The number of complexes from the following with no electrons in the $$t_2$$ orbital is $$TiCl_4, [MnO_4]^-, [FeO_4]^{2-}, [FeCl_4]^-, [CoCl_4]^{2-}$$

We need to find how many of the given complexes have no electrons in the $$t_2$$ orbitals.

Note: In a tetrahedral field, the d-orbitals split into $$e$$ (lower, $$d_{z^2}$$ and $$d_{x^2-y^2}$$) and $$t_2$$ (higher, $$d_{xy}$$, $$d_{xz}$$, $$d_{yz}$$). For the $$t_2$$ orbitals to have no electrons, the metal must have $$d^0$$ configuration or all electrons must be in the $$e$$ set (at most 4 electrons with 2 in each $$e$$ orbital).

1. $$TiCl_4$$: Ti is in +4 state. $$Ti^{4+}$$: [Ar], $$d^0$$. No d-electrons at all, so no electrons in $$t_2$$. Yes - no electrons in $$t_2$$.

2. $$[MnO_4]^-$$: Mn is in +7 state. $$Mn^{7+}$$: [Ar], $$d^0$$. No d-electrons, so no electrons in $$t_2$$. Yes - no electrons in $$t_2$$.

3. $$[FeO_4]^{2-}$$: Fe is in +6 state. $$Fe^{6+}$$: [Ar], $$d^2$$. In tetrahedral field, 2 electrons go into the lower $$e$$ orbitals. So $$t_2$$ has 0 electrons. Yes - no electrons in $$t_2$$.

4. $$[FeCl_4]^-$$: Fe is in +3 state. $$Fe^{3+}$$: [Ar] $$3d^5$$. In tetrahedral (weak field), electrons fill as $$e^2 t_2^3$$. The $$t_2$$ orbitals have 3 electrons. No - has electrons in $$t_2$$.

5. $$[CoCl_4]^{2-}$$: Co is in +2 state. $$Co^{2+}$$: [Ar] $$3d^7$$. In tetrahedral field, electrons fill as $$e^4 t_2^3$$. The $$t_2$$ orbitals have 3 electrons. No - has electrons in $$t_2$$.

The complexes with no electrons in $$t_2$$ orbitals are: $$TiCl_4$$, $$[MnO_4]^-$$, and $$[FeO_4]^{2-}$$.

The correct answer is Option (3): 3.