Question

The four quantum numbers of the outermost orbital of K (atomic no. $=19$) are

• A. $n=2,l=0,m=0,s=+\frac{1}{2}$
• B. $n=4,l=0,m=0,s=+\frac{1}{2}$
• C. $n=3,l=1,m=1,s=+\frac{1}{2}$
• D. $n=4,l=2,m=-1,s=+\frac{1}{2}$

Answer 1

The correct option is B $n=4,l=0,m=0,s=+\frac{1}{2}$

Electronic configuration of potassium K $\Rightarrow \left[Ar\right]4s^1$
The outermost orbital is $4s$
Hence the 4 quantum nos. are -

$n=4,l=0,m=0,s=+1/2$