Lanthanides and Actinides Questions and Answers

Answer: B

The coloured ion is Eu(III).

Lanthanide colours come from f-f transitions within the 4f shell. Eu(III) is 4f⁶, with several f-f transitions in the visible region, and its aqueous solutions are pale pink.

La(III) is 4f⁰ and Lu(III) is 4f¹⁴: with an empty or full f shell there are no f-f transitions, so both are colourless. Gd(III) is 4f⁷, a half-filled shell whose first excited state lies so high that its absorptions fall in the ultraviolet, so it too is colourless in water.

Answer: D

Lanthanoids form sulphides of the type \(Ln_{2}S_{3}\).

The characteristic oxidation state of the lanthanoids is +3. With sulphide (\(S^{2-}\)) that gives the formula \(Ln_{2}S_{3}\), formed when the metals are heated with sulphur.

\(LnS\), \(LnS_{2}\), \(LnS_{3}\) and \(Ln_{2}S\) would correspond to +2, +4, +6 and +1 states, which are not the typical state.

Answer: D

The lanthanoid is gadolinium (Z = 64).

Gadolinium has the configuration [Xe]4f⁷5d¹6s²: the extra stability of a half-filled 4f⁷ subshell makes one electron go into 5d rather than giving 4f⁸.

Europium (Z = 63) is [Xe]4f⁷6s², with no 5d electron. Neodymium is 4f⁴6s², samarium 4f⁶6s² and holmium 4f¹¹6s².

Answer: C

The lanthanoid with 4f⁷6s² is Europium (Eu, Z = 63).

Across the lanthanoids the 4f subshell fills one electron at a time: Nd [Xe]4f⁴6s², Pm 4f⁵6s², Sm 4f⁶6s², Eu 4f⁷6s². Europium reaches the stable half-filled 4f⁷ arrangement with no 5d electron.

Gadolinium (Z = 64) is the trap. It also has seven 4f electrons, but its extra electron goes into 5d rather than 4f, giving [Xe]4f⁷5d¹6s², which keeps the 4f set half-filled. So Gd does not have the configuration 4f⁷6s².

Answer: B

The incorrect statement is Actinoids are less reactive metals.

The actinoid metals are highly reactive, especially when finely divided. They react with boiling water to give the oxide and hydride, and combine with most non-metals at moderate temperatures.

The other statements are correct. The general oxidation state of the actinoids is +3, though many show higher states too. Hydrochloric acid attacks all the actinoid metals, while nitric acid affects them only slightly because a protective oxide layer forms. Their magnetic properties are more complex than those of the lanthanoids, because the 5f electrons are less shielded and interact more strongly with their surroundings.

Answer: E

The pair is Np and Pu.

Among the actinoids, the maximum oxidation state rises from thorium (+4) through protactinium (+5) and uranium (+6) to neptunium and plutonium, which reach +7, and then falls again. Americium reaches about +6, and the later actinoids are mostly +3.

Th, Pa and U cannot reach +7 because they do not have enough valence electrons, and Cm, Bk and Fm favour +3 (or +4 for Bk).

Answer: D

The correct order is \(\mathrm{Yb^{3+} \lt Pm^{3+} \lt Ce^{3+} \lt La^{3+}}\).

Across the lanthanoid series the ionic radius of the M³⁺ ions decreases steadily as atomic number increases. The added 4f electrons shield the growing nuclear charge poorly, so the outer electrons are pulled in: the lanthanoid contraction.

In order of atomic number, La (57), Ce (58), Pm (61), Yb (70), so La³⁺ is largest and Yb³⁺ smallest, with Ce³⁺ before Pm³⁺. The options starting with La³⁺ as the smallest run the trend backwards, and the '>' option puts Pm³⁺ above Ce³⁺.

Answer: E

The incorrect statement is the configuration \((n-1)f^{1-14}(n-1)d^{0-1}ns^{2}\).

In the f-block, the f orbitals being filled are two shells inside the outermost: for actinoids the configuration is \(5f^{1-14}6d^{0-1}7s^{2}\), that is \((n-2)f^{1-14}(n-1)d^{0-1}ns^{2}\) with \(n = 7\). Writing \((n-1)f\) puts the f electrons in the 6th shell, which is wrong.

The other statements are correct: f-block elements are those whose differentiating electron enters an f orbital; the lanthanoids Ce to Lu (58 to 71) and actinoids Th to Lr (90 to 103) make 14 + 14 = 28 elements; and all actinoids are radioactive.

Answer: C

The common oxidation state of the lanthanoids is +3.

Lanthanoids have the configuration 4fⁿ5d^0-16s². Losing the two 6s electrons and one 5d (or 4f) electron gives the +3 state, which is the most stable for all of them.

Some also show +2 or +4, usually when that gives an empty, half-filled or full f subshell: Eu²⁺ and Yb²⁺ (4f⁷, 4f¹⁴), Ce⁴⁺ (4f⁰). These are exceptions to the general +3, and +1 and +5 are not characteristic of the series.

Answer: B

They are comparable with those of Calcium.

The first and second ionization enthalpies of the lanthanoids are around 600 and 1100 to 1200 kJ mol^-1, similar to calcium's 590 and 1145. This is why the lanthanoids are reactive metals behaving much like calcium, losing electrons readily, while the third ionization enthalpy is what decides how stable the +3 state is.

Chromium, germanium and cadmium are not used as the comparison, and caesium's second ionization enthalpy is enormous because its second electron comes from a noble-gas core.