Question

Which set of quantum numbers is possible for the last electron of $M{g}^{+}$ ion?

• A. $n=3,l=2,m=0,s=+1/2$
• B. $n=2,l=3,m=0,s=+1/2$
• C. $n=1,l=0,m=0,s=+1/2$
• D. $n=3,l=0,m=0,s=+1/2$

Answer 1

Answer: (D)

$n=3,l=0,m=0,s=+1/2$

$M{g}^{+}⟶[Ne{]}_{10}3s^1$

So, principle quantum number $⟶n=3$

Azimuthal quantum number $⟶l=0$

Magnetic quantum number $⟶m=0$

Spin quantum number $⟶s=+\frac{1}{2}$

So, answer is

$n=3,l=0,m=0$ and $s=+\frac{1}{2}$