Q4. Give the hybridisation and shape of the central atom in XeF2 and ClF3.
XeF2: xenon has 8 valence electrons, 2 used in bonds, so 3 lone pairs; 2 + 3 = 5, sp3d; the lone pairs take the three equatorial positions, so the molecule is linear. ClF3: chlorine has 7 valence electrons, 3 used in bonds, so 2 lone pairs; 3 + 2 = 5, sp3d; both lone pairs are equatorial, so the molecule is T-shaped.
Q5. In propyne, CH3–C≡CH, which is shorter: a C–H bond of the CH3 group or the ≡C–H bond? Explain using s character.
The CH3 carbon is sp3 (25% s) and the ≡CH carbon is sp (50% s). More s character keeps the bonding electrons closer to the carbon nucleus, so the ≡C–H bond is shorter (about 106 pm against about 109 pm).
Q6. Give the hybridisation of each carbon in allene, CH2=C=CH2.
Each end carbon has three σ bonds (two C–H, one C–C): sp2. The middle carbon has two σ bonds and two π bonds: sp.
Q7. In PCl5, how many Cl–P–Cl angles are 90°, how many 120° and how many 180°?
Each of the 2 axial bonds is at 90° to each of the 3 equatorial bonds: 6 angles of 90°. The 3 equatorial bonds give 3 angles of 120°. The 2 axial bonds give 1 angle of 180°. In all, 6 + 3 + 1 = 10, the number of pairs among five bonds.
Q8. Which of the following are NOT required for hybridisation: (a) the orbitals belong to the valence shell; (b) the orbitals have almost equal energy; (c) an electron is promoted first; (d) only half-filled orbitals take part?
(c) and (d). Promotion is not essential, and filled orbitals can also be hybridised (NCERT page 121).