Question

If the spin magnetic moment of $X^{n+}$ ($Z=$ 26) is $\sqrt{24}$ B.M., then the 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)

$4,2$

Spin magnetic moment $=\sqrt{n(n+2)}=\sqrt{24}$ B.M.

$\therefore$ number of unpaired electron $=$ 4.

${}^{26}X:1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{6}4s^2$

Since, number of electrons unpaired $=4$, so from ${}^{26}X$, two electrons need to be removed.

$X^{2+}:1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^6$.