"Classification of Elements and Periodicity" — Practice

Question 1

Arrange $	ext{S}^{2-}$, $	ext{Cl}^-$, $	ext{K}^+$ and $	ext{Ca}^{2+}$ in order of increasing ionic radius, with reasoning.

Accepted Answer

All four have $18$ electrons, so the comparison is purely one of nuclear charge: $16<17<19<20$.

More protons pull the same eighteen electrons more tightly.

$	ext{Ca}^{2+}<	ext{K}^+<	ext{Cl}^-<	ext{S}^{2-}$

The two anions are larger than the two cations, as always for an isoelectronic series spanning the noble gas.

Question 2

Explain why the atomic radius of gallium ($135$ pm) is slightly smaller than that of aluminium ($143$ pm), despite gallium being in the next period.

Accepted Answer

Gallium is preceded by the ten first-row transition metals, so ten $3d$ electrons have been added.

$d$ orbitals shield poorly, so the ten added protons are not fully compensated and $Z_{eff}$ on gallium's $4p$ electron is unusually high.

That extra pull more than cancels the expected increase from the new shell, leaving gallium marginally smaller.

The same mechanism, one shell deeper, produces the lanthanoid contraction.

Question 3

The successive ionisation enthalpies of an element in kJ mol$^{-1}$ are $578$, $1817$, $2745$, $11577$ and $14842$. Identify the group and suggest the element.

Accepted Answer

The jump from the third to the fourth value is a factor of $4.2$, far larger than any other step.

Three electrons come off relatively easily and the fourth requires breaking into a noble-gas core, so the element is in **group 13**.

The magnitudes are consistent with aluminium.

Note that the answer follows from the pattern alone; no tabulated values need to be recalled.

Question 4

Explain why the first ionisation enthalpy of sulphur ($1000$) is lower than that of phosphorus ($1012$), even though sulphur has a higher nuclear charge.

Accepted Answer

Phosphorus is $3s^{2}3p^{3}$: a half-filled $p$ subshell with three parallel electrons and maximum exchange stabilisation.

Sulphur is $3s^{2}3p^{4}$: the fourth $p$ electron must pair with one already present.

Pairing repulsion in that doubly occupied orbital makes the electron easier to remove, which outweighs the small increase in nuclear charge.

The same argument explains why nitrogen exceeds oxygen in period 2.

Question 5

The second electron gain enthalpy of oxygen is $+780$ kJ mol$^{-1}$, yet metal oxides containing $	ext{O}^{2-}$ are common. Explain.

Accepted Answer

Forming $	ext{O}^{2-}$ from $	ext{O}^-$ requires forcing an electron onto an already negative species, which is strongly endothermic.

In isolation the doubly charged ion would therefore never form.

In a crystal, however, the doubly charged anion attracts the surrounding cations far more strongly than a singly charged one would, and the **lattice enthalpy** released more than pays the $780$ kJ mol$^{-1}$.

This is why oxide ions exist only in lattices and never in the gas phase.

Question 6

An element has an ionisation energy of $13.6$ eV and an electron affinity of $0.75$ eV. Compute its Mulliken electronegativity and compare it with that of an element having $17.4$ eV and $3.6$ eV.

Accepted Answer

First: $\chi_M=\dfrac{13.6+0.75}{2}=7.18$ eV

Second: $\chi_M=\dfrac{17.4+3.6}{2}=10.5$ eV

The second element is markedly more electronegative, because both of its atomic properties are larger rather than one compensating the other.

The first set of values corresponds to hydrogen and the second to a strongly electronegative element such as fluorine.

Question 7

Explain why $	ext{BeCl}_2$ is a polymeric solid that sublimes, while $	ext{CaCl}_2$ is a high-melting ionic lattice.

Accepted Answer

Polarising power scales as charge over the square of the radius. Beryllium(II) is doubly charged and very small; calcium(II) is doubly charged but much larger.

Beryllium(II) therefore polarises the chloride ion strongly, drawing electron density into the internuclear region and giving substantial covalent character.

$	ext{BeCl}_2$ is a chain polymer held by weak intermolecular forces and sublimes at about $520^{\circ}$C.

$	ext{CaCl}_2$ is essentially ionic and melts near $772^{\circ}$C with no covalent character worth speaking of.

Question 8

Arrange the following in order of increasing acidity of the terminal hydrogen: ethane, ethene, ethyne. Justify using electronegativity.

Accepted Answer

Ethane's carbon is $sp^{3}$ ($25\%$ $s$), ethene's is $sp^{2}$ ($33\%$), ethyne's is $sp$ ($50\%$).

Greater $s$ character holds the electron pair closer to the nucleus, so the carbanion left after deprotonation is more stable.

Acidity increases: ethane $<$ ethene $<$ ethyne, with $pK_a$ around $50$, $44$ and $25$.

The same atom in three environments spans twenty-five orders of magnitude in acidity, which is why electronegativity cannot be treated as a fixed elemental property.

Question 9

Give three chemical similarities between lithium and magnesium that distinguish lithium from the other alkali metals.

Accepted Answer

Both react directly with atmospheric nitrogen to form nitrides, which no other alkali metal does.

Both form hydroxides and carbonates that decompose on heating, whereas sodium and potassium carbonates are thermally stable.

Both form insoluble fluorides, carbonates and phosphates, unlike the freely soluble salts of the heavier alkali metals.

All three follow from lithium's unusually high charge density, comparable with that of the doubly charged but larger magnesium ion.

Question 10

Explain why lead(II) compounds are more stable than lead(IV) compounds, and predict whether $	ext{PbCl}_4$ or $	ext{PbCl}_2$ is the stronger oxidising agent.

Accepted Answer

Down group 14 the $ns^{2}$ pair becomes increasingly reluctant to participate in bonding, partly through relativistic contraction of the $6s$ orbital and partly through poor $4f$ and $5d$ shielding.

Lead therefore prefers $+2$, using only its $6p$ electrons.

$	ext{PbCl}_4$ is the stronger oxidising agent, since lead(IV) readily accepts two electrons to revert to the more stable lead(II).

It decomposes to $	ext{PbCl}_2$ and chlorine on gentle warming.

Question 11

Explain why zirconium and hafnium are chemically almost indistinguishable, and name one practical consequence.

Accepted Answer

Hafnium follows the fourteen lanthanoids, whose $4f$ electrons screen very poorly, so fourteen extra protons contract the atom by about $10$ pm.

That contraction almost exactly cancels the increase expected from adding a shell, leaving $r_{	ext{Zr}}=160$ pm against $r_{	ext{Hf}}=159$ pm.

Since size and charge govern chemistry, the two behave almost identically and always occur together in ores.

Separating them requires solvent extraction or ion exchange, which matters because zirconium is used in nuclear reactor cladding and hafnium, a strong neutron absorber, must be removed from it.

Question 12

Explain why nitrogen forms no pentahalide while phosphorus forms $	ext{PF}_5$ and $	ext{PCl}_5$ readily.

Accepted Answer

Nitrogen's valence shell contains only $2s$ and $2p$ orbitals — four in total — so it can form at most four bonds and cannot accommodate ten bonding electrons.

Phosphorus has energetically accessible $3d$ orbitals, allowing the octet to be expanded to ten electrons.

Nitrogen's small size would also make five halogens around it sterically impossible.

The same restriction explains why oxygen forms no hexafluoride and why fluorine, having no accessible $d$ orbitals at all, shows only the $-1$ oxidation state.

Practice — "Classification of Elements and Periodicity"

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