JEE Main 2026 — Classification of Elements and Periodicity in Properties Question with Solution

Statement I:
The third ionization energy ($I{E}_3$) corresponds to the removal of the third valence electron. Let us look at the electronic configurations of the divalent cations for the given metals:
$S{c}^{2+}$: $[Ar]3d^1$
$M{n}^{2+}$: $[Ar]3d^5$
$C{u}^{2+}$: $[Ar]3d^9$
$Z{n}^{2+}$: $[Ar]3d^{10}$

For $Sc$, removing the third electron involves taking it from the $3d^1$ state, which results in a highly stable noble gas core ($[Ar]$). Hence, $Sc$ has the lowest $I{E}_3$.
For $Zn$, the third electron must be removed from a completely filled, highly stable $3d^{10}$ subshell. Combined with its high effective nuclear charge ($Z_{eff}$) across the 3d series, $Zn$ requires the highest energy for the removal of the third electron.
Thus, Statement I is true.

Statement II:
The Crystal Field Stabilization Energy (CFSE) depends on the magnitude of the crystal field splitting energy (${\Delta }_o$), which in turn depends on the oxidation state of the central metal ion and the strength of the ligand.
1. Oxidation state: A higher oxidation state on the central metal ion leads to a larger ${\Delta }_o$. Therefore, the splitting for $C{o}^{3+}$ complexes is significantly greater than for $C{o}^{2+}$ complexes. This makes the CFSE of $[Co(H_{2}O{)}_6{]}^{3+}$ greater than that of $[Co(H_{2}O{)}_6{]}^{2+}$.
2. Nature of the ligand: Ethylenediamine ($en$) is a strong field bidentate ligand, whereas $H_{2}O$ is a weak field ligand. According to the spectrochemical series, $en$ causes a much larger splitting than $H_{2}O$. Thus, the ${\Delta }_o$ and CFSE for $[Co(en{)}_3{]}^{3+}$ are greater than those for $[Co(H_{2}O{)}_6{]}^{3+}$.

The overall order of CFSE is:
$[Co(H_{2}O{)}_6{]}^{2+}<[Co(H_{2}O{)}_6{]}^{3+}<[Co(en{)}_3{]}^{3+}$
Thus, Statement II is true.

Since both statements are correct, the correct option is (1).

Answer: Both Statement I and Statement II are true