change in basis of classification of elements from atomic weight to atomic number.

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**Sol.**Within a group, the atomic radius increases down the group. This is because a new energy shell (

*i.e.*, principal quantum number increases by unity) is added at each succeeding element while the number of electrons in the valence shell remains to be the same. In other words, the electrons in the valence shell of each succeeding element lie farther and farther away from the nucleus. As a result, the force of attraction of the nucleus for the valence electrons decreases and hence the atomic size increases. In contract, the atomic size decreases as we move from left to right in a period. This is because that within a period the outer electrons remains in the same shell but the nuclear charge increases by one unit at each succeeding element. Due to this increased nuclear charge, the attraction of the nucleus for the outer electrons increases and hence the atomic size decreases.

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**Sol.**Arranging the elements into different groups and periods in order of their increasing atomic numbers, we have, We know that the metallic character increases down a group and decreases along a period as we move from left to right. Therefore,

*Na*is the most metallic element, followed by

*Mg*and

*Si*, while

*P*is the least metallic element. Among and is more metallic than

*Be*. Therefore, the overall increasing order of metallic character is $P<S i<B e<M g<N a$

*s*-,

*p*-,

*d*– and

*f*-block elements :

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**Sol.**(i) $s$ -Block elements: $n s^{1-2}$ where $n=2-7$ (ii) $\quad p$ -Block elements: $n s^{2} n p^{1-6}$ where $n=2-6$ (iii) $d$ -Block elements: $(n-1) d^{1-10} n s^{0-2}$ where $n=4-7$ (iv) $f$ -Block elements: $(n-2) f^{0-14}(n-1) d^{0-1} n s^{2}$ where $n=6-7$

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**Sol.**(i) $\quad n=3$ suggest that the elements belongs to third period. since the last electron enters the $p$ -orbital, therefore, the given element is a $p$ -block element. Further since the valence shell contains $6(2+4)$ electrons, therefore, group number of the elements $=10+$ no. of electrons in the valence shell $=10+6=16$ The complete electronic configuration of the element is $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{4}$ and the element is $S($ sulphur). (ii) $\quad n=4$ suggest that the element lies in the $4^{\text {th }}$ Period. since the $d$ -orbitals are incomplete, therefore, it is $d$ -block element. The group number of the element $=$ no. of $d$ -electrons $+$ no. of $s$ electrons $=2+2=4 .$ Thus the elements lies in group 4 and $4^{\text {th }}$ period. The complete electronic configuration of the element is $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{2} 4 s^{2}$ and the element in Ti (Titanium). (iii) $\quad n=6$ means that the element lies in the sixth period. since the last electron goes to the $f$ -orbital, therefore, the element is a $f$ -block element. All $f$ -block elements lie in group 3. The complete electronic configuration of the element is $[\mathrm{Xe}\} 4 f^{7} 5 d^{1} 6 s^{2} .$ The atomic number of the element $54+7+1+2=64$ and the element $G d($ gadolinium).

*Z*= 117 and 120 have not yet been discovered. In which family group would you place these elements and also give electronic configuration in each case

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**Sol.**The electronic configuration of the element with $Z=117$ would be $[R n] 4 f^{14} 5 d^{10} 7 s^{2} 7 p^{5} .$ It has outermost $n s^{2} n p^{5}$ configuration and therefore, it belongs to halogen family or group 17 The electronic configuration of the element with $Z=120$ would be [Uuo] $8 s^{2} .$ It has outermost $n s^{2}$ configuration and therefore, belongs to alkaline earth metals family or group 2.

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**Sol.**Elements which have a strong tendency to lose electrons to form cations are called metals while those which have a strong tendency to accept electrons to form anions are called non metals. Thus, metals are strong reducing agents, they have low ionization enthalpies, have less negative electron gain enthalpies, low electronegativity, form basic oxides and ionic compounds. Non-metals, on the other hand, are strong oxidising agents, they have high ionization enthalpies, have high negative electron gain enthalpies, high electronegativity, form acidic oxides and covalent compounds.

*Z*=107 and

*Z*=109 have been made recently; element

*Z*=108 has not yet been made. Indicate the group in which you will place the above elements.

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**Sol.**The electronic configuration of these elements are : $Z=107 \quad[R n] 5 f^{14} 6 d^{5} 7 s^{2}$ $Z=108 \quad[R n] 5 f^{14} 6 d^{6} 7 s^{2}$ $Z=109 \quad[R b] 5 f^{14} 6 d^{7} 7 s^{2}$ These elements will be placed in $d$ -block in group $7^{\text {th }}, 8^{\text {th }}$ and $9^{\text {th }}$ respectively.

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**Sol.**– Among these elements, $N a$ is an alkali metal and has only one electron in its valence shell $\left(3 s^{1}\right) .$ Therefore, its $I E_{1}$ is very low. After the removal of one electron, it acquires neon configuration $i . e ., \quad\left(1 s^{2} 2 s^{2} 2 p^{6}\right) .$ Therefore, its $I E_{2}$ is expected to be very high. Consequently the difference in first and second ionisation enthalpies would be greater in case of $N a$

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**Sol.**Energy of the electron in ground state $=-2.18 \times 10^{-18} \mathrm{J}$ Energy required for removal of the electron $=0-\left(-2.18 \times 10^{-18} \mathrm{J}\right)=2.18 \times 10^{-18} \mathrm{J}$ This is the energy required to remove the electron from one atom of hydrogen in ground state. The energy required to remove electron from one mole $\left(6.022 \times 10^{23} \text { atoms }\right)$ will be $=2.18 \times 10^{-18} \times 6.022 \times 10^{23}=13.12 \times 10^{5} \mathrm{J} \mathrm{mol}^{-1}$

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**Sol.**We know that the metallic character increases down a group and decreases along a period as we move from left to right. Hence the order of increasing metallic character is $P<S i<B e<M g<N a$

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**Sol.**(i) Group I elements are largest in size than alkaline earth metals (Group II elements), therefore, There is less force of attraction between nucleus and valence electron, that is why their ionization energy is lower. (ii) Anions are largest than neutral atom because electrons are more than the protons, therefore, effective nuclear charge is less, therefore, distance between centre of nucleus and valence electrons is more.

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**Sol.**It will be more close to $575 \mathrm{kJ} \mathrm{mol}^{-1}$ because the value of Al should be lower than that of $M g .$ This is due to effective shielding of $3 p$ electrons from the nucleus by $3 s$ -electrons.

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**Sol.**The ionic radius of a cation is always smaller than the parent atom because the loss of one or more electrons increases the effective nuclear charge. As a result, the force of attraction of nucleus for the electrons increases and hence the ionic radii decreases. In contrast, the ionic radius of an anion is always larger than its parent atom because the addition of one or more electrons decreases the effective nuclear charge. As a result, the force of attraction of the nucleus for the electrons decreases and hence the ionic radii increases.

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**Sol.**Both electrons gain enthalpy and electronegativity refer to the tendency of the atom of an element to attract electrons. Whereas electron gain enthalpy refers to the tendency of an isolated gaseous atom to accept an additional electron to form a negative ion, electronegativity refers to the tendency of the atom of an element to attract the shared pair of electrons towards it in a covalent bond.

*IE*of oxygen is less than that of nitrogen ?

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**Sol.**(i) It belong to group 8 of periodic table because its electronic configuration is $[\mathrm{Ar}] 4 \mathrm{s}^{2} 3 d^{6}$ (ii) It is because nitrogen has stable electronic configuration i.e., half-filled $p$ -orbitals $\therefore$ its I.E. is more than that of oxygen.

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**Sol.**Arrange the elements $N a, M g$ and $S i$ into different groups and periods in order of their increasing atomic numbers, we have, since in case of $A l,$ a $3 p$ -electron is to be lost while in $M g,$ a $2 s$ electron is to be lost, therefore, value for $A l$ will be lower than that of $M g\left(737 k J \mathrm{mol}^{-1}\right)$ because of the effective shielding of the $3 p-$ electron from the nucleus by 3 s-electrons. Therefore, $\Delta_{i} H$ for $A l$ will be more close to $575 \mathrm{kJ} \mathrm{mol}^{-1}$

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**Sol.**– The size of the isoelectronic ions depends upon the nuclear charge $(Z) .$ As the nuclear charge increase the size decreases. For example, $F^{-}(+9)>N e(+10)>N a^{+}(+11) .$ Therefore, statement (i) is correct while all other statements are wrong.

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**Sol.**The elements of group 1 have only one electron in their respective valence shell and thus a strong tendency to lose this electron. The tendency to lose electrons. in turn, depends upon the ionization enthalpy. Since the ionization enthalpy decreases down the group, therefore, the reactivity of group 1 elements increases in the same order : $L i<N a<K<R b<C s .$ In contrast, the elements of group 17. have seven electrons in their respective valence shells and thus have a strong tendency to accept one more electron. The tendency to accept electrons, in turn depends upon their electrode potentials. since the electrode potential of group 17 elements decreases in the order $F(+2.87 V)>C l(+1.36 V), B r(1.08 V)$ and $I(+0.53),$ therefore, their reactivities also decrease in the same order: $F>C l>B r>I$

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**Sol.**(i) $\quad M g$ and $A l$ belong the third period. Across a period, atomic radii decreases due to increased nuclear charge. Therefore, atomic size of Al is smaller than that of $M g .$ (ii) Further, cations are smaller than their parent atoms. Therefore, $M g^{2+}$ is smaller than $M g$ and $A l^{3+}$ is smaller than $A l$ (iii) and are isoelectronic ions. Among isoelectronic ions, higher the tye charge, smaller the size. Therefore, ionic radius of $A P^{3+}$ is smaller than that of From the above discussion, it follows that $M g$ has the largest while has the smallest size.

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**Sol.**(i) Ionization enthalpy is the minimum amount of energy required to move the most loosely bound electrons from an isolated gaseous atom so as to convert it into a gaseous cation is called ionization enthalpy. The force with which an electron is attracted by the nucleus of an atom is appreciably affected by presence of other atoms within its molecule or in the neighbourhood. Therefore, for the purpose of determination of ionization enthalpy, it is essential that these interatomic forces of attraction should be minimum. Since in the gaseous state, the atoms are widely separated, therefore, these interatomic forces are minimum. Further since it is not possible to isolate a single atom for the purpose of determination of its ionization enthalpy, therefore, the interatomic distances are further reduced by carrying out the measurement at a low pressure of the gaseous atom. Due to these reasons, the term isolated gaseous atoms has been included in the definition of ionization enthalpy. (ii) Electron gain enthalpy is the energy released when an isolated gaseous atom in the ground state accepts an extra electron to form the gaseous negative ion. The term ground state here means that the atom must be present in the most stable state,

*i.e.*, the ground state. The reason being that when the isolated gaseous atoms is in the excited state, lesser amount of energy will be released when it gets converted into gaseous anions after the accepting an electron. Therefore, for comparison purposes, the electron gains enthalpy of gaseous atoms be determined in their respective most stable state,

*i.e.*, ground state.

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**Sol.**(i) Electronic configuration of elements

*A*,

*B*,

*C*,

*D*and

*E*are as (ii) Element $A$ receives the last electron in $2 p-$ orbital, therefore, it belongs to $p$ -block elements and its group number $=10+\mathrm{No.}$ of electrons in the valence shell $=10+7=17 .$ Further the period of the element $=$ No. of the principal quantum number of the valence shell $=2^{\mathrm{nd}}$ Element $B$ receives the last electron in $3 s$ -orbital, therefore, it belongs to the $s$ -block elements and its group number $=$ No. of electrons in the valence shell $=2 .$ Further the period of the element $=$ No. of the principal quantum number of the valence shell $=3^{\text {rd }}$ Element $C$ receives the last electron in the $3 d$ -orbital, therefore, it belongs to $d$ block elements and its group number $=$ No. of electrons in the penultimate shell and valence shell $=10+1$ $=11 .$ Further, the period of the element $=$ No. of principal quantum number of the valence shell $=4^{\text {th }}$ Element D receives its last electron in the 5 p-orbital, therefore, it belongs to $p$ -block elements and its group number $=10+\mathrm{No.}$ of electrons in the valence shell $=10+8=18 .$ Further, the period of the element $=$ No. of the principal quantum number of valence shell $=5^{\text {th }}$ Element $E$ receives its last electron in the $4 f$ -orbital, therefore, it belongs to $f$ -block elements. It may be noted here that the filling of 4 f-orbital occurs only when one electron has already entered $5 d$ -orbital. Therefore, element $E-$ belongs to $f-$ block elements and not to $d$ -block elements. since it belongs to lanthanide series therefore, as such it does not have any group number of its own but is usually considered to lie in group $3 .$ However, its period $=$ No. of the principal quantum number of the valence shell $=6^{\text {th }}$ (iii) Elements $A$ and $B$ are representative elements since their last electron enters $p-$ and $s$ -orbital respectively. Elements $C$ is a transition element since it receives its last electron in the $d$ – orbital. Element $D$ is a $p$ -block element with completely filled $s$ – and $p$ -orbitals of the valence shell. Such a type of $p-$ block element is called a noble gas. Element $E$ is an inner transition element since it receives its last electron in the f-orbital.

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**Sol.**(i) Lithium belongs to group 1 with a valence of 1 while oxygen belongs to group 16 with a valence of $2 .$ Hence, the formula of the compound is $L i_{2} O$. (ii) Magnesium belongs to group 2 with a valence of 2 while nitrogen belongs to group 5 with a valence of 3. Hence, the formula of compound is $M g_{3} N_{2}$ (iii) Aluminium belongs to group 13 with a valence of 3 and iodine belongs to group 17 with a valence of $1 .$ Hence the formula of the compound is $A l l_{3}$ (iv) Silicon belongs to group 14 with valence of 4 and oxygen belongs to group 16 with a valence of $2 .$ Hence, the formula of the compound is $S i O_{2}$ (v) Phosphorus belongs to group 15 and it has valence 3 or $5,$ while fluorine belongs to group 17 with a valence of 1 Hence the formula of the compound may be $P F_{3}$ or $P F_{5} .$ (vi) Element with atomic number 71 is a lanthanoid called Lutetium $(L u),$ its common value is $3 .$ Fluorine is a group 17 (halogen) element with a valence of $1 .$ Therefore, the formula of the compound formed would be $L u F_{3}$ (Lutetium fluoride).

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**Sol.**(i)

**Gain of electron.**When a neutral atom gains one electron to form an anion, its radius increases. The reason being that the number of electrons in the anion increases while its nuclear charge remains the same as the parent atom. Since the same nuclear charge now attracts greater number of electrons, therefore, force of attraction of the nucleus on the electrons of all the shells decreases (

*i.e.*, effective nuclear charge decreases) and hence the electron cloud expands. In other words, the distance between the centre of the nucleus and the last shell that contains electrons increases thereby increasing the ionic radius. Thus, (ii)

**Loss of electrons.**When a neutral atom loses one electron to form a cation, its atomic radius decreases. The reason being that the number of electrons in the cation decreases while its nuclear charge remains the same as the parent atom. Since the same nuclear charge now attracts lesser number of electrons, therefore, the force of attraction of the nucleus on the electrons of all the shells-increases (

*i.e.,*effective nuclear charges increases) and hence the size of cation decreases. Thus,

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**Sol.**On moving the group 13 from $B$ to $A l,$ the ionization enthalpy decreases as expected due to an increases in atomic size and screening effect which outweigh the effect of increased nuclear charge. However, $\Delta_{i} H$ of $G a$ is only slightly higher $\left(2 k J m o l^{-1}\right)$ than that of $A l$ while that of $T i$ is much higher than those of $A l, G a$ and $I n .$ These deviations can be explained as follows: Al follows immediately after $s$ -block elements while $G a$ and In follows after $d$ -block elements and $T i$ after $d$ – and $f$ – block elements. These extra $d-$ and $f-$ electrons do not shield (or screen) the outer shell-electrons from the nucleus very effectively. As a result, the valence electrons remain more tightly held by the nucleus and hence larger amount of energy is needed for their removal. This explains why $G a$ has higher ionization enthalpy than $A l .$ Further on moving down the group from $G a$ to $I n,$ the increased shielding effect (due to presence of additional $4 d$ -electrons ) outweighs the effect of increased nuclear charge $(49-31=18$ units) and hence the $\Delta_{i} H_{1}$ of $\operatorname{In}$ is lower than that of the $G a .$ Thereafter, the effect of increased nuclear charge $(81-49=32$ units ) outweigh the shielding effect due to the presence of additional 4 $f$ and $5 d$ electrons and hence the of $T l$ is higher than that of In.

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**Sol.**(i) The element $V$ has highest first ionization enthalpy $\left(\Delta_{i} H_{1}\right)$ and positive electron gain enthalpy $\left(\Delta_{e g} H\right)$ and hence it is the least reactive element. since inert gases have positive $\Delta_{e g} H,$ therefore, the element-V must be inert gas. The values of $\Delta_{i} H_{1}, \Delta_{i} H_{2}$ and $\Delta_{e g} H,$ match that of $H e .$ (ii) The element II which has the least first ionization enthalpy and a low negative electron gain enthalpy is the most reactive metal. The values of and match that of $K$ (potassium). (iii) The element III which has high first ionization enthalpy and a very high negative electrons gain enthalpy is the most reactive non-metal. The values of and match that of (fluorine). (iv) The element IV has a high negative electron gain enthalpy but not so high first ionization enthalpy. Therefore, it is the least reactive non-metal. the values of and match that of $I$ (Iodine). (v) The element VIhas low first ionization enthalpy but higher than that of alkalimetals. Therefore, it appears that the element is an alkaline each metal and hence will form binary halide of the formula (where $X=$ halogen). The values of and match that of (magnesium). $M g$ (vi) The element I has low first ionization but a very high second ionization enthalpy, therefore, it must be an alkali metal. since the metal forms a predominary stable covalent halide of the formula $(X=$ halogen), therefore, the alkali metal must be least reactive. athe values of and match that of Li (lithium).

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**Sol.**Ions of different elements which have the same number of electrons but different magnitude of the nuclear charge are called isoelectronic ions. (i) $\quad F^{-}$ has $10(9+1)$ electrons. Therefore, the species nitride ion, $N^{3-}(7+3) ;$ oxide ion; $O^{2-}(8+2),$ neon, $N e(10+0)$ sodium ion, $N a^{+}(11-1) ;$ magnesium ion, $M g^{2+}(12-2)$ aluminium ion, $A l^{3+}(13-3)$ etc. each one of which contain 10 electrons, are isoelectronic with it. (ii) $\quad$ Ar has 18 electrons. Therefore, the species phosphate ion, $P^{3-}(15+3),$ sulphide ion; $S^{-2}(16+2) ;$ chloride ion; $\mathrm{Cl}^{-}(17+1),$ potassium ion, $K^{+}(19-1),$ Calcium ion, $\mathrm{Ca}^{2+}(20-2),$ etc. each one of which contain 18 electrons, are isoelectronic with it. (iii) $M g^{2+}$ has $10(12-2)$ electrons, therefore, the species $N^{3-}, O^{2-}, F^{-}, N e, N a^{+}, A l^{3+}$ etc. each one of which contains 10 electrons, are isoelectronic with it. (iv) $\quad R b^{+}$ has $36(37-1)$ electrons. Therefore, the species bromide ion, $B r^{-}(35+1),$ krypton, $K r(36+0)$ and strontium $S r^{2+}(38-2)$ each one of which has 36 electrons, are isoelectronic with it.

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**Sol.**(i) 6, 6 (ii) 26, 30 (iii) 38, 50 (iv) 92, 143.

*cm*if the zinc atoms are arranged side by side lengthwise.

**[NCERT]**

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**Sol.**Charge carried by one electron $=1.6022 \times 10^{-19} \mathrm{C}$ $\therefore \quad$ Electrons present in particle carrying $2.5 \times 10^{-16} \mathrm{C}$ charge $=\frac{2.5 \times 10^{-16}}{-1.6022 \times 10^{-19}}=1560$

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**Sol.**(i) $2 s$ is closer to the nucleus than 3 s. Hence 2 swill experience larger effective nuclear charge. (ii) $4 d$ (iii) $3 p$ (for same reasons).

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**Sol.**Silicon has greater nuclear charge $(+14)$ than aluminium $(+13) .$ Hence, the unpaired $3 p$ electron in case of silicon will experience more effective nuclear charge.

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**Sol.**The increasing order of frequency is : Radiation from

*FM*radio < radiation from microwave oven < amber light from traffic signal < X-rays < cosmic rays from outer space.

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**Sol.**$E_{n}=\frac{E_{1}}{n^{2}} ; E_{5}=\frac{-2.17 \times 10^{-18}}{(5)^{2}} \mathrm{J} /$ atom $=-8.68 \times 10^{-20} \mathrm{J} /$ atom.

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**Sol.**$E=N h v=N h \frac{c}{\lambda} \quad\left(\because v=\frac{c}{\lambda}\right)$ $=\frac{\left(5.6 \times 10^{24}\right)\left(6.626 \times 10^{-34} \mathrm{J} \mathrm{s}\right)\left(3.0 \times 10^{8} \mathrm{ms}^{-1}\right)}{\left(337.1 \times 10^{-9} \mathrm{m}\right)}$ $=3.3 \times 10^{6} J$

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**Sol.**Quantization of energy means the energy of energy levels can have some specific values of energy and not all the values.

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**Sol.**$\mathrm{E}_{\mathrm{n}}=-\frac{1312 \mathrm{kJ} \mathrm{mol}^{-1}}{\mathrm{n}^{2}}$

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**Sol.**(a) Subshells in $n=4$ are \[ 4 s(l=0), 4 p(l=1), 4 d(l=2), 4 f(l=3) \] The number of electrons having $m_{s}=-\frac{1}{2}$ in $n=4$ will be 16

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**Sol.**It means that 4

*d*sub-shell has 6 electrons. 4 represents fourth energy shell and

*d*is a sub-shell and 6 electrons are present in

*d*orbitals of sub-shell.

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**Sol.**According to Heisenberg’s Uncertainty principle ‘the exact position of electron cannot be determined, i.e., the exact path of electron cannot be determined but we can determine the region or space where electron spends most of its time, and this region or space is called orbital.

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**Sol.**The energy of photon is sufficient to disturb a subatomic particle so that there is uncertainty in the measurement of position and momentum of the subatomic particle. However, the energy is insufficient to disturb a macroscopic object.

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**Sol.**No, because, velocity = 0 and position can be measured accurately.

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**Sol.**(i) n (ii) n (iii) l (iv) $m_{l}$

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**Sol.**It measures the electron probability density at a point in an atom.

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**Sol.**(i) $n=4, m_{\mathrm{s}}=+1 / 2=16$ electrons (ii) $n=3, l=0=2$

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**Sol.**(i) $2 s$ (ii) $3 p$ (iii) both have same energy (iv) both have same energy (v) $5 \mathrm{s}$

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**Sol.**$\mathrm{n}=3,1=2, \mathrm{m}_{1}=+2,+1,0,-1,-2$

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**Sol.**$1=2,$ and $\mathrm{m}_{1}=+2,+1,0,-1,-2$

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**Sol.**Number of electrons : $H_{2}^{+}=1 ; H_{2}=2 ; \quad O_{2}^{+}=15$

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**Sol.**(i) $L i$ (ii) $P \quad$ (iii) $S c$

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**Sol.**An atom will cover length equal to its diameter Diameter of carbon atom $=\frac{2.4}{2 \times 10^{8}}$ $=1.2 \times 10^{-8} \mathrm{cm}=1.2 \times 10^{-10} \mathrm{m}$ Radius of carbon atom $=\frac{1.2 \times 10^{-10}}{2}=0.6 \times 10^{-10} \mathrm{m}$ $=0.06 \times 10^{-9} \mathrm{m}=0.06 \mathrm{nm}$

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**Sol.**For violet light $v=7.50 \times 10^{14} \mathrm{Hz}$ For red light $v=4.00 \times 10^{14} \mathrm{Hz}$

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**Sol.**since the ion carries 3 units of positive charge, it will have 3 electrons less than the number of protons. Let number of electrons $=x$ No. of protons $=x+3$ No. of neutrons $=x+\frac{x \times 30.4}{100}$ $=x+0.304 x=1.304 x$ Now, No. of protons $+$ No. of neutrons $=56$ $x+3+1.304 x=56$ $2.304 x=53$ $x=\frac{53}{2.304}=23$ No. of electrons $=23,$ No. of protons $=23+3=26$ No. of neutrons $=56-26=30$

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**Sol.**We know, Mass number $=$ No. of protons $+$ No. of neutrons $=81$ i.e., $p+n=81$ Let number of protons $=x$ Number of neutrons $=x+\frac{x \times 31.7}{100}=1.317 x$ $\therefore \quad x+1.317 x=81$ $x=\frac{81}{2.317}=34.96=35$ Symbol $=\begin{array}{l}{81} \\ {35}\end{array} B r$

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**Sol.**Since the ion carries one unit of negative charge, it means that in the ion the number of electrons is one more than the number of protons. Total number of electrons and neutrons in the ion = 37 + 1 = 38 Let the number of electrons in the ion be =

*x*Number of neutrons in the $=\frac{x \times 111.1}{100}=1.111 x$ $=x+1.111 x=2.111 x$ $2.111 x=38$ \[ x=\frac{38}{2.111}=18 \] $\therefore$ Number of electrons in the ion $=18$ Number of protons in the ion $=18-1=17$ Thus, atomic number of the elements is 17 which corresponds to chlorine. $\therefore$ Symbol of the ion is $_{17}^{37} \mathrm{Cl}^{-}$

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**Sol.**$K . E .=\frac{1}{2} m v^{2}=\frac{1}{2} m_{e} v^{2}=h\left(v-v_{0}\right)$ $=\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(1.0 \times 10^{15} \mathrm{s}^{-1}-7.0 \times 10^{14} \mathrm{s}^{-1}\right)$ $=\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(10 \times 10^{14} \mathrm{s}^{-1}-7.0 \times 10^{14} \mathrm{s}^{-1}\right)$ $=\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(3.0 \times 10^{14} \mathrm{s}^{-1}\right)=1.988 \times 10^{-19} \mathrm{J}$

*kHz*. Calculate the wavelength of the electromagnetic radiation emitted by the transmitter. Which part of the electromagnetic spectrum does it belong ?

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**Sol.**The wavelength, $\lambda,$ is equal to $c / v$ where $^{t} c^{\prime}$ is the speed of electromagnetic radiation in vacuum and ‘ $v^{\prime}$ is the frequency. Substituting the given values, we have, $\lambda=\frac{3.00 \times 10^{8} \mathrm{ms}^{-1}}{1368 \mathrm{kHz}}$ $=\frac{3.00 \times 10^{8} \mathrm{ms}^{-1}}{1368 \times 10^{3} \mathrm{s}^{-1}}=219.3 \mathrm{m}$ This is a characteristic radiowave wavelength.

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**Sol.**$v=3.29 \times 10^{15}\left(\frac{1}{3^{2}}-\frac{1}{n^{2}}\right) s^{-1}$ $\lambda=1285 \mathrm{nm}$ $\therefore \quad v=\frac{c}{\lambda}=\frac{3.0 \times 10^{8} \mathrm{ms}^{-1}}{1285 \times 10^{-9} \mathrm{m}}$ $=2.33 \times 10^{14} s^{-1}$ $2.33 \times 10^{14}=3.29 \times 10^{15}\left(\frac{1}{9}-\frac{1}{n^{2}}\right)$ or $\quad \frac{1}{9}-\frac{1}{n^{2}}=\frac{2.33 \times 10^{14}}{3.29 \times 10^{15}}=0.0708$ $|-\frac{1}{n^{2}}=0.0708-\frac{1}{9}=-0.0403$ $\frac{1}{n^{2}}=0.0403$ or $n^{2}=24.82$ $\therefore \quad n=5$

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**Sol.**$h v=h \theta_{\circ}+\frac{1}{2} m v^{2} \quad$ since $v=0$ $h v=h v_{o}+0$ $h v=h v_{0} \Rightarrow v=v_{0}$ $v=\frac{c}{\lambda} \Rightarrow v=\frac{3 \times 10^{8} \mathrm{ms}^{-1}}{6800 \times 10^{-10} \mathrm{m}}=\frac{300}{86} \times 10^{14}=4.41 \times 10^{14} \mathrm{s}^{-1}$ $v=v_{0}=4.41 \times 10^{14} \mathrm{s}^{-1}$ $W_{o}=h v_{o}=6.626 \times 10^{-34} \mathrm{Js} \times 4.41 \times 10^{14} \mathrm{s}^{-1}=2.91 \times 10^{-19} \mathrm{J}$

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**Sol.**Energy of electron in nth orbit is $E_{n}=-\frac{13.595}{n^{2}} Z^{2} e V$ atom $^{-1}$ For first orbit of $H e^{+}, Z=2, n=1$ $E_{1}=-\frac{13.595 \times(2)^{2}}{1^{2}}=-54.380 e V$ Radius of nth orbit is, $r_{n}=\frac{0.529 n^{2}}{Z} A$

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**Sol.**The longest wavelength corresponds to minimum energy $(\Delta E)$ transition. For Balmer series this transition is from $n_{2}=3$ to $n_{1}=2$ $v=109677\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right] \mathrm{cm}^{-1}$ $=109677\left[\frac{1}{(2)^{2}}-\frac{1}{(3)^{2}}\right] \mathrm{cm}^{-1}$ $=15233 \mathrm{cm}^{-1}=1.5233 \times 10^{4} \mathrm{cm}^{-1}$ or $\quad=1.5233 \times 10^{6} \mathrm{m}^{-1}$

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**Sol.**According to Bohr postulate of angular momentum,. \[ m v r=n \frac{h}{2 \pi} \quad \text { or } \quad 2 \pi r=n \frac{h}{m v} \quad \ldots .(\mathrm{i}) \] Substituting this value in equation (i), we get $2 \pi r=n \lambda$ Thus, the circumference $(2 \pi r)$ of the Bohr orbit for hydrogen atoms is an integral multiple of de-Broglie wavelength.

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**Sol.**We know that $\lambda=\frac{h}{m v}$ $m=9.1 \times 10^{-31} \mathrm{kg}, h=6.63 \times 10^{-34} \mathrm{kg} \mathrm{m}^{2} \mathrm{s}^{-1}$ $v=1 \%$ of speed of light $=\frac{1 \times 3.0 \times 10^{8}}{100} m s^{-1}$ $=3.0 \times 10^{6} \mathrm{ms}^{-1}\left(\because \text { speed of light }=3.0 \times 10^{8} \mathrm{ms}^{-1}\right)$

**[NCERT]**

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**Sol.**$K . E .=\frac{1}{2} m v^{2}$ $\therefore v=\sqrt{\frac{2 K \cdot E}{m}}=\sqrt{\frac{2 \times 3.0 \times 10^{-25} \mathrm{J}}{9.1 \times 10^{-31} \mathrm{kg}}=812} \mathrm{ms}^{-1}$ $\left(1, J=1 k g m^{2} s^{-2}\right)$ By de-Broglie equation, $\lambda=\frac{h}{m v}=\frac{6.626 \times 10^{-34} \mathrm{Js}}{\left(9.1 \times 10^{-31} \mathrm{kg}\right)\left(812 \mathrm{ms}^{-1}\right)}$

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**Sol.**(i) $n=4, l=0,1,2,3,(4 \text { subshells, viz, } s, p, d \text { and } f)$ (ii) No. of orbitals in 4 th shell $=n^{2}=4^{2}=16$ Each orbital has one electron with $m_{s}=-1 / 2 .$ Hence, there will be 16 electrons with $m_{s}=-1 / 2$

**[NCERT]**

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**Sol.**Mass of wagon $=2000 \mathrm{kg}$ Uncertainty in position, $\Delta x=\pm 10 \mathrm{m}$ According to Heisenberg uncertainty principle $\Delta x \times \Delta p=\frac{h}{4 \pi}$ or $\Delta x \times \Delta v=\frac{h}{4 \pi m}$ or $\quad \Delta v=\frac{h}{4 \pi m \Delta x}$

**[NCERT]**

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**Sol.**$\Delta x=0.002 n m=2 \times 10^{-3} n m=2 \times 10^{-12} m$ $\Delta x \times \Delta p=\frac{h}{4 \pi}$ $\therefore \quad \Delta p=\frac{h}{4 \pi \Delta x}=\frac{6.626 \times 10^{-34} \mathrm{kg} m^{2} s^{-1}}{4 \times 3.14 \times\left(2 \times 10^{-12} \mathrm{m}\right)}$ $=2.638 \times 10^{-23} \mathrm{kg} \mathrm{ms}^{-1}$ Actual momentum $=\frac{h}{4 \pi \times 0.05 \mathrm{nm}}=\frac{h}{4 \pi \times 5 \times 10^{-11} \mathrm{m}}$ It cannot be defined as the actual magnitude of the momentum is smaller than the uncertainty.

*s*,

*p*,

*d*notations, describe the orbital with the following quantum numbers :

**[NCERT]**

**[NCERT]**

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**Sol.**For $n=4, l$ can have values $0,1,2,3 .$ Thus, there are four sub-shells in $n=4$ energy level. These four sub-shells are $4 s, 4 p, 4 d$ and $4 f$

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**Sol.**i) $\quad_{15} P=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p_{x}^{1} 3 p_{y}^{1} 3 p_{z}^{1}$ No. of unpaired electron $=3$ (ii) $\quad_{14} S i=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p_{x}^{1} 3 p_{y}^{1}$ No. of unpaired electron $=2$ (iii) $\quad_{24} C r=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{5} 4 s^{1}$ No. of unpaired electrons $=6$ (iv) $\quad 26 F e=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{6} 4 s^{2}$ No. of unpaired electrons $=4(\text { in } 3 d)$ (v) $\quad 36 K r=$ Noble gas. All orbitals are filled. Unpaired electron $=0$

**[NCERT]**

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**Sol.**No. of protons $=$ Number of electrons $=29$ Electronic configuration $=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1} 3 d^{10}$

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**Sol.**(a) The element corresponding to ( i ) is Lithium (Li). (b) This electronic configuration represents the same element in the excited state. (c) By supplying energy to the element when the electron jumps from the lower energy $2 s$ -orbital to the higher energy $3 s$ orbital. (d) It is easier to remove an electron from (ii) than from (i) since in the former case the electron is present in a $3 s$ -orbital which is away from the nucleus and hence is less strongly attracted by the nucleus than an electron in the $2 s$ -orbital.

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**Sol.**(i) $1 s^{2} 2 s^{2} 2 p_{x}^{2} 2 p_{y}^{2} 2 p_{z}^{1} .$ Atomic number is 9 and it is configuration of fluorine in ground state. (ii) $1 s^{2} 2 s^{1} 2 p_{x}^{1} 2 p_{y}^{1} 2 p_{z}^{1} .$ Atomic number is 6 and it is configuration of carbon in excited state. (iii) $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p_{x}^{1} 3 p_{y}^{1} .$ Atomic number is 14 and it is configuration of silicon in ground state. (iv) $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{1} 3 p_{x}^{1} 3 p_{y}^{1} 3 p_{z}^{1} 3 d^{1} .$ Atomic number is 15 and it is configuration of phosphorus in excited state. (v) $[A r] 3 d^{5} 4 s^{2} .$ Atomic number is 25 and it is configuration of manganese in ground state.

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**Sol.**The electronic configuration of the atom is : $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2}$ (i) Atomic number $=2+2+6+2=12$ (ii) Number of principal quantum numbers $=3$ (iii) Number of sub-levels $=4(1 s, 2 s, 2 p, 3 s)$ (iv) Number of $s$ -orbitals $=3(1 s, 2 s, 3 s)$ (v) Total number of $p$ -electrons $=6$

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**Sol.**(i) It is because half filled and completely filled orbitals are more stable. (ii) $4 s$ orbital has lower energy than $3 d$ orbital, therefore, it is filled before $3 d$ orbital.

**[NCERT]**

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**Sol.**(i) 1 molecule of methane contains $=6+4=10$ electrons, $6.022 \times 10^{23}$ molecules of methane $(1 \mathrm{mol})$ contain electrons $=$ $10 \times 6.022 \times 10^{23}=6.022 \times 10^{24}$ (ii) One atom of $^{14} C$ contains $=8$ neutrons, $6.022 \times 10^{23}$ atoms of carbon weight $=14 \mathrm{g}$ $14 g$ of $C$ contain $=6.022 \times 10^{23}$ atoms $=6.022 \times 10^{23} \times 8$ neutrons \[ 7.0 \times 10^{-3} g \text { of } C \text { contains } \] $=\frac{6.022 \times 10^{23} \times 8 \times 7.0 \times 10^{-3}}{14}=2.409 \times 10^{21}$ Mass of 1 neutron $=1.675 \times 10^{-27} \mathrm{kg}$ Mass of $2.409 \times 10^{21}$ neutrons $=1.675 \times 10^{-27} \times 2.409 \times 10^{21}$ $=4.035 \times 10^{-6} \mathrm{kg}$ (iii) No. of protons in 1 molecule of $N H_{3}$ \[ =7+1+1+1=10 \text { protons } \] $17 g$ of $N H_{3}$ contains $=6.022 \times 10^{23}$ molecules $=6.022 \times 10^{23} \times 10$ protons $=6.022 \times 10^{24}$ protons $34 \times 10^{-3} g$ of $N H_{3}$ contains $=\frac{6.022 \times 10^{24} \times 34 \times 10^{-3}}{17}=1.204 \times 10^{22}$ Mass of 1 proton $=1.67 \times 10^{-27} \mathrm{kg}$ Mass of $1.204 \times 10^{22}$ protons $=1.67 \times 10^{-27} \times 1.204 \times 10^{22}$ $=2.01 \times 10^{-5} \mathrm{kg}$ No answer with not change by changing temp. and pres.,

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**Sol.**– Energy of a photon of radiation of wavelength 300 nm will be $E=h v=h \frac{c}{\lambda}=\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(3.0 \times 10^{8} \mathrm{ms}^{-1}\right)}{\left(300 \times 10^{-9} \mathrm{m}\right)}$ \[=6.626 \times 10^{-19} \mathrm{J}\] $\therefore$ Energy of 1 mole of photons $=\left(6.626 \times 10^{-19} \mathrm{J}\right) \times\left(6.022 \times 10^{23} \mathrm{mol}^{-1}\right)$ $=3.99 \times 10^{5} \mathrm{J} \mathrm{mol}^{-1}$ As $\quad E=E_{0}+K . E .$ of photoelectrons emitted. $\therefore$ Miniumu energy $\left(E_{0}\right)$ required to remove 1 mole of electrons from sodium $=E-K . E$ $=(3.99-1.68) 10^{5} J m o l^{-1}$ $=2.31 \times 10^{5} \mathrm{J} \mathrm{mol}^{-1}$ $\therefore$ Minimum energy required to remove one electron.

**[NCERT]**

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**Sol.**Suppose threshold wavelength $=\lambda_{0} n m=\lambda_{0} \times 10^{-9} \mathrm{m}$ Then $h\left(v-v_{0}\right)=\frac{1}{2} m v^{2} \quad$ or $\quad h c\left(\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right)=\frac{1}{2} m v^{2}$ Substituting the given results of the three experiments, we get $\frac{h c}{10^{-9}}\left(\frac{1}{500}-\frac{1}{\lambda_{0}}\right)=\frac{1}{2} m\left(2.55 \times 10^{6}\right)^{2}$ $\frac{h c}{10^{-9}}\left(\frac{1}{450}-\frac{1}{\lambda_{0}}\right)=\frac{1}{2} m\left(4.35 \times 10^{6}\right)^{2}$ $\frac{h c}{10^{-9}}\left(\frac{1}{400}-\frac{1}{\lambda_{0}}\right)=\frac{1}{2} m\left(5.20 \times 10^{6}\right)^{2}$ Dividing equation (ii) by equation (i), we get $\frac{\lambda_{0}-450}{450 \lambda_{0}} \times \frac{500 \lambda_{0}}{\lambda-500}=\left(\frac{4.35}{2.55}\right)^{2}$ or $\quad \frac{\lambda_{0}-450}{\lambda_{0}-500}=\frac{450}{500}\left(\frac{4.35}{2.55}\right)^{2}=2.619$ or $\lambda_{0}-450=2.619 \lambda-1309.5$ or $\quad 1.619 \lambda_{0}=859.5 \quad \therefore \quad \lambda_{0}=531 \mathrm{nm}$ Substituting this value in equation (iii), we get $\frac{h \times\left(3 \times 10^{8}\right)}{10^{-9}}\left(\frac{1}{400}-\frac{1}{531}\right)=\frac{1}{2}\left(9.11 \times 10^{-31}\right)\left(5.20 \times 10^{6}\right)^{2}$ or $\quad h=6.66 \times 10^{-34} \mathrm{Js}$

*nm*

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**Sol.**For a spectral line, $\frac{1}{\lambda}=R Z^{2}\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]$ For Lyman line, $\frac{1}{\lambda_{\mathrm{Lyman}}}=R Z^{2}\left[\frac{1}{1^{2}}-\frac{1}{2^{2}}\right]=\frac{3 R Z^{2}}{4}$ or $\quad \lambda_{L y \operatorname{man}}=\frac{4}{3 R Z^{2}}$ For Balmer line, $\frac{1}{\lambda_{\text {Balmer }}}=R z^{2}\left[\frac{1}{2^{2}}-\frac{1}{3^{2}}\right]=\frac{5 R Z^{2}}{36}$ or $\quad \lambda_{\text {Balmer }}=\frac{36}{5 R Z^{2}}$ $\lambda_{\text {Balmer }}-\lambda_{\text {Lyman }}=\frac{36}{5 R Z^{2}}-\frac{4}{3 R Z^{2}}=59.3 \times 10^{-7}$ $=\frac{1}{R Z^{2}}\left[\frac{36}{5}-\frac{4}{3}\right]=59.3 \times 10^{-7}$ $=\frac{1}{109677.8 Z^{2}}\left[\frac{108-20}{15}\right]=59.3 \times 10^{-7}$ or $\quad Z^{2}=\frac{1}{109677.8} \times \frac{88}{15 \times 59.3 \times 10^{-7}}=9$ or $\quad Z=3 .$ This corresponds to $L i^{2+}$ ion.

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**Sol.**Velocity of electron is given as : $v=\frac{2 \pi e^{2}}{n h}$ Now, $e=4.8 \times 10^{-10}$ e.s.u., $n=3, h=6.63 \times 10^{-27}$ erg sec $\therefore \quad v=\frac{2 \times 22 \times\left(4.8 \times 10^{-10}\right)^{2}}{7 \times 3 \times 6.63 \times 10^{-27}}=7.27 \times 10^{7} \mathrm{cm} \mathrm{s}^{-1}$ No. of revolutions per second $=\frac{\text { Velocity }}{\text { Circumference }}=\frac{v}{2 \pi r}$ but $r=\frac{n^{2} h^{2}}{4 \pi^{2} m e^{2}}$ $\therefore$ No. of revolutions $=\frac{2 \pi e^{2}}{n h} \times \frac{4 \pi^{2} m e^{2}}{2 \pi n^{2} h^{2}}=\frac{4 \pi^{2} m e^{4}}{n^{3} h^{3}}$ $=\frac{4 \times\left(\frac{22}{7}\right)^{2} \times\left(9.1 \times 10^{-28}\right) \times\left(4.8 \times 10^{-10}\right)^{4}}{3^{3} \times\left(6.62 \times 10^{-27}\right)^{3}}$ $=2.42 \times 10^{14}$

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**Sol.**Angular momentum $(m v r)=n \frac{h}{2 \pi}$ $=4.22 \times 10^{-34} \mathrm{kg} \mathrm{m}^{2} \mathrm{s}^{-1}$ (Given) $\therefore \quad n=4.22 \times 10^{-34} \times \frac{2 \pi}{h}$ $=\frac{2 \times 4.22 \times 10^{-34} \times 3.14}{6.626 \times 10^{-34}}=4$ When the electron jumps from $n=4$ to $n=3,$ the wavelength of the spectral line can be calculated as follows: $\frac{1}{\lambda}=R_{H}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$ $=109,677 \mathrm{cm}^{-1}\left(\frac{1}{3^{2}}-\frac{1}{4^{2}}\right)$ $=109,677 \times\left(\frac{1}{9}-\frac{1}{16}\right)=109,677 \times \frac{7}{144} \mathrm{cm}^{-1}$ or $\lambda=\frac{144}{109,677 \times 7} \mathrm{cm}=1.88 \times 10^{-4} \mathrm{cm}$

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**Sol.**(i) $\quad C u^{+}(29) 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{0} 3 d^{10}$ (ii) $\quad m=1 m g \quad v=1 \% \times 3 \times 10^{8} m s^{-1}$ $m=10^{-6} \mathrm{kg} \quad v=\frac{1}{100} \times 3 \times 10^{8} \mathrm{ms}^{-1}=3 \times 10^{6} \mathrm{ms}^{-1}$ $\lambda=\frac{h}{m v}=\frac{6.63 \times 10^{-34} \mathrm{kg} m^{2} s^{-1}}{10^{-6} \mathrm{kg} \times 3 \times 10^{6} \mathrm{ms}^{-1}}=2.21 \times 10^{-34} \mathrm{m}$

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**Sol.**During the formation of cations, electrons are lost while in the formation of anions, electrons are added to the valence shell. The number of electrons added or lost is equal to the numerical value of the charge present on the ion. Following this general concept, we can write the electronic configurations of all the ions given in the question. (i) $\quad C u^{2+}=_{29} C u-2 e^{-}$ $=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{1}-2 e^{-}$ $=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{9}$ (ii) $\quad C r^{3+}=_{24} C r-3 e^{-}$ $=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{5} 4 s^{1}-3 e^{-}$ $=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{3}$ (iii) $\quad F e^{2+}=_{26} F e-2 e^{-}$ $=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{6} 4 s^{2}-2 e^{-}$ $=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{6}$ $=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{6}$ $F e^{3+}=26 F e-3 e^{-}$ $=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{5}$ (iv) $\quad H^{-}=_{1} H+1 e^{-}=1 s^{1}+1 e^{1}=1 s^{2}$ (v) $\quad S^{2-}=_{16} S+2 e^{-}$ $=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p_{x}^{2} 3 p_{y}^{1} 3 p_{z}^{1}+2 e^{-}$ $=1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p_{x}^{2} 3 p_{y}^{2} 3 p_{z}^{2}$

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**[NCERT]**

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**Sol.**The S.I. unit of mass is kilogram. Mass is defined as the amount of matter present in substance. The unit of mass kilogram is defined as equal to the mass of the international prototype of the kilogram.

*S.I.*base units using power of 10 notation. (example $\left.2.54 \mathrm{mm}=2.54 \times 10^{-3} \mathrm{m}\right)$ (i) $1.35 \mathrm{mm}$ (ii) 1 day (iii) $6.45 \mathrm{mL}$ (iv) $48 \mu g$ (microgram) (v) 0.0426 inches

**[NCERT]**

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**Sol.**(i) $1.35 m m=1.35 \times 10^{-3} \mathrm{m}$ (ii) 1 day $=24 \times 60 \times 60 s=86400 s=8.64 \times 10^{4} s$ (iii) $6.45 \times 10^{-6} \mathrm{m}^{3}\left[\because 1 \mathrm{mL}=10^{-6} \mathrm{m}^{3}\right]$ (iv) $48 \mu g=48 \times 10^{-6} g=4.8 \times 10^{-5} g$ 0.0426 inches (v) $\quad$ linch $=2.54 \times 10^{-2} \mathrm{m}$ \[ =2.54 \times 10^{-2} \mathrm{m} \times 0.0426=1.082 \times 10^{-3} \mathrm{m} \]

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**Sol.**These observations are in agreement with law of conservation of mass.

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**Sol.**

**Avogadro’s Law :**Equal volume of all gases under similar conditions of temperature and pressure contain equal number of molecules.

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**Sol.**$\mathrm{CO}_{2}, \quad \mathrm{CH}_{4}$

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**Sol.**The limiting reactant is that reactant which is present in minimum amount as required by the stoichiometry of the balanced equation. Since the limiting reactant is completely consumed during the reaction, all the stoichiometric calculations are done keeping the amount of limiting reactant in mind.

**[NCERT]**

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**Sol.**According to equation mole ratio of $H C l: M n O_{2}=4: 1$ and mass ratio is $36.5 \times 4: 87=146: 87$ $\therefore \quad H C l$ required in $g=\frac{146}{87} \times 5=8.39 \mathrm{g}$

**[NCERT]**

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**Sol.**Molarity of solution $\left(\mathrm{mol} L^{-1}\right)$ $=\frac{\text { Mass of solute }(g)}{\mathrm{M} . \text { mass }} \times \frac{1000}{V \mathrm{inmL}}$ Conc. of sugar $=\frac{20}{342} \times \frac{1000}{2000}=0.0292 \mathrm{mol} \mathrm{L}^{-1}$

**[NCERT]**

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**Sol.**$X_{\text {ethonol }}=0.040 ; X_{H_{2} O}=0.960$ Molality ( $m)$ $=\frac{1000 \times X_{\text {eth }}}{X_{H_{2} O} \times G M M_{\left(H_{2} O\right)}}=\frac{1000 \times 0.040}{0.96 \times 18}=2.315 m$

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**Sol.**$N_{2}(g)+3 H_{2}(g) \longrightarrow 2 N H_{3}(g)$ $6.023 \times 10^{23}$ molecules of $N_{2}$ react completely with $H_{2}$ to save $2 \times 6.023 \times 10^{23}$ molecules $=1.204 \times 10^{24}$ molecules.

**[NCERT]**

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**Sol.**Pressure is the force (i.e., weight) acting per unit area

**[NCERT]**

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**Sol.**$816 \times 0.02456=20.0$ Product rounded off to 3 significant figures because the least number of significant figures in this multiplication is three.

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**Sol.**(i) 5.608 (ii) 32.39 (iii) $1.790 \times 10^{3}$ (iv) $7.837 \times 10^{-3}$

**(i) $6.022 \times 10^{23}$**

**(ii) 5.359**

**(iii) 0.04597**

**(iv) 34.216**

**[NCERT]**

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**Sol.**(i) The last digit to be retained is 2 and the digit to be dropped is 2 which is less than $5 .$ The result will be expressed as $6.02 \times 10^{23}$ (ii) In this case the last digit to be retained is 5 and the digit to be dropped is $9,$ which is greater than $5 .$ Hence, the last digit to be retained is increased by one. The number will be written as 5.36. (iii) The three digits to be retained in this case are $4,5$ and $9 .$ The digit 7 is to be dropped which is greater than $5 .$ Hence, the last digit to be retained will therefore be increased by one. The number will be rounded off to three significant digits as 0.0460 (iv) In this case the digits 1 and 6 will be dropped. since, the digit following the last digit to be retained is $1,$ the last digit will be kept unchanged and the number is written as $34.2 .$

**[NCERT]**

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**Sol.**No. of atoms can be calculated as: (i) $1 g A u=\frac{1}{197} \times 6.022 \times 10^{23}=\frac{6.022}{197} \times 10^{23}$ (ii) $\quad 1 g N a=\frac{6.022}{23} \times 10^{23}$ (iii) $1 g L i=\frac{6.022 \times 10^{23}}{7}$ (iv) $1 g C l_{2}=\frac{2 \times 6.022 \times 10^{23}}{71}=\frac{6.022 \times 10^{23}}{35.5}$ It is clear that contains largest number of atoms.

**[NCERT]**

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**Sol.**(i)

**Average atomic mass :**It is defined as average of the mass of all of the atoms of an elements, e.g., average atomic mass of is 35.5

*u*. (ii)

**Mole**is defined as amount of substance that contains as many atoms, molecules and particles as there are atoms in exactly 0.012

*kg*of Carbon-12 isotope. (iii) Molar mass : It is mass of 1 mole of substance which contains $6.022 \times 10^{23}$ particles. (iv)

**Unit factor :**The factor which is used to convert one unit into another is called Unit factor. (v)

**Molarity :**It is defined as number of moles of solute dissolved per litre of solution. (vi)

**Precision and accuracy :**Precision means how closely the experimental measurements agree with another.

**Accuracy**means how close the experimental measurements and exact values are with each other.

**[NCERT]**

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**Sol.**$3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g)$ 2 moles of $H N O_{3}$ need 3 moles of $N O_{2}$ 7.33 mole of $\mathrm{HNO}_{3}$ need $\frac{3}{2} \times 7.33=\frac{21.99}{2}=10.995$ moles of $\mathrm{NO}_{2}$

**[NCERT]**

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**Sol.**Average atomic mass

**[NCERT]**

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**Sol.**$M=n \times \frac{1000}{\text { Vol. of solution in } \mathrm{cm}^{3}} \Rightarrow 0.50=n \times \frac{1000}{250}$ $\Rightarrow n=0.125 \mathrm{mol}$ Amount of $N a C l=n \times M \cdot w t .=0.125 \times 58.5=7.30 \mathrm{g}$

**[NCERT]**

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**Sol.**$32.9 \mathrm{g}$ of $\mathrm{K}$ combines with $67.1 \mathrm{g}$ of $\mathrm{Br}$ $3.60 \mathrm{g}$ of $\mathrm{K}$ combines with $\frac{67.1}{32.9} \times 3.60=\frac{241.56}{32.9}=7.342 \mathrm{g}$ since we have $6.4 \mathrm{g}$ of bromine, it means $B r_{2}$ is limiting reactant. Number of moles of $B r_{2}=\frac{6.4 \mathrm{g}}{160}=0.04$ moles $2 K+B r_{2} \longrightarrow 2 K B r$ mole of $B r_{2}$ reacts with 2 moles of Potassium to form $K B r$ 0.04 mole of $B r_{2}$ will react with $2 \times 0.04$ moles of $K$ to form $K B r$ $=0.08$ moles $=8 \times 10^{-2}$ moles of $K$

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**Sol.**(i) Number of atoms of $^{\prime} A r^{\prime}=52 \times 6.02 \times 10^{23}=3.132 \times 10^{25}$ atoms. (ii) 1 atom of He has a mass of 4 u. Number of atoms of ‘ $^{\prime} H e^{\prime}=\frac{52}{4}=13$ atoms. (iii) Number of moles of ‘ $\mathrm{He}^{\prime}=\frac{52}{4}=13$ moles $\therefore$ Number of atoms of $^{\prime} H e^{\prime}=13 \times 6.023 \times 10^{23}=7.826 \times 10^{24}$ atoms.

*g*of this complex.

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**Sol.**– Mass of magnesium in $5.00 \mathrm{g}$ of complex $=\frac{2.68}{100} \times 5.00=0.134 g$ No. of moles of $M g=\frac{0.134}{24}$ atoms of $M g=6.023 \times 10^{23} \times \frac{0.134}{24}=3.36 \times 10^{22}$ atoms

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**Sol.**(i) 1 mole of $C_{2} H_{6}$ contains 2 moles of carbon $\therefore \quad$ Number of moles of carbon in 3 moles of $C_{2} H_{6}=6$ (ii) 1 mole of $C_{2} H_{6}$ contain 6 mole atoms of hydrogen $\therefore \quad$ Number of moles of hydrogen atoms in 3 mole of $C_{2} H_{6}=3 \times 6=18$ (iii) 1 mole of $C_{2} H_{6}=6.022 \times 10^{23}$ molecules $\therefore \quad$ Number of molecules in 3 mole of $C_{2} H_{6}=3 \times 6.022 \times 10^{23}=1.807 \times 10^{24}$ molecules

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**Sol.**$-0.375 M$ aqueous solution means that $1000 \mathrm{mL}$ of the solution contain sodium acetate $=0.375$ mole $\therefore \quad 500 \quad \mathrm{mL}$ of the solution should contain sodium acetate $=\frac{0.375}{2}$ mole Molar mass of sodium acetate $=82.0245 \mathrm{g} \mathrm{mol}^{-1}$ $\therefore\ Mass of sodium acetate acquired $=\frac{0.375}{2}$ mole $\times 82.0245 \mathrm{g} \mathrm{mol}^{-1}=15.380 \mathrm{g}$

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**Sol.**I mole of $\mathrm{CuSO}_{4}$ contains 1 mole $(1 \mathrm{gatom})$ of $^{c} \mathrm{Cu}^{\prime}$ $\mathrm{CuSO}_{4}=63.5+32+4 \times 16=159.5 \mathrm{gmol}^{-1}$ Thus, $\mathrm{Cu}$ that can be obtained from $159.5 \mathrm{g}$ of $\mathrm{CuSO}_{4}=63.5 \mathrm{g}$ $\therefore \quad \mathrm{Cu}$ that can be obtained from $100 \mathrm{g}$ of $\mathrm{CuSO}_{4}=\frac{63.5}{159.5} \times 100 \mathrm{g}=39.81 \mathrm{g}$ Molar mass of

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**Sol.**(i) Molecular mass of $N_{2}=14 \times 2=28$ $28 g$ of $N_{2}$ occupy at S.T.P. $=22.4$ litres $\therefore 14 g$ of $N_{2}$ occupy at S.T.P. $=\frac{22.4}{28} \times 14=11.2$ litre. (ii) $6.023 \times 10^{23}$ molecules of $\mathrm{NH}_{3}$ occupy at S.T.P. $=22.4$ litres. $\therefore 6.023 \times 10^{22}$ molecules of $\mathrm{NH}_{3}$ occupy at S.T.P. $=\frac{22.4}{6.023 \times 10^{23}} \times 6.023 \times 10^{22}=2.24 L$ (iii) 1 mole of $S O_{2}$ occupies at S.T.P. $=22.4$ litres $\therefore 0.1 \mathrm{mol}$ of $\mathrm{SO}_{2}$ occupies at S.T.P. $=\frac{22.4}{1} \times 0.1=2.24 L$

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**Sol.**The formula of the compound is $F e_{2}\left(S O_{4}\right)_{3}$ The formula mass $=2 \times 56+3(32+64)=400 u$ $\therefore \quad$ Percentage of iron $=\frac{2 \times 56 \times 100}{400}=28 \%$ \[\begin{array}{l} {\text { Percentage of sulphur }=\frac{3 \times 32 \times 100}{400}=24 \%} \\ {\text { Percentage of oxygen }=\frac{3 \times 64 \times 100}{400}=48 \%} \end{array}\] [esquestion] Calculate the mass percent of different elements present in sodium sulphate $\left(N a_{2} S O_{4}\right)$. #tag#

**[NCERT]**

**[NCERT]**

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**Sol.**Hence, molecular formula is same as empirical formula, viz.. $\mathrm{Fe}_{2} \mathrm{O}_{3}$

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**Sol.**$Z n(s)+2 H C l(a q) \longrightarrow Z n C l_{2}(a q)+H_{2}(g)$ 1 mole of $Z n$ reacts with 2 moles of $H C l$ 0.3 mole of $Z n$ will react with $2 \times 0.30=0.6$ moles of $\mathrm{HCl}$ But we have only 0.52 moles of $H C l,$ therefore, $H C l$ is limiting reactant. 2 moles of $\mathrm{HCl}$ react with 1 mole of $\mathrm{Zn}$ to form 1 mole of $\mathrm{H}_{2}$. 0.52 moles of $\mathrm{HCl}$ react with $\frac{0.52}{2}$ moles of $\mathrm{Zn}$ to form $\frac{0.52}{2}=0.26$ moles of $H_{2}$

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**Sol.**According to the equation, 1 mole of Areacts with 1 mole of $B_{2}$ and 1 atom of Areacts with 1 molecule of $B_{2}$ (i) $\quad B$ is limiting reagent because 200 molecules of $B_{2}$ will react with 200 atoms of $A$ and 100 atoms will be left in excess. (ii) $\quad A$ (iii) Both will react completely because it is stoichiometric mixture. No limiting reagent. (iv) $\quad B$ (v) $A$

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**Sol.**$M=\frac{W_{B}}{M_{B}} \times \frac{1000}{\text { Vol. of solution in } \mathrm{cm}^{3}}=\frac{18.25}{40} \times \frac{1000}{200}$ $=2.28 \mathrm{mol} L^{-1}$

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**Sol.**Amount of $M g$ present in chlorophyll $=$ Mass of chlorophyll $\times \%$ of $M g$ $=2 g \times \frac{2.68}{100}=0.0536 \mathrm{g}$ of $\mathrm{Mg}$ $24 g$ of $M g$ contains $6.023 \times 10^{23}$ atoms $0.0536 g$ of $M g$ contains $\frac{6.023 \times 10^{23}}{24} \times 0.0536=1.345 \times 10^{21}$ atoms

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**Sol.**Molecular weight of $\mathrm{NaNO}_{3}=23+14+3 \times 16=23+14+48=85 g \mathrm{mol}^{-1}$ $M=\frac{W_{B}}{M_{B}} \times \frac{1000}{\text { Volume of solution in } m L}$ $=\frac{0.83}{85} \times \frac{1000}{50}=\frac{830}{4250}=\frac{83}{425}=0.1952 M$

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**Sol.**$M=\frac{W_{B}}{M_{B}} \times \frac{1000}{\text { Volume of solution }}$ $0.15=\frac{0.184}{40} \times \frac{1000}{\text { Volume of solution }}$ Volume of $\mathrm{NaOH}$ solution $=\frac{184}{40 \times 0.15}=\frac{184}{6}=30.66 \mathrm{mL}$

*gm*dihydrogen gas to yield ammonia ? Also calculate the amount of ammonia formed.

**[NCERT]**

**[NCERT]**

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**Sol.**Amount of methanol required $m(g)=\frac{\text { Molarity } \times \mathrm{M} . \text { Mass } \times V_{i n} m L}{1000}$ $=\frac{0.25 \times 32 \times 2500}{1000}=20 g$ since, density $=\frac{\text { Mass }}{\text { Volume }}$ $\therefore$ Volume of $\mathrm{CH}_{3} \mathrm{OH}$ required $=\frac{\text { Mass }}{\text { density }}=\frac{20}{0.793}=25.22 \mathrm{mL}$

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**Sol.**(i) Moles of $\mathrm{NaOH}$ in $50.0 \mathrm{mL}$ solution $=\frac{38}{40}$ $\therefore$ moles of $\mathrm{NaOH}$ in $1.0 \mathrm{L}$ solution \[ =\frac{38}{40} \times \frac{1000}{50}=19 M \] (ii) $1.0 L$ of solution contains moles $=0.15$ $27.0 \mathrm{mL}$ of solution contains moles $=\frac{0.15}{1000} \times 27=4.05 \times 10^{-3} \mathrm{moles}$

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**Sol.**(i) 15 ppm means 15 parts by mass of $\mathrm{CHCl}_{3}$ in $10^{6}$ parts by mass of water Hence, percent by mass \[ =\frac{15}{10^{6}} \times 100=1.5 \times 10^{-3} \% \] (ii) Moles of chloroform $=\frac{15}{119.5}=0.1255$ Hence, molality $(m)$ of $\mathrm{CHCl}_{3}=\frac{0.1255}{10^{6}} \times 10^{3}=1.25 \times 10^{-4} \mathrm{m}$

**[At. Wt. of $B a=137 u, S=3 u, O=16 u]$**

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**Sol.**$B a C l_{2}(a q)+N a_{2} S O_{4}(a q) \longrightarrow B a S O_{4}(s)+2 N a C l(a q)$ Number of moles of $N a_{2} \mathrm{SO}_{4}$ $=\left(\frac{500 m L}{1000 m L}\right) \times 0.25 m o l$ $=\frac{0.25}{2}=0.125 \mathrm{mol} \mathrm{Na}_{2} \mathrm{SO}_{4}$ Molar mass of $B a C l_{2}=208.2 \mathrm{g}$ Number of moles of $B a C l_{2}$ $=\frac{15}{208.2}=0.072 \mathrm{mol}$ of $\mathrm{BaCl}_{2}$ 0.072 moles of $B a C l_{2}$ reacts with 0.072 moles of $B a C l_{2}$ to form 0.072 mole of $B a S O_{4}$ Mass of $B a S O_{4}$ formed $=0.072 \times 233.4=16.80 g$ of $B a S O_{4}$

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**Sol.**(i) $\quad M=\frac{W_{B}}{M_{B}} \times \frac{1000}{\text { Volume of solution in } m L}$ \[ =\frac{0.38}{40} \times \frac{1000}{50}=0.19 \mathrm{mol} L^{-1} \] (ii) Law of multiple proportion states whenever two elements reacts to form two or more compounds, the ratio between different weights of one of the elements which combine with fixed weight of another is always simple, e.g., Nitrogen reacts with oxygen to form $\mathrm{NO}$ and $\mathrm{NO}_{2} \cdot$ In $\mathrm{NO}, 14 \mathrm{g}$ of $N$ reacts with $16 \mathrm{g}$ of $\mathrm{O}$ In $\mathrm{NO}_{2} 14 \mathrm{g}$ of $N$ reacts with $32 \mathrm{g}$ of $\mathrm{O} .$ The ratio between weights of oxygen which combine with fixed weight of nitrogen is $16: 32,$ i.e., 1: 2

*S.I.*units : (i) 93 million miles (this is the distance between the earth and the sun). (ii) 5 feet 2 inches (this is the average height of an Indian female). (iii) 100 miles per hour (this is the typical speed of Rajdhani Express). (v) $46^{\circ} \mathrm{C}$ (this is the peak summer temperature in Delhi). (vi) 150 pounds (this is the average weight of an Indian male).

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**Sol.**(i) The S.I. unit of distance is metre ( $m$ ) \[ \begin{array}{l} {1 \text { mile }=1.60 \text { kilometre }=1.60 \times 1000 \mathrm{m}} \\ {\text { Unit factor }=\frac{1.60 \times 1000(\mathrm{m})}{1 \mathrm{mile}}=\frac{1.6 \times 10^{3}(\mathrm{m})}{1 \mathrm{mile}}} \end{array} \] 93 millionmiles $=\frac{93 \times 10^{6}(\text { miles }) \times 1.6 \times 10^{3}(\mathrm{m})}{1 \text { mile }}$ $\therefore=93 \times 1.6 \times 10^{9} \mathrm{m}=148.8 \times 10^{9} \mathrm{m}=1.49 \times 10^{11} \mathrm{m}$

**(ii) 5 feet 2 inches $=62$ inches**\[ \begin{array}{l} {1 \text { inch }=2.54 \times 10^{-2} \mathrm{m}} \\ {\text { Unit factor }=\frac{2.54 \times 10^{-2}(\mathrm{m})}{1 \mathrm{inch}}} \\ {62 \text { inches }=\frac{62(\text { inches }) \times 2.54 \times 10^{-2} \mathrm{m}}{1 \mathrm{inch}}} \\ {=62 \times 2.54 \times 10^{-2} \mathrm{m}} \\ {=157.48 \times 10^{-2} \mathrm{m}=1.57 \mathrm{m}} \end{array} \] (iii) 1 mile $=1.60 \mathrm{km}=1.60 \times 10^{3} \mathrm{m}$ \[ \begin{array}{l} {\text { Unit factor }=\frac{1.60 \times 10^{3} \mathrm{m}}{1 \mathrm{mile}}} \\ {1 \mathrm{hr}=60 \times 60 \mathrm{s}=3.6 \times 10^{3} \mathrm{s}} \\ {\text { Unit factor }=\frac{3.6 \times 10^{3} \mathrm{s}}{1 \mathrm{hr}}} \end{array} \] $\therefore \quad$ Speed $=\frac{100 \text { miles }}{h r}$ $=\frac{100 \text { miles }}{h r} \times \frac{1.60 \times 10^{3} \mathrm{m}}{1 \mathrm{mile}} \times \frac{1 \mathrm{hr}}{3.6 \times 10^{3} \mathrm{s}}=44 \mathrm{ms}^{-1}$ (iv) $1 \dot{A}=10^{-10} \mathrm{m}$ \[ \text { Unit factor }=\frac{10^{-10} \mathrm{m}}{1(\hat{A})} \] $\therefore \quad 0.74 \mathrm{A}=\frac{0.74 \hat{A} \times 10^{-10}(m)}{1(\hat{A})}$ \[ =0.74 \times 10^{-10} \mathrm{m} \text { or }=7.4 \times 10^{-11} \mathrm{m} \] (v) $\quad 0^{\circ} \mathrm{C}=273.15 \mathrm{K}$ $\quad 46^{\circ} \mathrm{C}=273.15 \mathrm{K}+46 \mathrm{K}=319.15 \mathrm{K}$ (vi) 1 pound $=454 \times 10^{-3} \mathrm{kg}$ \[ \text { Unit factor }=\frac{454 \times 10^{-3}(k g)}{1(\text { pound })} \] $\begin{aligned} \therefore \quad 150 \text { pound }=\frac{150(\text { pound }) \times 454 \times 10^{-3}(k g)}{1(\text { pound })} & \\=& 150 \times 454 \times 10^{-3} \mathrm{kg}=68.1 \mathrm{kg} \end{aligned}$

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**Sol.**(i) Conversion factor $=\frac{\left(10^{-3} \mathrm{m}^{3}\right)}{(1 . L)}$ \[ 25 L=(25 L) \times \frac{\left(10^{-3} m^{3}\right)}{(1 L)}=2.5 \times 10^{-2} m^{3} \] (ii) Conversion factor= $=\frac{(1000 m g)}{(1 g)} \times \frac{(10 d L)}{(1 L)}$ \[ \begin{array}{l} {25 g L^{-1}=(25 g) \times \frac{\left(10^{3} m g\right)}{(1 g)} \times\left(1 L^{-1}\right) \times \frac{\left(10 d L^{-1}\right)}{\left(1 L^{-1}\right)}} \\ {=2.5 \times 10^{3} \mathrm{mg} d L^{-1}} \end{array} \] (iii) Conversion factor $=\frac{\left(10^{9} p m\right)}{(1 m m)} \times \frac{\left(10^{-6} \mu s\right)}{(1 s)}$ \[ \begin{array}{l} {1.54 \mathrm{mm} \mathrm{s}^{-1}=(1.54 \mathrm{mm}) \times \frac{\left(10^{9} \mathrm{pm}\right)}{(1 \mathrm{mm})}\left(1 \mathrm{s}^{-1}\right) \times \frac{\left(10^{-6} \mu \mathrm{s}^{-1}\right)}{\left(1 \mathrm{s}^{-1}\right)}} \\ {=1.54 \times 10^{3} \mathrm{pm} / \mathrm{s}^{-1}} \end{array} \] (iv) Conversion factor $=\frac{\left(10^{6} \mu g\right)}{(1 g)} \times \frac{\left(10^{4} \mu m\right)}{(1 \mathrm{cm})}$ $2.66 g \mathrm{cm}^{-3}=(2.66 g) \times \frac{\left(10^{6} \mu g\right)}{(1 g)} \times\left(\mathrm{cm}^{-3}\right) \frac{\left(10^{4} \mu m\right)^{-3}}{\left(1 \mathrm{cm}^{-3}\right)}$ $=2.66 \times\left(10^{6} \mu g\right) \times\left(10^{-12} \mu m^{-3}\right)$ \[ =2.66 \times 10^{-6} \mu g \mu m^{-3} \] (v) Conversion factor $=\frac{10^{3} m L}{(1 L)} \times \frac{(60 \times 60 \mathrm{s})}{1 h}$ $\quad 4.2 \mathrm{Lh}^{-2}=(4.2 \mathrm{L}) \times \frac{\left(10^{3} \mathrm{mL}\right)}{(1 \mathrm{L})} \times\left(h^{-2}\right) \times \frac{(60 \times 60 \mathrm{s})^{-2}}{\left(h^{-2}\right)}$ $=3.24 \times 10^{-4} \mathrm{mL} s^{-2}$

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**Sol.**(a) Fixing the mass of dinitrogen as 28

*g*, masses of dioxygen combined will be 32, 64, 32 and 80

*g*in the given four oxides. These are in the ratio 1 : 2 : 1 : 5 which is a simple number ratio. Hence, the given data obey the law of multiple proportion.

**(i) $11 \mathrm{g}$ of $\mathrm{CO}_{2}$**

**(ii) $3.01 \times 10^{22}$ molecules of $\mathrm{CO}_{2}$**

**(iii) 1.12 litre of $\mathrm{CO}_{2}$ at $\mathrm{S.T.P.}$**

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**Sol.**(i) Molecular mass of $C O_{2}$ \[ =(12+2 \times 16)=44 \] $44 \mathrm{g}$ of $\mathrm{CO}_{2}=1 \mathrm{mole}$ of $\mathrm{CO}_{2}$ $11 \mathrm{g}$ of $\mathrm{CO}_{2}=\frac{1}{44} \times 11=0.25 \mathrm{mol}$ (ii) $6.02 \times 10^{23}$ molecules of $\mathrm{CO}_{2}$ $=1$ mole of $C O_{2}$ $3.01 \times 10^{22}$ molecules of $\mathrm{CO}_{2}$ $=\frac{1}{6.02 \times 10^{23}} \times 3.01 \times 10^{22}=0.05 \mathrm{mol}$ (iii) 22.4 litres of $\mathrm{CO}_{2}$ at S.T.P. $=1$ mole of $C O_{2}$ 1.12 litres of $\mathrm{CO}_{2}$ at S.T.P. $=\frac{1}{22.4} \times 1.12=0.05 \mathrm{mol}$

*C*,

*H*,

*O*elements. A 4.24

*mg*sample of butyric acid is completely burnt in oxygen. It gives 8.45

*mg*of carbon dioxide and 3.46

*mg*of water. What is the mass percentage of each element ? Determine the empirical and molecular formula of butyric acid if molecular mass of butyric acid is determined to be 88

*u*.

**[NCERT]**

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**Sol.**Mass of carbon present in $8.45 \mathrm{mg}$ of $\mathrm{CO}_{2}$ $=\frac{8.45 \times 12}{44} m g=2.30 m g$ Percentage of carbon $=\frac{2.30 \times 100}{4.24}=54.24 \%$ Mass of hydrogen in $3.46 \mathrm{mg}$ of $\mathrm{H}_{2} \mathrm{O}$ $=\frac{3.46 \times 2}{18} m g=0.384 m g$ Percentage of hydrogen $=\frac{0.384 \times 100}{4.24}=9.05 \%$ Percentage of oxygen $=100-54.24-9.05=36.71 \%$

*g*of carbon dioxide and 0.690

*g*of water and no other products. 10.0

*L*of this welding gas at

*S.T.P.*is found to weigh 11.6

*g*. Calculate : (i) empirical formula, (ii) molar mass, and (iii) molecular formula of welding gas.

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**Sol.**

**(i) Calculation of empirical formula:**Mass of carbon in 3.38 g carbon dioxide $=\frac{12 \times 3.38}{44}=0.922 g$ Mass of hydrogen in 0.690

*g*of water $=\frac{2 \times 0.690}{18}=0.077 g$ The ratio by mass of $C$ and $H$ in the sample $=0.922: 0.077$ The mole ratio of $C$ and $H$ in the sample $=\frac{0.922}{12}: \frac{0.077}{1}=0.077: 0.077=1: 1$ Thus empirical formula of the gas is $C H$ (ii) Calculation of molar mass: $10.0 \mathrm{L}$ of gas at S.T.P. weighs $=11.6 \mathrm{g}$ $22.4 \mathrm{L}$ of gas at S.T.P. weighs $=\frac{11.6 \times 22.4}{10}=25.98 g$ Thus, molar mass of gas $=25.98 \mathrm{g} \mathrm{mol}^{-1}$ (iii) Calculation of molecular formula: $n=\frac{\text { Molecular mass }}{\text { Empirical formula mass }}=\frac{25.98}{13}=1.998 \approx 2$ Thus, molecular formula is $(C H)_{2}=C_{2} H_{2}$

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**Sol.**From the data given it is clear that if we have $100.0 \mathrm{g}$ of naphthalene, there are $93.71 \mathrm{g}$ of $\mathrm{C}$ and $6.29 \mathrm{g}$ of $\mathrm{H.Number}$ of moles of $\mathrm{C}$ and H in $100.0 \mathrm{g}$ sample of naphthalene are $93.71 g C\left(\frac{1 \mathrm{mol} C}{12.0 \mathrm{g} C}\right)=7.80 \mathrm{mol} \mathrm{C}$ $6.29 g H\left(\frac{1 \mathrm{mol} H}{1.00 g H}\right)=6.29 \mathrm{mol} \mathrm{H}$ Mole ratio $=\left(\frac{7.80 \mathrm{mol} \mathrm{C}}{6.29 \mathrm{mol} \mathrm{H}}\right)=\left(\frac{1.24 \mathrm{mol} \mathrm{C}}{1.00 \mathrm{mol} \mathrm{H}}\right)$ Changing the decimal fraction into whole number ratio of $C$ and $H,$ we know $1.24 \cong \frac{5}{4}$ Therefore, the ratio of $C$ to $H$ is Mole ratio $=\frac{5 / 4 \mathrm{mol} \mathrm{C}}{1 \mathrm{mol} \mathrm{H}}=\frac{5 \mathrm{mol} \mathrm{C}}{4 \mathrm{mol} \mathrm{H}}$ The ratio of carbon to hydrogen atoms in naphthalene is 5: 4 giving the empirical formula, $C_{5} H_{4} .$ The ratio of molecular mass to the empirical mass gives the factor with which the empirical formula is multiplied to obtain the molecular formula. $\frac{128 g / m o l \text { of napthalene }}{64.0 g / m o l \text { of } C_{5} H_{4}}=2.00$ Thus, molecular formula of naphthalene is given by empirical formula multiplied by $2,$ i.e. $\left(C_{5} H_{4}\right)_{2}$ or $C_{10} H_{8}$

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**Sol.**(i) Calculation of the weight of calcium bicarbonate present. Wt. of calcium bicarbonate present in 100 litres of well water $=16.2 \mathrm{g}$ Wt. of calcium bicarbonate present in $60,000$ litres of \[ \text { well water }=\frac{16.2}{100} \times 60000 g=9720 g \] (ii) Calculation of the quantity of lime required. The equation involved is:

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**Sol.**$10 \%$ (w/w) solution of glucose means that $10 g$ of glucose is present in $100 g$ of solution or in $90 g$ of water. (i) Calculation of molality \[ \begin{array}{l} {\text { Mass of glucose }=10 g} \\ {\text { Moles of glucose }=\frac{10}{180}=0.0556} \end{array} \] (Molar mass of glucose $=180$ ) Mass of water $=90 g$ Molality $=\frac{\text { Moles of glucose }}{\text { Mass of water }} \times 1000$ \[ =\frac{0.0556}{90} \times 1000=0.618 m \] (ii) Calculation of molarity Moles of glucose $=0.0556$ Volume of solution $=\frac{\text { Mass }}{\text { Density }}=\frac{100}{1.20}=83.3 \mathrm{mL}$ Molarity $=\frac{\text { Moles of glucose }}{\text { Vol. of solution }} \times 1000$ $=\frac{0.0556}{83.3} \times 1000=0.667 M$ (iii) Calculation of mole fraction of components \[ \begin{array}{l} {\text { Moles of glucose }=0.0556} \\ {\text { Moles of water }=\frac{90}{18}=5.0} \\ {\text { Total moles }=5.0+0.0556=5.0556} \end{array} \] Mole fraction of glucose $=\frac{0.0556}{5.0556}=0.011$ Mole fraction of water $=\frac{5.0}{5.0556}=0.989$

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**Sol.**Carbon dioxide is not poisonous.

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**Sol.**Solid carbon dioxide gas is called dry ice, solid carbon dioxide can sublime to the gaseous state without passing through the liquid state and therefore, it is called dry ice.

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**Sol.**The hybridisation of carbon in (i) $\mathrm{CO}_{3}^{2-}$ is $s p^{2}$ (ii) in diamond is $s p^{3}$ and in (iii) graphite is $\mathrm{sp}^{2}$

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**Sol.**Isotope of carbon used for radio carbon dating is 14 C. Isotope of carbon used for international standard for atomic mass is $12 \mathrm{C}$.

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**Sol.**Allotropy. “The existence of an element in two or more different forms which have different physical properties but same chemical properties, is called allotropy and the forms are called allotropes.” The allotropic forms of carbon are: (i) Crystallined forms – Diamond, graphite. (ii) Non-Crystallined forms – Charcoal, lamp-black.

**[NCERT]**

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**Sol.**Diamond is very hard crystalline solid because each carbon is linked to four other carbon atoms tetrahedrally by using $s p^{3}$ hybrid orbitals giving rise to a rigid three dimensional network of carbon atoms. It does not conduct electricity because it does not have free electrons. Graphite has structure in which each carbon is bonded to three other carbon atoms to form a hexagonal sheet. Each ‘CI atoms forms three sigma bonds by means of $s p^{2}$ hybrid orbitals. The unhybridised $p$ -orbitals of each carbon atoms having electrons overlaps with $p$ -orbitals of other carbon atoms to form $\pi$ -bonds. The hexagonal layers are held together by weak Vander Wall’s force of attraction, that is why it is soft and slippery and used as lubricant. It is conductor of electricity because -electrons within layer are free to move and this accounts for electrical conductivity.

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**Sol.**(i) 2 (ii) 2 (iii) 4 (iv) –1

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**Sol.**– Carbon cannot lose electrons to form $C^{4+}$ because the ionization energy required is very high. It cannot gain electrons to form $C^{4-}$ because it is energetically not favourable. Hence $C$ forms covalent compounds. Down the group, the $I . E .$ decreases. $P b$ being the last element has so low, I.E. that it can lose electrons to form ionic compounds.

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**Sol.**(i) Calcium carbide and Carbon monoxide are formed. \[ \mathrm{CaO}+3 \mathrm{C} \stackrel{\text { heat }}{\longrightarrow} \mathrm{CaC}_{2}+\mathrm{CO} \] (ii) Phosgene gas is formed. \[ \mathrm{CO}+\mathrm{Cl}_{2} \longrightarrow \mathrm{COCl}_{2} \] [Phosgene gas (Carbonyl Chloride)] (iii) They form glucose and starch.

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**Sol.**Compounds of electropositive elements with carbon are called carbides. There are three types of Carbides: (i) Ionic Carbide : Most electropositive metals form ionic carbide, e.g., calcium carbide. (ii) Covalent Carbide : Less electropositive metals and nonmetals form covalent carbides, e.g., silicon carbide, boron carbides. (iii) Interstitial Carbides : Those carbides which are formed by transition metals in which carbon atoms occupy interstitial voids, $e . g ., W C(\text { Tungsten Carbide })$

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**Sol.**(i) Carbon has the maximum tendency for catenation due to stronger $C-C \quad\left(353.5 \mathrm{kJ} \mathrm{mol}^{-1}\right)$ and hence it forms large number of alkanes. silicon, on the other hand, bonds has much lesser tendency for catenation due to weak $S i-S i \quad\left(200 k J \mathrm{mol}^{-1}\right)$ and hence it forms only few silanes (ii) In presence of conc. $H N O_{3}, A l$ becomes passive, due to a thin protective layer of its oxide $\left(A l_{2} O_{3}\right)$ which is formed on its surface which prevents the further action between the metal and the acid. Therefore $A l$ containers can be used for storing conc. However, it cannot be stored in zinc vessels because zinc reacts with $\mathrm{HNO}_{3}$. \[ \mathrm{Zn}+4 \mathrm{HNO}_{3} \longrightarrow \mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}+2 \mathrm{NO}_{2}+2 \mathrm{H}_{2} \mathrm{O} \] (iii) Heat evolved during burning of $M g$ is enough to decompose $N O$ to $\mathrm{N}_{2}$ and $\mathrm{O}_{2} .$ The $\mathrm{O}_{2}$ thus, produced support combustion of $M g .$ In contrast, the heat produced during burning of $S$ is not quite sufficient to decompose $N O .$ As a result, sulphur stops burning. (iv) The alkaline nature of aqueous solution is due to the hydrolysis of these salts. $\underset{\text { Bicarbonates }}{H C O_{3}^{-}}+H_{2} O \rightleftharpoons H_{2} C O_{3}+\overline{O H}$ \[ \underset{\text { Cattonates }}{C O_{3}^{2-}}+H_{2} O \rightleftharpoons H C O_{3}^{-}+\overline{O H} \]

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**Sol.**(i) Carbon is able to form $p \pi-p \pi$ bond with $O$ atom and constitute a stable non-polar molecule $O=C=O .$ Due to weak interparticle forces its boiling point is low and it is a gas at room temperature. Si$,$ on the other hand is not able to form bond with $O$ atoms because of its relatively large size. In order to complete it octet sillicon is linked to four $O$ atoms around it by sigma bonds and thus it constitutes network structure which is responsible for its solid state. (ii) In $\mathrm{SiCl}_{4},$ the silicon atom has empty 3 dorbitals in its valence shell which can accept an electron pair from $H_{2} O$ molecule resulting the co-ordinate bond. The co-ordinated molecule loses a molecule of $H C l$ and as a result, a chlorine atom gets replaced by an $O H$ group as shown. In the same way, all the four $C l$ atoms are replaced by $O H$ group and the molecule gets completely hydrolyse. Carbon atom, on the other hand does not have vacant $d$ -orbitals in 2nd level. Hence, it cannot accept electron pair form $H_{2} O$ molecule and therefore, $\mathrm{CCl}_{4}$ is not hydrolysed.

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**Sol.**Electronic configuration of $\mathrm{p}$ -block elements is $n s^{2} n p^{1-6}$

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**Sol.**The element having highest melting point is carbon (diamond).

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**Sol.**Amorphous and Crystalline.

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**Sol.**Tincal is an ore of boron. Its formula is $\mathrm{Na}_{2} \mathrm{B}_{4} \mathrm{O}_{10} \cdot 10 \mathrm{H}_{2} \mathrm{O}$

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**Sol.**Boron has very small size and has very high sum of three ionisation enthalpies $\left(I E+I E_{2}+I E_{3}\right)$ therefore it cannot lose its three electrons to form $B^{3+}$ ions.

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**Sol.**

**$2 \mathrm{BCl}_{3}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \frac{\mathrm{sun}}{_{1270 \mathrm{K}}} 2 \mathrm{B}(\mathrm{s})+6 \mathrm{HCl}(\mathrm{g})$**

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**Sol.**$B F_{3}$ is weaker lewis acid than $B C l_{3}$ because of more effecttive back bonding in case of $F$ due to smaller size than $C l$

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**Sol.**It is because in boric acid, boron does not have its octet complete. It accepts $O H^{-}$ from water.

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**Sol.**Boron does not have vacant d-orbitals as second shell is the outer most shell.

**[NCERT]**

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**Sol.**$B(O H)_{3}$ is a weak monobasic acid. It acts as a lewis acid by accepting electrons from a hydroxy ion in water. $B(O H)_{3}+2 H O H \longrightarrow\left[B(O H)_{4}\right]^{-}+H_{3} O^{+}$

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**Sol.**Sodium borohydride is formed $2 \mathrm{NaH}+\mathrm{B}_{2} \mathrm{H}_{6} \stackrel{\text { heat }}{\longrightarrow} 2 \mathrm{NaBH}_{4}$

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**Sol.**Pyrex glass is a glass which is heat resistant. It can with stand high temperature.

**[NCERT]**

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**Sol.**Boron can be obtained by reduction of Boron oxide with electropositive metal like

*Mg.*

**$B_{2} O_{3}(s)+3 M g(g) \longrightarrow 2 B(s)+3 M g O(s)$**

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**Sol.**Boric acid is $H_{3} B O_{3}$ Oxidation state of boron $=+3$ Boric acid is tribasic.

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**Sol.**Boron does not have vacant $d$ -orbitals in its valence shell. Therefore, it cannot expand its octet. Therefore, maximum covalency of boron cannot exceed and it does not form $B F_{6}^{3-}$ ion.

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**Sol.**Boron halides are electron deficient, therefore, they can form addition compounds with electron rich amines $e . g .$ $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N} \longrightarrow \mathrm{BF}_{3}$

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**Sol.**When boric acid is heated at $370 K .$ metaboric acid $\left(H B O_{2}\right)$ is formed. On heating strongly $B_{2} O_{3}$ is formed. (i) $\mathrm{H}_{3} \mathrm{BO}_{3} \stackrel{370 \mathrm{K}}{\longrightarrow} \mathrm{HBO}_{2}+\mathrm{H}_{2} \mathrm{O}$ (ii) $2 H B O_{2} \stackrel{\text { heat }}{\longrightarrow} B_{2} O_{3}+H_{2} O$

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**Sol.**$\mathrm{Na}_{2} \mathrm{B}_{4} \mathrm{O}_{7} \cdot 10 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{HCl}+5 \mathrm{H}_{2} \mathrm{O} \longrightarrow 2 \mathrm{NaCl}$ $+4 H_{2} B Q_{3}+12 H_{3} Q$

**[NCERT]**

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**Sol.**$=B-C l$ bond has dipole moment due to difference in electronegativity of $B$ and chlorine atom. $B C l_{3}$ on the contrary has a zero dipole moment because dipole moment is a vector quantity. The net resultant of the three bonds is zero due to its symmetrical planer triangular structure.

**[NCERT]**

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**Sol.**– In $B F_{3}$ the hybridisation of boron is $s p^{2}$ and hence the molecule has a planar triangular shape. In $B H_{4}^{-}$ the hybridisation of boron is $s p^{3}$ and hence it has a tetrahedral shape.

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**Sol.**Electron deficient compounds are those which have incomplete octet. $B C l_{3}$ is an electron deficient compound as it has sextet of electrons. On the contrary $\mathrm{SiCl}_{4}$ is not an electron deficient compound. However, due to the presence of vacant $d$ -orbitals in its valence shell it can extend its co-ordination number and form species like $S i F_{3}^{-}, S i F_{6}^{2-},$ etc.

**[NCERT]**

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**Sol.**– The boron family exhibits two oxidation states of $+3$ and $+1$ in their compounds on moving down the group from boron to thallium, the stability of lower oxidation state increases $i . e .,+1$ oxidation state becomes more prominent due to ” inert pair effect?

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**Sol.**$B C l_{3}$ has more stability than $T I C l_{3} .$ It is because boron due to its small size can exhibit only one oxidation state of $+3$ in its compounds. On the contrary. Thallium exhibits two oxidation states of $+3$ and $+1$ in its compounds. The lower oxidation state of $+1$ is more stable because of inert pair effect. To add further the increase in the magnitude of nuclear charge is not compensated by the increase in size as a result of which the $n s$ -electrons do not participate in bonding.

**[NCERT]**

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**Sol.**It is obtained by reacting $A I C I_{3}$ with $L i H$ in the presence of dry ether

**Uses :**It is a very good reducing agent used in many organic synthesis.

**[NCERT]**

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**Sol.**$\mathrm{GaCl}_{3}+\mathrm{NaOH} \longrightarrow \mathrm{Ga}(\mathrm{OH})_{3}+3 \mathrm{NaCl}$ $\mathrm{Ga}(\mathrm{OH})_{3}+\mathrm{NaOH} \longrightarrow \underset{\text { Souble Complex }}{\mathrm{NaGaO}_{2}}+2 \mathrm{H}_{2} \mathrm{O}$

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**Sol.**Uses of boron compounds : (i) The main use of borax and boric acid is in the preparation of heat resistant borosilicate glass such as pyrex glass. (ii) Borax is used as flux for soldering of metals and procelain enamels. (iii) Dilute aqueous solution of boric acid is used as mild antiseptic. (iv) Boron rods are used as control rods in nuclear reactors.

**[NCERT]**

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**Sol.**(i) $\quad 2 B+3 C l_{2} \stackrel{\text { heat }}{\longrightarrow} 2 B C l_{3}$ (ii) $\quad 4 B+3 O_{2} \stackrel{\text { heat }}{\longrightarrow} 2 B_{2} O_{3}$ (iii) $\quad 2 B+N_{2} \stackrel{\text { heat }}{\longrightarrow} 2 B N$

**[NCERT]**

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**Sol.**$B-F$ bond length in $B F_{3}$ is 130 pm because boron in is hybridised and has a planar triangular structure. As a result of deficiency of electrons on the boron atom the excessive electron density on the fluorine atoms (due to its small size and strong interelectronic repulsions) is pushed on to the boron atom $i . e .$ back donation or back bonding occurs. This introduces a partially double bond character shortening the bond. However, in the boron atom is hybridised with bond length of $143 \mathrm{pm}$

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**Sol.**Both $B F_{3}$ and $B B r_{3}$ are electron deficient molecules due to presence of only six electrons around boron atom. Therefore, these molecules act as Lewis acids due to their tendency to accept a pair of electrons. In , the 2 p-orbital of adjoining $F$ atom containing lone pair of electrons form effective bonding with empty $2 \mathrm{p}$ -orbital of boron atom. As a result, the lone pair of $F$ is donated to $B$ atom and hence the electron-deficiency ofboron decreases. In contrast in, the size of $B r$ atom is much bigger than the $F$ atom, hence donation of lone pair of electrons of $B r$ to $B$ does not occur to any significant extent. As a result, the electron-deficiency of $B$ is much higher in $\mathrm{BBr}_{3}$ than that in $\mathrm{BF}_{3},$ and hence $\mathrm{BBr}_{3}$ is a stronger Lewis acid than $\mathrm{BF}_{3}$

**[NCERT]**

**or**What is the action of heat on boric acid ?

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**Sol.**– Boric acid is $H_{3} B O_{3} .$ It can also be represented by $B(O H)_{3} .$ It has layer type structure in which $B O_{3}$ units are linked to one another through $H$ -bonds. The $H$ -atom constitutes covalent bond with one unit and $H$ -bond with other unit. Boric acid is a weak monobasic acid, it does not act as proton donor but as Lewis acid. $\mathrm{B}(\mathrm{OH})_{3}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{B}(\mathrm{OH})_{4}^{-}+\mathrm{H}^{+}$ Preparation (i) By the action of $\mathrm{SO}_{2}$ on boiling suspension of colemanite in water. \[ \mathrm{Ca}_{2} \mathrm{B}_{6} \mathrm{O}_{11}+2 \mathrm{SO}_{2}+9 \mathrm{H}_{2} \mathrm{O} \longrightarrow 2 \mathrm{CaSO}_{3}+6 \mathrm{H}_{3} \mathrm{BO}_{3} \] (ii) By acidifying aqueous solution of borax with mineral acids. \[ \mathrm{Na}_{2} \mathrm{B}_{4} \mathrm{O}_{7}+2 \mathrm{HCl}+5 \mathrm{H}_{2} \mathrm{O} \longrightarrow 2 \mathrm{NaCl}+4 \mathrm{H}_{3} \mathrm{BO}_{3} \] Effect of heat. Heating orthoboric acid above $370 K$ leads to partial removal of water to yield metaboric acid, $\left(\mathrm{HBO}_{2}\right) ;$ further heating yields.

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**Sol.**It is the point at which solid, liquid and vapour of a substance are in equilibrium with each other. For example, triple point of water is the temperature at which ice, water and its vapour coexist.

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**Sol.**HF molecules are associated with intermolecular $H-$ bonding, therefore, it is liquid whereas $H C l$ is gas because less Vander Waal’s forces of attraction.

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**Sol.**$(i)<100^{\circ} C(\text { ii })>100^{\circ} C$

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**Sol.**Heavier air will come down and lighter air goes up. Air at lower level is denser since it is compressed by mass of air above it.

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**Sol.**Solid $\mathrm{CO}_{2}$ It is because solid $\mathrm{CO}_{2}$ is directly converted into gaseous state (sublimes) and does not change into liquid, so it is called dry ice.

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**Sol.**Balloon will expand because rate of diffusion of $H_{2}$ is greater than that of $D_{2}$

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**Sol.**A straight line parallel to pressure axis.

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**Sol.**Charle’s Law.

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**Sol.**$\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}$ or $P_{2}=\frac{P_{1} V_{1} T_{2}}{T_{1} V_{2}}$ But $d \propto \frac{1}{V} .$ Hence $P_{2}=\frac{P_{1} T_{2}}{T_{1}}\left(\frac{d_{2}}{d_{1}}\right)$ $=\frac{760 \times 263}{273}\left(\frac{1}{10}\right)=73.2 \mathrm{mm}$

**[NCERT**]

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**Sol.**Number of years $=\frac{6.023 \times 10^{23}}{10^{10} \times 365 \times 24 \times 60 \times 60}=1,908,00$ years.

**[NCERT]**

**[NCERT]**

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**Sol.**$\frac{p V^{2} T^{2}}{n}=\frac{\left(N m^{-2}\right)\left(m^{3}\right)^{2}(K)^{2}}{m o l}$ $=N m^{4} K^{2} m o l^{-1}$

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**Sol.**$R=8.314 \mathrm{JK}^{-1} \mathrm{mol}^{-1}$

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**Sol.**$^{c} a^{\prime}$ is a measure of the magnitude of the intermolecular forces of attraction while $b$ is a measure of the effective size of the gas molecules.

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**Sol.**The air pressure decreases with increases in altitude. That is why jet aeroplane flying at high altitude need pressurization of the cabin so that partial pressure of oxygen is sufficient for breathing.

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**Sol.**Both have same molar mass $\left(=44 g m o l^{-1}\right)$. According to Graham’s law of diffusion, rates of diffusion of different gases are inversely proportional to the square root of their molar masses under same conditions of temperature and pressure.

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**Sol.**At $30^{\circ} \mathrm{C},$ kinetic energy depends only on absolute temperature and not on the identity of a gas.

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**Sol.**(i) Inter molecular forces between molecules are negligible. (ii) Molecules of a gas have negligible volumes.

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**Sol.**(i) High compressibility is due to large empty spaces between the molecules. (ii) Due to absence of attractive forces between molecules, the molecules of gases can easily separate from one another.

**[NCERT]**

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**Sol.**Higher the critical temperature, more easily the gas can be liquefied i.e. greater are the intermolecular forces of attraction. Therefore, $\mathrm{CO}_{2}$ has stronger intermolecular forces than $\mathrm{CH}_{4}$

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**Sol.**$(\mathrm{i}) \mathrm{pCH}_{4}=\frac{n R T}{\mathrm{V}}=\frac{3.2}{16} \times \frac{0.0821 \mathrm{L} \operatorname{atm} \mathrm{K}^{-1} \mathrm{mol}^{-1} \times 300 \mathrm{K}}{9 \mathrm{L}}$$p C H_{4}=\frac{0.2 \times 24.63}{9}=\frac{4.926}{9}=0.547 \mathrm{atm}$ $p C H_{4}=\frac{0.2 \times 24.63}{9}=\frac{4.926}{9}=0.547 \mathrm{atm}$ (ii) Boron and silicon are example of Covalent solids.

**Or**(i) $28 g$ of Nitrogen gas contains $2 \times 7 \times 6.023 \times 10^{23}$ electrons. $1+1$ $\begin{array}{llll}{1.4} & {g} & {\text { of }} & {\text { Nitrogen }} & {\text { gas }} & {\text { contains }}\end{array}$ $\frac{2 \times 7 \times 6.023 \times 10^{23}}{28} \times 1.4$ $=\frac{2 \times 7 \times 6.023 \times 10^{23}}{20}=\frac{84.32 \times 10^{23}}{20}$ $=4.2161 \times 10^{23}$ electrons (ii) $N H_{3}$ will diffuse faster $\frac{r_{N H_{3}}}{r_{H C l}}=\sqrt{\frac{36.5}{17}}=\sqrt{2.14}=1.46$ times faster. (iii) Urea is a crystalline solid therefore it has sharp melting point whereas glass does not have sharp melting point because it is amorphous, i.e., does not have regular three dimensional structure.

*cm*whereas the arm connected to the bulb reads 15.6

*cm*. If the barometric pressure is 743

*mm*mercury, what is the pressure of the bulb reads 15.6

*cm*. If the barometric pressure is 743

*mm*mercury, what is the pressure of the gas in bar ?

**[NCERT]**

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**Sol.**Pressure of gas $=$ Atmospheric pressure $+$ Difference between levels of $\mathrm{Hg}$ $=743 \mathrm{mm}+(43.7 \mathrm{cm}-15.6 \mathrm{cm})=74.3 \mathrm{cm}+28.1 \mathrm{cm}$ $P=102.4 \mathrm{cm} P$ in bar $=\frac{102.4}{76}=1.347 \mathrm{bar}$

**[NCERT]**

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**Sol.**$P V=n R T, \quad P=1$ atm, $\quad V=500 \mathrm{cm}^{3}, \quad n=?$ $R=82.1 \mathrm{atm} \mathrm{cm}^{3} \mathrm{K}^{-1} \mathrm{mol}^{-1}, T=300 \mathrm{K}$ $n=\frac{1 a t m, \times 500 \mathrm{cm}^{3}}{\left(82.1 \mathrm{atm} \mathrm{cm}^{3} \mathrm{K}^{-1} \mathrm{mol}^{-1}\right) \times(300 \mathrm{K})}=0.02 \mathrm{mol}$

**[NCERT]**

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**Sol.**$p V=n R T$ 1 bar $\times \frac{34.05}{1000} L=\frac{0.0625}{M} \times 0.083 \times 819 K$ $M=\frac{0.0625 \times 0.083 \times 819 K \times 1000}{34.05}=\frac{4248.5625}{34.05}$ $M=124.77 \mathrm{g} \mathrm{mol}^{-1}$

**Or**Charle’s plotted the volume against temperature in $^{\circ} C$. These plots when extraplotted intersect the temperature axis at the same point $-273^{\circ} \mathrm{C} .$ He concluded that all gases at this temperature could have zero volume and below this temperature volume would be negative. It shows $-273^{\circ} \mathrm{C}$ is lowest temperature attainable.

**[NCERT]**

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**Sol.**Since temperature and amount of gas remains constant, therefore, Boyle’s law is applicable.

*L*volume, upto what volume can the balloon be expanded ?

**[NCERT]**

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**Sol.**According to Boyle’s law, at constant temperature, Since balloon bursts at 0.2 bar pressure, the volume of the balloon should be less than 11.35

*L*.

**[NCERT]**

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**Sol.**Suppose volume of vessel $=V c m^{3}$ i.e., volume of air in the flask at $27^{\circ} \mathrm{C}=\mathrm{Vcm}^{3}$ $\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}, \quad$ i.e., $\quad \frac{V}{300}=\frac{V_{2}}{750} \quad$ or $\quad V_{2}=2.5 \mathrm{V}$ $\therefore$ Volume expelled $=2.5 \mathrm{V}-\mathrm{V}=1.5 \mathrm{V}$ Fraction of air expelled $=\frac{1.5 \mathrm{V}}{2.5 \mathrm{V}}=\frac{3}{5}$

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**Sol.**Pressure $=$ height $\times$ density $\times g$ Case (i). Pressure $=76 \mathrm{cm} \times 0.99 \mathrm{g} / \mathrm{cm}^{3} \times 981 \mathrm{cm} / \mathrm{s}^{2}$ $=7.38 \times 10^{4}$ dynes $\mathrm{cm}^{-2}$ $=0.073$ atm $\left(1 \mathrm{atm}=1.013 \times 10^{6} \text { dynes } \mathrm{cm}^{-2}\right)$ Case (ii). Pressure $=76 \mathrm{cm} \times 13.6 \mathrm{g} / \mathrm{cm}^{3} \times 981 \mathrm{cm} / \mathrm{s}^{2}$ $=1.013 \times 10^{6}$ dynes $\mathrm{cm}^{-2}=1 \mathrm{atm}$

**[NCERT]**

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**Sol.**According to Charles’ law $\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}$ $V_{1}=2 L$ $V_{2}=?$ $T_{1}=273+23.4=296.4 \mathrm{K} \quad T_{2}=273+26.1=299.1$ $\therefore V_{2}=\frac{V_{1} T_{2}}{T_{1}}=\frac{2 L \times 299.1 K}{296.4 K}=2.018 L$

**[NCERT]**

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**Sol.**Since the amount of gas and the pressure remains constant, Charles’ law is applicable. i.e. $\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}$ $V_{1}=800 \mathrm{mL}$ $V_{2}=?$ $T_{1}=273+27=300 K ; \quad T_{2}=273+47=320 K$ $\frac{800 m L}{300 K}=\frac{V_{2}}{320 K}$ or $\quad V_{2}=\frac{(800 \mathrm{mL})}{(300 \mathrm{K})} \times(320 \mathrm{K})=853.3 \mathrm{mL}$ Increase in volume of air $=853.3-800=53.3 \mathrm{mL}$

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**Sol.**since gas is confined in a cylinder, its volume will remain constan

**[NCERT]**

**[NCERT]**

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**Sol.**According to ideal gas equation: $p V=n R T$ $p=3.32$ bar, $V=5 d m^{3}, n=4.0 \mathrm{mol}$ $R=0.083$ bar $d m^{3} K^{-1} \mathrm{mol}^{-1}$ $T=\frac{p V}{n R}=\frac{3.32 \text { bar } \times 5 d m^{3}}{4.0 \mathrm{mol} \times 0.083 \text { bar } d m^{3} K^{-1} \mathrm{mol}^{-1}}=50 \mathrm{K}$

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**Sol.**According to ideal gas equation,

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**Sol.**Weight of liquid $=148-50=98 g$ Density of liquid $=0.98 g m l^{-1}$ $\therefore$ Volume of liquid $=\frac{98}{0.98}=100 \mathrm{ml}$ Weight of gas $=50.5-50.0=0.5 g$ Volume $=\frac{100}{1000} L, P=1$ atm, $T=300 K$ $p V=n R T$ or $p V=\frac{W}{M} R T$ $\frac{1 \times 100}{1000}=\frac{0.5}{M} \times 0.082 \times 300$ $\therefore M=123.15 g \mathrm{mol}^{-1}$

**[NCERT]**

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**Sol.**Calculation of density of $N_{2}$ at 5 bar and $0^{\circ} \mathrm{C}$ $d=\frac{P M}{R T}=\frac{5(\text { bar }) \times 28\left(g m o l^{-1}\right)}{0.0831\left(L \text { bar } K^{-1} \mathrm{mol}^{-1}\right) \times 273.15(K)}$ $=6.168 g L^{-1}$ Calculation of molar mass of gaseous oxide $M=\frac{d R T}{P}$ $=\frac{6.168\left(g L^{-1}\right) \times 0.0831\left(L \text { bar } K^{-1} \mathrm{mol}^{-1}\right) \times 273.15(K)}{2}$ $=70.0 \mathrm{g} \mathrm{mol}^{-1}$

**NCERT]**

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**Sol.**$p=\frac{n}{V} R T=\frac{w}{M} \frac{R T}{V}$ $p_{C H_{4}}=\left(\frac{3.2}{16} m o l\right) \frac{0.0821 d m^{3} a t m K^{-1} m o l^{-1} \times 300 K}{9 d m^{3}}$ $=0.55 \mathrm{atm}$ $p_{\mathrm{CO}_{2}}=\left(\frac{4.4}{44} \mathrm{mol}\right) \frac{0.0821 \mathrm{dm}^{3} \mathrm{atm} \mathrm{K}^{-1} \mathrm{mol}^{-1} \times 300 \mathrm{K}}{9 \mathrm{dm}^{3}}$ $=0.27 \mathrm{atm}$ $p_{\text {Total }}=0.55+0.27=0.82 \mathrm{atm}$

**[NCERT]**

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**Sol.**$p V=n R T$ $p \times 1=\frac{8}{32} \times 0.083 \times 300=\frac{24.9}{4}=6.225$ bar $p \times 1=\frac{4}{2} \times 0.083 \times 300=49.80 \mathrm{bar}$ Total Pressure $=6.225+49.80=56.025$ bar

**[NCERT]**

**[NCERT]**

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**Sol.**Pay load is the difference between the mass of displaced air and the mass of the balloon. Volume of ballon $=\frac{4}{3} \pi r^{3}$ Radius of balloon, $r=10 \mathrm{m}$ $V=\frac{4}{3} \times 3.14 \times(10)^{3}=4186.7 \mathrm{m}^{3}$ Mass of displaced air $=4186.7 \mathrm{m}^{3} \times 1.2 \mathrm{kg} \mathrm{m}^{-3}$ $=5024.04 \mathrm{kg}$ Moles of gas present $=\frac{p V}{R T}$ $=\frac{1.66 \times 4186.7 \times 10^{3}}{0.083 \times 300}=279.11 \times 10^{3} \mathrm{mol}$ Mass of helium present $=279.11 \times 10^{3} \times 4$ $=1116.44 \times 10^{3} g=1116.4 \mathrm{kg}$ Mass of filled balloon $=100+1116.4=1216.4 \mathrm{kg}$ Pay load $=$ mass of displaced air – Mass of balloon $=5024.4-1216.44=3807.6 \mathrm{kg}$

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**Sol.**Volume of balloon, $V=\frac{4}{3} \pi r^{3} r=10 m$ $\therefore V=\frac{4}{3} \times 3.14 \times(10)^{3}=4187 \mathrm{m}^{3}$ Mass of displaced air $=4187 \mathrm{m}^{3} \times 1.2 \mathrm{kg} \mathrm{m}^{-3}$ $=5024.44 \mathrm{kg}$ Moles of gas present, $n=\frac{P V}{R T}=\frac{1 \times 4187 \times 10^{3}}{0,082 \times 298}$ $=171.3 \times 10^{3} \mathrm{mol}$ Mass of He present $=171.3 \times 10^{3} \times 4$ $=685.3 \times 10^{3} g$ $=685.3 \mathrm{kg}$ Mass of filled balloon $=100+685.3=785.3 \mathrm{kg}$ Pay load $=$ Mass of displaced air – Mass of balloon $=5024.4-785.3=4239.1 \mathrm{kg}$

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**Sol.**$\left(p+\frac{a n^{2}}{V^{2}}\right)(V-n b)=n R T$ $n=2 \mathrm{mol}, V=4 L, p=11 \mathrm{atm}$ $T=300 K, b=0.05 L m o l^{-1}$ Substituting the values $\left(11+\frac{a \times 4}{16}\right)(4-2 \times 0.05)=2 \times 0.082 \times 300$ $\frac{176+4 a}{16} \times 3.9=49.2$ $(176+4 a) 3.9=49.2 \times 16$ $15.6 a=787.2-686.4$ $a=6.4616 \mathrm{atm} \mathrm{L}^{2} \mathrm{mol}^{-2}$

**[NCERT]**

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**Sol.**Mole of phosphorus vapour $(n)=\frac{P V}{R T}$ $=\frac{1(\text {bar}) \times 34.05 \times 10^{-3}(L)}{0.0831\left(\text {bar } L K^{-1} \mathrm{mol}^{-1}\right) \times 819.15(K)}=5.0 \times 10^{-4}$ Let molar mass of phosphorus be $M g m o l^{-1}$ $\therefore$ Mole of phosphorus vapour $=\frac{0.0625}{M}$ Now, $\frac{0.0625}{M}=5.0 \times 10^{-4}$ or $M=\frac{0.0625}{5.0 \times 10^{-4}}$ $=125 g \mathrm{mol}^{-1}$

**[NCERT]**

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**Sol.**The chemical reaction taking place is

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**Sol.**For a real gas, the plot of $P V_{m}$ vs $P$ can be of the type $A$ or $B$ but at the point of intercept, $P=0$ and at any low pressure, vander Waal’s equation reduce to ideal gas equation. $\mathrm{PV}=\mathrm{nRT}$ or $\mathrm{PV}_{\mathrm{m}}=\mathrm{RT}$ Hence, $y$ -intercept of graph will be $=R T$

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**Sol.**(i) For statement of the Graham’s law. Molar mass of: $\mathrm{CO}_{2}=44 \mathrm{u} ; \mathrm{SO}_{2}=64 \mathrm{u}$ and $\mathrm{NO}_{2}=46 \mathrm{u}$ As $r_{d i f f} \propto \frac{1}{\sqrt{M}},$ therefore, larger the molar mass lesser will be the rate of diffusion under similar condition. Thus, increasing order of rates of diffusion is $r_{S O_{2}}<r_{N O_{2}}<r_{C O_{2}}$ (ii) $V=\frac{n R T}{P}=\frac{0.30(\text { mol }) \times 0.0821\left(L \text { atm } K^{-1} \mathrm{mol}^{-1}\right) \times 333(\mathrm{K})}{0.821(\mathrm{atm})}$ $=9.99 L \approx 10.8 L$

**[NCERT]**

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**Sol.**$\frac{r_{N H_{3}}}{r_{H C l}}=\frac{l_{1}}{200-l_{1}}=\sqrt{\frac{M_{H C l}}{M_{N H_{3}}}}=\sqrt{\frac{36.5}{17}}$ $\frac{l_{1}}{200-l_{1}}=\sqrt{2.147}=1.465$ $l_{1}=293-1.465 l_{1} ; 2.465 l_{1}=293$ $l_{1}=\frac{293}{2.465}=118.88 \mathrm{cm}$ $200-l_{1}=200-118.88=81.12 \mathrm{cm}$ from $\mathrm{HCl}$ end.

*A*and

*B*diffuse through a porous pot in 20 and 10 seconds respectively. If the molar mass of

*A*be 80, find the molar mass of

*B*.

**[NCERT]**

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**Sol.**$\frac{r_{A}}{r_{B}}=\sqrt{\frac{M_{B}}{M_{A}}}$ or $\frac{t_{B}}{t_{A}}=\sqrt{\frac{M_{B}}{M_{A}}}$ $t_{A}=20$ sec, $t_{B}=10$ sec, $M_{A}=80, M_{B}=?$ $\frac{10}{20}=\sqrt{\frac{M_{B}}{80}}$ $M_{B}=\frac{80}{4}=20 g \mathrm{mol}^{-1}$

**[NCERT]**

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**Sol.**As the mixture $H_{2}$ and $O_{2}$ contains $20 \%$ by weight of hydrogen, therefore, if $H_{2}=20 g,$ then $O_{2}=80 g$ $n_{H_{2}}=\frac{20}{2}=10$ moles, $n_{O_{2}}=\frac{80}{32}=2.5$ moles $p_{H_{2}}=\frac{n_{H_{2}}}{n_{H_{2}}+n_{O_{2}}} \times P_{\text {total }}=\frac{10}{10+2.5} \times 1$ bar $=0.8$ bar.

**[NCERT]**

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**Sol.**Partial pressure of oxygen gas, $p=\frac{n R T}{V}$ $n=\frac{8}{32}$ mol, $\quad \mathrm{V}=1 \mathrm{dm}^{3}, T=300 \mathrm{K}$ $p_{\left(O_{2}\right)}=\frac{8 \times 0.083 \times 300}{32 \times 1}=6.225$ bar Partial pressure of hydrogen gas $p=\frac{n R T}{V}$ $n=\frac{4}{2}=2 m o l$ $p\left(H_{2}\right)=\frac{2 \times 0.083 \times 300}{1}=49.8 \mathrm{bar}$ Total pressure $=p_{\left(O_{2}\right)}+p_{\left(H_{2}\right)}$ $=6.225+49.8=56.025$ bar

**[NCERT]**

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**Sol.**Let the partial pressure of hydrogen be $P_{H_{2}}$ and the partial pressure of oxygen be $P_{O_{2}}$ The number of mole of hydrogen $\left(n_{1}\right)=\frac{4}{2}=2$ mole The number of mole of oxygen $\left(n_{2}\right)=\frac{8}{32}=0.25$ mole Now, applying ideal gas equation for each gas $p_{H_{2}} \times V=n_{1} R T$ $p_{H_{2}}=\frac{n_{1} R T}{V}=\frac{2 \times 0.083 \times 300}{1}=49.8$ bar Similarly, $p_{O_{2}} V=n_{2} R T$ $p_{O_{2}}=\frac{n_{2} R T}{V}=\frac{0.25 \times 0.083 \times 300}{1}=6.225 \mathrm{bar}$ Total pressure of gaseous mixture $=p_{H_{2}}+p_{O_{2}}=49.8+6.225=56.025$ bar.

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**Sol.**(i) Average kinetic energy is given as $E_{k}=\frac{3 n R T}{2}$ Here, $n=\frac{32}{16}=2 ; R=8.314 \mathrm{JK}^{-1} \mathrm{mol}^{-1} ; T=300 \mathrm{K}$ $\therefore E_{k}=\frac{3 \times 2 \times 8.314 \times 300}{2}=7482.6 \mathrm{J}$ (ii) Root mean square speed is given as $u_{r m s}=\sqrt{\frac{3 R T}{M}}$ Here use $R=8.314 \times 10^{7}$ ergs $K^{-1} \mathrm{mol}^{-1}$ to get speed in $\mathrm{cm} \mathrm{s}^{-1}$ $u_{m s}=\sqrt{\frac{3 \times 8.314 \times 10^{7} \times 300}{16}}=68385.85 \mathrm{cm} s^{-1}=68.38 \mathrm{ms}^{-1}$ (iii) Most probable speed $(\alpha)=\frac{u_{r m s}}{1.224}=\frac{683.8}{1.224}=558.7 \mathrm{ms}^{-1}$

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**Sol.**Volume occupied by $1 \mathrm{mol}$ of $\mathrm{O}_{3}$ at $20^{\circ} \mathrm{C}$ and $82 \mathrm{mm}$ pressure is calculated by applying general gas equation, $\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}$ $\therefore \quad V_{2}=\frac{P_{1} V_{1} T_{2}}{T_{1} P_{2}}=\frac{76 \times 22400 \times 293}{273 \times 82}=22281.92 \mathrm{cm}^{3}$ Now, $u=\sqrt{\frac{3 P V}{M}}$ Here, we use $P$ in dyne/cm$^{2}, P=82 \times 13.6 \times 981$ dyne/cm$^{2}$ $\therefore u=\sqrt{\frac{3 \times 82 \times 13.6 \times 981 \times 22281.92}{48}}=3.90 \times 10^{4} \mathrm{cm} \mathrm{s}^{-1}$

**[NCERT]**

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**Sol.**(i) The kinetic theory of gases assumes that pressure of gas is due to collision of gas molecules with the walls of the container. The more will be frequency of collision, more will be pressure. The reduction in volume of gas increases no. of molecules per unit volume to which pressure is directly proportional. Therefore, the volume of the gas is reduced if pressure is increased or pressure is inversely proportional to volume. $\frac{1}{2} m u^{2}=\frac{3}{2} k T$ $p=\frac{1}{3} \frac{N}{V} m u^{2}$ or $p=\frac{2}{3} \frac{N}{V} \times \frac{1}{2} m u^{2}=\frac{2}{3} \frac{N}{V} \times \frac{3}{2} k T$ $\Rightarrow p V=N k T \Rightarrow P \alpha \frac{1}{V}$ It can be seen that at a constant temperature for a fixed number of gas molecules, the pressure is inversely proportional to volume. (ii) $\frac{T_{1}}{T_{2}}=\frac{n_{1}}{n_{2}} \quad \begin{array}{cc}{T_{1}=27^{\circ} C+273} & {=300 K} \\ {T_{2}=477+273=} & {750 K}\end{array}$ $\frac{300 K}{750 K}=\frac{n_{1}}{n_{2}}, \quad \frac{n_{1}}{n_{2}}=\frac{2}{5} b v$ fraction of air escaped $=1-\frac{2}{5}=\frac{3}{5}$

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**Sol.**Average velocity $=\sqrt{\frac{8 R T}{\pi M}}$ Root mean square velocity $=\sqrt{\frac{3 R T}{M}}=\sqrt{\frac{3 R \times 300 K}{M}}$ For equal values, $\sqrt{\frac{8 R T}{\pi M}}=\sqrt{\frac{3 R \times 300}{M}}$ or $\quad \frac{8 R T}{\pi M}=\frac{3 R \times 300}{M}$ or $\frac{8 T}{\pi}=900$ or $\quad T=353.57 K=80.57^{\circ} \mathrm{C}$

*mL*at a pressure of 0.720 bar. It is subjected to an external pressure of 0.900 bar. What is the resulting volume of the gas ?

**[NCERT]**

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**Sol.**Boyle’s law is applicable as the amount and temperature are unaltered $p_{1} V_{1}=p_{2} V_{2}$ or $p_{1} / p_{2}=V_{2} / V_{1}$ ‘Substituting the values 0.720 bar/ 0.900 bar $=V_{2} / 200 \mathrm{mL}$ $V_{2}=\frac{720}{900} \times 200 m L=160 m L$ Boyle’s law is manifested in the working of many devices used in daily life such as cycle pump, aneroid barometer and tyre pressure gauge etc.

*g*of nitrogen gas.

**[NCERT]**

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**Sol.**Number of moles of nitrogen $=2.8 g / 28 g \mathrm{mol}^{-1}=0.1 \mathrm{mol}$ Number of nitrogen molecules $=0.1 \mathrm{mol} \times 6.022 \times 10^{23}$ $m o l^{-1}=6.022 \times 10^{22}$

**[NCERT]**

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**Sol.**$p V_{m}=R T$ or $p M / d=R T$ or $M=d R T / p$ $=\frac{1.29 \mathrm{kg} m^{-3} \times 8.314 \mathrm{NmK}^{-1} \mathrm{mol}^{-1} \times 273.15 \mathrm{K}}{1.0 \times 10^{5} \mathrm{Nm}^{-2}(\mathrm{or} P a)}$ $=\frac{1.29 \times 8.314 \times 273.15 k g m o l^{-1}}{1 \times 10^{5}}$ $=0.0293 \mathrm{kg} \mathrm{mol}^{-1}$ or molar mass is $29.3 \mathrm{g} \mathrm{mol}^{-1}$

**[NCERT]**

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**Sol.**$r_{N H_{3}} / r_{H C 1}=\left(M_{H C 1} / M_{N H_{3}}\right)^{1 / 2}$ $=(36.5 / 17)^{1 / 2}=1.46$ or $r_{N H_{3}}=1.46 r_{H C 1}$ Thus ammonia will diffuse 1.46 times faster than hydrogen chloride gas.

**[NCERT]**

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**Sol.**Partial pressure of sulphur dioxide, $p_{S O_{2}}=n R T / V$ $=\frac{0.25 \mathrm{mol} \times 8.314 \mathrm{JK}^{-1} \mathrm{mol}^{-1} \times 300 \mathrm{K}}{2.5 \times 10^{-3} \mathrm{m}^{3}}$ $=2.49 \times 10^{5} \mathrm{Nm}^{-2}=2.49 \times 10^{5} \mathrm{Pa}$ Similarly $p_{N_{2}}=2.49 \times 10^{5} \mathrm{Pa}$ Following Dalton’s law $p_{\text {Total }}=p_{N_{2}}+p_{S O_{2}}$ $=2.49 \times 10^{5} P a+2.49 \times 10^{5} P a=4.98 \times 10^{5} P a$

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**Sol.**As the vessel is open, pressure and volume remain constant. Thus, if $n_{1}$ moles are present at $T_{1}$ and $n_{2}$ moles are present at $T_{2},$ we can write $P V=n_{1} R T_{1} ; P V=n_{2} R T_{2}$ Hence, $n_{1} R T_{1}=n_{2} R T_{2}$ or $n_{1} T_{1}=n_{2} T_{2}$ or, $\quad \frac{n_{1}}{n_{2}}=\frac{T_{2}}{T_{1}}$ Suppose the no. of moles of air originally present $=n$ After heating, no. of moles of air expelled $=\frac{3}{5} n$ $\therefore$ No. of moles left after heating $=n-\frac{3}{5} n=\frac{2}{5} n$ Thus, $n_{1}=n, T_{1}=300 K ; n_{2}=\frac{2}{5} n, T_{2}=?$ $\frac{n}{\frac{2}{5} n}=\frac{T_{2}}{300}$ or $\frac{5}{2}=\frac{T_{2}}{300}$ or $, T_{2}=750 \mathrm{K}$ Alternatively, suppose the volume of the vessel $=V$ i.e. Volume of air initially at $27^{\circ} C=V$ Volume of air expelled $=\frac{3}{5} V$ $\therefore$ Volume of air left at $27^{\circ} \mathrm{C}=\frac{2}{5} \mathrm{V}$ However, on heating to $T^{\circ} K,$ it would become $=V$ As pressure remains constant, (vessel being open), $\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}$ i.e. $\frac{2 / 5 \mathrm{V}}{300 \mathrm{K}}=\frac{\mathrm{V}}{T_{2}}$ or $T_{2}=750 \mathrm{K}$

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**Sol.**Weight of $2.24 \mathrm{L}$ of $\mathrm{H}_{2}$ at $\mathrm{S} . T . P .=0.2 \mathrm{g}$ (Mol. mass of $\left.H_{2}=2\right)=0.1 \mathrm{mol}$ Weight of $1.12 \mathrm{L}$ of $D_{2}$ at $S . T . P .=0.2 \mathrm{g}$ (Mol. mass of $\left.D_{2}=4\right)=0.05 \mathrm{mol}$ As number of moles of two gases are different but

*V*and

*T*are same, therefore, their partial pressures will be different, i.e., in the ratio of their number of moles. Thus, $\frac{P_{H_{2}}}{P_{D_{2}}}=\frac{n_{H_{2}}}{n_{D_{2}}}=\frac{0.1}{0.5}=2$ Now, $D_{2}$ present in the first bulb $=0.1 g$ (Given) $D_{2}$ diffused into the second bulb $=0.2-0.1=0.1 g=0.56 L$ at $S . T . P .$ Now, $\frac{r_{H_{2}}}{r_{D_{2}}}=\frac{P_{H_{2}}}{P_{D_{2}}} \times \sqrt{\frac{M_{D_{2}}}{M_{H_{2}}}}$ Or, $\frac{v_{H_{2}}}{t} \times \frac{t}{v_{D_{2}}}=\frac{P_{H_{2}}}{P_{D_{2}}} \times \sqrt{\frac{M_{D_{2}}}{M_{H_{2}}}}$ $\frac{v_{H_{2}}}{t} \times \frac{t}{0.56 L}=2 \times \sqrt{\frac{4}{2}}$ or $\quad v_{H_{2}}=1.584 L=0.14 g$ of $H_{2}$ $\therefore$ Weight of the gases in 2 nd bulb $=0.10 g\left(D_{2}\right)+0.14 g\left(H_{2}\right)=0.24 g$ Hence, in the 2 nd bulb, $\%$ of $D_{2}$ by weight $=\frac{0.10}{0.24} \times 100=41.67 \%$ $\%$ of $H_{2}$ by weight $=100-41.67=58.33 \%$

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**Sol.**Weight of $L P G$ originally present $=29.0-14.8$ $=14.2 \mathrm{kg}$ Pressure $=2.5$ atm Weight of LPG present after use $=23.2-14.8$ $=8.4 \mathrm{kg}$ since volume of the cylinder is constant, applying since volume of the cylinder is constant, applying $p V=n R T$ $\frac{p_{1}}{p_{2}}=\frac{n_{1}}{n_{2}}=\frac{W_{1} / M}{W_{2} / M}=\frac{W_{1}}{W_{2}}$ $\frac{2.5}{p_{2}}=\frac{14.2}{8.4}$ $p_{2}=\frac{2.5 \times 8.4}{14.2}=1.48 \mathrm{atm}$ or Weight of used gas $=14.2-8.4=5.8 \mathrm{kg}$ Moles of gas $=\frac{5.8 \times 10^{3}}{58}=100$ moles Normal conditions $p=1$ atm; $T=273+27=300 K$ Volume of 100 moles of $L P G$ at 1 atm and $300 \mathrm{K}$ $V=\frac{n R T}{p}=\frac{100 \times 0.082 \times 300}{1}=2460$ litre $V=2.460 \mathrm{m}^{3}$

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**Sol.**Compressibility factor $Z=\frac{p V}{R T} ; \quad 0.5=\frac{100 \times V}{0.082 \times 273}$ $\therefore V=\frac{0.5 \times 0.082 \times 273}{100}=0.1119 L$ If volume of molecules is negligible i.e. $b$ is negligible vander Waals’ equation: $\left(p+\frac{a}{V^{2}}\right)(\mathrm{V})=R T$ or $p V=R T-a / V$ or $a=R T V-P V^{2}$ $=(0.082 \times 273 \times 0.1119)-(100 \times 0.1119 \times 0.1119)$ $=1.253 \mathrm{atm} L^{2} \mathrm{mol}^{-2}$

**Click Here for Detailed Chapter-wise Notes of Chemistry for Class 11th, JEE & NEET.**You can access free study material for all three subject’s Physics, Chemistry and Mathematics. Click Here for Detailed Notes of any chapter. eSaral provides you complete edge to prepare for Board and Competitive Exams like JEE, NEET, BITSAT, etc. We have transformed classroom in such a way that a student can study anytime anywhere. With the help of AI we have made the learning Personalized, adaptive and accessible for each and every one. Visit eSaral Website to download or view free study material for JEE & NEET. Also get to know about the strategies to Crack Exam in limited time period.

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**Sol.**A precipitate will be formed if ionic product > solubility product.

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**Sol.**Lewis acid is a species which is electron deficient or positively charged Lewis base is a species which is electron rich,

*i.e.,*has lone pair of electron or is negatively charged.

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**Sol.**Bronsted acid can donate $H^{+}$ where as bronsted base can accept.

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**Sol.**Yes, it is true, It is because Lewis bases are – velycharged or electron rich. They are bronsted bases also because they can accept $H^{+}$ easily.

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**Sol.**Acid $^{\prime} A^{\prime}$ with $\mathrm{pKa}=1.5$ is strongest acid, lower the value of $P K_{a}$ stronger will be the acid.

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**Sol.**Neutral: $\mathrm{NaCl}, \mathrm{KBr} \quad$ Acidic : $\mathrm{NH}_{4} \mathrm{NO}_{3}$ Basic : $\mathrm{NaCN}, \mathrm{NaNO}_{2}$ and $\mathrm{KF}$

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**Sol.**An acid-base which differ by a proton is called conjugate acid base pair. $\mathrm{NO}_{2}^{-}, \mathrm{HCN}, \mathrm{ClO}_{4}^{-}, \mathrm{HF}, \mathrm{H}_{2} \mathrm{O}(\mathrm{acid})$ or $\mathrm{O}^{2-}$ (base) $\mathrm{HCO}_{3}^{-}$ and $\mathrm{HS}^{-}$

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**Sol.**$p H=-\log \left[H^{+}\right]=-\log \left(3.8 \times 10^{-3}\right)$ $=-\log 3.8+3=3-0.5798=2.4202=2.42$

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**Sol.**$p H=-\log \left[H^{+}\right]$ or $\log \left[H^{+}\right]=-p H=-3.76=\overline{4} .24$ $\therefore\left[H^{+}\right]=$ Antilog $\overline{4} 24=1.738 \times 10^{-4} M=1.74 \times 10^{4} M$

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**Sol.**(i), ( iii), (iv)

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**Sol.**$\mathrm{Cod}+H_{2} \mathrm{O} \rightleftharpoons \mathrm{CodH}^{+}+\mathrm{OH}^{-}$ $p H=9.95 \therefore p O H=14-9.95=4.05$ i.e., $-\log \left[\mathrm{OH}^{-}\right]=4.05$ or $\log \left[O H^{-}\right]=-4.05=5.95$ or $\left[O H^{-}\right]=8.913 \times 10^{-5}$ $K_{b}=\frac{\left[\mathrm{Cod} H^{+}\right][\mathrm{OH}]}{[\mathrm{Cod}]}=\frac{[\mathrm{OH}]^{2}}{[\mathrm{Cod}}=\frac{\left(8.91 \times 10^{5}\right)^{2}}{5 \times 10^{-3}}=1.588 \times 10^{-6}$ $p K_{b}=-\log \left(1.588 \times 10^{-6}\right)=6-0.20=5.8$

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**Sol.**$p K_{a}=-\log K_{a}=-\log 1.8 \times 10^{-4}=3.74$ Buffer capacity is maximum near $p K_{a}$ of the acid $i . e .,$ at $p H=3.74 .$ In order to obtain a buffer solution of $p H=4.25$ we have $\log \frac{[\text { salt }]}{[\text { acid }]}=p H-p K_{a}=4.25-3.74=0.51$ $\frac{[\text { salt }]}{[\text { acid }]}=$ Antilog $0.51=3.24$ It means concentration of salt should be 3.24 times that of acid.

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**Sol.**HCNO ionizes as: $H C N O=H^{+}+C N O^{-}$ The dissociation constant $K_{a}=\frac{\left[H^{+}\right][C N O]}{[H C N O]} \Rightarrow\left[H^{+}\right]=\left[C N O^{-}\right]$ Now, $p H=2.34$ $-\log \left[H^{+}\right]=2.34 \Rightarrow \log \left[H^{+}\right]=-2.34=3.36$ $\left(H^{+}\right)=$ Antilog $(3.36)=4.57 \times 10^{-3} \mathrm{M}$ $\therefore\left[H^{+}\right]=\left[C N O^{-}\right]=4.57 \times 10^{-3} M$ and $[H C N O]=0.1 M$ $K_{a}=\frac{\left(4.57 \times 10^{-3}\right) \times\left(4.57 \times 10^{-3}\right)}{0.1}=2.09 \times 10^{-4}$ Degree of dissociation $=\alpha=\sqrt{\frac{K_{a}}{c}}$ $=\sqrt{\frac{2.09 \times 10^{-4}}{0.1}}=4.57 \times 10^{-2} \mathrm{or}$

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**Sol.**$5 m L$ of $0.5 M K O H=2.5$ millimoles $100 \mathrm{mL}$ of $0.05 \mathrm{M} \mathrm{HBr}=5$ milli moles $105 \mathrm{mL}$ of solution contains $5-2.5=2.5$ milli moles $=2.5 \times 10^{-3}$ moles of acid $1000 \mathrm{mL}$ of contains $=\frac{2.5 \times 10^{-3}}{105} \times 1000$ $=2.38 \times 10^{-2} \mathrm{mol} L^{-1}$ $p H=-\log 2.38 \times 10^{-2}=-\log 2.38-\log 10^{-2}$ $=-0.38+2.000=1.62$ Phenolpthalein and Thymolpt halein are suitable indicators.

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**Sol.**$50 \mathrm{mL}$ of $0.2 \mathrm{MNaOH}=50 \times 0.2=10 \mathrm{millimoles}$ of $\mathrm{NaOH}$ 10 millimoles of $N a O H$ will neutralize 10 millimoles of Benzoic acid. 10 millimoles of benzoic acid $=1.22 \mathrm{g}$ 1 mole $=1000$ millimoles $=\frac{1.22}{10} \times 1000=122 \mathrm{g} \mathrm{mol}^{-1}$ Molecular weight $=122 \mathrm{g} \mathrm{mol}^{-1}$

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**Sol.**When equal volume of $\mathrm{NalO}_{3} \& \mathrm{CuClO}_{3}$ are mixed, the concentration of each of the solution will become half. $\left[\mathrm{IO}_{3}^{-}\right]=\frac{1}{2} \times 0.002=0.001=10^{-3}$ $\left[\mathrm{Cu}^{2+}\right]=\frac{1}{2} \times 0.002=0.001=10^{-3}$ $\mathrm{Cu}\left(\mathrm{IO}_{3}\right)_{2} \mathrm{H}$ tyth $\quad \mathrm{Cu}^{2+}+2 \mathrm{IO}_{3}^{-}$ I.P (Ionic product) $=\left[C u^{2+}\right]\left[I O_{3}^{-}\right]^{2}=\left[10^{-3}\right]\left[10^{-3}\right]^{2}=10^{-9}$ since $I . P .\left(1 \times 10^{-9}\right)$ is less than $K_{s p}\left(7.4 \times 10^{-8}\right),$ therefore No precipitation will take place.

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**Sol.**$K_{s p}=6.3 \times 10^{-8}$ FeS $\mathrm{HHH} \quad \mathrm{Fe}_{‘^{\prime} \mathrm{S}}^{+2}+\mathrm{S}^{-2}$ $K_{s p}=6.3 \times 10^{-18}$ $K_{s p}=s \times s=s^{2}=6.3 \times 10^{-18}$ $s=2.51 \times 10^{-9}$ since volume of solution will become double on mixing equal volume of each solution. $\therefore$ concentration will become half on mixing max. concentration of each ion $=2 \times S$ $=2 \times 2.51 \times 10^{-9}=5.02 \times 10^{-9} \mathrm{m}$

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**Sol.**The relation between ionization constant of an acid and that of its conjugate base is $K_{a} \times K_{b}=K_{w}$ $\therefore \quad K_{b}=\frac{K_{w}}{K_{a}}$ The conjugate base of $H F$ is $F^{-}$ $K_{b\left(F^{-}\right)}=\frac{K_{w}}{K_{a}(H F)}=\frac{1 \times 10^{-14}}{6.8 \times 10^{-4}}=1.47 \times 10^{-11}$ The conjugate base of $H C O O H$ is $H C O O^{-}$ $K_{b}\left(H C O O^{-}\right)=\frac{K_{w}}{K_{a(H C O O H)}}=\frac{1 \times 10^{-14}}{1.8 \times 10^{-4}}=5.56 \times 10^{-11}$ The conjugate base of $H C N$ is $C N^{-}$ $K_{b\left(C N^{-}\right)}=\frac{K_{w}}{K_{a(H C N)}}=\frac{1 \times 10^{-14}}{4.8 \times 10^{-9}}=2.08 \times 10^{-6}$

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**Sol.**If $C$ is the initial concentration and $\alpha$ is the degree of ionization of $H F$ then