JEE Main 2024 — Coordination Compounds Question with Solution

To determine the atomic number of the metal based on the provided spin-only magnetic moment value, we use the formula for calculating the spin-only magnetic moment:

$\mu =\sqrt{n(n+2)}{\mu }_B$

where $\mu$ is the magnetic moment in Bohr magnetons (${\mu }_B$), and $n$ is the number of unpaired electrons.

Given that the spin-only magnetic moment value is $3.86\mathrm{BM}$, we can set this value equal to the formula to solve for $n$:

$3.86=\sqrt{n(n+2)}$

Squaring both sides gives:

$14.8996=n(n+2)$

Since $n$ must be an integer and this equation doesn't solve neatly for an integer $n$, we look for a value of $n$ that would give a product close to $14.8996$ when plugged into $\sqrt{n(n+2)}$.

Practically, we can test the square root values near $3.86$ for different $n$ to see which one gives a close match. Considering the usual spin-only magnetic moments for common numbers of unpaired electrons:

• $n=1$, $\mu =\sqrt{3}\approx 1.73{\mu }_B$


• $n=2$, $\mu =\sqrt{8}\approx 2.83{\mu }_B$


• $n=3$, $\mu =\sqrt{15}\approx 3.87{\mu }_B$


• $n=4$, $\mu =\sqrt{24}\approx 4.90{\mu }_B$

The value $n=3$ gives a magnetic moment of approximately $3.87{\mu }_B$, which is very close to the given value, $3.86{\mu }_B$. Therefore, the metal has 3 unpaired electrons in its d-orbital configuration.

In the +2 oxidation state, metals lose electrons from the s-orbital before the d-orbital. Given that the element is a first row transition metal, starting with scandium (Sc) at atomic number 21 which has an electronic configuration of $[Ar]3d^{1}4s^2$ in its neutral state, we can deduce the configurations:

• $Z=22$ for Titanium (Ti), which has a neutral configuration of $[Ar]3d^{2}4s^2$. In the +2 oxidation state, it would have an electronic configuration of $[Ar]3d^2$, resulting in 2 unpaired electrons.


• $Z=23$ for Vanadium (V), with a neutral configuration of $[Ar]3d^{3}4s^2$. In the +2 state, its configuration would be $[Ar]3d^3$, presenting 3 unpaired electrons, which matches our calculation.


• $Z=26$ for Iron (Fe), indicating a neutral configuration of $[Ar]3d^{6}4s^2$. In the +2 state, $[Ar]3d^6$, which would have 4 unpaired electrons, not matching the calculation.


• $Z=25$ for Manganese (Mn), with a neutral configuration of $[Ar]3d^{5}4s^2$. In the +2 state, $[Ar]3d^5$, indicating 5 unpaired electrons, which also does not match our calculation.

So, the atomic number of the metal with a +2 oxidation state and a spin-only magnetic moment value of $3.86\mathrm{BM}$ (corresponding to 3 unpaired electrons) is 23, which identifies the metal as Vanadium (V). Therefore,

Option B (23) is the correct answer.