Question
The total spin and magnetic moment for the atom with atomic number 24 are :
A.
$$ \pm 3,\sqrt {48} \,BM$$
B.
$$ \pm 3,\sqrt {35} \,BM$$
C.
$$ \pm \frac{3}{2},\sqrt {48} \,BM$$
D.
$$ \pm \frac{3}{2},\sqrt {35} \,BM$$
Answer :
$$ \pm 3,\sqrt {48} \,BM$$
Solution :
$$\eqalign{
& {\text{Atomic number}} = 24 \cr
& {\text{Electronic configuration is}}\,\,3{d^5},4{s^1} \cr
& {\text{No}}{\text{. of valence}}\,\,{e^{ - 1}}\left( s \right) = 6 \cr
& {\text{Total spin}}\, = \pm \frac{1}{2} \times 6 = \pm 3 \cr
& {\text{Magnetic moment}} \cr
& = \sqrt {n\left( {n + 2} \right)} \cr
& = \sqrt {6\left( {6 + 2} \right)} \cr
& = \sqrt {48} \,BM \cr} $$