Isomerism

The compound has two double bonds:

$ \mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_2 $

Only the central $\mathrm{C}=\mathrm{C}$ (second one) can show geometrical isomerism, as the terminal alkene ( $\mathrm{CH}=\mathrm{CH}_2$ ) lacks two different groups on one carbon.

So, only one C=C shows cis-trans isomerism $\rightarrow 2$ isomers

Although $ \mathrm{CH}_{3} \mathrm{C}=\mathrm{CH}-\mathrm{CH}=\mathrm{CH}_{2} $ has two double bond but one of them has two similar atoms (i.e. H -atom) therefore no geometrical isomerism is possible around the double bond. However each carbon atom of other double bond has two different atoms/groups and hence geometrical isomerism is possible around this double bond.

Hence, only two geometrical isomers are possible.

If two compounds have the same empirical formula but different molecular formulae, they must have different molecular weights. The empirical formula represents the simplest whole-number ratio of the elements in the compound, but it doesn't provide information about the actual number of atoms in a molecule (molecular formula). Because the molecular formula is a multiple of the empirical formula, compounds with different molecular formulae will have different molecular weights even if their empirical formula is the same.

To illustrate with an example, consider the empirical formula $CH_2$. This empirical formula can correspond to several different molecular formulae such as $C_2H_4$, $C_3H_6$, $C_4H_8$, and so on. Each of these molecular formulae has a different molecular weight, even though they share the same empirical formula.

Therefore, the correct option is:

Option B: different molecular weights

According to sequence rule, the priority order are

$\mathop {Cl}\limits_{(1)} $ > $\mathop {{C_2}{H_5}}\limits_{(2)} $ > $\mathop {C{H_3}}\limits_{(3)} $ > $\mathop H\limits_{(4)} $, so in R configuration.