Colour, Magnetism And Stability Of Co-ordination Compounds

ChemistryCoordination CompoundsFor JEE aspirants

The colour, magnetism and stability of coordination compounds all come from the d electrons of the metal and how tightly the ligands hold it. A complex looks coloured because a d-d transition absorbs part of visible light; it is paramagnetic when unpaired electrons remain; and its stability in solution is measured by the stability constant. This page explains the colour, magnetism and stability of coordination compounds with worked numbers, then the uses of complexes. These ideas are tested every year in JEE Main and NEET.

On this page1Why complexes are coloured2Colourless complexes3Magnetic moment4Stability constants5Factors6Applications7Solved examples

Key Formulas - Quick Reference

★ Must learn Colour seen = complementary colour of the light absorbed
★ Must learn Energy absorbed in a d-d transition: $Δ_{o} = h ν = \frac{h c}{λ}$ ; larger $Δ_{o}$ means a shorter wavelength is absorbed
No d-d transition, so colourless: $d^{0}$ ( $S c^{3 +}$ , $T i^{4 +}$ ) and $d^{10}$ ( $Z n^{2 +}$ , $C u^{+}$ , $C d^{2 +}$ ). $Mn O_{4}^{-}$ and $Cr O_{4}^{2 -}$ are coloured by charge transfer
★ Must learn Spin-only moment: $μ = n (n + 2)$ BM, giving 0, 1.73, 2.83, 3.87, 4.90, 5.92 for n = 0 to 5
★ Must learn Overall stability constant: $M + n L ⇌ ML_{n}$ , $β_{n} = \frac{[ ML _{n} ]}{[ M ] [ L ] ^{n}}$
Instability (dissociation) constant: $K_{i} = \frac{1}{β}$
★ Must learn Stepwise constants: $β_{n} = K_{1} K_{2} \dots K_{n}$ , $lo g β_{n} = \sum lo g K$ ; usually $K_{1} > K_{2} > K_{3} > \dots$
Stability rises with the charge density of the metal ion, the basicity of the ligand and chelation
Mond process: $Ni + 4CO$ → $Ni (CO)_{4}$ (330-350 K), decomposed at 450-470 K
Gold: $4Au$ + $8C N^{-}$ + $2 H_{2} O$ + $O_{2}$ → $4 [Au (CN)_{2}]^{-}$ + $4O H^{-}$

1. Colour of Coordination Compounds

1.1 Why Complexes Are Coloured

In a complex the d orbitals of the metal are split (see Bonding). If the $t_{2g}$ level holds an electron and the $e_{g}$ level has room, the electron can jump up by absorbing a photon whose energy equals $Δ_{o}$ . For most complexes this energy falls in the visible region, so part of white light is absorbed. This is a d-d transition.

The colour of a complex depends on the metal, its oxidation state and its ligands.
Coloured compounds absorb visible light. The colour we see is the mixture of the wavelengths that are not absorbed, that is, the complementary colour of the light absorbed.
A plot of how much light is absorbed at each wavelength is the absorption spectrum of the complex.
For example, $[Ti (H_{2} O)_{6}]^{3 +}$ has one d electron ( $t_{2g}^{1}$ ). It absorbs most strongly near 498 nm, in the blue-green region, and so appears violet.

Figure 1: A complex absorbs light whose energy equals

Δ_{o}

and shows the complementary colour.

[Ti (H_{2} O)_{6}]^{3 +}

absorbs near 498 nm and looks violet.

Figure 2: One d electron, one broad band. Its peak at about 498 nm is

Δ_{o}

for

[Ti (H_{2} O)_{6}]^{3 +}

; the unabsorbed ends of the spectrum give the violet colour.

1.2 Absorbed Colour and Observed Colour

The table (NCERT data) shows that a stronger ligand field shifts the absorption to shorter wavelength:

Complex	$λ$ absorbed (nm)	Colour absorbed	Colour observed
$[CoCl (N H_{3})_{5}]^{2 +}$	535	yellow	violet
$[Co (N H_{3})_{5} (H_{2} O)]^{3 +}$	500	blue-green	red
$[Co (N H_{3})_{6}]^{3 +}$	475	blue	yellow-orange
$[Co (CN)_{6}]^{3 -}$	310	ultraviolet	pale yellow
$[Cu (H_{2} O)_{4}]^{2 +}$	600	red	blue
$[Ti (H_{2} O)_{6}]^{3 +}$	498	blue-green	violet

Going from $C l^{-}$ to $H_{2} O$ to $N H_{3}$ to $C N^{-}$ around cobalt(III), $Δ_{o}$ rises and the wavelength absorbed falls from 535 nm to 310 nm, following the spectrochemical series. $[Co (CN)_{6}]^{3 -}$ absorbs mainly in the ultraviolet and so looks only pale yellow.

Figure 3:

Δ_{o} = \frac{N _{A} h c}{λ}

: as the ligand goes from

C l^{-}

C N^{-}

the absorbed wavelength falls from 535 to 310 nm and

Δ_{o}

rises from about 224 to 386 kJ/mol.

Exam Trick

Opposites on the wheel: violet-yellow, blue-orange, green-red. Find the colour absorbed, jump straight across the colour wheel, and you have the colour seen. $[Cu (H_{2} O)_{4}]^{2 +}$ absorbs red (600 nm), so it looks blue.

1.3 What Decides the Colour

The ligand. Pale blue $[Cu (H_{2} O)_{4}]^{2 +}$ turns deep blue $[Cu (N H_{3})_{4}]^{2 +}$ when aqueous ammonia is added. Adding ethane-1,2-diamine step by step to green $[Ni (H_{2} O)_{6}]^{2 +}$ changes it to pale blue, blue-purple and finally violet $[Ni (en)_{3}]^{2 +}$ .
The geometry. Pink octahedral $[Co (H_{2} O)_{6}]^{2 +}$ becomes blue tetrahedral $[CoC l_{4}]^{2 -}$ in concentrated HCl, because the splitting pattern changes.
The metal and its oxidation state, which fix the number of d electrons and the size of $Δ$ .
The presence of ligands at all. Anhydrous $CuS O_{4}$ is white; with no water ligands there is no splitting to cause a d-d transition. $CuS O_{4} \cdot 5 H_{2} O$ is blue.

Figure 4: The colour of a complex follows its ligands and its shape. Without ligands (anhydrous

CuS O_{4}

) there is no splitting and no colour.

1.4 When a Complex Is Colourless

A partly filled d subshell is usually needed for a complex to be coloured. $d^{0}$ ions ( $S c^{3 +}$ , $T i^{4 +}$ ) have no electron to promote, and $d^{10}$ ions ( $Z n^{2 +}$ , $C d^{2 +}$ , $C u^{+}$ , $A g^{+}$ ) have no empty d orbital to receive one, so their complexes are colourless. Important exceptions are $Mn O_{4}^{-}$ (purple), $Cr O_{4}^{2 -}$ (yellow) and $C r_{2} O_{7}^{2 -}$ (orange): the metal is $d^{0}$ , but the colour comes from a different process.

JEE Advanced

Charge transfer and pale $d^{5}$ complexes. In $Mn O_{4}^{-}$ an electron moves from an oxygen (ligand) orbital to the metal: a ligand-to-metal charge transfer. Such transitions are fully allowed, so the colour is very intense. d-d transitions are "forbidden" by the selection rules and are weak, which is why most complexes are only lightly coloured. In high spin $d^{5}$ complexes such as $[Mn (H_{2} O)_{6}]^{2 +}$ and $[Fe F_{6}]^{3 -}$ every d-d jump would also have to flip an electron spin, so they are almost colourless (very pale pink and colourless).

Gemstones. Ruby is aluminium oxide containing 0.5-1% $C r^{3 +}$ ions in octahedral sites; the splitting makes it absorb green-yellow light and look red. Emerald has $C r^{3 +}$ in the mineral beryl, where a slightly different crystal field makes it absorb differently and look green.

Key idea

Colour = complementary colour of the d-d absorption; a stronger field ligand shifts the absorption to a shorter wavelength.

2. Magnetic Properties

Many transition metal complexes are paramagnetic: they have unpaired electrons and are attracted into a magnetic field. Complexes with all electrons paired are diamagnetic and are weakly repelled. The effect is measured with a magnetic (Gouy) balance, and the result is expressed as the magnetic moment $μ$ in Bohr magnetons (BM). For most complexes of the 3d metals the orbital contribution is small, so the spin-only formula works well:

μ = n (n + 2) BM

Here n is the number of unpaired electrons. For example, $[Co (N H_{3})_{6}]^{3 +}$ has no unpaired electrons ( $μ = 0$ ) but $[Co F_{6}]^{3 -}$ has four ( $μ = 24 = 4.90$ BM).

Figure 5: Spin-only magnetic moments. For

n \geq 2

a quick estimate is

μ \approx n + 1

BM, so a measured value tells you n at a glance.

2.1 Using the Magnetic Moment

Since n follows from $μ$ , a single measurement reveals whether a complex is high spin or low spin, and so which orbitals were used in bonding (see Bonding). Pairs of complexes of the same metal ion show this clearly:

Complex	d electrons	Spin state	Unpaired	$μ$ (BM)	Hybridisation
$[Fe (CN)_{6}]^{3 -}$	$d^{5}$	low spin	1	1.73	$d^{2} sp^{3}$
$[Fe (H_{2} O)_{6}]^{3 +}$	$d^{5}$	high spin	5	5.92	$sp^{3} d^{2}$
$[Mn (CN)_{6}]^{3 -}$	$d^{4}$	low spin	2	2.83	$d^{2} sp^{3}$
$[MnC l_{6}]^{3 -}$	$d^{4}$	high spin	4	4.90	$sp^{3} d^{2}$
$[Co (C_{2} O_{4})_{3}]^{3 -}$	$d^{6}$	low spin	0	0.00	$d^{2} sp^{3}$
$[Co F_{6}]^{3 -}$	$d^{6}$	high spin	4	4.90	$sp^{3} d^{2}$

For four-coordinate complexes the magnetic moment decides the shape: $[Ni (CN)_{4}]^{2 -}$ is diamagnetic and square planar, while $[NiC l_{4}]^{2 -}$ has $μ = 2.83$ BM and is tetrahedral. Tetrahedral complexes are almost always high spin because $Δ_{t}$ is small.

Exam Trick

Read n straight from $μ$ . For n = 2 to 5 the spin-only value is close to n + 1: 2.83 (2), 3.87 (3), 4.90 (4), 5.92 (5). So a measured 3.9 BM means 3 unpaired electrons, and 1.73 BM means exactly 1.

Figure 6: One magnetic measurement tells you whether the ligand paired the electrons, and so which hybridisation and shape the complex has.

Quick Recall: tap to check

Spin-only moment of

[Fe (CN)_{6}]^{3 -}

1.73 BM: low spin

d^{5}

has 1 unpaired electron.

Why is anhydrous

CuS O_{4}

white?

With no ligands the d orbitals are not split, so there is no d-d transition.

Which absorbs the shorter wavelength,

[Co (N H_{3})_{6}]^{3 +}

[CoCl (N H_{3})_{5}]^{2 +}

[Co (N H_{3})_{6}]^{3 +}

(475 nm against 535 nm):

N H_{3}

gives the larger

Δ_{o}

Key idea

Count the unpaired electrons, then

μ = n (n + 2)

BM; for

n \geq 2

μ \approx n + 1

3. Stability of Complexes in Solution

3.1 Stability and Instability Constants

The stability of a complex in solution measures how strongly it resists breaking up or having its ligands replaced. Because dissociation is an equilibrium, stability is expressed by an equilibrium constant. For the dissociation of the tetraamminecopper(II) ion:

[Cu (N H_{3})_{4}]^{2 +} ⇌ C u^{2 +} + 4N H_{3}

K_{i} = \frac{[ C u ^{2 +} ] [ N H _{3} ] ^{4}}{[ [ Cu ( N H _{3} ) _{4} ] ^{2 +} ]}

$K_{i}$ is the instability constant. If instead we write the formation of the complex, the reverse reaction:

C u^{2 +} + 4N H_{3} ⇌ [Cu (N H_{3})_{4}]^{2 +}

K = β_{4} = \frac{[ [ Cu ( N H _{3} ) _{4} ] ^{2 +} ]}{[ C u ^{2 +} ] [ N H _{3} ] ^{4}} = 2.1 \times 1 0^{13}

This is the stability (formation) constant, and $K = 1/ K_{i}$ . The larger the stability constant, the more stable the complex in solution.

3.2 Stepwise and Overall Constants

A complex forms in steps, one ligand at a time, and each step has its own stepwise constant:

M + L ⇌ ML, K_{1} = \frac{[ ML ]}{[ M ] [ L ]} ML + L ⇌ M L_{2}, K_{2} = \frac{[ M L _{2} ]}{[ ML ] [ L ]}

and so on. The overall constant is their product, and its logarithm is their sum:

β_{n} = K_{1} \times K_{2} \times \dots \times K_{n}, lo g β_{n} = lo g K_{1} + lo g K_{2} + \dots + lo g K_{n}

Usually $K_{1} > K_{2} > K_{3} > K_{4}$ : as ligands are added there are fewer places left for the next one and more ligands that can leave, and a growing negative charge (for anionic ligands) repels the next ligand.

Figure 7: Stepwise constants fall steadily, and their logarithms add up to

lo g β_{4}

. Typical literature values give

β_{4} \approx 2.1 \times 1 0^{13}

, the value NCERT uses.

Exam Trick

Logs add, constants multiply. $lo g β_{n}$ is the sum of the stepwise $lo g K$ values, and $K_{i} = 1/ β$ , so $lo g K_{i} = - lo g β$ . A higher charge or a chelate always means a bigger $lo g β$ .

3.3 Stability Constants of Common Complexes

Figure 8: Approximate

lo g β

values. Compare pairs: Co(III) vs Co(II) (charge),

[Ni (en)_{3}]^{2 +}

[Ni (N H_{3})_{6}]^{2 +}

(chelation), Pb vs Ca with EDTA (metal), and cyanide vs chloride with Ag(I) (ligand).

Complex	Overall stability constant β (approx.)	log β
$[AgC l_{2}]^{-}$	$1.1 \times 1 0^{5}$	5.0
$[AgB r_{2}]^{-}$	$1.3 \times 1 0^{7}$	7.1
$[Ag (N H_{3})_{2}]^{+}$	$1.6 \times 1 0^{7}$	7.2
$[Ag (CN)_{2}]^{-}$	about $1 0^{20}$	20
$[Cu (N H_{3})_{4}]^{2 +}$	$2.1 \times 1 0^{13}$	13.3
$[Ni (N H_{3})_{6}]^{2 +}$	$4 \times 1 0^{8}$	8.6
$[Ni (en)_{3}]^{2 +}$	$2 \times 1 0^{18}$	18.3
$[Co (N H_{3})_{6}]^{2 +}$	about $1 0^{5}$	5.1
$[Co (N H_{3})_{6}]^{3 +}$	about $1 0^{34}$	34
$[Fe (CN)_{6}]^{4 -}$	about $1 0^{35}$	35
$[Fe (CN)_{6}]^{3 -}$	about $1 0^{43}$	43
$[Ca (EDTA)]^{2 -}$	$5 \times 1 0^{10}$	10.7
$[Pb (EDTA)]^{2 -}$	$1 \times 1 0^{18}$	18.0

Values vary slightly between data sources and with conditions; use them for comparison, not as exact data.

3.4 Factors Affecting Stability

(a) Nature of the central ion.

Charge density (charge ÷ radius): a smaller, more highly charged ion holds ligands more tightly. $[Co (N H_{3})_{6}]^{3 +}$ (log β about 34) is enormously more stable than $[Co (N H_{3})_{6}]^{2 +}$ (about 5).
Higher oxidation state gives more stable complexes with the same ligand: $[Fe (CN)_{6}]^{3 -}$ is more stable than $[Fe (CN)_{6}]^{4 -}$ .
Electronegativity: the more electronegative (electron-attracting) the metal ion, the stronger its bonds to ligands.

(b) Nature of the ligand.

Basicity: a more basic ligand donates its electron pair more readily and usually forms a more stable complex ( $N H_{3}$ beats $H_{2} O$ for most metal ions).
Chelation: chelating ligands form far more stable complexes than similar unidentate ligands (the chelate effect): $[Ni (en)_{3}]^{2 +}$ is about $1 0^{10}$ times more stable than $[Ni (N H_{3})_{6}]^{2 +}$ . Five- and six-membered chelate rings are the most stable.
Matching the metal: $C N^{-}$ and chelates such as EDTA form very stable complexes with most metals; soft ions such as $A g^{+}$ prefer soft ligands ( $I^{-}$ > $B r^{-}$ > $C l^{-}$ ).

Stable is not the same as inert. Thermodynamic stability (a large β) says where the equilibrium lies; kinetic lability says how fast ligands exchange. $[Ni (CN)_{4}]^{2 -}$ has a huge formation constant (about $1 0^{30}$ ) yet exchanges its cyanide ligands almost instantly (stable but labile). $[Co (N H_{3})_{6}]^{3 +}$ should decompose in acid, but does so only over days (unstable in acid but inert). For ions of the same charge, the stability of high spin complexes follows the Irving-Williams order: $M n^{2 +}$ < $F e^{2 +}$ < $C o^{2 +}$ < $N i^{2 +}$ < $C u^{2 +}$ > $Z n^{2 +}$ .

Thermodynamically stableLarge

β

: the equilibrium lies far on the side of the complex.

Kinetically inertSlow ligand exchange: says how fast, not how far. A stable complex can still be labile.

3.5 Evidence of Complex Formation in Solution

Change observed	Example
Solubility increases	AgCl dissolves in aqueous $N H_{3}$ as $[Ag (N H_{3})_{2}]^{+}$
Colour changes	pale blue $C u^{2 +}$ (aq) turns deep blue with $N H_{3}$
Ion tests disappear	$K_{4} [Fe (CN)_{6}]$ gives no test for $F e^{2 +}$ or $C N^{-}$
Conductivity and freezing point change	the number of free ions and particles in solution changes
pH changes	EDTA titrations release $H^{+}$ , so a buffer is needed
Magnetic moment changes	unpaired electrons are paired by strong ligands

Key idea

A large

β

means a stable complex: high charge density, basic ligands and chelation all raise it.

4. Importance and Applications of Coordination Compounds

Figure 9: Coordination compounds at work. Almost every use depends on one property: a stable complex that holds a metal ion tightly and selectively.

4.1 Analytical Chemistry

Qualitative analysis uses coloured complexes: a red precipitate with dimethylglyoxime for $N i^{2 +}$ , a blood-red colour with thiocyanate for $F e^{3 +}$ , a deep blue colour with ammonia for $C u^{2 +}$ , and Prussian blue with potassium ferrocyanide for $F e^{3 +}$ .
Hardness of water is found by titration with EDTA, which forms stable complexes with $C a^{2 +}$ and $M g^{2 +}$ . The difference in stability constants of the Ca and Mg complexes lets them be estimated separately.
Gravimetric analysis: nickel is weighed as its dimethylglyoxime complex.

4.2 Metallurgy

Gold and silver are extracted by dissolving the metal as a cyanide complex in the presence of air, then displacing it with zinc:

4Au + 8C N^{-} + 2 H_{2} O + O_{2} ⟶ 4 [Au (CN)_{2}]^{-} + 4O H^{-}

2 [Au (CN)_{2}]^{-} + Zn ⟶ [Zn (CN)_{4}]^{2 -} + 2Au

Impure nickel is purified by the Mond process: it reacts with carbon monoxide at 330-350 K to form volatile $[Ni (CO)_{4}]$ , which is decomposed at 450-470 K to give pure nickel.

Ni + 4CO ⇌ Ni (CO)_{4}

4.3 Biological Systems

Chlorophyll, the green pigment of plants, is a coordination compound of magnesium. Haemoglobin, the red oxygen carrier of blood, and myoglobin, which stores oxygen, are coordination compounds of iron. Vitamin $B_{12}$ (cyanocobalamin), the anti-pernicious anaemia factor, is a complex of cobalt. These contain large ring (porphyrin or corrin) ligands. Many enzymes, such as carboxypeptidase A and carbonic anhydrase, contain coordinated metal ions.

4.4 Medicine

Chelation therapy removes toxic metals: excess copper is removed with D-penicillamine and excess iron with desferrioxamine B. EDTA (as its calcium salt) treats lead poisoning.
Cisplatin, cis- $[PtC l_{2} (N H_{3})_{2}]$ , and related platinum complexes inhibit the growth of tumours.

4.5 Industry, Photography and Electroplating

Catalysis: Wilkinson's catalyst, $[RhCl (PP h_{3})_{3}]$ , is used for the hydrogenation of alkenes.
Photography: in fixing, unexposed silver bromide is dissolved by sodium thiosulphate (hypo) as a soluble complex.
Electroplating: $[Ag (CN)_{2}]^{-}$ and $[Au (CN)_{2}]^{-}$ keep the concentration of free metal ions very low, which gives a smooth, even coating.

AgBr + 2N a_{2} S_{2} O_{3} ⟶ N a_{3} [Ag (S_{2} O_{3})_{2}] + NaBr

Figure 10: The four properties on one page. Colour and magnetism come from the d electrons; stability comes from how tightly the ligands hold the metal.

Quick Recall: tap to check

How is

β_{4}

related to the stepwise constants?

β_{4} = K_{1} K_{2} K_{3} K_{4}

, so

lo g β_{4} = \sum lo g K

Why is

[Co (N H_{3})_{6}]^{3 +}

far more stable than

[Co (N H_{3})_{6}]^{2 +}

C o^{3 +}

has the higher charge density and binds ammonia more strongly.

Which metals are in chlorophyll and vitamin

B_{12}

Magnesium and cobalt.

5. Solved Examples

Solved Example 1

Which of the two is more stable and why:

K_{4} [Fe (CN)_{6}]

K_{3} [Fe (CN)_{6}]

Solution:

$K_{3} [Fe (CN)_{6}]$ is more stable (overall stability constant about $1 0^{43}$ , against about $1 0^{35}$ for the ferrocyanide ion). In $[Fe (CN)_{6}]^{3 -}$ iron is in the +3 state; the smaller, more highly charged $F e^{3 +}$ ion has a higher charge density and binds the cyanide ligands more strongly than $F e^{2 +}$ .

Trap: $[Fe (CN)_{6}]^{4 -}$ follows the EAN rule (36) and $[Fe (CN)_{6}]^{3 -}$ does not (35), but the EAN rule is not a test of stability in solution.

Solved Example 2

Cobalt(II) is stable in aqueous solution, but in the presence of strong field ligands and air it is oxidised to cobalt(III). Explain. (Atomic number of Co = 27.)

Solution:

$C o^{2 +}$ is $3 d^{7}$ . Water is a weak field ligand, so $[Co (H_{2} O)_{6}]^{2 +}$ is high spin and nothing drives oxidation.
With strong field ligands such as $N H_{3}$ the electrons pair up. In valence bond terms the seventh electron has to go into a higher (5s or 4d) orbital, from where it is easily lost.
In crystal field terms, $C o^{3 +}$ ( $d^{6}$ ) with strong ligands is low spin $t_{2g}^{6}$ with a very large stabilisation ( $- 2.4 Δ_{o}$ ), much more than $C o^{2 +}$ ( $t_{2g}^{6} e_{g}^{1}$ , $- 1.8 Δ_{o}$ ). So air readily oxidises it:

4 [Co (N H_{3})_{6}]^{2 +} + O_{2} + 2 H_{2} O ⟶ 4 [Co (N H_{3})_{6}]^{3 +} + 4O H^{-}

Solved Example 3

Some four-coordinate complexes of Ni(II) are diamagnetic while others are paramagnetic. Justify.

Solution:

$N i^{2 +}$ is $3 d^{8}$ with two unpaired electrons.

With strong field ligands ( $C N^{-}$ , dmg) the two electrons pair up, one 3d orbital becomes free and $dsp^{2}$ hybridisation gives a square planar, diamagnetic complex such as $[Ni (CN)_{4}]^{2 -}$ .
With weak field ligands ( $C l^{-}$ ) no pairing occurs, $sp^{3}$ hybrids are used and the complex is tetrahedral and paramagnetic with two unpaired electrons ( $μ = 2.83$ BM), such as $[NiC l_{4}]^{2 -}$ .

So the diamagnetic ones are square planar and the paramagnetic ones are tetrahedral.

Solved Example 4

The spin-only magnetic moment of cobalt in

Hg [Co (SCN)_{4}]

is
(A)

3

(B)

8

(C)

15

(D)

24

Solution:

Answer: (C). Mercury is +2, so the anion is $[Co (SCN)_{4}]^{2 -}$ and cobalt is +2 ( $3 d^{7}$ ). A four-coordinate Co(II) thiocyanate complex is tetrahedral and high spin: $e^{4} t_{2}^{3}$ , 3 unpaired electrons. $μ = 3 \times 5 = 15 = 3.87$ BM.

Solved Example 5

[Mn (CN)_{6}]^{3 -}

and

[MnC l_{6}]^{3 -}

both contain

M n^{3 +}

. Predict their numbers of unpaired electrons, magnetic moments and hybridisation.

Solution:

	$[Mn (CN)_{6}]^{3 -}$	$[MnC l_{6}]^{3 -}$
$M n^{3 +}$	$3 d^{4}$	$3 d^{4}$
Ligand	strong field: pairing, $t_{2g}^{4}$	weak field: $t_{2g}^{3} e_{g}^{1}$
Unpaired electrons	2	4
$μ$ (spin-only)	$8 = 2.83$ BM	$24 = 4.90$ BM
Hybridisation	$d^{2} sp^{3}$ (inner orbital)	$sp^{3} d^{2}$ (outer orbital)

Solved Example 6

Among

[Ti F_{6}]^{2 -}

[Co F_{6}]^{3 -}

C u_{2} C l_{2}

and

[NiC l_{4}]^{2 -}

, the colourless species are
(A)

[Co F_{6}]^{3 -}

and

[NiC l_{4}]^{2 -}

(B)

[Ti F_{6}]^{2 -}

and

[Co F_{6}]^{3 -}

(C)

[NiC l_{4}]^{2 -}

and

C u_{2} C l_{2}

(D)

[Ti F_{6}]^{2 -}

and

C u_{2} C l_{2}

Solution:

Answer: (D). $T i^{4 +}$ is $3 d^{0}$ and $C u^{+}$ is $3 d^{10}$ , so neither can show a d-d transition. $C o^{3 +}$ ( $d^{6}$ ) and $N i^{2 +}$ ( $d^{8}$ ) have partly filled d subshells and are coloured.

Solved Example 7

On adding excess HCl to a light pink aqueous solution of cobalt(II) chloride, a deep blue solution forms. This is due to
(A) octahedral

[Co (H_{2} O)_{6}]^{2 +}

changing to octahedral

[CoC l_{6}]^{4 -}

(B) octahedral

[Co (H_{2} O)_{6}]^{2 +}

changing to tetrahedral

[CoC l_{4}]^{2 -}

CoC l_{2}

Solution:

Answer: (B). The large chloride ions replace water and the geometry changes from octahedral to tetrahedral. The different splitting pattern (and the more intense absorption of tetrahedral complexes) gives the blue colour:

[Co (H_{2} O)_{6}]^{2 +} + 4C l^{-} ⇌ [CoC l_{4}]^{2 -} + 6 H_{2} O

Solved Example 8

Aqueous copper sulphate (blue) gives (a) a green precipitate with aqueous potassium fluoride and (b) a bright green solution with aqueous potassium chloride. Explain.

Solution:

In both cases the water ligands of the blue aqua ion are replaced by halide ions, and the new ligand field changes the colour:

[Cu (H_{2} O)_{4}]^{2 +} + 4 F^{-} ⟶ [Cu F_{4}]^{2 -} + 4 H_{2} O (green precipitate)

[Cu (H_{2} O)_{4}]^{2 +} + 4C l^{-} ⟶ [CuC l_{4}]^{2 -} + 4 H_{2} O (bright green solution)

Solved Example 9

F e^{3 +}

with excess thiocyanate gives a blood-red complex A, which turns into a colourless complex B with excess fluoride. Identify A and B, name them and find the spin-only magnetic moment of B.

Solution:

F e^{3 +} + 6SC N^{-} ⟶ [Fe (SCN)_{6}]^{3 -}

[Fe (SCN)_{6}]^{3 -} + 6 F^{-} ⟶ [Fe F_{6}]^{3 -} + 6SC N^{-}

A: $[Fe (SCN)_{6}]^{3 -}$ , hexathiocyanato-N-ferrate(III) ion (blood red). In dilute solution the red species is mainly $[Fe (SCN) (H_{2} O)_{5}]^{2 +}$ .
B: $[Fe F_{6}]^{3 -}$ , hexafluoridoferrate(III) ion; fluoride forms the more stable complex with the hard $F e^{3 +}$ ion.
$F e^{3 +}$ is $d^{5}$ and $F^{-}$ is weak field: 5 unpaired electrons, $μ = 35 = 5.92$ BM. B is colourless because every d-d jump of a high spin $d^{5}$ ion needs a spin flip.

Solved Example 10

Why does

[CuC l_{4}]^{2 -}

exist but

[Cu I_{4}]^{2 -}

does not?

Solution:

Iodide is a reducing agent and $C u^{2 +}$ is an oxidising agent. Instead of forming a complex, copper(II) oxidises iodide to iodine and is itself reduced to insoluble copper(I) iodide:

2C u^{2 +} + 4 I^{-} ⟶ 2CuI ↓ + I_{2}

Chloride is not oxidised by $C u^{2 +}$ , so $[CuC l_{4}]^{2 -}$ is stable.

Solved Example 11

The overall stability constant of

[Cu (N H_{3})_{4}]^{2 +}

2.1 \times 1 0^{13}

(NCERT value). (i) Find its instability constant. (ii) Find the concentration of free

C u^{2 +}

in a solution that is 0.10 M in the complex and 1.0 M in free

N H_{3}

Solution:

(i) $K_{i} = \frac{1}{β} = \frac{1}{2.1 \times 1 0 ^{13}} = 4.8 \times 1 0^{- 14}$

(ii) Rearranging the expression for $β$ :

[C u^{2 +}] = \frac{[ [ Cu ( N H _{3} ) _{4} ] ^{2 +} ]}{β [ N H _{3} ] ^{4}} = \frac{0.10}{2.1 \times 1 0 ^{13} \times ( 1.0 ) ^{4}} = 4.8 \times 1 0^{- 15} M

Almost all the copper is held in the complex, which is why the solution gives hardly any reactions of free $C u^{2 +}$ .

Solved Example 12

Calcium disodium EDTA is used to treat lead poisoning. Given

lo g K = 10.7

for

[Ca (EDTA)]^{2 -}

and 18.0 for

[Pb (EDTA)]^{2 -}

, explain how it works and find the equilibrium constant for the exchange.

Solution:

[Ca (EDTA)]^{2 -} + P b^{2 +} ⇌ [Pb (EDTA)]^{2 -} + C a^{2 +}

K = \frac{K _{Pb}}{K _{Ca}} = 1 0^{18.0 - 10.7} = 1 0^{7.3} \approx 2 \times 1 0^{7}

The equilibrium lies far to the right: lead replaces calcium, and the soluble lead complex is excreted in urine. The calcium salt is used (not free EDTA) so that the body's calcium is not removed.

Solved Example 13

Aluminium fluoride is insoluble in anhydrous HF but dissolves in the presence of KF. When

B F_{3}

is added to this solution,

Al F_{3}

precipitates again. Explain with equations.

Solution:

Liquid HF is covalent and hydrogen bonded, so it supplies almost no free $F^{-}$ ions. KF is ionic and supplies $F^{-}$ , which forms the stable complex $[Al F_{6}]^{3 -}$ :

Al F_{3} + 3KF ⟶ K_{3} [Al F_{6}]

$B F_{3}$ is a stronger fluoride acceptor; it takes the fluoride back as the more stable $[B F_{4}]^{-}$ , and $Al F_{3}$ precipitates:

K_{3} [Al F_{6}] + 3B F_{3} ⟶ Al F_{3} ↓ + 3KB F_{4}

Solved Example 14

In the extraction of gold: roasted ore +

C N^{-}

H_{2} O

O_{2}

→ [X] +

O H^{-}

, then [X] + Zn → [Y] + Au. X and Y are
(A)

[Au (CN)_{2}]^{-}

[Zn (CN)_{4}]^{2 -}

(B)

[Au (CN)_{4}]^{3 -}

[Zn (CN)_{4}]^{2 -}

(C)

[Au (CN)_{2}]^{-}

[Zn (CN)_{6}]^{4 -}

(D)

[Au (CN)_{4}]^{-}

[Zn (CN)_{4}]^{2 -}

Solution:

Answer: (A). Gold(I) forms the linear dicyanidoaurate(I) ion, and zinc, being more reactive, displaces gold and forms the tetrahedral tetracyanidozincate(II) ion (see the equations in section 4.2).

Practice Questions: Source Exercises 3 to 5

Identify the complexes expected to be coloured and explain: (a) $Ti (N O_{3})_{4}$ (b) $[Cu (NCC H_{3})_{4}]^{+} B F_{4}^{-}$ (c) $[Cr (N H_{3})_{6}]^{3 +} 3C l^{-}$ (d) $K_{3} [V F_{6}]$ Answer: (c) and (d). $T i^{4 +}$ ( $d^{0}$ ) and $C u^{+}$ ( $d^{10}$ ) cannot show d-d transitions; $C r^{3 +}$ ( $d^{3}$ ) and $V^{3 +}$ ( $d^{2}$ ) can.
Account for the following: $[Ni (CN)_{4}]^{2 -}$ is square planar and diamagnetic whereas $[NiC l_{4}]^{2 -}$ is tetrahedral and paramagnetic.Answer: $C N^{-}$ pairs the two 3d electrons of $N i^{2 +}$ , freeing a 3d orbital for $dsp^{2}$ ; weak $C l^{-}$ does not, so $sp^{3}$ with 2 unpaired electrons (see the nickel orbital diagrams on the Bonding page).
What coordination entity forms when excess aqueous KCN is added to aqueous copper sulphate? Why is no copper sulphide precipitated when $H_{2} S$ is then passed?Answer: $K_{3} [Cu (CN)_{4}]$ (tetracyanidocuprate(I)); $2CuS O_{4}$ + $10KCN$ → $2 K_{3} [Cu (CN)_{4}]$ + $2 K_{2} S O_{4}$ + $(CN)_{2}$ . The complex is so stable that too few free copper ions remain to precipitate a sulphide.

Practice Questions

Which compound is not coloured? (A) $N a_{2} [CuC l_{4}]$ (B) $N a_{2} [CdC l_{4}]$ (C) $K_{4} [Fe (CN)_{6}]$ (D) $K_{3} [Fe (CN)_{6}]$ Answer: (B): $C d^{2 +}$ is $d^{10}$ .
The colour of $[Ti (H_{2} O)_{6}]^{3 +}$ is due to (A) electron transfer between Ti atoms (B) water molecules (C) a d-d transition (D) molecular vibrationAnswer: (C)
Which ion has the highest paramagnetism? (A) $[Cr (H_{2} O)_{6}]^{3 +}$ (B) $[Fe (H_{2} O)_{6}]^{2 +}$ (C) $[Cu (H_{2} O)_{6}]^{2 +}$ (D) $[Zn (H_{2} O)_{6}]^{2 +}$ Answer: (B): 4 unpaired electrons.
Which is not paramagnetic? (A) $[Cu (N H_{3})_{4}] C l_{2}$ (B) $[Ag (N H_{3})_{2}] Cl$ (C) NO (D) $N O_{2}$ Answer: (B): $A g^{+}$ is $4 d^{10}$ .
How many d electrons does $[Cr (H_{2} O)_{6}]^{3 +}$ have?Answer: 3
Which is paramagnetic: potassium ferrocyanide, potassium ferricyanide, hexaamminecobalt(III) chloride, tetracarbonylnickel(0)?Answer: Potassium ferricyanide (1 unpaired electron).
A spin-only moment of 2.84 BM corresponds to which configuration? (A) $d^{4}$ in a strong field (B) $d^{2}$ in a weak field (C) $d^{3}$ in any field (D) $d^{5}$ in a strong fieldAnswer: (A) and (B) both: $t_{2g}^{4}$ and $t_{2g}^{2}$ each have 2 unpaired electrons.
What type of magnetism does $[Mn (H_{2} O)_{6}]^{2 +}$ show?Answer: Paramagnetism: high spin $d^{5}$ , 5 unpaired electrons, 5.92 BM.
$[Cu (N H_{3})_{4}]^{2 +}$ is coloured but $[Cu (CN)_{4}]^{3 -}$ is colourless. Why?Answer: Cu(II) is $d^{9}$ and can show d-d transitions; Cu(I) in the cyanide complex is $d^{10}$ .
$[Ti (H_{2} O)_{6}]^{3 +}$ is coloured but $[Sc (H_{2} O)_{6}]^{3 +}$ is colourless. Why?Answer: $T i^{3 +}$ is $d^{1}$ ; $S c^{3 +}$ is $d^{0}$ .
A complex absorbs light of 600 nm (red). What colour does it appear?Answer: Blue, like $[Cu (H_{2} O)_{4}]^{2 +}$ .
Which absorbs light of shorter wavelength, $[Co (N H_{3})_{6}]^{3 +}$ or $[Co (N H_{3})_{5} (H_{2} O)]^{3 +}$ ?Answer: $[Co (N H_{3})_{6}]^{3 +}$ (475 nm against 500 nm): $N H_{3}$ gives a larger $Δ_{o}$ than $H_{2} O$ .
The stability constant of $[Ag (N H_{3})_{2}]^{+}$ is $1.6 \times 1 0^{7}$ . Find its instability constant.Answer: $6.25 \times 1 0^{- 8}$
For a metal ion and ammonia, $lo g K_{1}$ to $lo g K_{4}$ are 4.0, 3.2, 2.7 and 1.8. Find $lo g β_{4}$ and $β_{4}$ .Answer: $lo g β_{4} = 11.7$ ; $β_{4} = 1 0^{11.7} \approx 5.0 \times 1 0^{11}$
Which is more stable, $[Co (N H_{3})_{6}]^{3 +}$ or $[Co (N H_{3})_{6}]^{2 +}$ , and why?Answer: $[Co (N H_{3})_{6}]^{3 +}$ : the higher charge density of $C o^{3 +}$ binds ammonia far more strongly.
True or false: the stability of a complex increases with the charge density of the metal ion.Answer: True
Dimethylglyoxime is used for the gravimetric estimation of ____ ions.Answer: $N i^{2 +}$
EDTA is used as a complexing agent in the ____ estimation of $C a^{2 +}$ , $M g^{2 +}$ and $Z n^{2 +}$ .Answer: complexometric (volumetric) titration
In silver electroplating, $K [Ag (CN)_{2}]$ is used instead of $AgN O_{3}$ because (A) a thin layer forms (B) more voltage is needed (C) all $A g^{+}$ is removed (D) the complex keeps the concentration of free $A g^{+}$ low, so deposition is slow and evenAnswer: (D)
Write the equation for the fixing of a photographic film with hypo.Answer: $AgBr$ + $2N a_{2} S_{2} O_{3}$ → $N a_{3} [Ag (S_{2} O_{3})_{2}]$ + $NaBr$
Briefly state the role of coordination compounds in (a) biological systems (b) analytical chemistry (c) medicine (d) metallurgy.Answer: (a) chlorophyll (Mg), haemoglobin (Fe), vitamin B12 (Co); (b) EDTA titrations, dmg for $N i^{2 +}$ ; (c) cisplatin, EDTA for lead poisoning; (d) cyanide extraction of Au and Ag, Mond process for Ni.
State the temperatures used in the Mond process.Answer: Nickel and CO combine at 330-350 K; $[Ni (CO)_{4}]$ decomposes at 450-470 K.

Common Mistakes to Avoid

Watch out

Saying a complex has the colour it absorbs. It shows the complementary colour: $[Cu (H_{2} O)_{4}]^{2 +}$ absorbs red and looks blue.
Thinking a stronger field ligand absorbs a longer wavelength. A larger $Δ$ means higher energy, so a shorter wavelength is absorbed.
Expecting $d^{0}$ or $d^{10}$ complexes to be coloured by d-d transitions, or calling $Mn O_{4}^{-}$ colourless; its colour comes from charge transfer.
Explaining colour with valence bond theory. VBT has no d-d transitions; use crystal field theory.
Putting the total number of d electrons into $n (n + 2)$ . Only unpaired electrons count: $[Fe (CN)_{6}]^{4 -}$ ( $d^{6}$ ) has $μ = 0$ .
Treating tetrahedral complexes as low spin. $Δ_{t}$ is small, so they are high spin.
Mixing up stability and instability constants. $K_{stability} = 1/ K_{i}$ , and a large stability constant means a stable complex.
Assuming later stepwise constants are larger. Usually $K_{1} > K_{2} > K_{3} > K_{4}$ .
Confusing thermodynamically stable with kinetically inert: $[Ni (CN)_{4}]^{2 -}$ is very stable but exchanges ligands rapidly.

Frequently Asked Questions

Why are most coordination compounds coloured?

The ligands split the metal d orbitals into two sets. An electron in the lower set can absorb visible light of energy equal to the gap and jump to the upper set, a d-d transition. The complex absorbs that colour from white light and shows the complementary colour, for example violet for hexaaquatitanium(III).

Why are zinc(II) and scandium(III) complexes colourless?

A d-d transition needs an electron in a lower d orbital and a vacancy in a higher one. Scandium(III) has no d electrons (d0) and zinc(II) has a full set (d10), so neither can absorb visible light this way. Their complexes are white or colourless unless a charge transfer band appears.

How is the spin-only magnetic moment of a complex calculated?

Find the number of unpaired electrons, n, from the metal's d count and whether the complex is high spin or low spin. Then use mu = square root of n(n+2) Bohr magnetons. One to five unpaired electrons give 1.73, 2.83, 3.87, 4.90 and 5.92 BM respectively.

What is the stability constant of a complex?

It is the equilibrium constant for forming the complex from the free metal ion and ligands, such as copper(II) and four ammonia molecules. A large value means the complex hardly dissociates. Its reciprocal is the instability constant, and the overall value is the product of the stepwise constants.

What factors affect the stability of a complex?

A smaller, more highly charged metal ion gives a more stable complex, as does a more basic ligand. Chelating ligands such as ethylenediamine and EDTA give much more stable complexes than unidentate ligands (the chelate effect), especially when they form five or six membered rings.

What are the main applications of coordination compounds?

They are used in chemical analysis (EDTA titrations, dimethylglyoxime for nickel), metallurgy (cyanide extraction of gold, the Mond process for nickel), medicine (cisplatin, chelation therapy for lead), catalysis (Wilkinson's catalyst), photography and electroplating. Chlorophyll, haemoglobin and vitamin B12 are natural examples.

What colour and magnetism questions are asked in JEE Main?

JEE Main commonly asks which complex is coloured or colourless, which has the highest magnetic moment, the spin-only moment of a given complex, and why a colour changes when a ligand or geometry changes. Knowing the d electron count and whether the ligand is strong or weak field answers almost all of them.

Which stability and application facts are important for NEET?

NEET often asks about the chelate effect, the use of EDTA in water hardness and lead poisoning, cisplatin as an anticancer drug, the metals in chlorophyll, haemoglobin and vitamin B12, and the cyanide and Mond processes. These come straight from the NCERT section on the importance of coordination compounds.

Previous year questions on Colour, Magnetism And Stability Of Co-ordination Compounds

70 questions from past papers, each with a step-by-step solution.

Show all 70 questions

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