Identify the geometry and crystal field splitting

All the given species are tetrahedral complexes (or oxyanions with tetrahedral coordination).

In a tetrahedral field, the $d$-orbitals split as: $e ext{ (lower)}, _2 ext{ (higher)}$

We need complexes with no electrons in the $t_2$ orbitals, i.e. all $d$ electrons, if any, must fit only in the lower $e$ set.

Since $e$ can hold a maximum of $4$ electrons, this means the metal must have configuration: $d^0, d^1, d^2, d^3, ext{ or } d^4$ in tetrahedral high-spin filling before $t_2$ starts getting occupied.

Find oxidation state and $d$-electron count for each complex

(i) $\mathrm{TiCl}_4$

Let oxidation state of Ti be $x$: $x + 4(-1)=0 \Rightarrow x=+4$ Ti: $Z=22$, ground state $[\mathrm{Ar}]3d^24s^2$

$\mathrm{Ti}^{4+} : d^0$

So $t_2$ has no electrons. ✅

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

Let oxidation state of Mn be $x$: $x + 4(-2) = -1 \Rightarrow x=+7$ Mn: $Z=25$, ground state $[\mathrm{Ar}]3d^54s^2$

$\mathrm{Mn}^{7+}: d^0$

So $t_2$ has no electrons. ✅

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

Let oxidation state of Fe be $x$: $x + 4(-2) = -2 \Rightarrow x=+6$ Fe: $Z=26$, ground state $[\mathrm{Ar}]3d^64s^2$

$\mathrm{Fe}^{6+}: d^2$

In tetrahedral splitting, the two electrons occupy the lower $e$ orbitals: $e^2 t_2^0$

So $t_2$ has no electrons. ✅

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

Let oxidation state of Fe be $x$: $x + 4(-1) = -1 \Rightarrow x=+3$ $\mathrm{Fe}^{3+}: d^5$

In tetrahedral field (generally high spin): $e^2 t_2^3$

So $t_2$ is occupied. ❌

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

Let oxidation state of Co be $x$: $x + 4(-1) = -2 \Rightarrow x=+2$ Co: $Z=27$, ground state $[\mathrm{Ar}]3d^74s^2$

$\mathrm{Co}^{2+}: d^7$

In tetrahedral field: $e^4 t_2^3$

So $t_2$ is occupied. ❌

Count complexes with no electrons in $t_2$

These are:

• $\mathrm{TiCl}_4$
• $[\mathrm{MnO}_4]^{-}$
• $[\mathrm{FeO}_4]^{2-}$

Hence, total number is: $3$

Compare with options

Option C: 3 is correct.