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Arrange the following compounds in decreasing order of acidity: $H_{2}O, ROH, HC \equiv CH$

Which of the following compounds do not have $sp^3$ carbon atom$(s)$?
$I$. Acetone
$II$. Acetic acid
$III$. Buta-$1,3$-diene
$IV$. Propyne
$V$. Naphthalene

In the reaction,the hybridisation states of carbon atoms $1, 2, 3, 4$ are:
$Br-CH=CH-Br \xrightarrow[Catalyst]{H_2} Br-CH_2-CH_2-Br$

The number of $sp^2$-hybridised carbon atoms in $HC\equiv C-CH_2-CO-CH_2-CH=CH_2$ is

The compound having both $sp$ and $sp^{2}$ hybridised carbon atoms is

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