What occurs in the following reaction?
$N H_{4}^{+} + H_{2} O ⇋ N H_{3} + H_{3} O^{+}$

Electron transfer

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Neutralization

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Double replacement

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Proton transfer

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Solution

The correct option is D Proton transfer
Option E is the correct answer.
$N H_{4}^{+}$ $+$ $H_{2} O$ $⇋$ $N H_{3}$ $+$ $H_{3} O^{+}$ .
In this reversible reaction, a proton or $H^{+}$ gets transferred from $N H_{4}^{+}$ to $H_{2} O$ . Thus, they form $N H_{3}$ and $H_{3} O^{+}$ , after the proton is transferred from $N H_{4}^{+}$ to $H_{2} O$ . Because after the removal of proton, $N H_{4}^{+}$ forms $N H_{3}$ , and Addition of that proton to $H_{2} O$ forms $H_{3} O^{+}$ .

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Similar questions

{N H}_{4}^{+} + H_{2} O ⇋ {N H}_{3} + H_{3} O^{+}

{N H}_{4}^{+} + H_{2} O ⇋ {N H}_{3} + H_{3} O^{+}

N H_{4}^{+} + 2 H_{2} O ⇌ N H_{4} O H + H_{3} O^{+}

K_{1} = 5.56 \times 10^{- 10}

H_{2} O + H_{2} O ⇌ H_{3} O^{+} + O H^{-}

K_{2} = 1 \times 10^{- 14}

K_{3}

N H_{4} O H ⇌ N H_{4}^{+} + O H^{-}

Which of the following conversions involve change in both shape and hybridisation?

a) CH4 -> C2H6

b) NH3 -> NH4+

c) BF3 -> BF4-

d) H2O ->H3O+

N H_{3} + H_{2} O \to N H_{4}^{+} + O H^{-}

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