Atomic number of $\mathrm{Cr}, \mathrm{Fe}$ and $\mathrm{Co}$ are 24, 26 and 27 respectively. Which of the…

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?
  1. $\left[\mathrm{Co}\left(\mathrm{NH}_3ight)_6ight]^{3+}$
  2. $\left[\mathrm{Co}(\mathrm{CN})_3ight]^{3+}$
  3. $\left[\mathrm{Fe}(\mathrm{CN})_6ight]^{4-}$
  4. $\left[\mathrm{Cr}(\mathrm{CN})_6ight]^{3-}$

Solution

$\left[\mathrm{Cr}(\mathrm{CN})_6ight]^{3-}$ 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 $\left[\mathrm{Cr}(\mathrm{CN})_6ight]^{3-}$ paramagnetic. ~

Asked in: JEE-TOPICTESTS-CHEMISTRY

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