Arrange the following compounds in decreasing order of acidic character of $C-H$ bonds.
$CH_2=CH_2, CH_3-CH_3, C_6H_6, CH\equiv CH$

(A) The acidic character of a $C-H$ bond depends on the $s$-character of the carbon atom involved in the bond.
Greater $s$-character leads to higher electronegativity of the carbon atom,which stabilizes the resulting conjugate base (carbanion) and increases acidity.
$1$. $CH\equiv CH$ ($sp$ hybridized,$50\% \ s$-character)
$2$. $CH_2=CH_2$ ($sp^2$ hybridized,$33.3\% \ s$-character)
$3$. $C_6H_6$ ($sp^2$ hybridized,but resonance stabilized)
$4$. $CH_3-CH_3$ ($sp^3$ hybridized,$25\% \ s$-character)
Therefore,the decreasing order of acidic character is: $CH\equiv CH > CH_2=CH_2 > C_6H_6 > CH_3-CH_3$.