Question

The magnetic moment of $M^{x+}$ (atomic number $=25$) is $\sqrt{15}$ . Then, the oxidation number $x$ of $M$ is:

• A. 4
• B. 3
• C. 2
• D. none of these

Answer 1

The correct option is C 2

$\sqrt{n(n+2)}=\sqrt{15}$

Squaring both sides

$\Rightarrow n(n+2)=15$

$\Rightarrow n^v+2n-15=0$

$\Rightarrow n^v+5n-3n-15=0$

$\Rightarrow n(n+5)-3(n+5)=0$

$\Rightarrow (n+5)(n-3)=0$

Solving we get,

$n=3$

Thus there are $3$ unpaired electrons in the element.

The configuration of an element with atomic number $2S$ is $\to$ ${1S}^2{2S}^2{2P}^6{3S}^2{3P}^6{4S}^2{3d}^5$. This configuration has electrons in the valence shell.

We have been given that the symbol of the element is $M^{x+}$.

So, oxidation number of $M=5-3$

$=2$