Question

Calcalculate spin and magnetic moment for nitrogen, chromium and kryton.

Answer 1

Spin-only magnetic moment is given by $\mu =\sqrt{n(n+2)}B.M$

where $n=no.ofunpairedelectrons$.

(1) For Nitrogen,

$N\left(7\right)=1s^{2}2s^{2}2p^3$

$n=7$

Spin -only magnetic moment$=\sqrt{n(n+2)}B.M=\sqrt{3\times 5}=3.873$

(2) For Chromium,

$Cr\left(24\right)=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}3d^5$

$n=6$

$\therefore \mu =\sqrt{6(6+2)}=\sqrt{48}=6.928B.M$

(3)For Krypton,

$Kr\left(36\right)=1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{2}3d^{10}4p^6$

$n=0$

$\therefore \mu =0B.M$