JEE Chemistry 2025 Question with Solution

Given: In the most stable structure of $\text{ClF}_3$, $n$ is the number of lone pairs present in the equatorial position. We must identify the ions having $n$ unpaired electrons.

Find: The correct option.

For $\text{ClF}_3$, chlorine has $7$ valence electrons and forms $3$ bonds with fluorine. Thus, there are $5$ electron pairs around chlorine, of which $3$ are bond pairs and $2$ are lone pairs. The electron pair geometry is trigonal bipyramidal and the molecule is T-shaped.

In a trigonal bipyramidal arrangement, lone pairs prefer equatorial positions to minimize repulsion. Therefore, both lone pairs occupy equatorial positions, so

Now compare the number of unpaired electrons in the given ions:

• $\text{V}^{3+} : [\text{Ar}]\,3d^2$, so it has $2$ unpaired electrons.
• $\text{Ti}^{3+} : [\text{Ar}]\,3d^1$, so it has $1$ unpaired electron.
• $\text{Cu}^{2+} : [\text{Ar}]\,3d^9$, so it has $1$ unpaired electron.
• $\text{Ni}^{2+} : [\text{Ar}]\,3d^8$, so it has $2$ unpaired electrons.
• $\text{Ti}^{2+} : [\text{Ar}]\,3d^2$, so it has $2$ unpaired electrons.

Thus, the ions with exactly $2$ unpaired electrons are $\text{V}^{3+}$, $\text{Ni}^{2+}$, and $\text{Ti}^{2+}$.

Therefore, the correct option is C: A, D and E Only.

The solution labels the correct option as A, but its working and final conclusion clearly give A, D and E only, which corresponds to option C in the listed options.

Given: $\text{ClF}_3$ has formula $AX_3E_2$.

Find: Which ions have the same number of unpaired electrons as the number of equatorial lone pairs in $\text{ClF}_3$.

Using VSEPR theory, $AX_3E_2$ corresponds to a trigonal bipyramidal electron arrangement with two lone pairs occupying equatorial positions. Hence,

Now evaluate each ion one by one:

1. $\text{V}^{3+}$: Vanadium is $[\text{Ar}]\,3d^3 4s^2$ in the neutral state, so $\text{V}^{3+} = [\text{Ar}]\,3d^2$. This gives $2$ unpaired electrons.
2. $\text{Ti}^{3+}$: Titanium is $[\text{Ar}]\,3d^2 4s^2$, so $\text{Ti}^{3+} = [\text{Ar}]\,3d^1$. This gives $1$ unpaired electron.
3. $\text{Cu}^{2+}$: Copper is $[\text{Ar}]\,3d^{10} 4s^1$, so $\text{Cu}^{2+} = [\text{Ar}]\,3d^9$. This gives $1$ unpaired electron.
4. $\text{Ni}^{2+}$: Nickel is $[\text{Ar}]\,3d^8 4s^2$, so $\text{Ni}^{2+} = [\text{Ar}]\,3d^8$. This gives $2$ unpaired electrons.
5. $\text{Ti}^{2+}$: $\text{Ti}^{2+} = [\text{Ar}]\,3d^2$. This gives $2$ unpaired electrons.

The matching ions are $\text{V}^{3+}$, $\text{Ni}^{2+}$ and $\text{Ti}^{2+}$, so the answer is A, D and E Only, which is option C.