"Coordination Compounds" — Practice

Question 1

For $\left[	ext{Co}\left(	ext{NH}_3ight)_6ight]^{3+}$ and $\left[	ext{CoF}_6ight]^{3-}$, give the spin state, number of unpaired electrons, magnetic moment, hybridisation and CFSE.

Accepted Answer

Both are $3d^{6}$.

$\left[	ext{Co}\left(	ext{NH}_3ight)_6ight]^{3+}$: ammonia is strong field, so **low spin** $t_{2g}^{6}$, zero unpaired, $\mu=0$, $d^{2}sp^{3}$ inner orbital.

$	ext{CFSE}=-0.4	imes6=-2.4\Delta_o$

$\left[	ext{CoF}_6ight]^{3-}$: fluoride is weak field, so **high spin** $t_{2g}^{4}e_g^{2}$, four unpaired, $\mu=\sqrt{24}=4.90$ BM, $sp^{3}d^{2}$ outer orbital.

$	ext{CFSE}=\left(-0.4	imes4+0.6	imes2ight)=-0.4\Delta_o$

Question 2

Explain why no octahedral $d^{4}$ to $d^{7}$ complex can be predicted as high or low spin without knowing the ligand, but $d^{1}$, $d^{2}$, $d^{3}$, $d^{8}$, $d^{9}$ and $d^{10}$ can.

Accepted Answer

For $d^{1}$ to $d^{3}$ the electrons occupy the three lower orbitals singly regardless, so no choice arises.

For $d^{8}$ to $d^{10}$ the lower set is already full and the remaining electrons must enter the upper set whatever the splitting.

Only between $d^{4}$ and $d^{7}$ is there a genuine alternative: place the next electron in the upper set at a cost of $\Delta_o$, or pair it in the lower set at a cost of $P$.

The ligand's field strength decides which is cheaper, so only those four configurations show two spin states.

Question 3

A complex absorbs light at $600$ nm. Find $\Delta_o$ in kJ mol$^{-1}$ and state the observed colour.

Accepted Answer

$\Delta_o=\dfrac{1240}{600}=2.07$ eV per ion

Per mole: $2.07	imes96.5=200$ kJ mol$^{-1}$

Absorption at $600$ nm is in the orange, so the complementary transmitted colour is **blue**.

A ligand higher in the spectrochemical series would raise $\Delta_o$, move the absorption to shorter wavelength and shift the observed colour towards orange or red.

Question 4

Write IUPAC names for $\left[	ext{Pt}\left(	ext{NH}_3ight)_4ight]\left[	ext{PtCl}_4ight]$ and $	ext{Na}_2\left[	ext{Ni}\left(	ext{CN}ight)_4ight]$.

Accepted Answer

The first has a complex cation and a complex anion, so both are named: **tetraammineplatinum(II) tetrachloridoplatinate(II)**.

Platinum is $+2$ in both, since ammonia is neutral and the four chlorides give $-4$ against a $2-$ charge.

The second is an anionic complex, so nickel takes the $-ate$ ending: **sodium tetracyanidonickelate(II)**.

Four cyanides give $-4$ against an overall $2-$, so nickel is $+2$.

Question 5

Two complexes have the empirical formula $	ext{CoBrSO}_4\cdot5	ext{NH}_3$. One gives a precipitate with barium chloride and the other with silver nitrate. Identify both and name the isomerism.

Accepted Answer

The one giving a barium sulphate precipitate has free sulphate outside the sphere: $\left[	ext{Co}\left(	ext{NH}_3ight)_5	ext{Br}ight]	ext{SO}_4$.

The one giving silver bromide has free bromide outside: $\left[	ext{Co}\left(	ext{NH}_3ight)_5	ext{SO}_4ight]	ext{Br}$.

This is **ionisation isomerism**, in which the two isomers differ only in which anion occupies the coordination sphere.

The test is decisive because bound ligands are not available to precipitate.

Question 6

Explain how conductivity measurements distinguish $\left[	ext{Co}\left(	ext{NH}_3ight)_6ight]	ext{Cl}_3$, $\left[	ext{Co}\left(	ext{NH}_3ight)_5	ext{Cl}ight]	ext{Cl}_2$ and $\left[	ext{Co}\left(	ext{NH}_3ight)_4	ext{Cl}_2ight]	ext{Cl}$.

Accepted Answer

The first gives four ions per formula unit: one complex cation and three chlorides.

The second gives three, and the third gives two.

Molar conductivity rises with the number of ions, so the three are readily distinguished by measurement alone.

Werner used precisely this method, together with silver nitrate precipitation, to establish that some chlorides are bound and some are free.

Question 7

Count all stereoisomers of $\left[	ext{Cr}\left(	ext{ox}ight)_3ight]^{3-}$ and of $\left[	ext{Co}\left(	ext{NH}_3ight)_3	ext{Cl}_3ight]$.

Accepted Answer

$\left[	ext{Cr}\left(	ext{ox}ight)_3ight]^{3-}$: three identical bidentate ligands, so **no geometrical isomers**. The propeller arrangement has no plane of symmetry, so it exists as a pair of enantiomers: **two** stereoisomers.

$\left[	ext{Co}\left(	ext{NH}_3ight)_3	ext{Cl}_3ight]$: the $	ext{MA}_3	ext{B}_3$ case, giving **facial** and **meridional** isomers.

Both have a plane of symmetry, so neither is optically active: **two** stereoisomers.

The counts coincide but for entirely different reasons.

Question 8

A four-coordinate complex $	ext{MA}_2	ext{B}_2$ is found to exist in two forms. What does this establish about its geometry, and why?

Accepted Answer

It establishes that the complex is **square planar**.

In a tetrahedron every pair of positions is equivalent, so there is exactly one way to place two of each ligand and no geometrical isomerism is possible.

In a square plane the two identical ligands can be adjacent (cis) or opposite (trans), giving two distinct compounds.

This argument is how the geometry of platinum(II) complexes was established, and it is the reason cisplatin and its inactive trans isomer both exist.

Question 9

Explain why EDTA forms exceptionally stable complexes and give two practical consequences.

Accepted Answer

EDTA is **hexadentate**, binding through two nitrogen and four oxygen atoms and forming five chelate rings simultaneously with a single metal ion.

Displacing six aqua ligands with one EDTA molecule releases six particles while consuming one, a very large entropy gain.

Practically, EDTA titration measures water hardness directly by complexing calcium and magnesium.

Medically, it is administered to treat lead poisoning, sequestering the lead as a soluble complex that is excreted.

Question 10

Apply the effective atomic number rule to predict the formula of the simplest stable carbonyl of vanadium ($Z=23$) and of molybdenum ($Z=42$).

Accepted Answer

Molybdenum: $42+2n=54$ (xenon) gives $n=6$, so $	ext{Mo}\left(	ext{CO}ight)_6$ — stable and well known.

Vanadium: $23+2n=36$ would need $n=6.5$, which is impossible.

Vanadium therefore forms $	ext{V}\left(	ext{CO}ight)_6$ with $35$ electrons, one short of krypton. It is a rare **odd-electron** carbonyl, is paramagnetic, and is unusually reactive.

It readily accepts an electron to give the stable $\left[	ext{V}\left(	ext{CO}ight)_6ight]^-$ anion, which does reach $36$.

Question 11

The carbon-oxygen stretching frequency is $2143$ cm$^{-1}$ in free carbon monoxide but $2060$ cm$^{-1}$ in nickel tetracarbonyl. Explain.

Accepted Answer

Carbon monoxide donates its carbon lone pair to nickel, which alone would barely change the carbon-oxygen bond.

The metal then back donates electron density from a filled $d$ orbital into the empty **antibonding** orbital of carbon monoxide.

Populating an antibonding orbital reduces the carbon-oxygen bond order, lengthening and weakening the bond, so its stretching frequency falls.

The size of the drop measures the extent of back donation, which is why infrared spectroscopy is the standard probe of carbonyl bonding.

Question 12

Explain the role of the metal ion in haemoglobin and why carbon monoxide is so much more toxic than its concentration would suggest.

Accepted Answer

Haemoglobin holds iron(II) in a porphyrin ring, which fixes the oxidation state and geometry and leaves one site free for a small ligand.

Oxygen binds there reversibly, which is essential: it must be released again in the tissues.

Carbon monoxide binds at the same site about $200$ times more strongly, because it accepts back donation from iron into its antibonding orbital far more effectively than oxygen does.

The resulting complex does not dissociate, so a small partial pressure disables a large fraction of the carrier permanently.

Practice — "Coordination Compounds"

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