Question

Hypothetically, if an electron can have three values of spin quantum numbers $(+\frac{1}{2},0$ and $-\frac{1}{2})$ then for $Pd$ (atomic number 46), the number of electrons that have the $(n+l)$ sum to be equal to 5 is:

• A. 27
• B. 24
• C. 18
• D. 16

Answer 1

The correct option is D 16

$n$ = principal quantum number,
$l$ = azimuthal quantum number
The electrons that have $n+l$ = 5, are those in $5s,4p$and $3d$ orbitals. Electronic configuration of $Pd$ is $1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}4d^{10}$
Three spin values would mean that each orbital would now have 3 electrons instead of two. The number of orbitals would be the same as $s=1,p=3,d=5$. The number of electrons in the sub shells would be 3, 9 and 15 respectively. The new configuration would be:
$1s^{3}2s^{3}2p^{9}3s^{3}3p^{9}3d^{15}4s^{3}4p^1$
The total number of electrons that satisfy the given conditions are $=15+1=16$.