1. Interpret the question

We need the number of acyclic structural isomers of pentene, and we must include geometrical isomers.

Pentene means an acyclic alkene with molecular formula: $C_5H_{10}$

So we list all open-chain alkene isomers with one double bond, then count any cis/trans (E/Z) isomers where possible.

2. List all acyclic structural isomers of $C_5H_{10}$ having one double bond

Possible carbon skeletons:

(i) Straight chain pentenes

1. Pent-1-ene

Structure: $CH_2=CH-CH_2-CH_2-CH_3$ This does not show geometrical isomerism because one double-bond carbon has two H atoms.

Count = $1$

2. Pent-2-ene

Structure: $CH_3-CH=CH-CH_2-CH_3$ Here each double-bond carbon has two different groups, so it shows geometrical isomerism:

• cis-pent-2-ene
• trans-pent-2-ene

Count = $2$

(ii) Branched chain alkenes

3. 2-Methylbut-1-ene

Structure: $CH_2=C(CH_3)-CH_2-CH_3$ No geometrical isomerism, since one double-bond carbon is $CH_2$.

Count = $1$

4. 3-Methylbut-1-ene

Structure: $CH_2=CH-CH(CH_3)-CH_3$ No geometrical isomerism, since one double-bond carbon is $CH_2$.

Count = $1$

5. 2-Methylbut-2-ene

Structure: $CH_3-C(CH_3)=CH-CH_3$ No geometrical isomerism, because one double-bond carbon has two identical groups ($CH_3, CH_3$).

Count = $1$

3. Total count

Adding all distinct acyclic isomers including geometrical isomers: $1 + 2 + 1 + 1 + 1 = 6$

4. Final answer

The number of acyclic structural isomers (including geometrical isomers) of pentene is: $\boxed{6}$

5. Comparison with stored correct answer

Stored correct answer = $6$

Our derived answer = $6$

So they agree.