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

We need to find the ions with $$n$$ unpaired electrons, where $$n$$ is the number of lone pairs in the equatorial position of the most stable structure of $$ClF_3$$.

In $$ClF_3$$, the central atom Cl has 7 valence electrons, three of which form bonds with F atoms, leaving two lone pairs (4 electrons). The resulting geometry is trigonal bipyramidal (sp$$^3$$d hybridization) with three F atoms and two lone pairs. Since lone pairs repel more strongly and occupy equatorial positions to minimize repulsion, the most stable T-shaped structure has both lone pairs equatorial. Hence $$n = 2$$.

Examining the given ions for two unpaired electrons: $$V^{3+}$$ has the configuration [Ar] 3d$$^2$$ and thus 2 unpaired electrons; $$Ti^{3+}$$ is [Ar] 3d$$^1$$ with 1 unpaired electron; $$Cu^{2+}$$ is [Ar] 3d$$^9$$ with 1 unpaired electron; $$Ni^{2+}$$ is [Ar] 3d$$^8$$ with 2 unpaired electrons (high‐spin); and $$Ti^{2+}$$ is [Ar] 3d$$^2$$ with 2 unpaired electrons.

Therefore, the ions with 2 unpaired electrons are $$V^{3+}$$, $$Ni^{2+}$$, and $$Ti^{2+}$$, corresponding to options A, D, and E. The correct answer is Option 2: A, D and E Only.