**66.** In which of the following complexes the CFSE, \( \Delta_0 \), will be equal to zero? - (1) \( [\ce{Fe(NH3)6}]Br_2 \) - (2) \( [\ce{Fe(en)3}]Cl_3 \) - (3) \( \ce{K4[Fe(CN)6]} \) - (4) \( \ce{K3[Fe(SCN)6]} \)

Key ideas to tackle the problem

1. Find the oxidation state & d-electron count

   • $\text{Fe}^{2+}$ ―> $d^6$
   • $\text{Fe}^{3+}$ ―> $d^5$

2. Judge ligand strength

   • Strong-field ligands (e.g. $\text{CN}^-$, $en$) usually produce low-spin complexes.
   • Weak-field ligands (e.g. $\text{H}_2\text{O}$, $\text{F}^-$, $\text{SCN}^-$ via S-bonding) give high-spin complexes.

3. Recall CFSE for an octahedral field

   $\text{CFSE} = \left[\text{number in } t_{2g}\right] \times (-0.4\Delta_0) + \left[\text{number in } e_g\right] \times (+0.6\Delta_0)$

4. When can CFSE be zero?

   • $d^0$ (empty)
   • High-spin $d^5$ (3 in $t_{2g}$, 2 in $e_g$)
   • $d^{10}$ (completely filled)

5. Match each option

   • Decide oxidation state & ligand field strength.
   • Check if electron configuration matches one of the ‘zero-CFSE’ cases.

Only the complex that supplies a high-spin $d^5$ iron ion will do the trick.

Imagine a playground with two kinds of slides

1. Low Slides (called t2g slots) – kids need less energy to sit here.
2. High Slides (called eg slots) – kids need more energy to reach here.

Every iron ‘kid’ brings a certain number of toys (electrons). Where the kids sit decides if they save energy (a negative score) or spend extra energy (a positive score).

• Sitting on a low slide saves $0.4$ points each.
• Sitting on a high slide costs $0.6$ points each.

Add all the points, and you get the Crystal-Field Stabilisation Energy (CFSE).

If the total points saved exactly cancel the points spent, the CFSE is zero – nothing gained, nothing lost.

So we only have to spot the iron ‘kid’ arrangement that ends up with no net score.

The magic arrangement is when iron has 5 scattered toys in a weak field (called ‘high-spin $d^5$’), because:

$3( -0.4 \Delta_0 ) + 2( +0.6 \Delta_0 ) = 0$

Among the given choices, only the complex with a weak ligand around an $\text{Fe}^{3+}$ ion gives this high-spin $d^5$ arrangement.

Step-by-step solution

1. Identify oxidation state and $d$ count

   Complex	Outside counter-ion	Oxidation state of Fe	$d$ electrons
   $[\text{Fe}(\text{NH}_3)_6]\,\text{Br}_2$	$2 \times \text{Br}^-$	$+2$	$d^6$
   $[\text{Fe}(en)_3]\,\text{Cl}_3$	$3 \times \text{Cl}^-$	$+3$	$d^5$
   $\text{K}_4[\text{Fe}(\text{CN})_6]$	$4 \times \text{K}^+$	$+2$	$d^6$
   $\text{K}_3[\text{Fe}(\text{SCN})_6]$	$3 \times \text{K}^+$	$+3$	$d^5$

2. Assess ligand field strength and spin state

   • $\text{NH}_3$: moderate field → usually high spin for $\text{Fe}^{2+}$.
   • $en$ (ethylenediamine): stronger than $\text{NH}_3$ → low spin for $\text{Fe}^{3+}$.
   • $\text{CN}^-$: very strong field → low spin for $\text{Fe}^{2+}$.
   • $\text{SCN}^-$ (S-bound): weak field → high spin for $\text{Fe}^{3+}$.

3. Work out electron distribution & CFSE

   • Option 1 – high-spin $d^6$: $t_{2g}^4 e_g^2$
   $\text{CFSE} = 4(-0.4\Delta_0) + 2(+0.6\Delta_0) = -0.4\Delta_0$

   • Option 2 – low-spin $d^5$: $t_{2g}^5$
   $\text{CFSE} = 5(-0.4\Delta_0) = -2.0\Delta_0$

   • Option 3 – low-spin $d^6$: $t_{2g}^6$
   $\text{CFSE} = 6(-0.4\Delta_0) = -2.4\Delta_0$

   • Option 4 – high-spin $d^5$: $t_{2g}^3 e_g^2$
   $\text{CFSE} = 3(-0.4\Delta_0) + 2(+0.6\Delta_0) = 0$

4. Conclusion

The only complex with CFSE = 0 is

$\boxed{\text{K}_3[\text{Fe}(\text{SCN})_6]}$

Hence, answer (4).