"p-Block Elements" — Practice

Question 1

Explain why $\mathrm{Tl^+}$ is more stable than $\mathrm{Tl^{3+}}$ while boron shows only the $+3$ state.

Accepted Answer

This is the inert pair effect. In thallium the $6s^2$ pair is held very tightly, because the intervening 4f and 5d electrons shield the nucleus poorly and raise $Z_{	ext{eff}}$, while the bonds thallium would form using those electrons are weak owing to its large size, so promotion costs more than the two extra bonds repay. Boron is small, has no intervening d or f electrons and forms short strong bonds, so using all three valence electrons is comfortably worthwhile. $\mathrm{Tl^{3+}}$ is correspondingly a strong oxidising agent.

Question 2

Gallium has a smaller atomic radius than aluminium despite lying below it. Explain.

Accepted Answer

Because the ten 3d electrons that fill immediately before gallium shield the nuclear charge very poorly. Going from aluminium to gallium adds a shell, which would increase the radius, but it also adds eighteen protons whose pull is only weakly screened by those diffuse d electrons. The contraction from the extra nuclear charge slightly outweighs the expansion from the extra shell, giving 135 pm against aluminium's 143. The same poor shielding disturbs the ionisation enthalpy trend in groups 13 and 14.

Question 3

Arrange $\mathrm{BF_3}$, $\mathrm{BCl_3}$, $\mathrm{BBr_3}$ and $\mathrm{BI_3}$ in increasing order of Lewis acid strength and justify the order.

Accepted Answer

The order is $\mathrm{BF_3 < BCl_3 < BBr_3 < BI_3}$, the reverse of the electronegativity prediction. Boron has an empty 2p orbital, and a filled halogen p orbital can donate density into it by sideways overlap. Fluorine's 2p orbital is almost the same size as boron's, so back-bonding is efficient and largely relieves the electron deficiency, making $\mathrm{BF_3}$ the weakest acceptor. Chlorine, bromine and iodine use progressively larger and more mismatched orbitals, so back-donation weakens and Lewis acidity rises.

Question 4

Explain why carbon dioxide is a gas while silicon dioxide melts above 1700 degrees Celsius.

Accepted Answer

Carbon is small enough for its 2p orbitals to overlap sideways with oxygen's 2p orbitals, so it forms two strong double bonds and exists as discrete linear $\mathrm{O=C=O}$ molecules held together only by weak dispersion forces. Silicon is much larger, so sideways overlap between its 3p orbitals and oxygen's 2p orbitals is ineffective and no useful double bond forms. Silicon therefore forms four single bonds to four different oxygens, each bridging onward, giving a giant covalent network that can only be melted by breaking covalent bonds throughout the solid.

Question 5

Why does carbon catenate essentially without limit while lead barely catenates at all?

Accepted Answer

Because the element-to-element bond enthalpy falls sharply down the group as atoms grow and their orbitals overlap less effectively: $\mathrm{C-C}$ is 348 kJ mol$^{-1}$, $\mathrm{Si-Si}$ 297, and by lead the bond is too weak to sustain a chain. Larger atoms also present their bonds more openly to attack, so heavier chains are kinetically vulnerable even where thermodynamically possible. For silicon there is a further and larger effect: $\mathrm{Si-O}$ at 452 outcompetes $\mathrm{Si-Si}$ by 155 kJ mol$^{-1}$, so silicon chains cannot survive in an oxygen-containing world.

Question 6

Explain why nitrogen is unreactive enough to make up most of the atmosphere while phosphorus ignites in air.

Accepted Answer

Nitrogen is small enough for effective sideways p overlap, so $\mathrm{N_2}$ has a triple bond of 941 kJ mol$^{-1}$, one of the strongest known, and almost nothing at ordinary temperatures can supply that. Phosphorus is too large for such overlap, so its triple bond would be worth only 490 while three single bonds are worth $3 	imes 201 = 603$, and it forms $\mathrm{P_4}$ tetrahedra instead. The 60 degree bond angles in that tetrahedron are severely strained, which is why white phosphorus ignites spontaneously in air and is stored under water.

Question 7

Account for the fall in bond angle from 107 degrees in $\mathrm{NH_3}$ to about 91 degrees in $\mathrm{SbH_3}$.

Accepted Answer

Nitrogen's 2s and 2p orbitals are close in energy, so $sp^3$ hybridisation is easy and the angle is near tetrahedral, compressed slightly by the lone pair. Down the group the s-to-p energy gap widens, hybridisation becomes progressively less favourable, and the central atom bonds using nearly pure p orbitals, three of which are mutually perpendicular and give an angle close to 90 degrees. Note that almost the entire fall happens in the first step, 107 to 93.6, because after phosphorus there is nothing left to lose. The lone pair correspondingly sits mostly in the s orbital, which is why basicity falls.

Question 8

Why does the acidity of group 16 hydrides increase from $\mathrm{H_2O}$ to $\mathrm{H_2Te}$ when oxygen is the most electronegative of the four?

Accepted Answer

Because the controlling factors are bond strength and anion stability, not bond polarity. Acidity requires the hydrogen to leave as a proton, and the bond holding it weakens steadily as the central atom grows and overlaps hydrogen's orbital less effectively, while the resulting anion grows and spreads its charge more thinly. The $\mathrm{p}K_a$ values run 15.7, 7.0, 3.9 and 2.6, so $\mathrm{H_2Te}$ is roughly $10^{13}$ times the acid water is. Electronegativity predicts the opposite order, which is why quoting it here is a standard error.

Question 9

Fluorine has a lower bond dissociation enthalpy than chlorine yet is the more reactive halogen. Explain both observations.

Accepted Answer

The $\mathrm{F-F}$ bond is very short because the atoms are small, so the three lone pairs on each atom are forced close together and repel strongly, weakening the bond to 155 kJ mol$^{-1}$ against chlorine's 242. Reactivity depends on the whole energy balance of a reaction and not on the bond broken alone. Fluorine's weak bond is cheap to break, and the bonds fluorine then forms are exceptionally strong because of its small size and high electronegativity. Both halves of the balance favour reaction, so the two facts are consistent rather than contradictory.

Question 10

Predict the shape and hybridisation of $\mathrm{XeF_4}$ and $\mathrm{XeO_3}$, showing your reasoning.

Accepted Answer

Xenon has eight valence electrons. In $\mathrm{XeF_4}$ four are used in bonds, leaving two lone pairs, so the steric number is 6, the electron geometry is octahedral and the hybridisation is $sp^3d^2$; the two lone pairs take opposite axial positions to be as far apart as possible, leaving a square planar molecule. In $\mathrm{XeO_3}$ the three double bonds to oxygen use six electrons, leaving one lone pair, so the steric number is 4, the hybridisation is $sp^3$ and the molecule is trigonal pyramidal.

Question 11

Why do xenon and radon form compounds while helium, neon and argon do not?

Accepted Answer

Because ionisation enthalpy falls steadily down group 18 as the atoms grow and their outer electrons sit further out and better shielded. Xenon's value of 1170 kJ mol$^{-1}$ is marginally below the 1175 of an oxygen molecule, which $\mathrm{PtF_6}$ was already known to ionise, and that comparison is exactly what led Bartlett to the first xenon compound in 1962. Helium at 2372 and neon at 2081 are beyond any available reagent. Radon reacts too, being lower still, but its radioactivity makes the chemistry difficult to study.

Question 12

Arrange $\mathrm{Na_2O}$, $\mathrm{Al_2O_3}$, $\mathrm{P_4O_{10}}$ and $\mathrm{Cl_2O_7}$ by acidic character, and state how the acidity of an oxide changes with the oxidation state of the element.

Accepted Answer

Acidity increases in the order $\mathrm{Na_2O < Al_2O_3 < P_4O_{10} < Cl_2O_7}$. Metal oxides are basic, non-metal oxides acidic, and aluminium sits at the metalloid boundary and gives an amphoteric oxide reacting with both acids and alkalis. For a single element in several oxidation states, the higher the state the more acidic the oxide, because each additional oxygen delocalises more of the anion's charge: the chlorine oxoacids run $\mathrm{p}K_a$ 7.5, 2.0, $-1$ and $-10$ across $\mathrm{HOCl}$ to $\mathrm{HClO_4}$, roughly five units per oxygen added.

Practice — "p-Block Elements"

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