QJEEM24Chf8d0: JEE Main 2024 Chemistry - Class 12 JEEM24Chf8d0 Solution: The number of complexes from the following with no electrons in …

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 ( $\mathrm{MnO_4}^-$ ) 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 $\mathrm{FeO_4}^{2-}$ : 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 $\mathrm{FeCl_4}^-$ : 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 $\mathrm{CoCl_4}^{2-}$ : 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