Q.1. ICl is more or less reactive than Cl2? Why?
Ans: ICl is more reactive than Cl2.
It is due to lower bond dissociation energy due to longer bond length and more polarity in ICl than Cl2.
Q.2. Why are NH3, glucose, and alcohol soluble in water, although they are covalent compounds?
Ans: They can form H-bonds with water because they are polar covalent compounds.
Q.3. On the basis of VSEPR theory, predict the shapes of the following molecules and ions:
(i) PH3
(ii) NH3
(iii) NH2–
(iv) H3O+
Ans:
(i) PH3 has 3 bond pairs, one lone pair, it has pyramidal shape.
(ii) NH3 has 3 bond pairs and one lone pair, it has pyramidal shape.
(iii) NH2- has 2 bond pairs and 2 lone pairs, therefore, it has V-shape or bent molecule.
(iv) H3O+ has 3 bond pairs and one lone pair, therefore, it is pyramidal.
Q.4. Compare the dipole moment of the compounds in each of the following sets:
(i) CHCl3, CCl4
(ii) CF4, SF4
(iii) BF3, BCl3
(iv) CO2, SO2
Ans:
(i) CHCl3 has higher dipole moment than CCl4.
(ii) SF4 has higher dipole moment than CF4.
(iii) BF3 and BCl3 have zero dipole moment or bent molecule.
(iv) SO2 has higher dipole moment than CO2.
Q.5. Which of the following has highest lattice energy and why?
CsF, CsCl, CsBr, CsI
Ans: CsF has highest lattice energy because 'F' is smallest in size and is more electronegative, therefore, it has maximum ionic character and maximum force of attraction, hence highest lattice energy.
Q.6. Why is dipole moment of HF (1.98 D) higher than that of HCl (1.03 D)?
Ans: HF is more polar as compared to HCl because 'F' is more electronegative than Cl. Greater the difference in electronegativity, more will be polarity, higher will be dipole moment.
Q.7. Calculate the number of σ and π–electrons in 0.1 mole of vinyl cyanide
(CH2=CH–C≡N)
Ans: In one molecule of vinyl cyanide there are 6 σ bonds and 3 π bonds, i.e. total 9 bonds, i.e. total 18 electrons.
l mole of vinyl cyanide will have 18 × 6.023 × 1023 electrons.
0.1 mole of vinyl cyanide will have 18 × 6.023 × 1023 × 0.1 = 10.8414 × 1023
= 1.08414 × 1024 electrons
Q.8. Give electron dot structure for carbon suboxide, C3O2. Write its structural formula.
Ans:
Q.9. AlF3 is a high melting solid whereas SiF4 is a gas. Explain.
Ans: AlF3 is an ionic compound and there is strong force of attraction between Al3+ and F- ions.
Q.10. Arrange the following in decreasing order of ionic characters of the bond and give reason:
NaF, NaBr, NaCl, NaI
Ans: NaF > NaCl > NaBr > NaI
Greater the difference in electronegativity, more will be ionic character.
Q.11. Explain the shapes of the following on the basis of VSEPR theory:
(i) BeCl2
(ii) PH4+
(iii) PF5
(iv) SF6
Ans:
(i) BeCl2 has 2 bond pairs of electrons, therefore, it has linear shape.
(ii) PH4+ has 4 bond pairs of electrons, it is tetrahedral.
(iii) PF5 has 5 bond pairs of electrons, it is trigonal bipyramidal.
(iv) SF6 has 6 bond pairs of electrons, therefore, it has octahedral shape.
Q.12. How do you express the bond strength in terms of bond order?
Ans: The higher bond order, more will be stability and smaller will be bond length.
Q.13. Define octet rule. Write its significance and limitations.
Ans:
Octet Rule: Every element tries to acquire 8 electrons in its outermost orbit by losing or gaining or sharing electrons.
Significance: Most of the molecules are formed by following octet rule e.g. O2, N2, Cl2, Br2 etc. It is useful for understanding most of the organic compounds and it applies mainly to the second period elements of the periodic table.
Limitations:
· Hydrogen gains one electron to fill its shell and attain Helium gas configuration (1s2). In this case, octet is not completed to achieve a stable configuration.
· In some compounds, the number of electrons surrounding the central atom is less than 8.
· In some compounds, the number of electrons surrounding the central atom is more than 8. E.g., In PF5, SF6, H2SO4, there is expanded octet or super octet, i.e. 10, 12 and 14 electrons respectively after sharing.
· In NO, NO2, octet rule is not satisfied. They are odd electron molecules.
· The octet rule fails to explain the formation of compounds of noble gases especially xenon and krypton such as XeF2, XeF6, etc.
· Octet rule theory does not account for the shape of the molecules.
· Octet rule theory cannot explain the relative stability of the molecule in terms of the energy.
Q.14. Write the favourable factors for the formation of ionic bond.
Ans:
(i) Lower Ionisation Enthalpy i.e., Larger metal ion
(ii) High Electron gain enthalpy, i.e., Smaller non-metallic ion
(iii) High lattice enthalpy of the compound formed.
Q.15. Explain why PCl5 is trigonal bipyramidal whereas IF5 is square pyramidal.
Ans: PCl5 has 5 bonded pairs of electrons, therefore, it is trigonal bipyramidal. It has sp3d hybridisation. IF5 has 5 bonded pairs and 1 lone pair of electrons, therefore, it is square pyramidal. It has sp3d2 hybridisation.
Q.16. Define hybridisation. Explain the structure of C2H2 with orbital diagram.
Ans: Hybridisation is process of intermixing of atomic orbitals of slightly different energies which give rise to hybridised orbitals having exactly equal energy, identical shape and more stability.
In H-C≡C-H each 'C' is 'sp' hybridised, therefore, it has linear shape.
Q.17. Give reason for the following:
(i) BF3 has a zero–dipole moment although the B–F bonds are polar.
(ii) All carbon to oxygen bonds in CO32– are equivalent.
Ans:
(i) Due to planar structure, individual dipoles get cancelled.
(ii) It is due to resonance.
Q.18. Give two differences between σ and π–bonds.
Ans:
|
Sigma bond |
Pi bond |
|
1. It is formed by overlap of orbitals along internuclear axis. |
1. It is formed by sideways overlapping of orbitals. |
|
2. It is formed by overlapping of s-s orbital, s-p orbital and p-p orbital. |
2. It is formed by overlap of p-p and p-,d orbitals. |
|
3. The electron cloud of a sigma bond is symmetrical about the internuclear axis and the overlapping is quite large, thus sigma bonds are strong. |
3. As the electron clouds overlap above and below the plane of internuclear axis, the overlapping is small, thus pi bonds are weak. |
|
4. Free rotation about a sigma bond is possible. |
4. Free rotation about a pi bond is not possible, since on rotating the overlapping vanishes and the bond breaks. |
|
5. Bond length is longer. |
5. Bond length is shorter. |
Q.19. Why the structure of NH3 molecule is pyramidal?
Ans: NH3 is pyramidal due to the presence of lone pair of electrons on nitrogen.
Q.20. (i) Describe the hybridisation in case of PCl5.
(ii) Deduce the shape of SF4 molecule on the basis of VSEPR theory.
Ans:
(i) Electronic configuration of P(15): 1s2 2s2 2p6 3s2 3p3
P(15) in excited state
sp3d hybridisation
It has sp3d hybridisation
(ii) In SF4, there are four bonded pair of electrons and one lone pair of electrons. It has see-saw shape so as to have minimum repulsion. It has see-saw shape.
Q.21. Bond angle in NH3 is more than in H2O. Justify.
Ans: Both NH3 and H2O are sp3 hybridised but there is only one lone pair present on N in NH3 and two lone pairs on O of H2O. Since lone pair-lone pair repulsion is greater than lone pair-bond pair and bond pair-bond pair repulsions; two lone pairs on oxygen push the bond pairs closer than one lone pair on nitrogen. This leads to smaller bond angle in H2O than in NH3.
Q.22. Which hybrid orbitals are used by underlined carbon in the following molecules?
(i) CH3CHO
(ii) CH3–CH=CH2
Ans:
(i) sp2
(ii) sp2
Q.23. (i) Explain the structure of CO32– ion in terms of resonance.
(ii) Which out of NH3 and NF3 has higher dipole moment and why?
Ans:
(i) In the resonance structure of CO32– ion, two C-O bonds are represented by single bonds with negative charge on oxygen and one of the C-O bonds is a double bond. All C-O bonds in CO32– are equivalent and hence CO32– is represented as the resonance hybrid of the canonical forms I, II and III below:
(ii) Dipole moment of NH3 is higher than that of NF3. This is because in NH3, the orbital dipole due to lone pair on N is in the same direction as the dipole of N-H bond. On the other hand, in NF3 direction of orbital dipole and bond dipole is opposite because of F being more electronegative than N.
Q.24. (i) How many sigma and pi bonds are there in the following molecule.
CH2=CH–CH2–C≡CH
(ii) Which type of hybrid orbitals are used by the second carbon atom in the following molecule.
CH≡C–CH2–CH=CH2
Ans:
(i) 10 σ and 3 π bonds
(ii) sp2 hybridisation. (while numbering the C-atoms, double bond is preferred over triple bond if they are equidistant)
5 4 3 2 1
CH≡C–CH2–CH=CH2
Q.25. You are given the electronic configuration of A, B, C, D and E:
A: ls2 2s2 2p6 3s2
B: 1s2 2s2 2p6 3s1
C: ls2 2s2 2pl
D: 1s2 2s2 2p5
E: ls2 2s2 2p6
Write the empirical formula for the substance containing:
(i) A and D
(ii) B and D
(iii) Only D
(iv) Only E
Ans:
(i) AD2
(ii) BD
(iii) D2
(iv) E
Q.26. Give shapes of
(i) NH4+
(ii) CO32–
(iii) BeF3–
(iv) SO42–
Ans:
(i) NH4+ is tetrahedral
(ii) CO32– is trigonal planar
(iii) BeF3– is trigonal planar
(iv) SO42– is tetrahedral.
Q.27. Why is BeCl2 linear whereas SnCl2 angular molecule?
Ans: BeCl2 is linear due to sp hybridisation. In SnCl2, there is sp2 hybridisation with a lone pair of electrons, hence it is bent molecule.
Q.28. Consider each of the following in solid state:
(i) Methane
(ii) Cesium chloride
(iii) Germanium,
(iv) Lithium
(v) Argon
(vi) Ice
State which would be an example of
(a) high melting, network solid
(b) a non–conducting solid which becomes a good conductor in the molten state
(c) a solid with high electrical and thermal conductivity
(d) a low melting solid held together by van der Waal's forces
(e) a solid in which hydrogen exists
Ans:
(i) Germanium
(ii) Caesium chloride
(iii) Lithium
(iv) Argon
(v) Ice
Q.29. Which end of ICl will be positive and which will be negative and why? Is it covalent or ionic?
Ans: I will be positive, Cl will be negative because Cl is more electronegative than I. It is a covalent compound.
Q.30. Write molecular orbital configuration of O2. Predict its magnetic behaviour and. Calculate its bond order.
Ans: O2(16), M.O. (configuration):
It is paramagnetic in nature due to presence of two unpaired electrons.
Bond Order = 
Q.31. Explain the following observations:
(i) CO2 and SO2 are not isostructural.
(ii) BF3 and NF3 are not isostructural.
(iii) BH4– and NH4+ are isostructural.
(iv) N2 has higher bond order than NO.
(v) In NO+ and N2, the bond order is same.
Ans:
(i) CO2 is linear due to sp hybridisation, SO2 is bent molecule due to sp2 hybridisation and one lone pair of electrons.
(ii) BF3 is planar due to sp2 hybridisation, NF3 is pyramidal due to sp3 hybridisation and one lone pair of electrons.
(iii) BH4– and NH4+ both are sp3 hybridised, tetrahedral in shape, therefore, isolable, i.e., have same structure.
(iv) N2 has bond order 3 because
Bond order = 
NO has 15 electrons, bond order is 2.5,
Bond order = 
N2 has one electron less in antibonding orbital than NO.
(v) NO+ and N2 have same number of bonding and antibonding electrons.
Q.32. N2 molecule has a greater bond dissociation energy than N2+ ion whereas O2 molecule has a lower bond dissociation energy than O2+ ion. Explain in terms of M.O. theory.
Ans:
Bond order =
Bond order = 
Since N2 has higher bond order than N2+, therefore N2 has higher bond dissociation energy than N2+.
Molecular orbital configuration of O2 is
Bond order of O2 =
Molecular orbital, configuration of O2+ is
Bond order of O2+ =
O2+ has higher bond order than O2, therefore, it has high bond dissociation energy than O2.
Q.33. Compare the relative stability of following species and indicate their magnetic properties
O2, O2+, O2– (superoxide), O22– (peroxide)
Ans: Molecular orbital configuration of O2 is
Bond order of O2 =
Molecular orbital, configuration of O2+ is
Bond order of O2+ =
Molecular orbital configuration of O2– is
Bond order of O2– = 
Molecular orbital configuration of O22–is
Bond order of O22– = 
Greater the bond order, greater is the bond dissociation energy and hence greater is the stability of species.
O2+ > O2 > O2– > O22–
On the basis of the presence of unpaired electrons, their magnetic nature is inferred as
O2 = paramagnetic
O2+ = paramagnetic
O2– = paramagnetic
O22– = diamagnetic
Q.34. Compare the relative stability of the following species on the basis of molecular orbital theory and indicate their magnetic properties: O2+, O2–.
Ans: Molecular orbital configuration of O2 is
Hence molecular orbital, configuration of O2+ is
Bond order of O2+ =
It is paramagnetic.
Molecular orbital configuration of O2– is
Bond order of O2– =
It is paramagnetic.
Since, the bond order of O2+ is greater than O2– therefore, O2+ is more stable than O2–.
Q.35. Draw molecular orbital energy level diagram for N2+. Calculate its bond order and explain its magnetic characteristics.
Ans:
Bond order =
Molecular orbital diagram of N2+.
It is paramagnetic due to presence of one unpaired electron.
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