Question

Magnetic moment of $Cr(Z=24),M{n}^{+}(Z=25)$ and $F{e}^{2+}(Z=26)$ are x, y, z. They are in order:

• A. $x<y<z$
• B. $x=y<z$
• C. $z<x=y$
• D. $x=y=z$

Answer 1

The correct option is C $z<x=y$

The expression for the magnetic moment is $\mu =\sqrt{N(N+2)}BM$

Where N= number of unpaired electrons.
The outer electronic configuration of $Cr\left(24\right)$ is $\left[Ar\right]3d^{5}4s^1$
It contains 6 unpaired electrons.
The expression for the magnetic moment is
$\mu =\sqrt{48}Bm=x$.
The outer electronic configuration of $M{n}^{+}\left(25\right)$ is $\left[Ar\right]3d^{5}4s^1$
It contains 6 unpaired electrons.
The expression for the magnetic moment is
$\mu =\sqrt{48}Bm=y$.
The outer electronic configuration of $F{e}^{2+}\left(26\right)$ is $\left[Ar\right]3d^{6}4s^0$
It contains 4 unpaired electrons.
The expression for the magnetic moment is
$\mu =\sqrt{24}Bm=z$.
Hence, $z<x=y$.
The option c is the correct answer.