CBSE Sample Papers For Class 11 Chemistry Mock Paper 1

Q1. The molar heat capacity of water at constant pressure, C_p is 75 J K^-1 mol^-1. When 10 kJ of heat supplied to 1 kg of water which is free to expand, the increase in the temperature of water is

Opt:

4.8 K

2.4 K

3.2 K

10 K

Ans: 2.4 K

Q2. The central carbon atom in  is

Opt: sp hybridized. sp² hybridized. sp³ hybridized. Unhybridized.

Ans: sp hybridized.
Q3. 0.3780 gm of an organic chloro compound gave 0.5740 gm of silver chloride in Carius estimation. Calculate the percentage of chlorine present in the compound.

Opt:

37.56 gm

39.52 gm

40.00 gm

37.03 gm

Ans: 37.56 gm

Q4. Reverse reaction in chemical equilibrium is favoured by

Opt: increasing the concentration of the reactant. removal of at least one of the products at regular intervals. increasing the concentration of one or more of the product. addition of catalyst.

Ans: increasing the concentration of one or more of the product.
Q5. At 350 K, K_p for the reaction given below is 3.0 x 10¹⁰ bar^-1 at equilibrium

${\displaystyle 2{\mathrm{N}}_{2}(\mathrm{g})+{\mathrm{O}}_{2}(\mathrm{g})\stackrel{}{\rightleftharpoons }2{\mathrm{N}}_{2}\mathrm{O}(\mathrm{g})}$

2N2​(g)+O2​(g)⇌​2N2​O(g)

The value of K_c at this temperature will be

a. 7.4 x 10¹¹ L mol^-1

b. 8715 x 10¹⁰ L mol^-1

c. 0.08 L mol^-1

d. 8.715 x 10¹¹ L mol^-1

Or

The solubility product of BaCl₂ is 3.2 10^-9. Its solubility in mol L^-1 will be

a. 4 x 10^-3

b. 3.2 x 10^-9

c. 1 x 10^-3

d. 1 x 10^-9

Ans: d. 8.715 x 10¹¹ L mol^-1

Or

c. 1 x 10^-3

Q6. An organic compound contains 69% carbon and 4.8% hydrogen. Calculate the mass of carbon dioxide produced when 0.20 gm of this substance is subjected to complete combustion.

Opt:

0.864 gm.

0.506 gm.

0.854 gm.

0.654 gm.

Ans: 0.506 gm.

Q7. The first and second electron affinities of oxygen are
Opt:

negative and positive respectively. positive and negative respectively. both positive. both negative.
Ans: negative and positive respectively.

Q8. 40gm of MgO has been mixed with 126 gm of

$HNO_3$

HNO3​   , the amount of

$Mg\left(NO_3\right)_2$

Mg(NO3​)2​    formed is

Opt:

4g.

64g.

82 g.

148g

Ans: 148g

Q9. What will be the heat of reaction for the following reaction? Will the reaction be exothermic or endothermic?

Fe₂O3(S) + 3H₂(g) → 2Fe(S) + 3H₂O (I)

Δ,H°(H₂O,I) = 285.83 kJ mol^-1Δ,H°(Fe₂O₃,S) = -824.2 kJ mol^-1

Opt:

-824.2 kJ mol^-1, exothermic

-33.3 kJ mol^-1, exothermic

+33.3 kJ mol^-1, endothermic

+824.2 kJ mol^-1, endothermic

Ans: -33.3 kJ mol^-1, exothermic

Q10. Argon is isoelectronic with

Opt: Ca²⁺. Ba²⁺. Al³⁺. F^–.

Ans: Ca²⁺.
Q11. Electrons can absorb photon but cannot emit the photon when present in

Opt:

2p orbital

1s orbital

3d orbital

4f orbital

Ans: 1s orbital

Q12. The most stable alkali metal hydride amongst the following is
a. LiH.
b. NaH.
c. KH.
d. RbH.

Or

In water-gas shift reaction, is employed for the manufacture of
a. carbon monoxide.
b. dinitrogen.
c. dihydrogen.
d. dioxygen.

Ans:

a. LiH.The hydride of lithium is more stable than hydrides of other alkali metals because of very small size and high polarizing power of litium.

Or

C. dihydrogen.

In water-gas shift reaction, carbon monoxide reacts with steam in the presence of iron-chromate as catalyst to form carbon dioxide and dihydrogen.

Q13. For the hypothetical reactions, the equilibrium constant (K) values are given

The equilibrium constant (K) for the reaction is

Opt:

24.

39.

160.

188.

Ans: 160.

Q14. The increasing order of stability of primary, secondary, tertiary and benzylic carbocation is

Opt:

primary < secondary < benzylic < tertiary. Primary < benzylic < secondary < tertiary. Primary < secondary < tertiary < benzylic.  Benzylic  <  Primary < secondary  < tertiary.

Ans: Primary < secondary < tertiary < benzylic.
Q15. In the question, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below the question.

Assertion (A): Phenols show +R effect.

Reason (R): The oxygen atom of alcohols contains lone pair of electrons.
Opt:

Both assertion (A) and reason (R) are true, and the reason (R) is the correct explanation of the assertion (A).​

​Both assertion (A) and reason (R) are true, but the reason (R) is not the correct explanation of the assertion (A).​

​Assertion (A) is true, but the reason (R) is false.​

​Assertion (A) is false, but the reason (R) is true.​

Ans: Both assertion (A) and reason (R) are true, and the reason (R) is the correct explanation of the assertion (A).​

Q16. The molecule having same hybridisation like that of XeF₆ ­is

Opt:

SF₆.

PF₅.

IF₇.

XeOF₄.

Ans: IF₇.

Q17.

${\displaystyle \begin{array}{l}\mathrm{T}\mathrm{h}\mathrm{e}\mathrm{o}\mathrm{x}\mathrm{i}\mathrm{d}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{s}\mathrm{t}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{o}\mathrm{f}\mathrm{s}\mathrm{u}\mathrm{l}\mathrm{p}\mathrm{h}\mathrm{u}\mathrm{r}\mathrm{i}\mathrm{n}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{a}\mathrm{n}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{s}\\ {\mathrm{S}\mathrm{O}}_3^{2-<!--\; -\; -->},{\mathrm{S}\mathrm{O}}_4^{2-<!--\; -\; -->},{\mathrm{S}}_2{\mathrm{O}}_4^{2-<!--\; -\; -->},{\mathrm{S}}_2{\mathrm{O}}_6^{2-<!--\; -\; -->}\mathrm{i}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{o}\mathrm{r}\mathrm{d}\mathrm{e}\mathrm{r}\mathrm{o}\mathrm{f}\end{array}}$

Opt:

S 2 O 4 2 −<!-- − --> > S 2 O 6 2 −<!-- − --> > S O 4 2 −<!-- − --> > S O 3 2 −<!-- − -->

S 2 O 6 2 −<!-- − --> > S O 3 2 −<!-- − --> > S 2 O 4 2 −<!-- − --> > S O 4 2 −<!-- − -->

S O 4 2 −<!-- − --> > S 2 O 6 2 −<!-- − --> > S O 3 2 −<!-- − --> > S 2 O 4 2 −<!-- − -->

S O 3 2 −<!-- − --> > S O 4 2 −<!-- − --> > S 2 O 4 2 −<!-- − --> > S 2 O 6 2 −<!-- − -->

Ans:

S O 4 2 −<!-- − --> > S 2 O 6 2 −<!-- − --> > S O 3 2 −<!-- − --> > S 2 O 4 2 −<!-- − -->

Q18. The increasing order of the size of an iodine atom/ ion is

Opt:

I> I^– > I⁺. I > I⁺ > I^–. I⁺ > I^– > I. I^– > I > I⁺.

Ans: I^– > I > I⁺.
Q19. For the reaction

2A_(g) + B_(g) → 2D_(g)

∆U^θ = –10.5 kJ and ∆S^θ= –44.1 JK^–1.

Calculate ∆G^θ for the reaction, and predict whether the reaction may occur spontaneously.

Ans: For the given reaction,

2  A_(g)  + B_(g)  → 2D_(g)

∆n_g  = 2  –  (3)

= –1  mole

Substituting  the value of ∆U^θ  in  the expression  of  ∆H:

∆H^θ  = ∆U^θ  +  ∆n_gRT

= (–10.5  kJ)  –  (–1)  (8.314  × 10–3 kJ  K^–1 mol^–1)  (298  K)

= –10.5  kJ  –  2.48  kJ

∆H^θ  = –12.98  kJ

Substituting  the values of  ∆H^θ  and  ∆S^θ  in  the expression  of  ∆G^θ:

∆G^θ  = ∆H^θ  –  T∆S^θ

= –12.98  kJ  –  (298  K)  (–44.1  J  K^–1)

= –12.98  kJ  + 13.14  kJ

∆G^θ  = + 0.16  kJ

Since ∆G^θ  for the  reaction  is positive, the reaction  will  not  occur  spontaneously.

Q20. Find out which of the following molecules does not exist –

(i) Be₂

(ii)C₂

Or

What is the total number of sigma and pi bonds in the following molecules?

C₂H₂ (b) C₂H₄

Ans: The molecule having zero bond order will not exist.

Or

A single bond is a result of the axial overlap of bonding orbitals and forms a sigma bond. A multiple bond (double or triple bond) is always formed as a result of the sidewise overlap of orbitals. A double bond is a combination of one sigma bond and one pi-bond. A triple bond is a combination of two pi-bonds and one sigma bond.

Structure of C₂H₂ can be represented as:

There are three sigma and two pi-bonds in acetylene molecule.

The structure of C₂H₄ can be represented as:

There are five sigma bonds and one pi-bond in the ethylene molecule.

Q21. Explain Wurtz reaction with an example. Where is it used?

Or
Draw the Newman’s projection of ethane.

Ans:

Alkyl halides on treatment with sodium metal in dry ether solution give higher alkanes. This reaction is known as Wurtz reaction.

 

This reaction is used for the preparation of higher alkanes containing even number of carbon atoms.

Or

Q22. Calculate the mass of sodium acetate (CH₃ COONa) required to make 500 ml of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol^-1

Ans:

0.375 M aqueous solution of sodium acetate
≡ 1000 mL of solution containing 0.375 moles of sodium acetate
∴ Number of moles of sodium acetate in 500 mL

= 0.375 1000 x500 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhcba9y8qqasFr0xc9YqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOqaaabaaaaaaaaapeGaeyypa0ZaaSaaa8aabaWdbiaaicdacaGGUaGaaG4maiaaiEdacaaI1aaapaqaa8qacaaIXaGaaGimaiaaicdacaaIWaaaaGqaciaa=HhacaaI1aGaaGimaiaaicdaaaa@4882@

= 0.1875 mole
Molar mass of sodium acetate = 82.0245 g mol^-1 (Given)
∴ Required mass of sodium acetate = (82.0245 g mol^-1) (0.1875 mole)
= 15.38 g

Q23. Explain why CCl₄ has a zero dipole moment although C-Cl bonds are polar.

Ans:

Dipole moment is a vector quantity.

The individual C-Cl bonds are polar due to the difference in electronegativities of carbon and chlorine atoms. Since, the CCl₄ molecule has regular geometry, the four bond moments cancel each other and give a net resultant as zero.

Q24. Write the balanced chemical equation for the following reaction:-
1) Iron reacts with dilute HNO₃.
2) Potassium permanganate is added to a hot solution of manganous sulphate.

Ans:

1)  With cold dilute HNO₃ iron produces ferrous nitrate and ammonium nitrate
2Fe + 10HNO₃  4Fe(NO₃)₂ + NH₄NO₃ + 3H₂O

2) In this case pottasium permangnate oxidises MnSO₄ to MnO₂

2KMnO₄ + 3MnSO₄ + 2H₂O  5MnO₂ + K₂SO₄ + 2H₂SO₄

Q25. Explian the following
(i)Entropy of pure crystalline substance is zero at 0 K.
(ii)All exothermic reactions are not spontaneous.

Ans:(i) Pure substances have perfectly orderly arrangement of constituent particles.Hence, there is no disorder.That’s why the entropy of pure crystalline substance is zero at 0 K.
(ii)ΔG =ΔH-TΔs
For a reaction to be spontaneous ΔG must be negative. If reaction is exothermic, ΔH is negative. Thus, it is magnitude of TΔS which decides whether the exothermic reaction will be spontaneous or not.

Q26. Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g mL^-1 and the mass per cent of nitric acid in it being 69%.

Ans:

Mass percent of nitric acid in the sample = 69 % [Given]

Thus, 100 g of nitric acid contains 69 g of nitric acid by mass.

Molar mass of nitric acid (HNO)₃

= {1 + 14 + 3(16)} g mol^–1

= 1 + 14 + 48

= 63 g mol^–1

∴ Number of moles in 69 g of HNO₃

= 69 63 g mo l −1 =1.095 mol Volume of 100g of nitric acid solution = Mass of Solution density of solution = 100 g 1.41 g m L −1 =70.92 mL ≡ 70.92 × 10 −3 L Concentration of nitric acid = 1.095 mole 70.92 × 10 −3 L = 15.44 mol/L ∴Concentration of nitric acid = 15.44 mol/LMathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhcba9y8qqasFr0xc9YqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOabaeqabaaeaaaaaaaaa8qacqGH9aqpdaWcaaWdaeaapeGaaGOnaiaaiMdaa8aabaWdbiaaiAdacaaIZaacbiGaa8hOaiaa=DgacaWFGcGaa8xBaiaa=9gacaWFSbWdamaaCaaaleqabaWdbiabgkHiTiaaigdaaaaaaaGcbaWdaiabg2da9iaabgdacaGGUaGaaGimaiaabMdacaqG1aGaaeiiaiaab2gacaqGVbGaaeiBaaqaaiaabAfacaqGVbGaaeiBaiaabwhacaqGTbGaaeyzaiaabccacaqGVbGaaeOzaiaabccacaqGXaGaaGimaiaaicdacaqGNbGaaeiiaiaab+gacaqGMbGaaeiiaiaab6gacaqGPbGaaeiDaiaabkhacaqGPbGaae4yaiaabccacaqGHbGaae4yaiaabMgacaqGKbGaaeiiaiaabohacaqGVbGaaeiBaiaabwhacaqG0bGaaeyAaiaab+gacaqGUbaabaWdbiabg2da9maalaaapaqaa8qacaWFnbGaa8xyaiaa=nhacaWFZbGaa8hOaiaa=9gacaWFMbGaa8hOaiaa=nfacaWFVbGaa8hBaiaa=vhacaWF0bGaa8xAaiaa=9gacaWFUbaapaqaa8qacaWFKbGaa8xzaiaa=5gacaWFZbGaa8xAaiaa=rhacaWF5bGaa8hOaiaa=9gacaWFMbGaa8hOaiaa=nhacaWFVbGaa8hBaiaa=vhacaWF0bGaa8xAaiaa=9gacaWFUbaaaaqaaiabg2da9maalaaapaqaa8qacaaIXaGaaGimaiaaicdacaWFGcGaa83zaaWdaeaapeGaaGymaiaac6cacaaI0aGaaGymaiaa=bkacaWFNbGaa8hOaiaa=1gacaWFmbWdamaaCaaaleqabaWdbiabgkHiTiaaigdaaaaaaaGcbaGaeyypa0JaaG4naiaaicdacaGGUaGaaGyoaiaaikdacaWFGcGaa8xBaiaa=XeacaWFGcGaeyyyIORaa8hOaiaaiEdacaaIWaGaaiOlaiaaiMdacaaIYaGaa8hOaiabgEna0kaaigdacaaIWaWdamaaCaaaleqabaWdbiabgkHiTiaaiodaaaGccaWFmbGaa8hOaaqaa8aacaqGdbGaae4Baiaab6gacaqGJbGaaeyzaiaab6gacaqG0bGaaeOCaiaabggacaqG0bGaaeyAaiaab+gacaqGUbGaaeiiaiaab+gacaqGMbGaaeiiaiaab6gacaqGPbGaaeiDaiaabkhacaqGPbGaae4yaiaabccacaqGHbGaae4yaiaabMgacaqGKbaabaWdbiabg2da9iaa=bkadaWcaaWdaeaapeGaaGymaiaac6cacaaIWaGaaGyoaiaaiwdacaWFGcGaa8xBaiaa=9gacaWFSbGaa8xzaaWdaeaapeGaaG4naiaaicdacaGGUaGaaGyoaiaaikdacaWFGcGaey41aqRaaGymaiaaicdapaWaaWbaaSqabeaapeGaeyOeI0IaaG4maaaakiaa=XeaaaaabaWdaiabg2da9iaabccacaqGXaGaaeynaiaac6cacaqG0aGaaeinaiaabccacaqGTbGaae4BaiaabYgacaGGVaGaaeitaaqaa8qacqGH0icxpaGaae4qaiaab+gacaqGUbGaae4yaiaabwgacaqGUbGaaeiDaiaabkhacaqGHbGaaeiDaiaabMgacaqGVbGaaeOBaiaabccacaqGVbGaaeOzaiaabccacaqGUbGaaeyAaiaabshacaqGYbGaaeyAaiaabogacaqGGaGaaeyyaiaabogacaqGPbGaaeizaiaabccacqGH9aqpcaqGGaGaaeymaiaabwdacaGGUaGaaeinaiaabsdacaqGGaGaaeyBaiaab+gacaqGSbGaai4laiaabYeaaaaa@1B36@

Q27. Write a Short note on Electromeric Effect and Hyperconjucation.

Ans:

“Electromeric Effect:   This is the complete transfer of shared pair of  electrons to one of the atoms joined by a multiple bond on the demand of an attacking reagent. It is a temporary effect. It can be divided into two parts:

Positive Electromeric Effect (+E)

Negative Electromeric Effect (-E)

 Hyperconjucation:    It is electronic interaction occurring between a double bond in a compound and a Carbon- Hydrogen single bond in attached alkyl group. Under certain Circumstances, the double bond and single C-H bond appear to behave rather like conjugated double bond. Greater the number of alkyl group, greater will be the hyperconjugation.

Q28. What are the  various  factors due  to  which  the ionization  enthalpy of the main group  elements tends to decrease  down  a  group?

Ans: The factors responsible for the ionization enthalpy of the main group elements to decrease down a  group  are  listed  below:

(i)  Increase  in  the  atomic  size  of  elements:  As  we  move  down  a  group,  the  number  of shells increases.  As  a  result,  the  atomic  size  also  increases  gradually  on  moving  down  a group.  As  the  distance  of  the  valence  electrons  from  the  nucleus  increases,  the  electrons are  not  held  very  strongly.  Thus, they can be removed easily.  Hence, on moving down a group, ionization energy decreases.

(ii)  Increase  in  the  shielding  effect:  The  number  of  inner  shells  of  electrons  increases  on moving  down  a  group.  Therefore, the shielding of the valence electrons from the nucleus by  the  inner  core  electrons  increases  down  a  group.  As  a result,  the  valence  electrons are  not  held  very  tightly  by  the  nucleus.  Hence, the energy required to remove a  valence electron  decreases  down  a  group.

Q29.

Give condensed and bond line structural formulas and identify the functional group(s) present, if any, for:

(a) 2, 2, 4-Trimethylpentane

(b) 2-Hydroxy-1, 2, 3-propanetricarboxylic acid

(c) Hexanedial

Opt:

Condensed structure:

a) CH₃C(CH₃)₂CH₂CH(CH₃)CH₃

b) HOOCCH₂C(OH)(COOH)CH₂COOH

c) OHC(CH₂)₄CHO

Bond line formula:

Ans:

Mass percent of nitric acid in the sample = 69 % [Given]

Thus, 100 g of nitric acid contains 69 g of nitric acid by mass.

Molar mass of nitric acid (HNO)₃

= {1 + 14 + 3(16)} g mol^–1

= 1 + 14 + 48

= 63 g mol^–1

∴ Number of moles in 69 g of HNO₃

= 69 63 g mo l −1 =1.095 mol Volume of 100g of nitric acid solution = Mass of Solution density of solution = 100 g 1.41 g m L −1 =70.92 mL ≡ 70.92 × 10 −3 L Concentration of nitric acid = 1.095 mole 70.92 × 10 −3 L = 15.44 mol/L ∴Concentration of nitric acid = 15.44 mol/LMathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8wrps0lbbf9q8WrFfeuY=Hhcba9y8qqasFr0xc9YqFj0dXdbba91qpepGe9FjuP0=is0dXdbba9pGe9xq=Jbba9suk9fr=xfr=xfrpeWZqaaeaabaGaaiaacaqabeaadaabauaaaOabaeqabaaeaaaaaaaaa8qacqGH9aqpdaWcaaWdaeaapeGaaGOnaiaaiMdaa8aabaWdbiaaiAdacaaIZaacbiGaa8hOaiaa=DgacaWFGcGaa8xBaiaa=9gacaWFSbWdamaaCaaaleqabaWdbiabgkHiTiaaigdaaaaaaaGcbaWdaiabg2da9iaabgdacaGGUaGaaGimaiaabMdacaqG1aGaaeiiaiaab2gacaqGVbGaaeiBaaqaaiaabAfacaqGVbGaaeiBaiaabwhacaqGTbGaaeyzaiaabccacaqGVbGaaeOzaiaabccacaqGXaGaaGimaiaaicdacaqGNbGaaeiiaiaab+gacaqGMbGaaeiiaiaab6gacaqGPbGaaeiDaiaabkhacaqGPbGaae4yaiaabccacaqGHbGaae4yaiaabMgacaqGKbGaaeiiaiaabohacaqGVbGaaeiBaiaabwhacaqG0bGaaeyAaiaab+gacaqGUbaabaWdbiabg2da9maalaaapaqaa8qacaWFnbGaa8xyaiaa=nhacaWFZbGaa8hOaiaa=9gacaWFMbGaa8hOaiaa=nfacaWFVbGaa8hBaiaa=vhacaWF0bGaa8xAaiaa=9gacaWFUbaapaqaa8qacaWFKbGaa8xzaiaa=5gacaWFZbGaa8xAaiaa=rhacaWF5bGaa8hOaiaa=9gacaWFMbGaa8hOaiaa=nhacaWFVbGaa8hBaiaa=vhacaWF0bGaa8xAaiaa=9gacaWFUbaaaaqaaiabg2da9maalaaapaqaa8qacaaIXaGaaGimaiaaicdacaWFGcGaa83zaaWdaeaapeGaaGymaiaac6cacaaI0aGaaGymaiaa=bkacaWFNbGaa8hOaiaa=1gacaWFmbWdamaaCaaaleqabaWdbiabgkHiTiaaigdaaaaaaaGcbaGaeyypa0JaaG4naiaaicdacaGGUaGaaGyoaiaaikdacaWFGcGaa8xBaiaa=XeacaWFGcGaeyyyIORaa8hOaiaaiEdacaaIWaGaaiOlaiaaiMdacaaIYaGaa8hOaiabgEna0kaaigdacaaIWaWdamaaCaaaleqabaWdbiabgkHiTiaaiodaaaGccaWFmbGaa8hOaaqaa8aacaqGdbGaae4Baiaab6gacaqGJbGaaeyzaiaab6gacaqG0bGaaeOCaiaabggacaqG0bGaaeyAaiaab+gacaqGUbGaaeiiaiaab+gacaqGMbGaaeiiaiaab6gacaqGPbGaaeiDaiaabkhacaqGPbGaae4yaiaabccacaqGHbGaae4yaiaabMgacaqGKbaabaWdbiabg2da9iaa=bkadaWcaaWdaeaapeGaaGymaiaac6cacaaIWaGaaGyoaiaaiwdacaWFGcGaa8xBaiaa=9gacaWFSbGaa8xzaaWdaeaapeGaaG4naiaaicdacaGGUaGaaGyoaiaaikdacaWFGcGaey41aqRaaGymaiaaicdapaWaaWbaaSqabeaapeGaeyOeI0IaaG4maaaakiaa=XeaaaaabaWdaiabg2da9iaabccacaqGXaGaaeynaiaac6cacaqG0aGaaeinaiaabccacaqGTbGaae4BaiaabYgacaGGVaGaaeitaaqaa8qacqGH0icxpaGaae4qaiaab+gacaqGUbGaae4yaiaabwgacaqGUbGaaeiDaiaabkhacaqGHbGaaeiDaiaabMgacaqGVbGaaeOBaiaabccacaqGVbGaaeOzaiaabccacaqGUbGaaeyAaiaabshacaqGYbGaaeyAaiaabogacaqGGaGaaeyyaiaabogacaqGPbGaaeizaiaabccacqGH9aqpcaqGGaGaaeymaiaabwdacaGGUaGaaeinaiaabsdacaqGGaGaaeyBaiaab+gacaqGSbGaai4laiaabYeaaaaa@1B36@

Q30. Write a Short note on Electromeric Effect and Hyperconjucation.

Ans:

“Electromeric Effect:   This is the complete transfer of shared pair of  electrons to one of the atoms joined by a multiple bond on the demand of an attacking reagent. It is a temporary effect. It can be divided into two parts:

Positive Electromeric Effect (+E)

Negative Electromeric Effect (-E)

 Hyperconjucation:    It is electronic interaction occurring between a double bond in a compound and a Carbon- Hydrogen single bond in attached alkyl group. Under certain Circumstances, the double bond and single C-H bond appear to behave rather like conjugated double bond. Greater the number of alkyl group, greater will be the hyperconjugation.

Q31. What are the  various  factors due  to  which  the ionization  enthalpy of the main group  elements tends to decrease  down  a  group?

Ans:

The factors responsible for the ionization enthalpy of the main group elements to decrease down a  group  are  listed  below:

(i)  Increase  in  the  atomic  size  of  elements:  As  we  move  down  a  group,  the  number  of shells increases.  As  a  result,  the  atomic  size  also  increases  gradually  on  moving  down  a group.  As  the  distance  of  the  valence  electrons  from  the  nucleus  increases,  the  electrons are  not  held  very  strongly.  Thus, they can be removed easily.  Hence, on moving down a group, ionization energy decreases.

(ii)  Increase  in  the  shielding  effect:  The  number  of  inner  shells  of  electrons  increases  on moving  down  a  group.  Therefore, the shielding of the valence electrons from the nucleus by  the  inner  core  electrons  increases  down  a  group.  As  a result,  the  valence  electrons are  not  held  very  tightly  by  the  nucleus.  Hence, the energy required to remove a  valence electron  decreases  down  a  group.

Q32. (i)Write the electronic configurations of the following ions: (a) H^– (b) Na⁺(c) O^2–(d) F^–

(ii) What are the atomic numbers of elements whose outermost electrons are represented by

(a) 3s¹ (b) 2p³ and (c) 3p⁵?

(iii)Which atoms are indicated by the following configurations?

(a) [He] 2s¹

(b) [Ne] 3s²3p³

(c) [Ar] 4s²3d¹.

Or

Indicate the number of unpaired electrons in: (a) P, (b) Si, (c) Cr, (d) Fe and (e) Kr.

Ans:

(i) (a) H^– ion:

The electronic configuration of H atom is 1s¹.

A negative charge on hydrogen indicates the gain of electron by it.

Electronic configuration of H^– = 1s²

(b) Na⁺ ion:

The electronic configuration of Na atom is 1s²2s²2p⁶3s¹.

A positive charge on the species indicates the loss of an electron by it.

Electronic configuration of Na⁺ =1s² 2s² 2p⁶ 3s⁰

=1s² 2s² 2p⁶

(c)O^2– ion:

The electronic configuration of O atom is 1s² 2s² 2p⁴

A dinegative charge on the species indicates that two electrons are gained by it.

Electronic configuration of O^2–ion = 1s² 2s² 2p⁶

(d)F^–ion:

The electronic configuration of F atom is 1s² 2s² 2p⁵.

A negative charge on the species indicates the gain of an electron by it.

Electron configuration of F^– ion = 1s² 2s² 2p⁶

(ii) (a) 3s¹:

Completing the electron configuration of the element as = 1s² 2s² 2p⁶ 3s¹

Number of electrons present in the atom of the element = 2 + 2 + 6 + 1 = 11

Atomic number of the element = 11

(b)2p³:

Completing the electron configuration of the element as

= 1s²2s²2p³

Number of electrons present in the atom of the element = 2 + 2 + 3 = 7

Atomic number of the element = 7

(c) 3p⁵:

Completing the electron configuration of the element as

= 1s²2s²2p⁵

Number of electrons present in the atom of the element = 2 + 2 + 5 = 9

Atomic number of the element = 9

(iii) (a)[He] 2s¹:

The electronic configuration of the element is

[He] 2s¹= 1s²2s¹.

Atomic number of the element = 3

The element is lithium (Li).

b)[Ne] 3s²3p³ :

The electronic configuration of the element is

[Ne]3s²3p³ = 1s²2s²2p⁶3s²3p³

Atomic number of the element = 15

Hence, the element with the electronic configuration

[Ne] 3s²3p³ is phosphorus (P).

(c)[Ar] 4s²3d¹:

The electronic configuration of the element is [Ar]4s²3d¹= 1s²2s²2p⁶3s²3p⁶4s²3d¹.

Atomic number of the element = 21

Hence, the element with the electronic configuration

[Ar] 4s²3d¹ is scandium (Sc).

Or

(a) Phosphorus (P):

Atomic number = 15

The electronic configuration of P is: 1s²2s²2p⁶3s²3p³

The orbital picture of P can be represented as:

Phosphorus has three unpaired electrons.

(b) Silicon (Si):

Atomic number = 14

The electronic configuration of Si is: 1s²2s²2p⁶3s²3p²

The orbital picture of Si can be represented as:

Silicon has two unpaired electrons.

(c) Chromium (Cr):

Atomic number= 24

The electronic configuration of Cr is:

1s²2s²2p⁶3s²3p⁶4s¹3d⁵

The orbital picture of Cr can be represented as:

Chromium has six unpaired electrons.

(d) Iron (Fe):

Atomic number= 26

The electronic configuration is:

1s²2s²2p⁶3s²3p⁶4s²3d⁶

The orbital picture of Fe can be represented as:

Iron has four unpaired electrons.

(e) Krypton (Kr):

Atomic number= 36

The electronic configuration is:

1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶

The orbital picture of Kr can be represented as:

Krypton has no unpaired electrons, all orbitals are fully occupied.

Q33. What are structural isomers. Explain different types of structural isomers with examples.

Ans:

The compounds which have the same molecular formula but different structural formula are called structural isomers and this phenomenon is known as structural isomerism. There are five types of structural isomerism:
(i) Chain isomerism:- When two or more compounds have similar molecular formula but different carbon skeletons, they are referred to as chain isomers and the phenomenon is termed as chain isomerism.

For example, C₄H₁₀ represents two compounds.

    

(ii) Position isomerism:-
When two or more compounds have similar molecular formula but they differ in the position of the functional group, C-C bond or substituent group are called position isomers and the phenomenon is termed as position isomerism.
Example:-

(iii) Functional group isomerism :- When two or more compounds have similar molecular formula but different functional groups in the molecule are called functional group isomers and the phenomenon is termed as functional group isomerism.

Example:-

              

(iv) Metamerism:-The compounds which have the same molecular formula but different number of carbon atoms (or alkyl groups) on either side of the functional group are called metamers and this phenomenon is known as metamerism.

Example:-

     

(v) Tautomerism:- It is a special kind of functional group isomerism in which the isomers differ in the arrangement of atoms but they exist in dynamic equilibrium with each other and this phenomenon is termed as tautomerism.

Example:-