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Atomic number of $\mathrm{Cr}, \mathrm{Fe}$ and $\mathrm{Co}$ are 24, 26 and 27 respectively. Which of the following inner orbital octahedral complexes are paramagnetic?
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Verified Answer
The correct answer is:
$\left[\mathrm{Cr}(\mathrm{CN})_6\right]^{3-}$
$\left[\mathrm{Cr}(\mathrm{CN})_6\right]^{3-}$
Oxidation state of $\mathrm{Cr}: x+(-1) 6=-3$
$$
\begin{aligned}
x & =+3 \\
\mathrm{Cr} & =[\mathrm{Ar}] 4 s^1 3 d^5 \\
\mathrm{Cr}^{3+} & =[\mathrm{Ar}] 4 s^0 3 d^3
\end{aligned}
$$
Since, $\mathrm{CN}$ is a strong field ligand, it will cause pairing of electrons in $\mathrm{Cr}^{3+}$,
So, there is one unpaired electron in $3 d$-orbital of $\mathrm{Cr}^{3+}$ which makes the compound $\left[\mathrm{Cr}(\mathrm{CN})_6\right]^{3-}$ paramagnetic.
Oxidation state of $\mathrm{Cr}: x+(-1) 6=-3$
$$
\begin{aligned}
x & =+3 \\
\mathrm{Cr} & =[\mathrm{Ar}] 4 s^1 3 d^5 \\
\mathrm{Cr}^{3+} & =[\mathrm{Ar}] 4 s^0 3 d^3
\end{aligned}
$$
Since, $\mathrm{CN}$ is a strong field ligand, it will cause pairing of electrons in $\mathrm{Cr}^{3+}$,
So, there is one unpaired electron in $3 d$-orbital of $\mathrm{Cr}^{3+}$ which makes the compound $\left[\mathrm{Cr}(\mathrm{CN})_6\right]^{3-}$ paramagnetic.
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