Q. $N H_{3}$ and $H_{2} O$ form $N H_{4} O H$ by

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A

electrovalent bond

32%

B

covalent bond

29%

C

coordinate bond

37%

D

none of these

2%

Solution:

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