Question

Spin multiplicity of nitrogen in ground state will be ___.

• A. $4$
• B. $3$
• C. $5$
• D. $0$

Answer 1

The correct option is A $4$

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

$\left[111\right]$

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

spin multiplicity=2s+1=2$(3\times \frac{1}{2})+1$=4