Question

Pick out the correct statement with respect \([Mn(CN)_6]^{3–}\) :

• A. It is sp³d² hybridised and octahedral
• B. It is sp³d² hybridised and tetrahedral
• C. It is d²sp³ hybridised and octahedral
• D. It is dsp² hybridised and square planar

Answer 1

The Correct Option is C

To determine the correct hybridization and geometry for the coordination complex \([Mn(CN)_6]^{3–}\), we need to follow these steps:

1.  Determine the oxidation state of the central metal ion, \(Mn\), in the complex:
   • Cyanide (\(CN^-\)) is a monodentate ligand with a charge of -1.
   • Let the oxidation state of manganese be \(x\).
   • The charge of the entire complex is -3.
   • Setting up the equation: \(x + 6(-1) = -3\).
   • Solving gives \(x = +3\).
2. Determine the electronic configuration of \(Mn^{3+}\):
   • The electronic configuration of neutral manganese (\(Mn\)) is \([Ar] 3d^5 4s^2\).
   • For \(Mn^{3+}\), remove three electrons: Two from the 4s orbital and one from the 3d orbital.
   • Electronic configuration of \(Mn^{3+}\) is \([Ar] 3d^4\).
3. Analyze the influence of cyanide ligands:
   • Cyanide (\(CN^-\)) is a strong field ligand that causes pairing of electrons in the 3d orbitals.
   • After pairing, the \(3d\) orbitals are configured as \(t_{2g}^4 e_g^0\).
4. Determine the hybridization of \(Mn^{3+}\):
   • After pairing, there are two empty d orbitals available (d₂), suitable for hybridization.
   • The configuration becomes \(t_{2g}^4 (d_{x^2-y^2} d_{z^2})\), preparing for \(d^2sp^3\) hybridization.
   • The resulting hybridization is \(d^2sp^3\), forming an octahedral geometry as six ligands fit perfectly with this configuration.

After evaluating the options, the correct statement for \([Mn(CN)_6]^{3–}\) is:

• It is \(d^2sp^3\) hybridised and octahedral.

This is because cyanide is a strong field ligand that induces pairing in the 3d orbitals of \(Mn^{3+}\), facilitating the \(d^2sp^3\) hybridization required for an octahedral geometry.