Question

How many maximum possible sets of quantum numbers are possible for the 6th electron of $Fe$?

• A. 1
• B. 2
• C. 6
• D. 10

Answer 1

The correct option is C 6

$Fe\left(26\right):1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{6}4s^2$
The 6th electron belongs to the $2p$ orbital.
$2p{e}^{-},n=2,l=1,m=+1,0,-1,s=\pm \frac{1}{2}$
The different sets of quantum numbers possible:
$\Rightarrow n=2,l=1,m=+1,m_s=\mp \frac{1}{2}$
$n=2,l=1,m=-1,m_s=\pm \frac{1}{2}$
$n=2,l=1,m=0,m_s=\pm \frac{1}{2}$
The answer is 6.