The calculated magnetic moment of $F e^{n +}$ ion is 4.90 BM. The sum of the number of unpaired electrons present in $3 d$ subshell of $F e^{n +}$ ion and the value of n is

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Solution

We know that, $μ = \sqrt{n (n + 2)} = 4.90 B M$
`\therefore` `n=4`
The ground state electronic configuration of iron is $[A r] 3 d^{6} 4 s^{2}$
The electronic configuration of $F e^{2 +}$ is $[A r] 3 d^{6} 4 s^{0}$
The number of unpaired electrons in the $3 d$ subshell is `4`
So, the sum of unpaired electrons in the $3 d$ subshell and the oxidation state is = 4+4= 8

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