KCET · Chemistry · Coordination Compounds
What will be the value of \(x\) in \(\mathrm{Fe}^{x+}\), if the magnetic moment, \(\mu=\sqrt{24} \mathrm{BM}\) ?
- A 3
- B 0
- C 1
- D 2
Answer & Solution
Correct Answer
(D) 2
Step-by-step Solution
Detailed explanation
Given that, magnetic moment, \(\mu=\sqrt{24}\) and \(\mu=\sqrt{n(n+2)}\) where, \(n=\) unpaired electrons
\[
\begin{gathered}
\sqrt{24}=\sqrt{n(n+2)} \\
24=n(n+2)=n^2+2 n \\
n^2+2 n-24=0 \Rightarrow n^2+6 n-4 n-24=0 \\
n(n+6)-4(n+6)=0 \Rightarrow(n-4)(n+6)=0 \\
n=4,-6
\end{gathered}
\]
\(\therefore\) Number of unpaired electrons \(=4\)
If oxidation of \(\mathrm{F}\) is +2 then it has 4 unpaired electrons.
\[
\begin{gathered}
\sqrt{24}=\sqrt{n(n+2)} \\
24=n(n+2)=n^2+2 n \\
n^2+2 n-24=0 \Rightarrow n^2+6 n-4 n-24=0 \\
n(n+6)-4(n+6)=0 \Rightarrow(n-4)(n+6)=0 \\
n=4,-6
\end{gathered}
\]
\(\therefore\) Number of unpaired electrons \(=4\)
If oxidation of \(\mathrm{F}\) is +2 then it has 4 unpaired electrons.
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