In $H S^{-}, I^{-}, R - N H_{2}, N H_{3}$ , the order of proton accepting tendency will be:

I^{-} > N H_{3} > R - N H_{2} > H S^{-}

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N H_{3} > R - N H_{2} > H S^{-} > I^{-}

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R - N H_{2} > N H_{3} > H S^{-} > I^{-}

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H S^{-} > R - N H_{2} > N H_{3} > I^{-}

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Solution

The correct option is C $R - N H_{2} > N H_{3} > H S^{-} > I^{-}$
The order of proton accepting tendency will be
$R - N H_{2} > N H_{3} > H S^{-} > I^{-}$
Iodide ion has least ability to accept proton as the negative charge is spread on bigger iodide ion. This results in lower charge density. Alkyl amine has greater tendency to accept proton than ammonia due to +I effect of alkyl group.

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