JEE Main 2024 — Coordination Compounds Question with Solution

Identifying the electronic configuration and geometry :

The condition "no electrons in the $t_2$ orbital" typically applies to a tetrahedral crystal field splitting pattern. In a tetrahedral field:

The $d$-orbitals split into two sets: $e$ (lower energy, 2 orbitals) and $t_2$ (higher energy, 3 orbitals).

Electrons fill the lower-energy $e$ set before occupying the $t_2$ set.

Thus, to have no electrons in $t_2$, either:

The metal has a $d^0$ configuration (no $d$-electrons at all), or

The $d$-electron count is so low that all electrons can fit into the $e$ orbitals without needing to occupy $t_2$.

Analyzing each complex :

(i) ${\mathrm{TiCl}}_4$ :

Ti in TiCl₄ is in the +4 oxidation state (since TiCl₄ is neutral and each Cl is -1).

Ti (Z = 22) neutral : $3d^{2}4s^2$. As Ti⁴⁺, it loses all 4 valence electrons (2 from 4s and 2 from 3d), resulting in $d^0$.

With $d^0$, there are no electrons to occupy $t_2$ orbitals.

No electrons in $t_2$.

(ii) $[{\mathrm{MnO}}_4{]}^{-}$ (permanganate) :

Mn in permanganate (${\mathrm{Mn}{\mathrm{O}}_4}^{-}$) is in the +7 oxidation state.

Mn (Z = 25) neutral is $3d^{5}4s^2$. Mn⁷⁺ means removing all 7 valence electrons, leaving $d^0$.

With $d^0$, no electrons in $t_2$.

No electrons in $t_2$.

(iii) $[{\mathrm{FeO}}_4{]}^{2-}$ (ferrate) :

Fe in ${\mathrm{Fe}{\mathrm{O}}_4}^{2-}$: Oxygen contributes -8 total. The ion is -2. Thus, Fe + (-8) = -2 → Fe = +6 oxidation state.

Fe (Z = 26) neutral is $3d^{6}4s^2$. Fe⁶⁺ means removing 6 electrons from the valence shell. After losing 2 from 4s and 4 from 3d, we get $d^2$.

In a tetrahedral field, $d^2$ fills the lower $e$ orbitals (2 electrons into e orbitals).

No need to occupy $t_2$ orbitals since both electrons fit into e.

No electrons in $t_2$.

(iv) $[{\mathrm{FeCl}}_4{]}^{-}$ :

Fe in ${\mathrm{FeC}{\mathrm{l}}_4}^{-}$: Cl total charge = -4, complex = -1, so Fe = +3.

Fe(III) is $d^5$.

In a tetrahedral field, with 5 $d$-electrons, after filling the 2 $e$ orbitals, we have 3 more electrons that must go into $t_2$.

Has electrons in $t_2$.

(v) $[{\mathrm{CoCl}}_4{]}^{2-}$:

Co in ${\mathrm{CoC}{\mathrm{l}}_4}^{2-}$ : Cl total = -4, complex = -2, so Co = +2.

Co(II) is $d^7$.

For $d^7$, even after filling the $e$ set (2 orbitals), we have 5 more electrons left, which must occupy the $t_2$ orbitals.

Has electrons in $t_2$.

Counting the complexes with no electrons in $t_2$ :

TiCl₄: No electrons in $t_2$.

[MnO₄]⁻: No electrons in $t_2$.

[FeO₄]²⁻: No electrons in $t_2$.

[FeCl₄]⁻: Has electrons in $t_2$.

[CoCl₄]²⁻: Has electrons in $t_2$.

Total with no electrons in $t_2$ = 3.

Answer :

3