Place the following in the correct order of acidity.

$\underset{\text{I}}{\text{CH}\equiv \text{C - C}{\text{H}}_2-\text{COOH}};\underset{\text{II}}{\text{C}{\text{H}}_{\text{2}}\text{= CH - C}{\text{H}}_{\text{2}}\text{COOH}}\text{;}\underset{\text{III}}{\text{C}{\text{H}}_{\text{3}}\text{C}{\text{H}}_{\text{2}}\text{C}{\text{H}}_{\text{2}}\text{COOH}}$

The correct option is A $I>II>III$
$\left(I\right)$ contains alkyne as well as carboxyl group, the alkyne having $sp$ hybridized $C$-the centre has high $e^{-}$ density which stabilizes the $-CO{O}^{-}$ group. Hence $H^{+}$ is easily releasable.

$\left(II\right)$ contains alkene$⟶$$s{p}^2$ hybridized centre$⟶$slightly less electronegative than $\left(I\right)$, hence less acidic compared to $\left(I\right)$

$\left(III\right)$ has an only carboxyl group. Does not have any alkyne or alkene to stabilize the $CO{O}^{-}$ group. Hence least acidic.

Hence the correct option is $A$