Q.Write the coordination isomer of $[Cu(NH_3)_4][PtCl_4]$.

Geometrical Isomerism: The Intuition

Imagine you have two friends standing on opposite sides of a door. If the door is open, they can walk around and swap places easily — there's no real difference between who is on the left and who is on the right. But if the door is locked shut, they are stuck. One is permanently on the left side, the other on the right. That locked door creates two distinct arrangements: Friend A on the left, Friend B on the right versus Friend A on the right, Friend B on the left.

That locked door is the key idea behind geometrical isomerism.

In chemistry, molecules are three-dimensional. Atoms connected by a single bond can rotate freely — like an open door. But a double bond (or a ring structure) locks the atoms in place. If you have two different groups attached to each carbon of a double bond, you get two distinct spatial arrangements that cannot interconvert without breaking the bond. These are geometrical isomers (also called cis-trans or E-Z isomers).

The Precise Conditions

For a molecule to show geometrical isomerism, it must satisfy two conditions simultaneously:

Let's unpack each.

Condition 1: Restricted Rotation

A single bond ($C-C$) allows free rotation — the atoms spin around the bond axis like a wheel. So no geometrical isomers exist there. A double bond ($C=C$) has a pi ($\pi$) bond that locks the molecule flat. Rotation would break the $\pi$ bond, which requires a lot of energy (about 250–270 kJ/mol). At room temperature, this rotation simply does not happen.

Rings (like cyclopropane, cyclobutane, etc.) also restrict rotation because the ring is a closed loop — atoms cannot rotate past each other without breaking the ring.

Condition 2: Two Different Substituents on Each End

This is the "different groups" rule. Look at each carbon of the double bond (or each ring carbon involved). If both carbons have two different groups attached, geometrical isomers exist. If even one carbon has two identical groups, there is only one possible arrangement.

How to Check: A Step-by-Step Method

Take any molecule with a double bond. Follow these steps:

1. Identify the double bond (or ring). Mark the two carbon atoms involved.
2. List the two groups attached to the first carbon. Are they different from each other? If yes, proceed. If no → no geometrical isomerism.
3. List the two groups attached to the second carbon. Are they different from each other? If yes → geometrical isomerism exists. If no → no geometrical isomerism.

Examples to Cement the Idea

Example 1: But-2-ene ($CH_3CH=CHCH_3$)

• Carbon 1 of the double bond: attached to $CH_3$ and $H$ → different ✓
• Carbon 2 of the double bond: attached to $CH_3$ and $H$ → different ✓

Result: Two geometrical isomers exist — cis (both methyl groups on the same side) and trans (methyl groups on opposite sides).

Example 2: 1,2-Dichloroethene ($ClCH=CHCl$)

• Carbon 1: attached to $Cl$ and $H$ → different ✓
• Carbon 2: attached to $Cl$ and $H$ → different ✓

Result: cis and trans isomers exist.

Example 3: 1,1-Dichloroethene ($Cl_2C=CH_2$)

• Carbon 1: attached to $Cl$ and $Cl$ → identical ✗ …