AIPMT PRELIMS 2004 Chemistry Crystal Field Theory MCQ Question

To solve the question regarding the number of unpaired electrons in the complex ion $[Mn(H_2O)_6]^{2+}$, we need to apply concepts from coordination chemistry, specifically Crystal Field Theory (CFT).

Step 1: Determine the oxidation state of manganese in $[Mn(H_2O)_6]^{2+}$

The oxidation state of manganese in this complex can be determined as follows:

$$\text{Oxidation state of Mn} + 6 \times \text{(Oxidation state of H}_2\text{O}) = +2$$

Since water ($H_2O$) is a neutral ligand, its oxidation state is 0. Thus, we have:

$$\text{Oxidation state of Mn} = +2$$

Step 2: Determine the electronic configuration of $Mn^{2+}$

The atomic number of manganese (Mn) is 25. The ground state electronic configuration of neutral manganese is:

$$[Ar] \, 4s^2 \, 3d^5$$

When manganese loses two electrons to form $Mn^{2+}$, the electrons are removed first from the $4s$ orbital. Therefore, the electronic configuration of $Mn^{2+}$ is:

$$[Ar] \, 3d^5$$

Step 3: Apply Crystal Field Theory

In the case of $[Mn(H_2O)_6]^{2+}$, which is a hexaaquamanganese(II) complex, we treat $H_2O$ as a weak field ligand. According to CFT, weak field ligands do not cause significant splitting of the $d$ orbitals; hence, they do not favor pairing of electrons.

For octahedral complexes, the five $d$ orbitals split into two sets due to the presence of ligands:

• $t_{2g}$ (lower energy): consists of three orbitals
• $e_g$ (higher energy): consists of two orbitals

For $Mn^{2+}$ with a $3d^5$ configuration and using water as a weak field ligand, the electrons are placed in the $d$ orbitals as follows:

• Each of the five $d$ electrons will occupy the $t_{2g}$ and $e_g$ orbitals singly before any pairing occurs.

This results in the following electron distribution:

• $t_{2g}^3 e_g^2$ (with all five electrons unpaired)

Step 4: Count the number of unpaired electrons

Since there are five electrons in total and none of them are paired, the number of unpaired electrons in $[Mn(H_2O)_6]^{2+}$ is:

$$\text{Number of unpaired electrons} = 5$$

Conclusion

Thus, the correct answer is:

A) Five

Clarification of Other Options:

• B) Two: Incorrect, as this would imply that three electrons are paired, which does not happen with weak field ligands.
• C) Four: Also incorrect for the same reason; it suggests that one electron is paired.
• D) Three: Incorrect; this implies that two electrons are paired, which does not occur with $H_2O$ as a weak field ligand.

In summary, the weak field nature of $H_2O$ leads to the preservation of all five unpaired electrons in the $Mn^{2+}$ ion in the complex $[Mn(H_2O)_6]^{2+}$.