How many geometrical isomers are possible in the following two alkenes? (I) CH3–CH=CH–CH=CH–CH3 (II) CH3–CH=CH–CH=CH–Cl

This molecule has two double bonds. Let's label them as $\text{C}_2=\text{C}_3$ and $\text{C}_4=\text{C}_5$.

• For $\text{C}_2=\text{C}_3$: The $\text{C}_2$ is attached to $\text{CH}_3$ and $\text{H}$. The $\text{C}_3$ is attached to $\text{H}$ and $\text{CH}=\text{CH}-\text{CH}_3$. Both carbons have two different groups, so this double bond can exhibit cis-trans isomerism.
• For $\text{C}_4=\text{C}_5$: The $\text{C}_4$ is attached to $\text{CH}_3$ and $\text{H}$. The $\text{C}_5$ is attached to $\text{H}$ and $\text{CH}=\text{CH}-\text{CH}_3$. Both carbons have two different groups, so this double bond can exhibit cis-trans isomerism. This molecule is symmetrical. The general formula for the number of geometrical isomers for a symmetrical polyene with $n$ double bonds is $2^{n-1} + 2^{(n/2)-1}$ if $n$ is even. Here, $n=2$. Number of isomers $= 2^{2-1} + 2^{(2/2)-1} = 2^1 + 2^{1-1} = 2^1 + 2^0 = 2 + 1 = 3$. The three isomers are:

1. cis, cis
2. trans, trans
3. cis, trans (which is identical to trans, cis due to symmetry)

This molecule also has two double bonds. Let's label them as $\text{C}_2=\text{C}_3$ and $\text{C}_4=\text{C}_5$.

• For $\text{C}_2=\text{C}_3$: The $\text{C}_2$ is attached to $\text{CH}_3$ and $\text{H}$. The $\text{C}_3$ is attached to $\text{H}$ and $\text{CH}=\text{CH}-\text{Cl}$. Both carbons have two different groups, so this double bond can exhibit cis-trans isomerism.
• For $\text{C}_4=\text{C}_5$: The $\text{C}_4$ is attached to $\text{H}$ and $\text{CH}=\text{CH}-\text{CH}_3$. The $\text{C}_5$ is attached to $\text{H}$ and $\text{Cl}$. Both carbons have two different groups, so this double bond can exhibit cis-trans isomerism. This molecule is unsymmetrical. For an unsymmetrical polyene with $n$ double bonds, the number of geometrical isomers is $2^n$. Here, $n=2$. Number of isomers $= 2^2 = 4$. The four isomers are:

1. cis, cis
2. cis, trans
3. trans, cis
4. trans, trans

Molecule (i) has 3 geometrical isomers.
Molecule (ii) has 4 geometrical isomers.
The correct option should be 3 and 4.

The final answer is $\boxed{\text{(4) 3 and 4}}$.