3 free questions from this chapter (the full set has 30). Marking: +4 for correct · −1 for incorrect · 0 for unattempted.
Question 1Level 3 · AdvancedSingle correct
Phosphorus forms the mixed halides PCl3F2 and PCl2F3, both with trigonal bipyramidal geometry. Which one of the following molecules has zero dipole moment?
APCl2F3
BClF3
CSF4
DPCl3F2
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Answer: D
Key idea (Bent's rule): in a trigonal bipyramid the more electronegative ligands occupy the axial positions, which use orbitals of lower s character.
In PCl3F2 both F are axial (their bond dipoles cancel at 180°) and the three Cl are equatorial at 120° (they cancel too), so μ=0.
In PCl2F3 two F are axial and cancel, but the equatorial set is one F and two Cl, which do not cancel, so μ=0.
SF4 (see-saw) and ClF3 (T-shaped) have equatorial lone pairs and are polar.
Trap: treating both mixed halides as 'symmetric TBP' and calling both non-polar, or placing Cl axial.
Exam pattern: JEE Main – dipole moment from VSEPR shape and positions of ligands/lone pairsQuestion 2Level 3 · AdvancedSingle correct
Arrange the bond angles in H2O, OF2, Cl2O and H2S in decreasing order.
AH2O>OF2>H2S>Cl2O
BCl2O>H2O>OF2>H2S
COF2>H2O>Cl2O>H2S
DH2O>Cl2O>OF2>H2S
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Answer: B
All four are AX2E2 (bent), so the differences come from bond-pair position and size of X.
OF2: highly electronegative F pulls bond pairs away from O, bp-bp repulsion falls, angle ≈ 103° (smaller than H2O, ≈ 104.5°).
Cl2O: the two large Cl atoms crowd each other (and O–Cl has some multiple-bond character), opening the angle to ≈ 111°, the largest.
H2S: S uses almost pure p orbitals, angle ≈ 92°, the smallest.
Order: Cl2O>H2O>OF2>H2S. Trap: assuming Cl, being less electronegative than F but more than H, puts Cl2O between OF2 and H2O.
Exam pattern: JEE Main – bond angle order using electronegativity of central/terminal atoms and steric sizeQuestion 3Level 3 · AdvancedSingle correct
Which pair of species has the same bond order but different magnetic behaviour (one paramagnetic, the other diamagnetic)?
ACN− and NO+
BO22− and F2
CC2 and O2
DN2+ and N2−
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Answer: C
C2 (8 valence e, s–p mixing): σ2s2σ2s∗2π2p4, bond order 2, all paired, diamagnetic.
O2 (12 valence e): …π2p4π2p∗2 with the two π∗ electrons unpaired, bond order 2, paramagnetic. So (C) fits.
(A) CN−, NO+: both 10 valence e, BO 3, both diamagnetic. (B) O22−, F2: both 14 e, BO 1, both diamagnetic.
(D) N2+ (9 e) and N2− (11 e): both BO 2.5 and both have one unpaired electron, both paramagnetic.
Trap: filling C2 without s–p mixing (σ2p below π2p) wrongly gives two unpaired electrons.
Exam pattern: JEE Main – MOT bond order and magnetic character of diatomic species