Questions - Answers
1. Explain why the bond angle in water molecule is 104.5°?
A: O in H2O undergoes sp3 hybridisation. O has two lone pairs of electrons in its outermost shell. Due to lone pair - lone pair repulsion, bond angle decreases from 109° 28' to 104.5°.
2. Why the boiling point of H2O is greater than HF?
A: Each H2O molecule can form 4, HF can form 2 Hydrogen bonds. In vapour state
H2O exists as isolated molecules. But HF exists as poly molecular.
3. Mention the types of bonds present in NH4Cl.
A:
1, 2, 3 - Covalent bonds
4 - dative bond
5 - ionic bond
4. Chloride ion is more stable than Cl atom. Why?
A: Cl has 7 electrons (3s2 4p5), Cl - has octet (3s2 3p6) in its valence shell. So Cl- is more stable.
5. Which of the two cations Ca+2 or Zn+2 is more stable and why?
A: Ca+2 has stable octet (2, 8, 8) configuration where as Zn+2 has pseudo inert gas (2, 8, 18) configuration in the valency shell. Ca+2 has octet configuration and is more stable than Zn+2.
6. How many sigma and pi bonds are present in a) C2H2 and b) C2H4
A: 
C2H2 has 3σ and 2Π bonds
C2H4 has 5σ and 1Π bond.
7. Even though Nitrogen in Ammonia is in sp3 hybridization the bond angle deviate from 109° 28'. Explain.
A: 
Due to presence of one lone pair of electrons in one sp3 hybrid orbital of Nitrogen, there will be lone pair-bond pair repulsion. Hence bond angle decreases from 109° 28' to 107°
8. Even though both NH3 and NF3 are pyramidal, NH3 has a higher dipole moment compared to NF3. Why?
A: 
NH3 & NF3 are pyramidal and having lone pair of electrons each. In NH3, the orbital dipole due to lone pair and dipole moment of N-H bonds are in the same direction. So NH3 has higher dipole moment. Where as in the case of NF3, they are in the opposite direction, which causes lower dipole moment.
9. Define dipole moment.
A: The product of the charge (Q) and the distance between the centres of positive and negative charges (r).
μ = Q.r; Units: Debye
1 Debye = 3.34 × 10-30 C. m (Coulomb meters)
10. Give two applications of dipole moment.
A: → C is isomers have dipole moment while trans isomers have zero dipole moment.
→ μ is useful in calculating % of ionic character of polar covalent bonds.
11. What is hydrogen bond? Explain the different types of hydrogen bonds with examples.
A: The weak electrostatic force of attraction between the hydrogen atom of one molecule and highly electronegative atom like F, O, N of the same or different molecule.
Inter molecular hydrogen bond: Hydrogen bond formed between H and electronegative atom of two different molecules of same compounds or different compounds.
e.g.: HF, H2O, NH3
Intra molecular hydrogen bond: Hydrogen bond formed between H and electronegative atom of the same molecule.
e.g.: O - Nitro phenol, O - Hydroxy Benzaldehyde.
12. Explain molecular orbital theory of Nitrogen molecule. Calculate its bond order and explain its magnetic property.
A:
N N2 N
Atomic Orbital Molecular Orbital Atomic Orbital
M.O. Configuration of N2 = 
Bond order of N2 = 
(Nb - Na) = (10 - 4) = 3
As all the orbitals are paired with electrons, it is diamagnetic.
13. Explain molecular orbital theory of Oxygen molecule. Calculate its bond order and explain its magnetic property.
A:
O O2 O
Atomic Orbital Molecular Orbital Atomic Orbital
M.O. Configuration of
Bond order of O2 = (Nb - Na) = (10 - 6) = 2
As 2 orbitals are having unpaired electrons, O2 is paramagnetic.
14. Give main postulates of V S E P R theory.
A: To explain the shapes of simple covalent molecules. Sidgwick and Powell, proposed a simple theory " Valency shell electron pair repulsion theory [VSEPR].
The main postulates of this theory are:
1) Central atom may have lone pairs or bond pairs in its valency shell.
2) Lone pairs occupy more space and bond pairs occupy less space.
3) Electron pairs around the central atom orient themselves, so that the repulsive forces are minimum between them.
4) The magnitude of repulsions between bond pairs depend on the electronegativity difference between central atom and other atoms (Bonded).
5) Order of repulsive forces: L.P. - L.P. > L.P. - B.P. - B.P.
6) The shape of the molecule depends on the no. of electron pairs present in the valence shell of the central atom.
7) If the no. of bond pairs are 2, 3, 4, 5, 6; the shapes of the molecules are linear, plane triangle, tetrahedral, trigonal bipyramidal, octahedral respectively.
15. Explain sp3d hybridisation with one example.
A: sp3d hybridisation: One s, three p, one d orbitals of same valence shell of an atom intermix together to give 5 identical sp3d H.O.'S is known as sp3d hybridisation. e.g.: PCl5
P = 1s2 2s2 2p6 3s1 3px1 3py1 3pz1 3d1 (1st excited state) 
One 3s, three 3p and one 3d atomic orbitals of valence shell of Phosphorous overlap to give 5 identical sp3 d hybrid orbitals. These 5 H.O's overlap with p orbitals of 5 Cl atoms. The shape of the molecule is trigonal bipyramidal, bond angle in the plane is 120°, perpendicular to plane is 90°. % of s & d is 20% each, % of p is 60%.
16. Explain the type of hybridisation involved in SF6.
A: sp3d2 hybridisation: One s, three p, two d orbitals of same valence shell of an atom intermix together to give 6 identical sp3d2 H.O.'s is known as sp3d2 hybridisation.
e.g.: SF6 S = 1s2 2s2 2p6 3s1 3px1 3py1 3pz1 3d1 3d1 (2nd excited state)
One 3s, three 3p, one 3d orbitals of same valence shell of 's' intermix to give 6 identical sp3d2 hybrid orbitals. The shape of the molecule is octa hedral, bond angles are 90°, 90°. % of s character is 16.66%, d is 33.33%, p is 50%.
17. Define Hybridisation of atomic orbitals. Explain sp, sp2, sp3 hybridisation with one example each.
A: The Process of intermixing of atomic orbitals (of almost equal energy) to give equal number of identical orbitals is known as hybridisation
sp hybridisation:
The hybridisation in which one 's' and one 'p' orbitals of same valence shell of an atom intermix together to give two identical sp hybrid orbitals.
e.g.: BeCl2
Be = 1s2 2s1 2px1 (1st excited state)
One 2s and one 2p orbitals of Be intermix together to give two identical sp hybrid orbitals. These two H.O'S overlap with p orbitals of 2 Cl atoms. The shape of the molecule is linear so its bond angle is 180°. % of s & p characters are 50% each.
sp2 hybridisation:
The hybridisation in which one 's' and two 'p' orbitals of same valence shell of an atom intermix together to give 3 identical sp2 hybrid orbitals.
e.g.: BCl3
B = 1s2 2s1 2px1 2py1 (1st excited state)
One 2s and two 2p orbitals of B inter mix together to give three identical sp2 hybrid orbitals. Three H.O' over lap with p orbitals of 3 Cl atoms. Molecule gets trigonal planar, bond angle is 120°, % of s is 33.33%, p is 66.67%.
sp3 hybridisation:
The hybridisation in which one s and three p orbitals of same valence shell of an atom intermix together to give 4 identical sp3 hybrid orbitals. e.g.: CH4
C = 1s2 2s1 2px1 2py1 2pz1 (1st excited state)
One 2s and three 2p orbitals of valence shell of carbon intermix together to give 4 identical sp3 hybrid orbitals. These 4 Orbitals overlap with s orbitals of 4 H atoms. The shape of the molecule is tetrahedral, bond angle is 109o 28', % of s is 25%, p is 75%.
