Question

Magnetic moment of $X^{n+}(Z=26)$ is $\sqrt{24}BM$. 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

The correct option is A $4,2$

$Magneticmovement$ $=\sqrt{n(n+2)}=\sqrt{24}=n(n+2)=24n=4$

$Here,n=numberofunpairedelectron=4Z=26⟶FeSinceithas4unpairedelectron,itwillbe{Fe}^{24}Whoseelectronicconfigrationis\left(Ar\right){3d}^{6}Hencein{X}^{n+}⟶X=Fen=2$