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