1. Find the number of unpaired electrons from magnetic moment

For spin-only magnetic moment,

where $n$ is the number of unpaired electrons.

Given:

Now check:

So, the complex has

unpaired electrons.

2. Electronic configuration of a $d^6$ ion

The metal ion is $d^6$.

We need a geometry that gives 4 unpaired electrons.

3. Check octahedral case

For octahedral splitting:

• Lower set: $t_{2g}$
• Upper set: $e_g$

For a $d^6$ ion:

• High spin octahedral: $t_{2g}^4 e_g^2$ → 4 unpaired electrons
• Low spin octahedral: $t_{2g}^6 e_g^0$ → 0 unpaired electrons

Since aqua is a weak field ligand, the octahedral aqua complex would be high spin. Thus configuration is:

which indeed has 4 unpaired electrons.

Now compute CFSE:

• each $t_{2g}$ electron contributes $-0.4\Delta_o$
• each $e_g$ electron contributes $+0.6\Delta_o$

So,

There is one paired electron pair, so pairing contribution is

Hence total:

This does not match option B or D.

4. Check tetrahedral case

For tetrahedral splitting:

• Lower set: $e$
• Upper set: $t_2$

Tetrahedral complexes are almost always high spin. For $d^6$ tetrahedral configuration:

This gives 4 unpaired electrons, consistent with the magnetic moment.

Now compute CFSE in tetrahedral field:

• each $e$ electron contributes $-0.6\Delta_t$
• each $t_2$ electron contributes $+0.4\Delta_t$

Thus,

There is one paired electron in $e^3$, but CFSE option is usually written separately unless pairing energy is explicitly added. So the correct matching expression is:

This matches Option C.

5. Evaluate options

• A: tetrahedral and $-1.6\Delta_t + 1P$ ❌ Incorrect CFSE
• B: octahedral and $-2.4\Delta_o + 2P$ ❌ Incorrect for $d^6$ high-spin aqua complex
• C: tetrahedral and $-0.6\Delta_t$ ✅ Correct
• D: octahedral and $-1.6\Delta_o$ ❌ Incorrect CFSE

6. Final answer

The complex is tetrahedral with CFSE

So the correct option is C.