Question

In case of nitrogen, if ${\text{M}}_1$ represents spin multiplicity if Hund's rule is followed and ${\text{M}}_2$ represents spin multiplicity if only Hund's Rule is violated then the value of $\frac{{\text{M}}_{\text{1}}}{{\text{M}}_{\text{2}}}$ will be:

Answer 1

Hund's rule-: Pairing of electron's takes place only after all the available degenerate orbitals are filled with one electron each.

spin multiplicity$=n=2;$s=total spin

nitrogen$\Rightarrow$$(z=7)$ $\Rightarrow$ $1s^{2}2s^2{2p}_x^1{2p}_4^1{2p}_z^1$

$\left[111\right]$

S=3$\times \frac{1}{2}$

spin multiplicity,n=2(3)$\left(\frac{1}{2}\right)+1=4$

$\Rightarrow M_1=4$

nitrogen$\Rightarrow (z=7)\Rightarrow 1s^{2}2s^{2}2p^3$

if hund's rule not followed

S=$\frac{1}{2}-\frac{1}{2}+\frac{1}{2}=v_2$

$n=2(\frac{1}{2}+1)=2$

$\Rightarrow M_2=2$

value of $\left(\frac{M_1}{M_2}\right)=\frac{4}{2}=2$