"Coordination Compounds" — Practice

Question 1

A compound of composition $\mathrm{CoCl_3 \cdot 4NH_3}$ precipitates one mole of silver chloride per mole with excess silver nitrate. Deduce its formula and name it.

Accepted Answer

One mole of chloride is precipitated, so only one chloride sits outside the coordination sphere as a free ion and the other two are bonded to cobalt. With a coordination number of six, the sphere holds four ammonias and two chlorides, giving $[\mathrm{Co(NH_3)_4Cl_2}]\mathrm{Cl}$. The single outer chloride means the complex ion carries $+1$, and with four neutral ammonias and two chlorides inside, cobalt must be $+3$. The name is tetraamminedichloridocobalt(III) chloride.

Question 2

Name $\mathrm{K_4[Fe(CN)_6]}$ and $[\mathrm{Cr(H_2O)_4Cl_2}]\mathrm{Cl}$.

Accepted Answer

In the first, potassium is the cation and named first; the ligand is cyanido and there are six; the complex is an anion, so iron takes its Latin stem and the -ate suffix. Four potassium ions give $4+$ and six cyanides give $6-$, so iron is $+2$, giving potassium hexacyanidoferrate(II). In the second the complex is a cation, so chromium keeps its English name; ligands are alphabetised as aqua before chlorido, and chromium is $+3$, giving tetraaquadichloridochromium(III) chloride.

Question 3

Explain the chelate effect and state why five- and six-membered rings are preferred.

Accepted Answer

A chelate complex is far more stable than a comparable complex of monodentate ligands with the same donor atoms: $\log \beta$ is 18.3 for $[\mathrm{Ni(en)_3}]^{2+}$ against 8.6 for $[\mathrm{Ni(NH_3)_6}]^{2+}$. The cause is almost entirely entropy, since the bonds formed are nearly identical: three bidentate ligands displacing six waters turn four particles into seven, whereas six ammonias displacing six waters leave the count unchanged. Five- and six-membered rings close without appreciable bond angle strain, whereas smaller rings are strained and larger ones are entropically costly to close.

Question 4

How many stereoisomers does $[\mathrm{Co(en)_2Cl_2}]^+$ have, and which of them are optically active?

Accepted Answer

The complex is octahedral with two bidentate ethane-1,2-diamine ligands and two chlorides, which may be adjacent or opposite, giving cis and trans geometrical isomers. The trans isomer has a plane of symmetry through the two chlorides, so it is superimposable on its mirror image and optically inactive. The cis isomer has no such plane and exists as a non-superimposable pair of enantiomers. There are therefore three stereoisomers in total, of which only the cis pair is optically active.

Question 5

Distinguish ionisation isomerism from coordination isomerism, with an example of each.

Accepted Answer

Ionisation isomerism swaps a ligand inside the coordination sphere with the counter ion outside it, so the two isomers give different ions in solution: pentaamminesulphatocobalt(III) bromide and pentaamminebromidocobalt(III) sulphate are the standard pair, one giving a positive test for bromide and the other for sulphate. Coordination isomerism requires both the cation and the anion to be complexes and swaps ligands between them, as in $[\mathrm{Co(NH_3)_6}][\mathrm{Cr(CN)_6}]$ against $[\mathrm{Cr(NH_3)_6}][\mathrm{Co(CN)_6}]$.

Question 6

Why do tetrahedral complexes never show geometrical isomerism while square planar ones do?

Accepted Answer

Because geometrical isomerism requires positions distinguishable as adjacent or opposite. In a tetrahedron every vertex is adjacent to every other vertex, all inter-ligand angles being the same 109.5 degrees, so there is no pair of positions that could be called trans and no second arrangement to find. In a square planar complex two positions are at 90 degrees and one is at 180, so a pair of identical ligands can be either adjacent or opposite, giving genuine cis and trans isomers. Cis-platin and trans-platin are exactly that case.

Question 7

Using valence bond theory, account for the fact that $[\mathrm{Fe(CN)_6}]^{3-}$ has one unpaired electron while $[\mathrm{FeF_6}]^{3-}$ has five.

Accepted Answer

Iron is $+3$ in both, so the configuration is $3d^5$. Cyanide is a strong field ligand and forces the five d electrons to pair as far as possible into three orbitals, leaving one unpaired and vacating two 3d orbitals; these combine with 4s and 4p to give $d^2sp^3$ hybridisation, an inner orbital octahedral complex with a moment of 1.73 BM. Fluoride is a weak field ligand and forces no pairing, so all five electrons stay unpaired and the metal must use 4d orbitals, giving $sp^3d^2$ hybridisation and a moment of 5.92 BM.

Question 8

State three limitations of valence bond theory as applied to coordination compounds.

Accepted Answer

First, it gives no account of colour at all, because it contains no energy gap that a photon could match. Second, it cannot explain why one ligand forces electron pairing and another does not; it can only label them strong and weak field after the experimental result is known. Third, it offers no way to rank ligands and makes no quantitative prediction of any property. Crystal field theory repairs all three by introducing a single measurable quantity, the splitting energy $\Delta_o$, whose size decides spin state and colour together.

Question 9

Explain why $[\mathrm{Co(NH_3)_6}]^{3+}$ is diamagnetic while $[\mathrm{CoF_6}]^{3-}$ is paramagnetic with four unpaired electrons.

Accepted Answer

Cobalt is $+3$ in both, giving $d^6$. Ammonia lies high in the spectrochemical series, so $\Delta_o$ exceeds the pairing energy and all six electrons occupy the lower set as $t_{2g}^6 e_g^0$, leaving none unpaired and making the complex diamagnetic. Fluoride lies low in the series, so $\Delta_o$ is smaller than $P$ and the electrons spread out as $t_{2g}^4 e_g^2$, leaving four unpaired and giving a moment of 4.90 BM. The metal ion is identical in both cases; only the splitting changed.

Question 10

Calculate the crystal field stabilisation energy for high spin and low spin $d^6$ octahedral complexes and comment on the difference.

Accepted Answer

Low spin $d^6$ is $t_{2g}^6 e_g^0$, so the CFSE is $-0.4 	imes 6 = -2.4\Delta_o$. High spin $d^6$ is $t_{2g}^4 e_g^2$, giving $-0.4(4) + 0.6(2) = -1.6 + 1.2 = -0.4\Delta_o$. The low spin arrangement is stabilised six times as strongly, which is why cobalt(III) forms low spin complexes with essentially every ligand stronger than fluoride. The pairing energy partly offsets this, but not enough once the ligand is a strong one.

Question 11

Explain why tetrahedral complexes are always high spin.

Accepted Answer

Because $\Delta_t = 	frac{4}{9}\Delta_o$ for the same metal and ligands. Two effects cause this: there are four ligands rather than six, and none of them points directly at any d orbital, so the electrostatic perturbation is much weaker. The resulting gap is essentially never larger than the pairing energy, so electrons always prefer to occupy the upper set singly rather than pair in the lower set. This means no low spin tetrahedral complex needs to be considered, which removes a whole branch of case analysis.

Question 12

Give one application of coordination chemistry in qualitative analysis, one in metallurgy and one in medicine.

Accepted Answer

In qualitative analysis, silver chloride dissolves in aqueous ammonia as the soluble diamminesilver(I) ion, which distinguishes silver from lead and mercury chlorides. In metallurgy, gold and silver are extracted by cyanide leaching, dissolving as soluble cyanido complexes and later displaced by zinc, and nickel is purified by the Mond process through volatile $\mathrm{Ni(CO)_4}$. In medicine, EDTA chelates lead for excretion in lead poisoning, given as the calcium salt so that it does not strip calcium indiscriminately, and cis-platin is a widely used anticancer drug whose trans isomer is inactive.

Practice — "Coordination Compounds"

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