Transition Elements: Chemistry 9701 (Cambridge International AS & A Level)

Zinc, $\text{Zn}$ ($Z=30$), is a d-block element, but unlike titanium to copper it is not classified as a transition element.

Which statement correctly explains this, in terms of electron configuration?

This question concerns the electron configurations of chromium ($\text{Cr}$, $Z=24$) and iron ($\text{Fe}$, $Z=26$) and their ions.

(a) Give the full electron configuration of a ground-state chromium atom, and explain why it is $[\text{Ar}]3d^54s^1$ rather than the configuration $[\text{Ar}]3d^44s^2$ that a simple extension of the vanadium-to-manganese trend would predict. [3]

(b) Give the full electron configurations of the $\text{Fe}^{2+}$ ion and the $\text{Fe}^{3+}$ ion. [2]

(c) State the definition of a transition element, and use your answer to (b) to explain why iron satisfies this definition. [2]

1,2-diaminoethane, $\text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2$ (commonly abbreviated "en"), is a bidentate ligand. In aqueous solution, nickel(II) ions normally exist as the octahedral complex $[\text{Ni}(\text{H}_2\text{O})_6]^{2+}$. Adding excess en to this solution replaces all six water ligands, forming $[\text{Ni}(\text{en})_3]^{2+}$:

$[\text{Ni}(\text{H}_2\text{O})_6]^{2+}(aq) + 3\text{en}(aq) \rightleftharpoons [\text{Ni}(\text{en})_3]^{2+}(aq) + 6\text{H}_2\text{O}(l)$

(a) State what is meant by the term bidentate ligand, and use this to explain why three molecules of en (rather than six) are needed to replace the six water ligands in this ligand-exchange reaction. [2]

(b) State the coordination number of nickel, and the shape, of the $[\text{Ni}(\text{en})_3]^{2+}$ ion. [1]

(c) $[\text{Ni}(\text{en})_3]^{2+}$ exists as two non-superimposable mirror-image forms (it shows optical isomerism), whereas $[\text{Ni}(\text{NH}_3)_6]^{2+}$, which also contains six monodentate nitrogen-donor ligands, does not. Suggest a reason for this difference. [2]

(d) Write an expression for the stability constant, $K_{stab}$, of the ligand-exchange reaction shown above, and state, with a reason, whether you would expect $K_{stab}$ for this reaction to be large or small. [3]

Standard electrode potentials for two relevant half-reactions are:

$\text{Cr}_2\text{O}_7^{2-}(aq) + 14\text{H}^+(aq) + 6e^- \rightleftharpoons 2\text{Cr}^{3+}(aq) + 7\text{H}_2\text{O}(l) \qquad E^{\ominus} = +1.33\ \text{V}$ $\text{Fe}^{3+}(aq) + e^- \rightleftharpoons \text{Fe}^{2+}(aq) \qquad E^{\ominus} = +0.77\ \text{V}$

(a) Use these standard electrode potentials to show that the reaction between $\text{Cr}_2\text{O}_7^{2-}(aq)$ and $\text{Fe}^{2+}(aq)$ in acidic solution is thermodynamically feasible, and calculate the standard cell potential, $E^{\ominus}_{cell}$, for this reaction. [2]

(b) Combine the two half-equations to write the overall ionic equation for this reaction. [2]

(c) A geologist analyses a $5.00\ \text{g}$ sample of crushed iron ore to determine its iron content. The sample is dissolved completely in excess dilute sulfuric acid and, by passing the resulting solution through a reducing column, all the iron present is converted to $\text{Fe}^{2+}(aq)$; this solution is then made up to exactly $250\ \text{cm}^3$ in a volumetric flask. A $25.0\ \text{cm}^3$ portion of this solution is titrated against $0.0150\ \text{mol dm}^{-3}$ standardised potassium dichromate(VI), using a few drops of sodium diphenylamine sulfonate as an indicator; $20.00\ \text{cm}^3$ is required to reach the end-point (a sharp colour change from green to violet). Calculate the number of moles of $\text{Cr}_2\text{O}_7^{2-}(aq)$ used, and hence the concentration, in $\text{mol dm}^{-3}$, of $\text{Fe}^{2+}(aq)$ in the $25.0\ \text{cm}^3$ sample. [3]

(d) Hence calculate the percentage, by mass, of iron in the original ore sample. [molar mass of $\text{Fe}=55.8\ \text{g mol}^{-1}$] [2]

Aqueous copper(II) sulfate contains the blue complex ion $[\text{Cu}(\text{H}_2\text{O})_6]^{2+}$. Adding excess concentrated hydrochloric acid converts this into the yellow-green complex ion $[\text{CuCl}_4]^{2-}$.

Which statement correctly compares the two complex ions and accounts for the colour change?

Cobalt forms stable complex ions in both the $+2$ and $+3$ oxidation states. The relevant standard electrode potentials, all at $298\ \text{K}$, are:

$[\text{Co(H}_2\text{O)}_6]^{3+}(aq) + e^- \rightleftharpoons [\text{Co(H}_2\text{O)}_6]^{2+}(aq) \qquad E^{\ominus} = +1.82\ \text{V}$ $\text{O}_2(g) + 4\text{H}^+(aq) + 4e^- \rightleftharpoons 2\text{H}_2\text{O}(l) \qquad E^{\ominus} = +1.23\ \text{V}$ $[\text{Co(NH}_3\text{)}_6]^{3+}(aq) + e^- \rightleftharpoons [\text{Co(NH}_3\text{)}_6]^{2+}(aq) \qquad E^{\ominus} = +0.10\ \text{V}$ $\text{O}_2(g) + 2\text{H}_2\text{O}(l) + 4e^- \rightleftharpoons 4\text{OH}^-(aq) \qquad E^{\ominus} = +0.40\ \text{V}$

(a) Use the first two electrode potentials to show that $[\text{Co(H}_2\text{O)}_6]^{3+}(aq)$ cannot exist for long in aqueous solution, and construct the overall balanced ionic equation for its decomposition. [3]

(b) In ammoniacal solution, atmospheric oxygen readily oxidises $[\text{Co(NH}_3\text{)}_6]^{2+}(aq)$ to $[\text{Co(NH}_3\text{)}_6]^{3+}(aq)$. Use the third and fourth electrode potentials to show that this reaction is feasible, and construct the overall balanced ionic equation. [2]

(c) The presence of $\text{NH}_3$ ligands, rather than $\text{H}_2\text{O}$ ligands, lowers the standard electrode potential for the $\text{Co}^{3+}/\text{Co}^{2+}$ couple from $+1.82\ \text{V}$ to $+0.10\ \text{V}$. Explain, in terms of the bonding between the ligand and the two cobalt ions, why this ligand substitution stabilises the $+3$ oxidation state relative to the $+2$ oxidation state. [2]

(d) State the coordination number and shape of the $[\text{Co(NH}_3\text{)}_6]^{3+}$ ion. [1]

Platinum forms a neutral square-planar complex known as cisplatin, $\text{cis-}[\text{Pt(NH}_3)_2\text{Cl}_2]$, which is used clinically as an anticancer drug. Its geometric isomer, $\text{trans-}[\text{Pt(NH}_3)_2\text{Cl}_2]$, has the identical molecular formula but is clinically ineffective.

(a) State the oxidation state of platinum in this complex, and use it, together with the charges on the ligands present, to show that the complex is electrically neutral overall. [2]

(b) State the coordination number of platinum in this complex and its shape. [1]

(c) Describe, in words, how the arrangement of the two $\text{NH}_3$ ligands and two $\text{Cl}^-$ ligands differs between the cis and trans isomers, and explain why this type of stereoisomerism could not occur if the complex were tetrahedral rather than square planar. [3]

(d) State the type of stereoisomerism shown by cis- and trans-$[\text{Pt(NH}_3)_2\text{Cl}_2]$, and suggest why these two isomers, despite having identical molecular formulae, can show very different biological activity. [2]

Solid chromium(III) chloride, $\text{CrCl}_3$, dissolves in water to form hydrated complex ions. In one such complex ion, the $\text{Cr}^{3+}$ ion is bonded directly to two chloride ligands and four water ligands; the remaining chloride ions present in the original solid are simple counter-ions, not bonded directly to chromium.

Which row correctly gives the formula, the coordination number of chromium, and the shape of this complex ion?

Ammonium vanadate(V), $\text{NH}_4\text{VO}_3$, dissolves in dilute sulfuric acid to give a yellow solution containing the $\text{VO}_2^+$ ion. Adding excess zinc powder to this acidified solution reduces vanadium successively through a series of distinctly coloured oxidation states, ending at the pale violet $\text{V}^{2+}(aq)$ ion.

Standard electrode potentials for the relevant half-reactions are:

$\text{VO}_2^+(aq) + 2\text{H}^+(aq) + e^- \rightleftharpoons \text{VO}^{2+}(aq) + \text{H}_2\text{O}(l) \qquad E^{\ominus} = +1.00\ \text{V}$ $\text{VO}^{2+}(aq) + 2\text{H}^+(aq) + e^- \rightleftharpoons \text{V}^{3+}(aq) + \text{H}_2\text{O}(l) \qquad E^{\ominus} = +0.34\ \text{V}$ $\text{V}^{3+}(aq) + e^- \rightleftharpoons \text{V}^{2+}(aq) \qquad E^{\ominus} = -0.26\ \text{V}$ $\text{Zn}^{2+}(aq) + 2e^- \rightleftharpoons \text{Zn}(s) \qquad E^{\ominus} = -0.76\ \text{V}$

(a) State the oxidation state of vanadium in each of $\text{VO}_2^+$, $\text{VO}^{2+}$, $\text{V}^{3+}$ and $\text{V}^{2+}$. [2]

(b) State the colour of each of these four vanadium species, in the order they would be observed as the zinc reduction proceeds from $\text{VO}_2^+$ to $\text{V}^{2+}$. [2]

(c) By combining the first half-equation above with the reverse of the zinc half-equation, construct the overall balanced ionic equation for the reduction of $\text{VO}_2^+(aq)$ to $\text{VO}^{2+}(aq)$ by zinc, and calculate $E^{\ominus}_{cell}$ for this reaction. [3]

(d) Explain, using the standard electrode potentials given, why zinc is able to reduce vanadium all the way down to $\text{V}^{2+}(aq)$, whereas a weaker reducing agent (with a standard electrode potential between $-0.26\ \text{V}$ and $+0.34\ \text{V}$) might succeed in producing $\text{VO}^{2+}(aq)$ or $\text{V}^{3+}(aq)$ but fail to reduce $\text{V}^{3+}(aq)$ further to $\text{V}^{2+}(aq)$. [2]

$\text{EDTA}^{4-}$ is a polydentate ligand with six donor atoms, two nitrogen atoms and four oxygen atoms, one from each of its four carboxylate groups. It forms very stable 1:1 complexes with many metal ions, including $\text{Pb}^{2+}$, and is used medically to treat heavy-metal poisoning.

Which row correctly gives the number of coordinate bonds a single $\text{EDTA}^{4-}$ ion forms with one $\text{Pb}^{2+}$ ion, the resulting coordination number of lead, and the overall charge of the complex ion formed?