The number of complexes from the following with no electrons in the $t_2$ orbital is ______.

$$\mathrm{TiCl}_4,\left[\mathrm{MnO}_4\right]^{-},\left[\mathrm{FeO}_4\right]^{2-},\left[\mathrm{FeCl}_4\right]^{-},\left[\mathrm{CoCl}_4\right]^{2-}$$

<p><p><strong>Identifying the electronic configuration and geometry :</strong></p> <p>The condition "no electrons in the $ t_2 $ orbital" typically applies to a tetrahedral crystal field splitting pattern. In a tetrahedral field:</p></p> <p><p>The $ d $-orbitals split into two sets: $ e $ (lower energy, 2 orbitals) and $ t_2 $ (higher energy, 3 orbitals).</p></p> <p><p>Electrons fill the lower-energy $ e $ set before occupying the $ t_2 $ set.</p> <p>Thus, to have no electrons in $ t_2 $, either:</p></p> <p><p>The metal has a $ d^0 $ configuration (no $ d $-electrons at all), or</p></p> <p><p>The $ d $-electron count is so low that all electrons can fit into the $ e $ orbitals without needing to occupy $ t_2 $.</p></p> <p><p><strong>Analyzing each complex :</strong></p> <p><strong>(i) $\mathrm{TiCl}_4$ :</strong></p></p> <p><p>Ti in TiCl₄ is in the +4 oxidation state (since TiCl₄ is neutral and each Cl is -1).</p></p> <p><p>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 $.</p></p> <p><p>With $ d^0 $, there are no electrons to occupy $ t_2 $ orbitals.</p></p> <p><p><strong>No electrons in $ t_2 $</strong>.</p> <p><strong>(ii) $[\mathrm{MnO}_4]^{-}$ (permanganate) :</strong></p></p> <p><p>Mn in permanganate ($\mathrm{MnO_4}^-$) is in the +7 oxidation state.</p></p> <p><p>Mn (Z = 25) neutral is $ 3d^5 4s^2 $. Mn⁷⁺ means removing all 7 valence electrons, leaving $ d^0 $.</p></p> <p><p>With $ d^0 $, no electrons in $ t_2 $.</p></p> <p><p><strong>No electrons in $ t_2 $</strong>.</p> <p><strong>(iii) $[\mathrm{FeO}_4]^{2-}$ (ferrate) :</strong></p></p> <p><p>Fe in $\mathrm{FeO_4}^{2-}$: Oxygen contributes -8 total. The ion is -2. Thus, Fe + (-8) = -2 → Fe = +6 oxidation state.</p></p> <p><p>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 $.</p></p> <p><p>In a tetrahedral field, $ d^2 $ fills the lower $ e $ orbitals (2 electrons into e orbitals).</p></p> <p><p>No need to occupy $ t_2 $ orbitals since both electrons fit into e.</p></p> <p><p><strong>No electrons in $ t_2 $</strong>.</p> <p><strong>(iv) $[\mathrm{FeCl}_4]^{-}$</strong> :</p></p> <p><p>Fe in $\mathrm{FeCl_4}^- $: Cl total charge = -4, complex = -1, so Fe = +3.</p></p> <p><p>Fe(III) is $ d^5 $.</p></p> <p><p>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 $.</p></p> <p><p><strong>Has electrons in $ t_2 $</strong>.</p> <p><strong>(v) $[\mathrm{CoCl}_4]^{2-}$</strong>:</p></p> <p><p>Co in $\mathrm{CoCl_4}^{2-}$ : Cl total = -4, complex = -2, so Co = +2.</p></p> <p><p>Co(II) is $ d^7 $.</p></p> <p><p>For $ d^7 $, even after filling the $ e $ set (2 orbitals), we have 5 more electrons left, which must occupy the $ t_2 $ orbitals.</p></p> <p><p><strong>Has electrons in $ t_2 $</strong>.</p></p> <p><strong>Counting the complexes with no electrons in $ t_2 $ :</strong></p> <p><p>TiCl₄: No electrons in $ t_2 $.</p></p> <p><p>[MnO₄]⁻: No electrons in $ t_2 $.</p></p> <p><p>[FeO₄]²⁻: No electrons in $ t_2 $.</p></p> <p><p>[FeCl₄]⁻: Has electrons in $ t_2 $.</p></p> <p><p>[CoCl₄]²⁻: Has electrons in $ t_2 $.</p></p> <p><strong>Total with no electrons in $ t_2 $ = 3.</strong></p> <p><strong>Answer :</strong></p> <p>3</p>