Question

How many unpaired electrons are there in $\mathrm{V}{\left[{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{+3}$?

Answer 1

Electronic configuration:

1. It helps to determine electron placement, whether an atom or compound contains a lone pair or not, magnetic characteristics, and so on.
2. $\mathrm{V}{\left[{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{+3}$ is a charged compound so for oxidation number of the compound, it will be equal to the charge on the compound.
3. Also water is a neutral ligand so it will not have any charge on it. Therefore, the charge on $\mathrm{V}=+3$
4. Electronic configuration of $\mathrm{V}=1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}4{\mathrm{s}}^{2}3{\mathrm{d}}^3$ and a plus sign indicates it will lose three electrons i.e. two from $4\mathrm{s}$ orbital and one from $3\mathrm{d}$ orbital.
5. ${\mathrm{V}}^{+3}=1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^{6}3{\mathrm{d}}^2$

Therefore, the number of unpaired electrons in $\mathrm{V}{\left[{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{+3}$ is $2$