JEE Main 2025 — Structure of Atom Question with 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 $=5.92\mathrm{BM}$ $\begin{aligned} & \mathrm{K}_4\left[\mathrm{Fe}(\mathrm{OH})_6\right] \\ & \mathrm{Fe}^{2+} \Rightarrow \mathrm{OH}^{-} \mathrm{(WFL)} \\ & \mathrm{Fe}^{2+} \Rightarrow 3 \mathrm{~d}^6=\mathrm{t}_{2 \mathrm{~g}}^4 \mathrm{e}_{\mathrm{g}}^2 \\ & \mathrm{n}=4 \end{aligned}$ ${\mu }_{\text{spin only }}=4.90\mathrm{BM}$