JEE Main 2024 — D And F Block Elements Question with Solution

In order to determine the spin-only magnetic moment value of the species with the least oxidizing ability, we first need to analyze their oxidation states and electron configurations.

1. $${\mathrm{VO}}_2^{+}$$:

For vanadium in $${\mathrm{VO}}_2^{+}$$, the oxidation state is +5. The electronic configuration of V is $$[Ar]3d^{3}4s^2$$. Thus, in +5 oxidation state, vanadium will have zero d-electrons (since it loses 5 electrons) and, consequently, $${\mathrm{VO}}_2^{+}$$ is diamagnetic (since zero unpaired electrons).

2. $${\mathrm{MnO}}_4^{-}$$:

For manganese in $${\mathrm{MnO}}_4^{-}$$, the oxidation state is +7. The electronic configuration of Mn is $$[Ar]3d^{5}4s^2$$. Thus, in +7 oxidation state, manganese will have zero d-electrons (since it loses 7 electrons) and, consequently, $${\mathrm{MnO}}_4^{-}$$ is also diamagnetic (since zero unpaired electrons).

3. $${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}$$:

For chromium in $${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}$$, we have two chromium atoms. Each chromium is in the +6 oxidation state. The electronic configuration of Cr is $$[Ar]3d^{5}4s^1$$. Thus, in +6 oxidation state, each chromium will have zero d-electrons (since it loses 6 electrons) and, consequently, $${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}$$ is also diamagnetic (since zero unpaired electrons).

From these observations, we note that all the species we analyzed are diamagnetic. However, to identify the species with the least oxidizing ability, we look at their standard reduction potentials (E° values).

Typically, the oxidizing ability increases with increasing positive E° values. Given the nature of these species, we can infer that:

• $${\mathrm{MnO}}_4^{-}$$ is a very strong oxidizing agent (very high E° value).
• $${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}$$ is also a strong oxidizing agent (high but less than $${\mathrm{MnO}}_4^{-}$$).
• $${\mathrm{VO}}_2^{+}$$ has a moderate oxidizing ability (lesser E° value than both $${\mathrm{MnO}}_4^{-}$$ and $${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}$$).

Therefore, among the given species, $${\mathrm{VO}}_2^{+}$$ has the least oxidizing ability.

As mentioned previously, $${\mathrm{VO}}_2^{+}$$ has zero unpaired electrons, making it diamagnetic. The spin-only magnetic moment is given by the formula:

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

where $$n$$ is the number of unpaired electrons.

For $${\mathrm{VO}}_2^{+}$$, $$n=0$$, thus:

$$\mu =\sqrt{0(0+2)}=0\text{BM}$$

Given the options, the nearest integer value of the spin-only magnetic moment for the species with the least oxidizing ability (which is $${\mathrm{VO}}_2^{+}$$) is indeed:

0 BM