Molecular Structure and Bonding Theory Questions and Answers

Answer: A

The molecule with zero dipole moment is \(\mathrm{BF_{3}}\).

BF₃ is trigonal planar, with the three B-F bonds at 120° to one another. Each bond is polar, but three equal bond dipoles arranged symmetrically in a plane add up to zero.

The others have lone pairs or unequal bonds that leave a net dipole. NH₃ and NF₃ are pyramidal, with a lone pair on nitrogen. SO₂ is bent, with a lone pair on sulphur. CH₂Cl₂ is tetrahedral but with two different kinds of bond, so the C-Cl and C-H dipoles do not cancel.

Answer: D

The molecule with an expanded octet is \(SF_{6}\).

Sulphur in \(SF_{6}\) forms six bonds, so it carries 12 electrons in its valence shell. This is possible for a period-3 element, which has 3d orbitals available.

\(BCl_{3}\) and \(BeH_{2}\) have an incomplete octet (6 and 4 electrons on the central atom). \(NO\) and \(NO_{2}\) are odd-electron molecules, the third type of exception to the octet rule, not expanded octets.

Answer: C

The molecule is \(ClF_{3}\).

Chlorine has 7 valence electrons; three are used in bonds with fluorine, leaving four as two lone pairs. Five electron pairs give a trigonal bipyramidal arrangement, and the molecule is T-shaped.

\(NH_{3}\) has 3 bond pairs and 1 lone pair; \(H_{2}O\) has 2 and 2; \(SF_{4}\) has 4 and 1; \(SO_{2}\) has 2 sigma bonds and 1 lone pair on sulphur.

The lone pairs occupy equatorial positions of the bipyramid, where they repel bonds least, which is why \(ClF_{3}\) is T-shaped rather than trigonal planar.

Answer: E

The correct statement is that CO₂ contains two C=O bonds and two lone pairs on each oxygen atom.

In the Lewis structure O=C=O, carbon uses its four valence electrons in two double bonds. Each oxygen shares two of its six valence electrons in the double bond and keeps four as two lone pairs, completing its octet. The molecule is linear.

A structure with one C-O single bond would leave carbon or oxygen without an octet unless formal charges are introduced, and a C≡O triple bond would also need formal charges; neither is the accepted structure. The options with only one lone pair per oxygen leave oxygen short of its octet.

Answer: D

The molecule is Xenon (IV) fluoride, XeF₄.

Xenon has 8 valence electrons. Four form bonds with fluorine and the other four make two lone pairs, giving six electron pairs (sp³d², octahedral arrangement). The two lone pairs sit opposite each other, leaving the four fluorines in a square plane.

CF₄ and SiF₄ are tetrahedral with no lone pair on the central atom. Platinum(IV) and nickel(IV) fluorides are not simple molecules with two lone pairs on the central atom, so they do not fit the description.

Answer: C

The shortest bond is in HF.

Bond length depends mainly on the sizes of the bonded atoms. Hydrogen is the smallest atom, so the H-F bond is very short, about 92 pm.

The others are longer: O2 about 121 pm (a double bond, but between two larger atoms), F2 about 142 pm, Cl2 about 199 pm and Br2 about 228 pm, increasing as the halogen atoms get larger down the group.

Answer: E

The molecule with an expanded octet is sulphur hexafluoride.

In SF6 the sulphur atom forms six bonds, so it has twelve electrons in its valence shell, more than eight. Elements of period 3 and below can do this because they have vacant d orbitals available in the valence shell.

Ozone and water obey the octet rule. Nitric oxide (NO) and nitrogen dioxide (NO2) are odd-electron molecules, where nitrogen has fewer than eight electrons, which is a different exception to the octet rule, not an expanded octet.

Answer: E

The hybridisation of phosphorus in PCl5 is sp³d.

Phosphorus has five valence electrons. To form five bonds it promotes one 3s electron to a 3d orbital and mixes one s, three p and one d orbital, all from the third shell, into five sp³d hybrids pointing to the corners of a trigonal bipyramid.

sp³d² and d²sp³ give six hybrid orbitals (octahedral), and sp²d² is not a standard set. dsp³ is the label used when an inner (n-1)d orbital takes part; phosphorus uses a d orbital of the same shell, so the textbook label is sp³d.

Answer: B

The molecule with the highest dipole moment is hydrogen fluoride.

Among the hydrogen halides, the dipole moment follows the electronegativity difference between H and the halogen. Fluorine is the most electronegative element, so H-F is the most polar bond: about 1.9 D, against about 1.0 D for HCl, 0.8 D for HBr and 0.4 D for HI.

Carbon dioxide is linear and symmetric (O=C=O), so its two bond dipoles cancel and its net dipole moment is zero.

Answer: C

The pair with almost the same bond length is C=O and N=O.

Standard tables give about 121 pm for C=O and about 122 pm for N=O, a difference of only 1 pm.

The other pairs differ more: C-H (about 107 pm) and O-H (about 96 pm); C=C (133 pm) and C=N (about 128 pm); C-C (154 pm) and C-N (about 147 pm); C-O (about 143 pm) and N-O (about 136 pm).