$\mathrm{Cr}(\mathrm{CO})_x$, find the value of $x$ ?
$\mathrm{Cr}(\mathrm{CO})_x$, find the value of $x$ ?
4
6
2
1
Solution
Here, $\mathrm{Cr}$ has zero oxidation state, as has zero oxidation state in the compound.
So, at zero oxidation state the electronic configuration of $\mathrm{Cr}$ is $[\mathrm{Ar}] 4 s^1, 3 d^5$
that means $\mathrm{Cr}$ has 6 valencies, as $\mathrm{Cr}$ in this state is stable due to half-filled orbitals. Now, for 6 valencies, $6 \mathrm{CO}$ is required. Therefore, $x=6$ The compound is $\mathrm{Cr}(\mathrm{CO})_6$.
Hence, the correct option is (2).