Question

The magnetic moment of a transition metal of 3d series is 6.92BM. Its electronic configuration will be
(A)- $3{{d}^{4}}\,4{{s}^{2}}$
(B)- $3{{d}^{5}}\,4{{s}^{1}}$
(C)- $3{{d}^{10}}$
(D)- $3{{d}^{5}}\,4{{s}^{0}}$

Answer 1

Hint:To answer this question we should know what the formula of magnetic moment of a metal is.
The formula for the spin-only magnetic moment is given by:
\[\mu \,=\,\sqrt{n(n+1)}\]
Where, the number of unpaired electrons in a metal is given by n.

Complete step by step solution:
Let’s look at the given options:
In option one the configuration is given as $3{{d}^{4}}\,4{{s}^{2}}$. In this configuration the number of unpaired electrons is 4.
Now, we will calculate the magnetic moment for n=4 and then match our solution with the given value
\[\mu \,=\,\sqrt{4(4+1)}\]
\[\mu \,=\,\sqrt{4\times 5}\,=\,4.47\].
The calculated value is not close to the given value. So, option (A) is incorrect.
In option (B) the electronic configuration is given as $3{{d}^{5}}\,4{{s}^{1}}$. The number of unpaired electrons is 6.
Therefore, $\mu \,=\,\sqrt{6(6+1)}$
\[\mu \,=\,\sqrt{6\times 7}\,=\,6.48\]
The calculated value is close to the given value
So, option (B) is the correct answer.
In option (C) the electronic configuration is $3{{d}^{10}}$. The number of unpaired electrons is 0.
Therefore, $\mu $= 0
In option (D) the electronic configuration is $3{{d}^{5}}\,4{{s}^{0}}$. The number of unpaired electrons is 5.
Therefore, $\mu \,=\,\sqrt{5(5+1)}$
\[\mu \,=\,\sqrt{5\times 6}\,=\,5.477\]
Which is not close to the value given in the question. So, option (D) is incorrect.
Hence, the answer to the given question is option (B)

Note:
While answering this question the d orbital configuration must be kept in mind for some metals like copper which show anomalous electronic configuration. Students should only take the unpaired electrons for calculation. The full form of BM is Bohr Magneton.