The number of acyclic structural isomers (including geometrical isomers) for pentene are ___.

For pentene ($$\text{C}_5\text{H}_{10}$$), we need to count all acyclic structural isomers including geometrical (cis-trans) isomers.

The possible structures are: (1) pent-1-ene: $$\text{CH}_2{=}\text{CH-CH}_2\text{-CH}_2\text{-CH}_3$$, (2) trans-pent-2-ene: $$\text{CH}_3\text{-CH}{=}\text{CH-CH}_2\text{-CH}_3$$, (3) cis-pent-2-ene, (4) 2-methylbut-1-ene: $$\text{CH}_2{=}\text{C(CH}_3\text{)-CH}_2\text{-CH}_3$$, (5) 3-methylbut-1-ene: $$\text{CH}_2{=}\text{CH-CH(CH}_3\text{)}_2$$, and (6) 2-methylbut-2-ene: $$\text{CH}_3\text{-C(CH}_3\text{)}{=}\text{CH-CH}_3$$.

Pent-1-ene has no geometrical isomers because one carbon of the double bond bears two hydrogen atoms. Pent-2-ene has both cis and trans forms since each doubly-bonded carbon has two different groups. 2-methylbut-1-ene has no geometrical isomers (terminal $$=\text{CH}_2$$). 3-methylbut-1-ene has no geometrical isomers (terminal $$=\text{CH}_2$$). 2-methylbut-2-ene has no geometrical isomers because one carbon bears two methyl groups.

The total count is $$6$$.