An element 'y' present in its ground state, the value of principal quantum number and azimuthal quantum number of the last electron of element 'y' is n = 3, and $\mathcal{l}$ = 1 and spin multiplicity for a given element is 4. 

Select incorrect statements based on the above information.

$\text{ Spin multiplicity }=\left|\right(2s+1\left)\right|=4\text{, indicates, three }{e}^{-1}\mathrm{s}\text{, with same spin, i.e.,}$

$\left|2\left(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\right)+1\right|=4$

$n=3,l=1\Rightarrow 3p,\text{ with }3e^{-1}\text{ s with same spin, }\Rightarrow 3p^3$

$\text{ Element ' }y\text{ ' electronic configuration }=1s^2,2s^{2}2p^6,3s^2,3p^3$

Thus element 'y' - Phosphorous (Z = 15)(3rd period, 15 group) 

$m_{\mathrm{mm}}=\sqrt{n(n+2)}=\sqrt{3(3+2)}=\sqrt{15}\mathrm{BM}$