The type of hybridization and the magnetic property of $\left[\mathrm{MnCl}_6\right]^{3-}$ are,

<p>Determine the oxidation state of manganese in $\left[\mathrm{MnCl}_6\right]^{3-}:$ </p> <p><p>Chloride (Cl⁻) has a charge of -1. </p></p> <p><p>With six chloride ions, the total charge contributed by the ligands is $6(-1) = -6.$ </p></p> <p><p>Let the oxidation state of Mn be $x.$ Then: </p> <p>$x - 6 = -3 \quad \Longrightarrow \quad x = +3.$ </p></p> <p><p>So, manganese is in the +3 oxidation state.</p></p> <p>Find the d-electron count for Mn(III): </p> <p><p>The neutral manganese atom has the electronic configuration $[\mathrm{Ar}]\, 3d^5 4s^2.$ </p></p> <p><p>Removing three electrons (first from the 4s, then from the 3d) gives: </p> <p>$\mathrm{Mn}^{3+} : [\mathrm{Ar}]\, 3d^4.$ </p></p> <p><p>Therefore, Mn(III) is a $d^4$ system.</p></p> <p>Predict the spin state in an octahedral complex: </p> <p><p>Chloride (Cl⁻) is a weak-field ligand, meaning it produces a relatively small crystal field splitting ($\Delta_o$). </p></p> <p><p>When $\Delta_o$ is small, the pairing energy is higher than $\Delta_o,$ so electrons prefer to remain unpaired. </p></p> <p><p>For a $d^4$ configuration in a high-spin octahedral complex, the electrons will be distributed as: </p> <p><p>Three electrons in the three $t_{2g}$ orbitals (one in each) </p></p> <p><p>One electron in one of the $e_g$ orbitals </p></p></p> <p><p>This results in a total of 4 unpaired electrons, making the complex paramagnetic.</p></p> <p>Determine the type of hybridization: </p> <p><p>In octahedral complexes, the two common hybridizations are: </p> <p><p>$d^2sp^3$ (inner orbital complex), typically seen in low-spin complexes where inner 3d orbitals are available because the electrons are paired. </p></p> <p><p>$sp^3d^2$ (outer orbital complex), seen in high-spin complexes where the inner $3d$ orbitals are occupied by unpaired electrons. </p></p></p> <p><p>Since $\left[\mathrm{MnCl}_6\right]^{3-}$ is high-spin (because of Cl⁻ being a weak-field ligand), the inner 3d orbitals are not available for hybridization. </p></p> <p><p>Therefore, the complex uses the outer orbitals (the 4s, 4p, and 4d orbitals), leading to an $sp^3d^2$ hybridization.</p></p> <p>Conclusion: </p> <p><p>The hybridization of $\left[\mathrm{MnCl}_6\right]^{3-}$ is $sp^3d^2.$ </p></p> <p><p>It is paramagnetic with four unpaired electrons.</p></p> <p>Thus, the correct answer is:</p> <p>Option A: $sp^3d^2$, paramagnetic with four unpaired electrons.</p>