QJEEM24Ch0121: JEE Main 2024 Chemistry - Class 12 JEEM24Ch0121 Solution: A first row transition metal in its +2 oxidation state has a…

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.