Question

The magnetic moment of ${Co^2+}$ in square planar complex is:
A. 3.87 B.M.
B. 1.73 B.M.
C. 4.87 B.M.
D. 5.87 B.M.

Answer 1

Hint: The magnetic moment is the magnetic strength and orientation of magnet or the other objects that produce a magnetic field. The magnetic field of a magnetic dipole is proportional to its magnetic dipole moment. Dipole moment can be calculated by using formula: $\mu = \sqrt {n(n + 2)} $ B.M, n is the number of unpaired electrons, B.M is Bohr Magneton.

Complete step by step answer:
The formula to calculate the magnetic moment is: $\mu = \sqrt {n(n + 2)} $B.M.
The Electronic configuration of Co in +2 oxidation state Co+2 = [Ar] 3d74S0
Box diagram for valence electrons is
Hence the number of unpaired electrons is 1, ie n=1. By substituting the value of n in the formula to calculate magnetic moment
$\mu = \sqrt {n(n + 2)} \\
   = \sqrt {1(1 + 2)} \\
   = \sqrt 3 \\
   = 1.73B.M. \\ $

Hence the option B is correct.

Note:
The product of charge and distance between the two ends is called dipole moment. In square planar complexes the central metal cation is either ${dsp^2}$ or ${sp^2}d$ hybridized. If strong ligands are present the electron pairing takes place then the shape of the complex will led to the formation of square planar and magnetic moment will be zero. If weak ligands are present the electron pairing does not take place then the shape of the complex will lead to the formation of tetrahedral and it has a magnetic moment depending on the number of unpaired electrons.