3 free questions from this chapter (the full set has 30). Marking: Multi-correct: +4 all correct, +1 per correct option if none wrong, −2 if any wrong · Numerical: +4 / 0 · Matching: +3 / −1.
Question 1Level 2 · ExamSingle correct
Take the internuclear axis as the z axis for two atoms A and B. Which one of the following pairs of atomic orbitals (first orbital on A, second on B) gives zero net overlap, so that no bond can form between them? Exactly one option is correct.
A2s on A and 2pz on B
B2pz on A and 3dz2 on B
C2py on A and 3dx2−y2 on B
D2px on A and 3dxz on B
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Answer: C
Net overlap is zero when the product of the two orbitals is odd under a symmetry operation that leaves the A–B axis unchanged.
(A) 2s and 2pz: both are symmetric about the z axis; head-on σ overlap, non-zero.
(B) 2pz and 3dz2: both have their large lobes along z; σ overlap, non-zero.
(C) 2py and 3dx2−y2: reflect through the xz plane (y→−y). py changes sign, dx2−y2 does not, so the product is odd and the positive and negative regions cancel exactly. Zero overlap.
(D) 2px and 3dxz: both change sign under x→−x, so their product is even; this is sideways (π) overlap, non-zero.
(A numerical integration with Slater-type functions gives overlaps of about 1.8, 1.8, 0.0 and 0.7 in arbitrary units, matching this reasoning.)
Answer: (C).
Difficulty: HardQuestion 2Level 3 · AdvancedSingle correct
Which one of the following dianions has a bond order of 2 and is paramagnetic? Use the usual valence MO energy order: for B, C and N species (s–p mixing) σ2s<σ2s∗<π2p<σ2p<π2p∗<σ2p∗; for O, F and Ne species σ2s<σ2s∗<σ2p<π2p<π2p∗<σ2p∗. Exactly one option is correct.
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Answer: A
Count valence electrons (n) and fill the MOs.
N22−: n=12: σ2s2σ2s∗2π2p4σ2p2π2p∗2. Bond order =(8−4)/2=2; the two π∗ electrons sit singly in the degenerate pair (Hund's rule), so 2 unpaired electrons: paramagnetic.
B22−: n=8: σ2s2σ2s∗2π2p4. Bond order 2, but the π2p set is full: diamagnetic. This is the trap: it has the right bond order but the wrong magnetism, because with s–p mixing the last 4 electrons fill π2p completely.
C22−: n=10, bond order 3, diamagnetic.
O22−: n=14, bond order 1, diamagnetic.
Answer: (A).
Difficulty: Very hardQuestion 3Level 3 · AdvancedSingle correct
In XeOF4 the Xe atom is bonded to one O atom (Xe=O) and four F atoms. In which one of the following species does the central atom have both the same hybridisation and the same VSEPR shape as Xe in XeOF4? Exactly one option is correct.
A[ClF4]−
B[IF4]+
C[SbF5]2−
D[XeF3]+
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Answer: C
XeOF4: Xe has 8 valence electrons; Xe=O uses 2 and four Xe–F use 4, leaving 2 electrons = 1 lone pair. Steric number 5+1=6: sp3d2, square pyramidal.
(A) [ClF4]−: 7+1=8 electrons; 4 bonds, 2 lone pairs; steric number 6: sp3d2 but square planar. Same hybridisation, different shape.
(B) [IF4]+: 7−1=6 electrons; 4 bonds, 1 lone pair; steric number 5: sp3d, see-saw.
(C) [SbF5]2−: 5+2=7 electrons; 5 bonds, 1 lone pair; steric number 6: sp3d2, square pyramidal. Matches.
(D) [XeF3]+: 8−1=7 electrons; 3 bonds, 2 lone pairs; steric number 5: sp3d, T-shaped.
Answer: (C).
Difficulty: Very hard