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JEE Main 202529 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
29 January
Shift
Evening Shift
Subject
Chemistry
Chapter
Structure of Atom

JEE Main 2025 · 29 January · Evening Shift · Chemistry · Structure of Atom

The calculated spin-only magnetic moments of K_3[Fe(OH)_6] and...

The calculated spin-only magnetic moments of K3[Fe(OH)6]\mathrm{K}_3\left[\mathrm{Fe}(\mathrm{OH})_6\right] and K4[Fe(OH)6]\mathrm{K}_4\left[\mathrm{Fe}(\mathrm{OH})_6\right] respectively are :

Options

  1. A3.87 and 4.90 B.M.
  2. B4.90 and 5.92 B.M.
  3. C4.90 and 4.90 B.M.
  4. D5.92 and 4.90 B.M.

Correct answer

D. 5.92 and 4.90 B.M.

Step-by-step solution

K3[Fe(OH)6]Fe3+⇒3 d5Fe3+ with  OH−(WFL)=t293eg2\begin{aligned} & \mathrm{K}_3\left[\mathrm{Fe}(\mathrm{OH})_6\right] \\ & \mathrm{Fe}^{3+} \Rightarrow 3 \mathrm{~d}^5 \\ & \mathrm{Fe}^{3+} \text{\text{ with } } \mathrm{OH}^{-}(\mathrm{WFL}) \\ & =\mathrm{t}_{29}{ }^3 \mathrm{e}_{\mathrm{g}}{ }^2 \end{aligned} Number of unpaired electron ( nn ) =5μ=5 \mu spin only =5.92BM=5.92 \mathrm{BM}K4[Fe(OH)6]Fe2+⇒OH−(WFL)Fe2+⇒3 d6=t2 g4eg2n=4\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}μspin only =4.90BM\mu_{\text{spin only }}=4.90 \mathrm{BM}
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