Consider 'n' is the number of lone pair of electrons present in the equatorial position of the most stable structure of ClF₃$\mathrm{ClF}_3$ . The ions from the following with 'n' number of unpaired electrons are : A. V3 +$\mathrm{V}^{3 + }$ B. Ti³⁺$\mathrm{Ti}^{3+}$ C. Cu2 +$\mathrm{Cu}^{2 + }$ D. Ni²⁺$\mathrm{Ni}^{2+}$ E. Ti²⁺$\mathrm{Ti}^{2+}$ Choose the correct answer from the options given below:

Step 1: Determine 'n'

ClF₃$\mathrm{ClF}_3$ has a central Chlorine atom with 7 valence electrons, bound to 3 Fluorine atoms. This leaves 2 lone pairs. The geometry is trigonal bipyramidal (T-shaped molecule). In its most stable geometry, both lone pairs lie in the equatorial plane to minimize lone pair-bonding pair repulsions. Thus, n = 2$n = 2$.

Step 2: Find ions with 2 unpaired electrons

Let us compute the number of unpaired electrons for each configuration:

• A. V³⁺$\mathrm{V}^{3+}$: [Ar] 3d² arrow 2$[\mathrm{Ar}] 3d^2 \rightarrow 2$ unpaired electrons.
• B. Ti³⁺$\mathrm{Ti}^{3+}$: [Ar] 3d¹ arrow 1$[\mathrm{Ar}] 3d^1 \rightarrow 1$ unpaired electron.
• C. Cu²⁺$\mathrm{Cu}^{2+}$: [Ar] 3d⁹ arrow 1$[\mathrm{Ar}] 3d^9 \rightarrow 1$ unpaired electron.
• D. Ni²⁺$\mathrm{Ni}^{2+}$: [Ar] 3d⁸ arrow 2$[\mathrm{Ar}] 3d^8 \rightarrow 2$ unpaired electrons.
• E. Ti²⁺$\mathrm{Ti}^{2+}$: [Ar] 3d² arrow 2$[\mathrm{Ar}] 3d^2 \rightarrow 2$ unpaired electrons.

Thus, A, D, and E have exactly n=2$n=2$ unpaired electrons.

Pattern Recognition

Sees: Number of equatorial lone pairs linked to unpaired electrons. Shortcut: Remember ClF₃$\mathrm{ClF}_3$ is T-shaped with 2 equatorial lone pairs. Look for d²$d^2$ or d⁸$d^8$ configurations among the transition metal ions.

Chapter Mix

Class 11 Chemistry: Chemical Bonding and Molecular Structure Class 12 Chemistry: The d-and f-Block Elements