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Valence Bond Theory of Coordination Compounds - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:35 AM | #JEE Main

Quick Facts

• Magnetic Moment(On the Basis of VBT) is considered one the most difficult concept.

• 66 Questions around this concept.

Solve by difficulty

Which of the following has a square planar geometry?

Which statement is incorrect with respect to bonding between two carbon atoms as per the VBT?

Consider the following complex ions

\begin{aligned} & \mathrm{P}=\left[\mathrm{FeF}_6\right]^{3-} \\ & \mathrm{Q}=\left[\mathrm{V}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+} \\ & \mathrm{R}=\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}\end{aligned}

The correct order of the complex ions, according to their spin only magnetic moment values in (B.M.) is :

Select the option with correct property -

Concepts Covered - 5

Valence Bond Theory

According to this theory, the metal atom or ion under the influence of ligands can use its (n-1)d or nd orbitals along with its ns and np for hybridisation to yield a set of equivalent orbitals of definite geometry such as octahedral, tetrahedral, square planar and so on. These hybridised orbitals are allowed to overlap with ligand orbitals that can donate electron pairs for bonding. The different types of hybridisation and their respective shapes are given below

 Coordination Number Type of Hybridisation Shape 4 sp3 Tetrahedral 4 dsp2 Square Planar 5 sp3d Trigonal Bipyramidal 6 sp3d2 Octahedral 6 d2sp3 Octahedral

Analysis of Complex Compound on the Basis of VBT

Let us consider the case of $[Ni(CN)_4]^{2-}$and try to predict the hybridisation of this complex

Here nickel is in +2 oxidation state and the ion has a valence electronic configuration $3d^8 4s^0$.

In the presence of Cyanide ions, the electrons will be paired up and the hybridisation of Ni in the complex will be $dsp^2$ as shown below

Each of the hybridised orbitals receives a pair of electrons from a cyanide ion. The compound is diamagnetic as evident from the absence of unpaired electron.

Exceptional Case of Hybridisation(VBT)

[Cu(NH3)4]2+ is an exception for determining the hybridisation. On the experimental basis, it has been found that its geometry is square planar, thus atleast one d orbital is compulsory.

The electronic configuration of [Cu(NH3)4]2+ is as follows:

But in this case, the last electron can easily be removed and Cu2+ will become Cu3+, which does not exist in reality.

Thus to explain its square planar geometry, Huggins proposed sp2d hybridisation that this last unpaired electron remains in the 3d orbital and the hybridisation is done by 4orbital, 2 4p orbitals and 1 4d orbital. The third 4p orbital does not participate in hybridisation. Thus, in this way, its hybridisation is sp2d and the geometry is square planar.

Magnetic Moment(On the Basis of VBT)

The magnetic moment of coordination compounds can be measured by the magnetic susceptibility experiments. The results can be used to obtain information about the number of unpaired electrons and hence structures adopted by metal complexes.
A critical study of the magnetic data of coordination compounds of metals of the first transition series reveals some complications. For metal ions with upto three electrons in the d orbitals, like Ti3+ (d1); V3+(d2); Cr3+(d3); two vacant d orbitals are available for octahedral hybridisation with 4s and 4p orbitals. The magnetic behaviour of these free ions and their coordination entities is similar. When more than three 3d electrons are present, the required pair of 3d orbitals for octahedral hybridisation is not directly available (as a consequence of Hund’s rule). Thus, for d4(Cr2+, Mn3+), d5(Mn2+, Fe3+), d6(Fe2+, Co3+) cases, a vacant pair of d orbitals results only by pairing of 3d electrons which leaves two, one and zero unpaired electrons, respectively.
The magnetic data agree with maximum spin pairing in many cases, especially with coordination compounds containing d6 ions. However, with species containing d4 and dions there are complications. [Mn(CN)6]3– has magnetic moment of two unpaired electrons while [MnCl6]3– has a paramagnetic moment of four unpaired electrons. [Fe(CN)6]3– has magnetic moment of a single unpaired electron while [FeF6]3– has a paramagnetic moment of five unpaired electrons. [CoF6]3– is paramagnetic with four unpaired electrons while [Co(C2O4)3]3– is diamagnetic. This apparent anomaly is explained by valence bond theory in terms of formation of inner orbital and outer orbital coordination entities. [Mn(CN)6]3–, [Fe(CN)6]3– and [Co(C2O4)3]3– are inner orbital complexes involving d2sp3 hybridisation, the former two complexes are paramagnetic and the latter diamagnetic. On the other hand, [MnCl6]3–, [FeF6]3– and [CoF6]3– are outer orbital complexes involving sp3d2 hybridisation and are paramagnetic corresponding to four, five and four unpaired electrons.

Limitations of VBT

While the Valence Bond theory, to a larger extent, explains the formation, structures and magnetic behaviour of coordination compounds, it has some shortcomings which are listed below:
(i) It involves a number of assumptions.
(ii) It does not give quantitative interpretation of magnetic data.
(iii) It does not explain the colour exhibited by coordination compounds.
(iv) It does not give a quantitative interpretation of the thermodynamic or kinetic stabilities of coordination compounds.
(v) It does not make exact predictions regarding the tetrahedral and square planar structures of 4-coordinate complexes.
(vi) It does not distinguish between weak and strong ligands.

Study it with Videos

Valence Bond Theory
Analysis of Complex Compound on the Basis of VBT
Exceptional Case of Hybridisation(VBT)

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Books

Reference Books

Valence Bond Theory

Chemistry Part I Textbook for Class XII

Page No. : 254

Line : 31

Magnetic Moment(On the Basis of VBT)

Chemistry Part I Textbook for Class XII

Page No. : 256

Line : 22