The calculated spin-only magnetic moments of $\mathrm{K}_3\left[\mathrm{Fe}(\mathrm{OH})_6\right]$ and…

The calculated spin-only magnetic moments of $\mathrm{K}_3\left[\mathrm{Fe}(\mathrm{OH})_6\right]$ and $\mathrm{K}_4\left[\mathrm{Fe}(\mathrm{OH})_6\right]$ respectively are :
  1. 3.87 and 4.90 B.M.
  2. 4.90 and 5.92 B.M.
  3. 4.90 and 4.90 B.M.
  4. 5.92 and 4.90 B.M.

Solution

$\begin{aligned}
& \mathrm{K}_3\left[\mathrm{Fe}(\mathrm{OH})_6\right] \\
& \mathrm{Fe}^{3+} \Rightarrow 3 \mathrm{~d}^5 \\
& \mathrm{Fe}^{3+} \text { with } \mathrm{OH}^{-}(\mathrm{WFL}) \\
& =\mathrm{t}_{29}{ }^3 \mathrm{e}_{\mathrm{g}}{ }^2
\end{aligned}$
Number of unpaired electron ( $n$) $=5$
$\mu$ spin only $=5.92 \mathrm{BM}$
$\begin{aligned}
& \mathrm{K}_4\left[\mathrm{Fe}(\mathrm{OH})_6\right] \\
& \mathrm{Fe}^{2+} \Rightarrow \mathrm{OH}^{-} \mathrm{WFL} \\
& \mathrm{Fe}^{2+} \Rightarrow 3 \mathrm{~d}^6=\mathrm{t}_{2 \mathrm{~g}}^4 \mathrm{e}_{\mathrm{g}}^2 \\
& \mathrm{n}=4
\end{aligned}$
$\mu_{\text {spin only }}=4.90 \mathrm{BM}$

Asked in: JEE Main 2025 (29 Jan Shift 2)

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