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JEE Main 2026 Preparation: Question Papers, Solutions, Mock Tests & Strategy Unacademy » JEE Study Material » Chemistry » Properties of Coordination Compounds

Properties of Coordination Compounds

Coordination compounds are formed when a transition metal covalently bonds with neutral and anionic ligands. These coordination compounds notes talk about their properties and examples.

Table of Content
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Coordination compounds are a class of chemical compounds in which there is a central metal atom that is bonded to several ligands simultaneously. The ligands may be anionic or neutral and are only rarely cationic. Their bonds are directional in nature, and their magnetic and optical properties are remarkable. These properties of coordination compounds can be explained by their bonding and structure. 

There are several types of coordination compounds with unique properties. Several theories have been proposed over time concerning coordination compounds that help explain it.

Valence Bond Theory (VBT)

To understand the properties of coordination compounds, let’s first understand how these coordination compounds are formed. The Valence Bond Theory can be used to explain their formation. 

As per the VBT, the metal can bind with ligands using its (n-1)d, ns, np or ns, np, nd orbitals. Ligands donate their electron pair to these hybrid orbitals and then yield definite geometries such as octahedral, square planar, tetrahedral, and so on. These vacant hybridised orbitals of the metal overlap with the orbitals of the ligand. 

Here is a table for your reference:

Coordination Number

Type of hybridisation

Geometry

Examples

4

sp3

Tetrahedral

[NiCl4]2-

4

dsp2

Square Planar

[Ni(CN)4]2-

5

sp3d

Triangular Bipyramidal

PCl5

6

sp3d2 or d2sp3

Octahedral

[Fe(CN)6]3-

Coordination Compounds Examples

Let us understand the concept of coordination compounds through certain examples.

For coordination number 4:

Let us take the example of [NiCl4]2-

Ni28 has the electronic configuration of [Ar] 4s2 3d8 

Ni2+ has the electronic configuration of [Ar] 4s0 3d8 

d-orbital can accommodate 10 electrons, but in Ni2+ only 8 electrons are there in the 3d-orbital. 4s and 4p have zero electrons. In [NiCl4]2- one 4s and three 4p orbitals hybridise and form four equivalent hybrid orbitals with tetrahedral geometry. Four chloride ions donate their electron density into the hybrid orbitals.

In [Ni(CN)4]2- the two unpaired electrons in the 3d orbital pair up due to the strong ligand effect of the cyanide ion. Now, one d orbital is vacant. 1, 3d orbital, 1, 4s orbital, and 2, 4p orbitals hybridise and form an equivalent hybrid dsp2 orbital with square planar geometry.

Whether the compound forms dsp2 or sp3, hybridisation depends upon the availability of vacant orbital and the ligand. For example, Zinc exists in the form of Zn2+ which has an electron configuration of [Ar] 4s0 3d10. When it binds with four ligands, it has no choice but to adopt sp3 hybridisation because no d-orbital is vacant.

However, when there is a vacant d-orbital, then the type of hybridisation formed depends upon the strength of the ligand (as in the case of nickel complexes discussed above).

For coordination number 4:

Let us take the example of [Co(NH3)6]3+

Co27 has the electronic configuration of [Ar] 4s2 3d7 

Co3+ has the electronic configuration of [Ar] 4s0 3d6 

Ammonia ligand, in this case, behaves as a strong field ligand and pairs up the unpaired electrons in 3d-orbital. As a result, two 3d orbitals, one 4d orbital, and three 4p orbitals hybridise to yield d2sp3 hybridised orbitals. 

Let us take the example of [CoF6]3-

Co27 has the electronic configuration of [Ar] 4s2 3d7 

Co3+ has the electronic configuration of [Ar] 4s0 3d6 

Fluoride is a weak field ligand, i.e., it cannot pair up the unpaired electrons in the 3d-orbital. As a result, all 3d orbitals are occupied and not available for hybridisation. One 4d orbital, three 4p orbitals, and two 4d orbitals hybridise to yield sp3d2 hybridised orbitals in which fluoride donates its electron pairs. An outer orbital, high spin, and a paramagnetic complex are formed. 

Properties of coordination compounds

The hybridisation that a coordination complex of a given coordination number would adopt depends majorly on the availability of vacant orbitals and the field strength of the ligand. The field strength of ligands is determined by the spectrochemical series.

The colour and magnetic properties of coordination compounds can be predicted by their bonding. A coordination compound can be diamagnetic or paramagnetic depending upon the presence of unpaired electrons in it. 

The presence of unpaired electrons makes the compound paramagnetic (as in the case of [NiCl4]2- and [CoF6]3-. If no unpaired electrons are present, then the compound is diamagnetic. 

Formula to calculate magnetic moment = [n(n+1)]½

where n is the number of unpaired electrons. 

The unit is BM (Bohr’s Magnetron).

The colour in different types of coordination compounds occurs due to d→ d transitions. The five d-orbitals are not equal in energy under the influence of a ligand (as explained by crystal field theory). Rather, they are split into 3 t2g and 2 eg orbitals. The colour is observed only when the d-orbital is partially filled (d1 to d9). 

Conclusion

VBT explains the hybridisation and geometry of the compound as seen in the above coordination compounds examples. Magnetic properties of coordination compounds can also be explained using VBT. To understand the reason for different colours in the compounds, we refer to Crystal Field Theory. Majorly, d→ d electron transition imparts colour to the compound as can be seen in the above coordination compounds notes.

faq

Frequently asked questions

Get answers to the most common queries related to the JEE Examination Preparation.

Why is CuSO4.5H2O blue, but it becomes white on heating?

Ans. CuSO4.5H...Read full

Why is the Ruby gemstone red coloured?

Ans. Ruby is Al2O...Read full

Why does a compound that absorbs red light appear blue?

Ans. The relationship between the colour of the complex and the colour of the absorbed light is alw...Read full

Calculate the magnetic moment of [MnBr4]2–.

Ans: In [MnBr...Read full

What is the IUPAC name for [Co(NH3)5(CO3)]Cl?

Ans :Pentaamminecarbonatocobalt(III) chloride Charge on carbonate: -2...Read full

Ans. CuSO4.5H2O is a coordination compound that exists as  [Cu(H2O)4]SO4.H2O. The d-orbitals of the copper ion splits into two energy levels under the influence of the water ligand, which results in d→ d electron transition, imparting the compound blue colour. Upon heating, these water molecules get lost, making such a phenomenon impossible. Thus, the compound appears colourless.

 

Ans. Ruby is Al2O3 containing a small percentage of Cr3+ ions which has a configuration of 3d3. These chromium ions that are present in the alumina lattice form octahedral complexes. d→ d electron transitions occur, giving rise to red colour.

 

Ans. The relationship between the colour of the complex and the colour of the absorbed light is always complimentary. For example, if a compound absorbs red colour, its actual colour will complement red, i.e. blue. Similarly, compounds absorbing yellow colour would appear violet. 

Ans: In [MnBr4]2–, Mn exists as Mn2+ having an electronic configuration of d5. Since Br– is a weak field ligand, the pairing of electrons does not take place. 

The number of unpaired electrons = 5.

So, magnetic moment = [5(5+1)]½ = 5.9 BM

Ans :Pentaamminecarbonatocobalt(III) chloride

Charge on carbonate: -2

Charge on the coordination sphere which is balanced by chloride: +1

Ammonia is neutral, so the charge on Co is +3.

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