Question
In $$H{S^ - },{I^ - },RN{H_2},N{H_3}$$ order of proton accepting tendency will be
A.
$${I^ - } > N{H_3} > RN{H_2} > H{S^ - }$$
B.
$$N{H_3} > RN{H_2} > H{S^ - } > {I^ - }$$
C.
$$RN{H_2} > N{H_3} > H{S^ - } > {I^ - }$$
D.
$$H{S^ - } > RN{H_2} > N{H_3} > {I^ - }$$
Answer :
$$RN{H_2} > N{H_3} > H{S^ - } > {I^ - }$$
Solution :
Basic strength $$ \propto $$ rate of accepting a proton.
In $$R - \ddot N{H_2},N - $$ has lone pair of electron which increases the intensity due to electron releasing $$R$$ - group and increases the tendency to donate lone pair of electrons to $${H^ + }.$$ Secondly as the size of the $$ion$$ increases, there is less attraction for $${H^ + }$$ and form weaker bond with $$H-atom$$ and less basic. The order of the given series is
$$RN{H_2} > N{H_3} > H{S^ - } > {I^ - }$$