JEE MAIN - Chemistry (2022 - 30th June Morning Shift - No. 20)
Spin only magnetic moment ($$\mu$$s) of $${K_3}[Fe{(CN)_6}]$$ is ____________ B.M.
(Nearest integer)
Answer
2
Explanation
$\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$
$\mathrm{CN}^{-}$ have $-1$ oxidation State.
Let oxidation state of $\mathrm{Fe}$ in the complex is $x$.
So, $x-6=-3 \Rightarrow x=+3$
Now, $\mathrm{CN}^{-}$ion is a strong field ligand.
For $Fe^{3+}: -3d^54s^0$_30th_June_Morning_Shift_en_20_1.png)
Unpaired electron $=1$
$\mu_{\mathrm{B}}=\sqrt{n(n+2)}=\sqrt{1 \times(1+2)}=1.7$
$\mu_{\mathrm{B}}=2$ B.M
$\mathrm{CN}^{-}$ have $-1$ oxidation State.
Let oxidation state of $\mathrm{Fe}$ in the complex is $x$.
So, $x-6=-3 \Rightarrow x=+3$
Now, $\mathrm{CN}^{-}$ion is a strong field ligand.
For $Fe^{3+}: -3d^54s^0$
Unpaired electron $=1$
$\mu_{\mathrm{B}}=\sqrt{n(n+2)}=\sqrt{1 \times(1+2)}=1.7$
$\mu_{\mathrm{B}}=2$ B.M
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