Q.Draw structures of geometrical isomers of $[Fe(NH_3)_2(CN)_4]^{-}$

The key idea is that geometrical isomerism in coordination compounds arises when ligands of two different types can occupy positions that are either adjacent (cis) or opposite (trans) to each other.

Reasoning:

1. The complex $[Fe(NH_3)_2(CN)_4]^{-}$ has a coordination number of 6, so it is octahedral.
2. There are two identical ammine ($NH_3$) ligands and four identical cyano ($CN^-$) ligands.
3. The two $NH_3$ ligands can be placed either at 90° to each other (adjacent, cis) or at 180° to each other (opposite, trans). The four $CN^-$ ligands occupy the remaining positions.

The two geometrical isomers are:

• cis isomer: The two $NH_3$ groups are adjacent.
• trans isomer: The two $NH_3$ groups are opposite each other.

Type of complex. Six ligands around Fe make it octahedral; with two $\text{NH}_3$ (A) and four $\text{CN}^-$ (B) it is an $MA_2B_4$ system, which gives exactly two geometrical isomers (no optical isomerism, since both forms have a plane of symmetry).

cis isomer — the two $\text{NH}_3$ groups occupy adjacent corners ($90^\circ$):

          NH3
           |
   H3N --- Fe --- CN
           |
           CN

(The remaining two $\text{CN}^-$ lie in front of and behind the plane. The two $\text{NH}_3$ groups share an edge of the octahedron.)

trans isomer — the two $\text{NH}_3$ groups occupy opposite (axial) corners ($180^\circ$), with the four $\text{CN}^-$ in the equatorial plane:

          NH3
           |
    NC --- Fe --- CN
           |
          NH3

(The remaining two $\text{CN}^-$ lie in front of and behind the plane; all four $\text{CN}^-$ are coplanar.)

Method: Ligand-Position Enumeration (cis/trans) for Geometrical Isomerism in Octahedral Complexes

Why this method?

Geometrical (cis-trans) isomerism in octahedral complexes arises when two identical ligands can be placed adjacent (cis) or opposite (trans) to each other. The test is positional: two arrangements are distinct isomers when no rotation of the whole complex converts one into the other — symmetry planes only matter later, for deciding optical activity.

Step-by-step solution for $[Fe(NH_3)_2(CN)_4]^{-}$

Step 1: Identify the coordination number and geometry

• Fe is the central metal.
• Ligands: 2 $\text{NH}_3$ (neutral) and 4 $\text{CN}^-$ (anionic).
• Total coordination number = 6 → octahedral geometry.

Step 2: Determine the formula type

This is a $[MA_2B_4]$ type complex (where A = $\text{NH}_3$, B = $\text{CN}^-$).

For such complexes, geometrical isomerism is possible.

Step 3: Draw the two possible arrangements

• Cis isomer: Both $\text{NH}_3$ ligands are adjacent (at 90° to each other).
• Trans isomer: Both $\text{NH}_3$ ligands are opposite (at 180° to each other).

Step 4: Check that the two arrangements are genuinely distinct

The two arrangements cannot be interconverted by any rotation of the octahedron, so they are distinct geometrical isomers. (Note: both forms retain a plane of symmetry, so neither is chiral — there is no optical isomerism here.)

Final Answer

Two geometrical isomers exist:

1. Cis isomer — both $\text{NH}_3$ groups on adjacent positions.
2. Trans isomer — both $\text{NH}_3$ groups on opposite positions.

Key point: No rotation converts the cis arrangement into the trans one — they are distinct, stable isomers. Both retain a plane of symmetry, so neither is optically active.

Quick Exam Tip

For $[M(A)_2(B)_4]$ type complexes:

• Cis → the 2 identical A ligands at 90°.
• Trans → the 2 identical A ligands at 180°.
• Total geometrical isomers = 2 (cis and trans) — and since both forms have a plane of symmetry, there is no optical isomerism.

Here are the most common mistakes students make when tackling geometrical isomerism for coordination compounds like $[Fe(NH_3)_2(CN)_4]^{-}$, along with clear strategies to avoid them.

Mistake 1: Forgetting to Determine the Coordination Number and Geometry First

• The Mistake: Students jump straight to drawing isomers without first identifying that the complex is octahedral (coordination number 6). They might incorrectly treat it as square planar or tetrahedral.
• Why it happens: The formula $[Fe(NH_3)_2(CN)_4]^{-}$ has 6 ligands, but the presence of only two types of ligands ($NH_3$ and $CN$) can trick students into thinking it’s simpler.
• How to avoid: Always start by counting the total number of ligands attached to the metal. Here, $2 + 4 = 6$ — six ligands mean an octahedral geometry. (Iron here is Fe$^{3+}$, i.e. $d^5$ — see Mistake 5's charge calculation.) Write this down before drawing anything.

Mistake 2: Confusing Geometrical Isomerism with Optical Isomerism

• The Mistake: Students often try to draw non-superimposable mirror images (optical isomers) for this complex, even though it is not chiral.
• Why it happens: The complex has two identical ligands ($NH_3$) and four identical ligands ($CN$). In an octahedral $[Ma_2b_4]$ type complex, the only possible geometrical isomers are cis and trans. There is no chiral center.
• How to avoid: Memorize the condition: For geometrical isomerism in octahedral complexes, you need at least two different types of ligands. For $[Ma_2b_4]$, only cis and trans exist. Do not draw mirror images unless the complex is $[M(AA)_3]$ or $[Ma_2b_2c_2]$ type.

Mistake 3: Drawing Only One Isomer (Usually the Trans)

• The Mistake: Students draw only the trans isomer (with $NH_3$ groups opposite each other) and forget the cis isomer (with $NH_3$ groups adjacent).
• Why it happens: The trans isomer looks more symmetric and is often the first one that comes to mind.
• How to avoid: Systematically place the two identical ligands ($NH_3$) in all possible distinct positions. In an octahedron:
  • Trans: $NH_3$ at positions 1 and 6 (opposite ends of an axis).
  • Cis: $NH_3$ at positions 1 and 2 (adjacent corners).
  • Any other arrangement is equivalent to one of these two by rotation.

Mistake 4: Drawing Incorrect Bond Angles or Ligand Positions

• The Mistake: Students draw the cis isomer with the two $NH_3$ groups at a 90° angle but then place the $CN$ groups incorrectly, making the structure look like a square planar complex or with distorted octahedral geometry.
• Why it happens: Lack of practice in drawing 3D octahedral structures on paper.
• How to avoid: Use the standard wedge-dash notation:
  • Draw a square plane (four $CN$ groups in the cis isomer; two $CN$ and two $NH_3$ in the trans isomer).
  • Place the remaining two ligands above and below the plane (axial positions).
  • For the cis isomer: the two $NH_3$ groups are on adjacent corners of the square plane.
  • For the trans isomer: the two $NH_3$ groups are at 180° — e.g. both on the axial positions. (An axial + equatorial pair is only 90° apart — that is cis, not trans.)

Mistake 5: Forgetting the Charge and Oxidation State

• The Mistake: Students draw the complex without considering the charge, leading to incorrect ligand placement or ignoring that $CN^{-}$ is a strong field ligand.
• Why it happens: The focus is only on the shape.
• How to avoid: Always note the overall charge ($-1$). Here, $Fe$ is in the $+3$ oxidation state (since $NH_3$ is neutral and $CN^{-}$ is $-1$ each: $x + 0 + 4(-1) = -1 \Rightarrow x - 4 = -1 \Rightarrow x = +3$). This affects the electronic configuration and stability of isomers, but for drawing geometrical isomers, it ensures you use the correct number of each ligand.

Correct Structures (for reference)

Trans isomer:

• Two $NH_3$ groups are opposite each other (e.g., axial positions).
• Four $CN$ groups occupy the equatorial square plane.

Cis isomer:

• Two $NH_3$ groups are adjacent (e.g., both in the equatorial plane, next to each other).
• Four $CN$ groups occupy the remaining four positions (two in the plane, two axial).

Key takeaway: For $[Fe(NH_3)_2(CN)_4]^{-}$, there are exactly two geometrical isomers: cis and trans. No optical isomers exist. Always verify the coordination number and ligand arrangement before drawing.