Number of unpaired electrons ${\mathrm{Mn}}^{+7}=a$

Number of d-subshell electrons $\mathrm{Cr}=b$

Number of f-subshell electrons in  $\mathrm{Hf}$ 

(Atomic no. = 72) = c 

Find the value of  $c-b+a$.

The electronic configuration of $M{n}^{+7}$ is,
$1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 4d^5$

From the electronic configuration, it is observed all the subshell are filled. There is no unpaired electron in $M{n}^{+7}$ the ion. Hence, a=0.

The chromium is in the 4s orbit and contains one electron. Due to the d-five orbital's subshells, each of which contains one electron, the atom gains extra stability by having a half-filled configuration, or 3d-orbital. Hence, b=5

Number of f-subshell electrons in $\mathrm{Hf}$

The calculation is,
=c-b+a
=14-5+0
=9