Consider that d⁶ metal ion (M²⁺) forms a complex with aqua ligands, and the spin only magnetic moment of the complex is…

🧠 μ = 4.90 BM Implies 4 Unpaired Electrons

$\mu = \sqrt{n(n+2)}$. For μ = 4.90: $4.90 = \sqrt{n(n+2)} \Rightarrow n(n+2) = 24 \Rightarrow n = 4$

🗺️ d⁶ With 4 Unpaired

For d⁶ M(II), 4 unpaired electrons matches:

• Octahedral HS d⁶ → $t_{2g}^4 e_g^2$ → 4 unpaired ✓
• Tetrahedral d⁶ → $e^3 t_2^3$ → 4 unpaired ✓ (Td is always HS for first-row TM)

Both could give 4 unpaired. Need additional reasoning.

🗺️ Aqua Ligand Strength

H₂O is weak/intermediate field. For d⁶ M(II) with H₂O:

• Octahedral HS d⁶: $t_{2g}^4 e_g^2$, CFSE = $-0.4\Delta_o$ + pairing penalty $P$ in $t_{2g}^4$ pair → $-0.4\Delta_o + P$ (1 extra pair). Or just $-0.4\Delta_o$ if pairing ignored.
• Tetrahedral d⁶: $e^3 t_2^3$, CFSE = $3(-0.6\Delta_t) + 3(+0.4\Delta_t) = -1.8\Delta_t + 1.2\Delta_t = -0.6\Delta_t$.

🗺️ Match the Options

Looking at options:

• Octahedral and $-2.4\Delta_o + 2P$: would need 6 paired electrons (LS) — μ=0, not 4.90 BM. ✗
• Tetrahedral and $-0.6\Delta_t$: matches Td d⁶ exactly. ✓
• Octahedral and $-1.6\Delta_o$: would be Td d⁵ ($e^2 t_2^3$ = $2(-0.6) + 3(+0.4) = 0$) or some other config — doesn't match d⁶ HS oct CFSE.
• Tetrahedral and $-1.6\Delta_t + 1P$: doesn't match standard Td d⁶ formula.

So the answer is (2) Tetrahedral and $-0.6\Delta_t$.

⚡ Td CFSE Coefficients

Td d-orbital splitting is inverted vs Oct: $e$ is lower (coeff $-0.6\Delta_t$), $t_2$ is upper (coeff $+0.4\Delta_t$).

⚠️ Td Always HS

For first-row tetrahedral complexes, $\Delta_t \approx (4/9)\Delta_o$ — too small to overcome pairing. So Td is universally HS for first-row TMs.

$\boxed{\text{Answer: (2) tetrahedral and } -0.6\Delta_t}$