Main Group Elements Questions and Answers

Answer: C

N(SiH₃)₃ is trigonal planar.

Unlike N(CH₃)₃, which is pyramidal, trisilylamine has a planar NSi₃ skeleton. The classic explanation is that the nitrogen lone pair is delocalised into vacant orbitals on silicon (described as pπ-dπ bonding, now often as negative hyperconjugation into Si-H σ* orbitals), which favours an sp²-like nitrogen. The molecule is correspondingly a very weak base.

Trigonal pyramidal is the trimethylamine answer, the trap here. Tetrahedral would need four groups on nitrogen, and linear is impossible for three substituents.

Answer: C

The order is ClO₄⁻ < ClO₃⁻ < ClO₂⁻ ≈ ClO⁻: oxidising tendency rises as the oxidation number of chlorine falls.

Chlorine is +7, +5, +3 and +1 in the four anions. Perchlorate, with chlorine buried among four oxygens in a symmetrical tetrahedron, is a notoriously sluggish oxidant in dilute solution. Chlorate oxidises more readily, and chlorite and hypochlorite, with fewer oxygens shielding the chlorine, are the strongest and fastest oxidants of the series and comparable to each other. In acid the couples HClO₂/HClO and HClO/Cl₂ have almost the same potential, about 1.6 V, which is why the last two are written as roughly equal.

The order with '>' throughout runs the trend backwards. Putting the approximate equality between perchlorate and chlorate, or between chlorate and chlorite, ignores how much more reactive each lower oxoanion is than the one above it.

Answer: B

XeOF₅⁻ is a pentagonal pyramid.

Count the electron domains on xenon. Xe brings 8 valence electrons and the negative charge adds 1. Five Xe-F bonds use 5 of them and the Xe=O double bond 2, leaving 2 electrons, one lone pair. With six bonding domains and one lone pair, VSEPR gives seven domains in a pentagonal bipyramid: the five fluorines form the pentagon, and the oxygen and the lone pair occupy the two axial positions.

Ignoring the lone pair, the atoms form a pentagonal pyramid with oxygen at the apex (C_5v), which is the structure found for this anion.

A trigonal bipyramid or square pyramid has only five positions, and an octahedron six; none holds six atoms plus a lone pair.

Answer: C

The metal is lithium, which forms Li₃N.

Lithium reacts directly with nitrogen, even at room temperature, to give the ruby-red nitride:

\[ 6\mathrm{Li} + \mathrm{N_{2}} \rightarrow 2\mathrm{Li_{3}N} \]

This is one of the ways lithium resembles magnesium (the diagonal relationship), which also burns in nitrogen to Mg₃N₂. The very small Li⁺ ion gives Li₃N a high lattice energy, enough to pay for breaking the N≡N bond and forming N³⁻.

Sodium, potassium and rubidium have larger, less polarizing cations, and their nitrides are not formed this way and are unstable, so the heavier alkali metals do not form M₃N on heating in nitrogen.

Answer: D

The set is HgO, KO₂, BaO₂.

Mercury(II) oxide decomposes on heating to mercury and oxygen (2HgO → 2Hg + O₂), the route by which Priestley first obtained oxygen. Barium peroxide gives up oxygen when heated strongly (2BaO₂ → 2BaO + O₂), the basis of the old Brin process. Potassium superoxide also releases O₂ on heating, decomposing to K₂O₂ or K₂O and oxygen.

The other sets contain normal oxides of reactive metals, CaO, BaO and Na₂O, which are thermally very stable and do not release oxygen on heating, so every set containing one of them is wrong.

Answer: A

The distorted octahedron is [IF₆]⁻.

All four ions have six bonding pairs and one lone pair on the central atom (AX₆E). Whether the lone pair distorts the octahedron depends on the size of the central atom and the crowding of the ligands. Iodine is large and fluorine small, so in IF₆⁻ the lone pair is stereochemically active and the structure is distorted, as in XeF₆.

In [SeBr₆]²⁻ and [TeCl₆]²⁻ the large halide ligands crowd the central atom and the lone pair stays in a spherical s orbital, giving regular octahedra. In [BrF₆]⁻ the small bromine centre is crowded by six fluorines, and its lone pair is also stereochemically inactive, so it too is a regular octahedron.

Answer: C

The incorrect statement is the one about making XeF₆ from SbF₅ and XeF₅⁺.

The reaction runs the other way: XeF₆ is a fluoride ion donor, and with the strong fluoride acceptor SbF₅ it forms the salt [XeF₅]⁺[SbF₆]⁻. XeF₅⁺ is a cation and is not a fluoride acceptor that would combine with SbF₅ to make XeF₆.

The other statements are correct. Xenon has six bonding pairs and one lone pair, 14 electrons in its valence shell. In the solid and liquid, XeF₆ associates into tetramers (and hexamers) built from XeF₅⁺ and F⁻ units. With silica, XeF₆ exchanges fluorine for oxygen, forming SiF₄ and oxyfluorides that end in XeO₃, which is why it attacks glass.

Answer: B

Among the options, the rate of water exchange increases as the coordination number increases.

For the alkali metal ions, water exchange is fastest for the largest ions: it speeds up down the group from Li⁺ to Cs⁺. Going down the group the ions grow larger, their surface charge density falls, the ion-water attraction weakens, and the number of water molecules they can hold rises from about four for Li⁺ to eight or more for Cs⁺. So faster exchange goes together with a higher coordination number.

The other three options all describe the opposite trend: a smaller ion, a higher surface charge density and a more electronegative ion all hold water more tightly and exchange it more slowly.

Answer: D

The correct statement is the one saying \(\text{NH}_{3}\) has the higher bond angle.

The H-N-H angle in NH₃ is about 107°, while the F-N-F angle in NF₃ is about 102°. Fluorine is much more electronegative than hydrogen, so the N-F bonding pairs are drawn towards fluorine and away from nitrogen. They repel each other less, and the lone pair on nitrogen squeezes them closer together.

The same pull of electron density towards fluorine also makes the nitrogen lone pair less available, so NF₃ is far less basic than NH₃. The three options claiming a larger angle for NF₃ are wrong on the basic fact, so only the last option fits, although the underlying cause of both observations is the electronegativity of fluorine.

Answer: B

The acids with S-S bonds are \(\text{H}_{2}\text{S}_{2}\text{O}_{3}\), \(\text{H}_{2}\text{S}_{2}\text{O}_{4}\) and \(\text{H}_{2}\text{S}_{2}\text{O}_{6}\).

Thiosulphuric acid (H₂S₂O₃) has one sulphur bonded directly to the central sulphur. Dithionous acid (H₂S₂O₄) and dithionic acid (H₂S₂O₆) are both built of two SO_x units joined by an S-S bond.

Disulphuric (pyrosulphuric) acid, H₂S₂O₇, joins its two sulphur atoms through an oxygen bridge (S-O-S), and peroxodisulphuric acid, H₂S₂O₈, through a peroxo bridge (S-O-O-S). Every option containing either of them is wrong.