If an iron(III) complex with the formula [Fe(NH3)x(CN)y]- has no electron in its eg orbital, then the value of x + y is:

🧠 No $e_g$ Electrons → LS Fe(III) d⁵

Fe(III) is d⁵. With no $e_g$ electrons, the configuration must be low-spin $t_{2g}^5 e_g^0$. This requires a strong-field ligand environment.

🗺️ CN⁻ Strong Field, NH₃ Moderate

For Fe(III) to be LS, the ligand field must be strong enough. CN⁻ is strong-field; NH₃ is moderate. A pure CN⁻ environment ($[\mathrm{Fe(CN)_6}]^{3-}$) is definitely LS.

But the formula is $[\mathrm{Fe(NH_3)_x (CN)_y}]^-$ — mixed.

🗺️ Charge Balance

Fe(III) charge = +3. NH₃ charge = 0. CN⁻ charge = -1. Overall complex charge = -1.

$+3 + 0 \cdot x + (-1) \cdot y = -1$ $3 - y = -1 \Rightarrow y = 4$

🗺️ Coordination Number = 6

$x + y = 6$ (octahedral). With $y = 4$: $x = 2$.

So formula: $[\mathrm{Fe(NH_3)_2 (CN)_4}]^-$. With 4 CN⁻ in the inner sphere, the field is sufficiently strong to force LS Fe(III) d⁵ → $t_{2g}^5 e_g^0$. ✓

$x + y = 2 + 4 = 6$

⚡ Charge Balance Strategy

When given complex formula with unknown ligand counts, set up two equations:

1. Charge balance.
2. Coordination number constraint (typically 6 for octahedral, 4 for sq pl/Td).

⚠️ Mixed Ligand Field Strength

A mixed NH₃/CN⁻ environment is strong enough for LS Fe(III) when CN⁻ count is high enough (≥ 3 or 4 typically).

$\boxed{\text{Answer: (3) 6}}$