column I	column II
(a)	No. of electrons in Na (11) having m = 0	(p)	7
(b)	No. of electrons in S (16)  having $(\mathrm{n} + \mathrm{l}) =3$	(q)	15
(c)	No. of maximum possible electrons having  $s=+\frac{1}{2}$ spin in $\mathrm{Cr}\left(24\right)$	(r)	8
		(s)	12

2 from 1s, 2 from 2s, 2 from 2p, and 1 from 3s add up to 7 electrons when m = 0 in the sodium atom.

Now we have,

${ }_{16}\mathrm{S}=1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^4$

Here $\mathrm{n}+\mathrm{l}=3$ for $2{\mathrm{p}}^6$  and $3{\mathrm{s}}^2$  electrons which is for 8 electrons.

The transition metal chromium has an atomic number of 24. No. of maximum possible electrons When the value of s is positive 1/2, it equals 15 in.