Question

If there are three possible values $(-1/2,0,+1/2)$ for the spin quantum, then electronic configuration of $K\left(19\right)$ will be:

• A. $1s^3,2s^{3}2p^9,3s^{3}3p^1$
• B. $1s^2,2s^{2}2p^6,3s^{2}3p^6,4s^1$
• C. $1s^2,2s^{2}2p^9,3s^{2}3p^4$
• D. None of these

Answer 1

The correct option is A $1s^3,2s^{3}2p^9,3s^{3}3p^1$

If there are three possible values $(-1/2,0,+1/2)$ for the spin quantum, then electronic configuration of K(19) will be
$1s^3,2s^{3}2p^9,3s^{3}3p^1$
1s, 2s and 3s orbitals will have 3 electrons each instead of 2 electrons.
2p and 3p orbitals will have 9 electrons each instead of 6 electrons.