How is $N H_{3}$ $NH_3$ acting in the reaction $H^{+} + : N H_{3} \to {[H : N H_{3}]}_{3}^{+}$ $H^++:NH_3 ->[H:NH_3]^+$ according to the Lewis definition?

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How is $N H_{3}$ $NH_3$ acting in the reaction $H^{+} + : N H_{3} \to {[H : N H_{3}]}_{3}^{+}$ $H^++:NH_3 ->[H:NH_3]^+$ according to the Lewis definition?

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Answer 1 · 2015-01-04T00:22:06

$N H_{3}$ $NH_3$ is acting as an electron donor, so by definition it is a Lewis base.

The lone pair of elctrons from the $N$ $N$ -atom is (mainly) donated to, are shared with, the proton ( $H^{+}$ $H^+$ ), after which all four $H$ $H$ 's fill up a nice tetrahedron around the $N$ $N$ .

The representation of ${[H : N H_{3}]}_{3}^{+}$ $[H:NH_3]^+$ is not really correct, as all $H$ $H$ -atoms are completely equal, and the electron deficit (the overall +) is shared around the whole ${[N H_{4}]}_{4}^{+}$ $[NH_4]^+$ ammonium -ion.

If the ammonium-ion is dissociated (e.g. in hot alkalic surroundings), there is no telling which of the four $H$ $H$ 's will leave.

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How is $N H_{3}$ $NH_3$ acting in the reaction $H^{+} + : N H_{3} \to {[H : N H_{3}]}_{3}^{+}$ $H^++:NH_3 ->[H:NH_3]^+$ according to the Lewis definition?