Molecular Structure and Bonding Theory Questions and Answers

Answer: D

The π-bonding orbital is \(c_{1}\phi_{x}^{C} + c_{2}\phi_{x}^{O}\).

With the molecule along the z axis, a π orbital is built from p orbitals perpendicular to the bond, that is, 2p_x (or 2p_y), and only orbitals of the same symmetry about the axis can combine. So the π MO combines 2p_x on carbon with 2p_x on oxygen, in phase for the bonding combination.

2p_z on carbon with 2p_z on oxygen point along the bond and give a σ orbital, not a π orbital. A 2p_z with a 2p_x on the other atom has zero net overlap by symmetry: one is symmetric and the other antisymmetric under reflection through the yz plane, so the two mixed options describe no bond at all.

Answer: B

The order of polarity is BF₃ < NF₃ < NH₃.

BF₃ is trigonal planar, so its three B-F bond dipoles cancel and its dipole moment is zero. NH₃ and NF₃ are both pyramidal with a lone pair on nitrogen. In NH₃ the N-H bond dipoles point towards nitrogen, the same way as the lone-pair dipole, so they add: about 1.47 D. In NF₃ the N-F bond dipoles point away from nitrogen, against the lone-pair dipole, so they largely cancel: about 0.23 D.

Any order placing NF₃ above NH₃ forgets that opposition, and any order not putting BF₃ first ignores its symmetry.

Answer: A

The shapes are T-shape, T-shape, linear and trigonal bipyramidal.

• ClF₃: three bonds and two lone pairs on Cl, five electron pairs in a trigonal bipyramid with the lone pairs equatorial, leaving a T shape.
• XeOF₂: Xe forms one double bond to O and two bonds to F and keeps two lone pairs; three bonded positions and two equatorial lone pairs again give a T shape.
• N₃⁻: 16 valence electrons, isoelectronic with CO₂, so linear.
• XeO₃F₂: five bonded groups and no lone pair on Xe, so trigonal bipyramidal.

Calling ClF₃ or XeOF₂ trigonal planar ignores their lone pairs, azide is not bent, and XeO₃F₂ has no lone pair to make it square pyramidal.

Answer: D

The statement that is always correct is \(0 \le S \le 1\).

The 1s functions are real, positive everywhere and normalised. Their overlap integral is therefore a real number that cannot be negative, and by the Schwarz inequality it cannot exceed the product of the norms, 1. It equals 1 only when the two nuclei coincide and approaches 0 as they move far apart.

\(S = 1\) holds only at zero separation and \(S = 0\) only at infinite separation, so neither is always true. The overlap of two real functions is real, so it is never an imaginary constant.

Answer: A

The ligand group orbital is \(\frac{1}{\sqrt{12}}(2\sigma_1+2\sigma_2-\sigma_3-\sigma_4-\sigma_5-\sigma_6)\).

In \(O_{h}\) the \(d_{z^{2}}\) orbital belongs to e_g. Its LGO must copy the orbital's shape: large positive lobes along +z and -z, where \(\sigma_{1}\) and \(\sigma_{2}\) point, and the smaller negative collar in the xy plane, where \(\sigma_{3}\) to \(\sigma_{6}\) point. So \(\sigma_{1}\) and \(\sigma_{2}\) take the same positive weight, twice that of the four equatorial orbitals, which take a negative sign. The normalising factor is

\[ \frac{1}{\sqrt{4 + 4 + 1 + 1 + 1 + 1}} = \frac{1}{\sqrt{12}} \]

The all-positive combination \(\frac{1}{\sqrt{6}}(\sigma_{1} + \dots + \sigma_{6})\) is the a_1g LGO that matches the 4s orbital. The \(2\sigma_{1} - 2\sigma_{2}\) option changes sign between +z and -z, which a gerade d orbital never does. The \(\frac{1}{\sqrt{6}}\) option with equal weights of opposite sign is not a symmetry-adapted orbital at all: it is a mixture of the a_1g and e_g combinations.

Answer: C

Sulphur in SF₄ is sp\(^{3}\)d hybridised.

Sulphur has six valence electrons. Four form bonds to fluorine and the remaining two make one lone pair, giving five electron pairs around the central atom. Five pairs need five hybrid orbitals, sp³d, arranged as a trigonal bipyramid; the lone pair occupies an equatorial position, so the molecule has a see-saw shape.

sp³ would allow only four pairs (as in SCl₂ with two lone pairs, or a tetrahedral molecule), dsp² gives a square planar set of four, and sp³d² six pairs, as in SF₆.

Answer: D

NO⁺ has bond order three and is diamagnetic.

NO⁺ has 10 valence electrons (5 from N, 6 from O, minus 1), the same as N₂ and CO. Filling the molecular orbitals:

\[ \begin{aligned} &\sigma_{2s}^{2}\,\sigma_{2s}^{*2} \\ &\pi_{2p}^{4}\,\sigma_{2p}^{2} \end{aligned} \]

\[ \text{bond order} = \frac{8 - 2}{2} = 3 \]

All electrons are paired, so the ion is diamagnetic. Neutral NO has one more electron, in a \(\pi^{*}\) orbital, giving bond order 2.5 and paramagnetism; options with bond order two or paramagnetism describe neither species correctly.

Answer: A

The smallest bond angle is in PH₃, about 93°.

The measured angles are roughly NH₃ 107°, NF₃ 102°, PF₃ 98° and PH₃ 93°.

Going down the group from N to P, the central atom becomes larger and less electronegative. Its bonding pairs sit farther from the nucleus and repel each other less, and the bonds use orbitals of nearly pure p character, so the angle closes toward 90°. That puts both phosphorus compounds below both nitrogen compounds.

Within the nitrogen pair, fluorine pulls the bonding pairs away from N, reducing their mutual repulsion, so NF₃ has a smaller angle than NH₃. Between the phosphorus compounds, PH₃ is the smaller; PF₃ is opened slightly by the bulk of the fluorine atoms and by their lone pairs interacting with phosphorus.

Answer: A

The bond order of \(\mathrm{N_{2}^{2-}}\) is 2.

N₂ has 10 valence electrons; the dianion has 12. Filling the molecular orbitals in the order used for N₂:

\[ \begin{aligned} &\sigma_{2s}^{2}\,\sigma_{2s}^{*2}\,\pi_{2p}^{4} \\ &\sigma_{2p}^{2}\,\pi_{2p}^{*2} \end{aligned} \]

That is 8 bonding and 4 antibonding electrons, so

\[ \text{bond order} = \frac{8 - 4}{2} = 2 \]

3 is the bond order of neutral N₂, and 2.5 is that of N₂⁻, where only one electron has entered \(\pi^{*}\). Each extra electron goes into an antibonding orbital and lowers the bond order by half, so 3.5 is impossible here.

Answer: A

The correct match is i-d, ii-c, iii-b, iv-a.

Count the electron domains (bond pairs plus lone pairs) on the central atom; that number fixes the hybridization.

• I₃⁻: the central I has 2 bonds and 3 lone pairs, 5 domains, so sp³d (linear ion, lone pairs equatorial).
• PCl₃: 3 bonds and 1 lone pair, 4 domains, so sp³ (pyramidal).
• BF₃: 3 bonds and no lone pair, so sp² (trigonal planar).
• CO₂: two double bonds, 2 domains, so sp (linear).

The other options give I₃⁻ sp (counting only its two bonds), or swap PCl₃ with BF₃, or give CO₂ a trigonal hybrid; each fails one of these counts.