An aqueous solution containing $V^{x+}$ ion has magnetic moment equal to $\sqrt{15\text{BM}}$ Therefore, the $x$ is$:$
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Let $n$ be the number of unpaired electrons.
Magnetic moment $=\sqrt{\text{n(n+2)}}=\sqrt{15}$
$\Rightarrow n = 3$
Atomic number of Vanadium is $23.$
So, it must loose all its $4s$ electrons to get $3$ unpaired $d$ electrons.
$\therefore$ value of $x$ will be $2$
Magnetic moment $=\sqrt{\text{n(n+2)}}=\sqrt{15}$
$\Rightarrow n = 3$
Atomic number of Vanadium is $23.$
So, it must loose all its $4s$ electrons to get $3$ unpaired $d$ electrons.
$\therefore$ value of $x$ will be $2$
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