The type of isomerism not possible with the molecular formula ${\mathrm{C}}_4{\mathrm{H}}_{10}\mathrm{O}$ is

The existence of ${\mathrm{CH}}_3-{\mathrm{CH}}_2-{\mathrm{CH}}_2-{\mathrm{CH}}_2\mathrm{OH}$ and $\mathrm{CH}{\left({\mathrm{CH}}_3\right)}_2-{\mathrm{CH}}_2\mathrm{OH}$ show that chain isomerism is possible for ${\mathrm{C}}_4{\mathrm{H}}_{10}\mathrm{O}$.
Functional isomerism is possible in ${\mathrm{C}}_4{\mathrm{H}}_{10}\mathrm{O}$ as the alcohol group could shift constructively keeping the carbon chain similar.

To show geometrical it must be either be  unsaturated or cyclic.