Arrange the following compounds in decreasing | vedclass

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(B) The correct order of acidity is: $H_{2}O > ROH > HC \equiv CH$.$1$. $H_{2}O$: The oxygen atom is highly electronegative,making the $O-H$ bond high

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$H_{2}O > HC \equiv CH > ROH$

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(B) The correct order of acidity is: $H_{2}O > ROH > HC \equiv CH$.$1$. $H_{2}O$: The oxygen atom is highly electronegative,making the $O-H$ bond highly polar and the resulting hydroxide ion $(OH^-)$ relatively stable.$2$. $ROH$: The alkyl group $(R)$ is electron-donating ($+I$ effect),which destabilizes the alkoxide ion $(RO^-)$ formed after the loss of a proton,making alcohols less acidic than water.$3$. $HC \equiv CH$: The terminal hydrogen is attached to an $sp$-hybridized carbon. Although $sp$-carbon is more electronegative than $sp^3$ or $sp^2$ carbons,it is still less electronegative than oxygen. Therefore,ethyne is the least acidic among the three.

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

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