Question

Magnetic moment of $X^{n+}(Z=26)$is $\sqrt{24}$ B.M. Hence number of unpaired electrons and value of n respectively are :

• A. $4,2$
• B. $2,4$
• C. $3,1$
• D. $0,2$

Answer 1

Answer: (A)

$4,2$

Option 'A" is correct.
${\mu }_{eff}=\sqrt{n^{\prime }(n^{\prime }+2)}BM=\sqrt{4(4+2)}BM=\sqrt{24}BM$,
there are four unpaired electrons(n').