Question

The type of hybridisation and magnetic property of the complex $[MnCl_6]^{3-}$, respectively, are :

• A. d$^2$sp$^3$ and paramagnetic
• B. sp$^3$d$^2$ and diamagnetic
• C. sp$^3$d$^2$ and paramagnetic
• D. d$^2$sp$^3$ and diamagnetic

Hint:

Remember the spectrochemical series to identify weak and strong field ligands. Halides (like $Cl^-, Br^-, I^-$) are generally weak field ligands, leading to high-spin complexes. Ligands with C or N donor atoms (like $CN^-, CO, NH_3$) are often strong field, leading to low-spin complexes.

Answer 1

The Correct Option is C

Step 1: Determine the oxidation state of the central metal ion, Manganese (Mn).
Let the oxidation state of Mn be x. The charge on each chloride ligand is -1. The overall charge of the complex is -3.
$x + 6(-1) = -3 \implies x - 6 = -3 \implies x = +3$.
So, we have Mn$^{3+}$.
Step 2: Write the electronic configuration of Mn$^{3+}$.
The atomic number of Mn is 25. Its configuration is $[Ar] 3d^5 4s^2$.
For Mn$^{3+}$, we remove three electrons (two from 4s, one from 3d).
The configuration of Mn$^{3+}$ is $[Ar] 3d^4$.
Step 3: Consider the ligand and determine if the complex is high-spin or low-spin.
The ligand is $Cl^-$, which is a weak field ligand. Weak field ligands do not cause pairing of electrons in the d-orbitals. Therefore, this will be a high-spin complex.
Step 4: Determine hybridization and magnetic properties.
The $3d^4$ configuration for a high-spin octahedral complex will have 4 unpaired electrons:
$3d$: [$\uparrow$][$\uparrow$][$\uparrow$][$\uparrow$][ ]
For bonding with six $Cl^-$ ligands, we need six empty hybrid orbitals. Since this is a high-spin complex, the inner 3d orbitals are not available. The complex will use the outer orbitals: one 4s, three 4p, and two 4d orbitals.
Hybridisation: $sp^3d^2$ (outer orbital complex).
Magnetic property: Since there are 4 unpaired electrons, the complex is strongly paramagnetic.
The correct combination is $sp^3d^2$ and paramagnetic.