1. Condition for geometrical isomerism

Geometrical isomerism about a double bond is possible only when each carbon of the $C=C$ bond has two different substituents.

If any one of the double-bonded carbons has two identical groups, then geometrical isomerism is not possible.

2. Check each option

Option A: $1,2$-dichloro-$1$-pentene

Structure: $\mathrm{CHCl=C(Cl)-CH_2-CH_2-CH_3}$

• Carbon-1 of double bond has: $\mathrm{H}$ and $\mathrm{Cl}$
• Carbon-2 of double bond has: $\mathrm{Cl}$ and $\mathrm{C_3H_7}$

Both double-bonded carbons have two different groups.

$\Rightarrow$ Shows geometrical isomerism.

Option B: $1,3$-dichloro-$2$-pentene

Structure: $\mathrm{CH_2Cl-CH=C(Cl)-CH_2-CH_3}$

• Carbon-2 of double bond has: $\mathrm{H}$ and $\mathrm{CH_2Cl}$
• Carbon-3 of double bond has: $\mathrm{Cl}$ and $\mathrm{C_2H_5}$

Again, both carbons have two different substituents.

$\Rightarrow$ Shows geometrical isomerism.

Option C: $1,1$-dichloro-$1$-pentene

Structure: $\mathrm{CCl_2=CH-CH_2-CH_2-CH_3}$

• Carbon-1 of double bond has: $\mathrm{Cl}$ and $\mathrm{Cl}$
• Carbon-2 of double bond has: $\mathrm{H}$ and $\mathrm{C_3H_7}$

Here, one double-bond carbon has two identical substituents ($\mathrm{Cl}$ and $\mathrm{Cl}$).

$\Rightarrow$ Does not show geometrical isomerism.

Option D: $1,4$-dichloro-$2$-pentene

Structure: $\mathrm{CH_2Cl-CH=CH-CHCl-CH_3}$

• Carbon-2 of double bond has: $\mathrm{H}$ and $\mathrm{CH_2Cl}$
• Carbon-3 of double bond has: $\mathrm{H}$ and $\mathrm{CHClCH_3}$

Both double-bonded carbons have two different groups.

$\Rightarrow$ Shows geometrical isomerism.

3. Conclusion

The compound that does not show geometrical isomerism is: $\boxed{\text{C: }1,1\text{-dichloro-}1\text{-pentene}}$

4. Comparison with stored correct answer

Stored correct answer: C

Derived answer: C

They match.