A) \[\left( \underset{\begin{smallmatrix} | \\ H \end{smallmatrix}}{\overset{\begin{smallmatrix} H \\ | \end{smallmatrix}}{\mathop{H-N\to H}}}\, \right)\]
B) \[\left( \underset{\begin{smallmatrix} | \\ H \end{smallmatrix}}{\overset{\begin{smallmatrix} H \\ | \end{smallmatrix}}{\mathop{H-N\leftarrow H}}}\, \right)\]
C) \[\left( \underset{\begin{smallmatrix} \uparrow \\ H \end{smallmatrix}}{\overset{\begin{smallmatrix} H \\ \downarrow \end{smallmatrix}}{\mathop{H\to N-H}}}\, \right)\]
D) \[\left( \underset{\begin{smallmatrix} \downarrow \\ H \end{smallmatrix}}{\overset{\begin{smallmatrix} H \\ \uparrow \end{smallmatrix}}{\mathop{H\leftarrow N\to H}}}\, \right)\]
Correct Answer: A
Solution :
The correct representation of\[NH_{4}^{+}\]ion is \[\left[ H-\underset{\begin{smallmatrix} | \\ H \end{smallmatrix}}{\overset{\begin{smallmatrix} H \\ | \end{smallmatrix}}{\mathop{N}}}\,\to H \right]\] This is because nitrogen\[(N{{H}_{3}})\]contains a lone pair of electron and donates it to\[{{H}^{+}}\]ion through coordinate bond.
You need to login to perform this action.
You will be redirected in
3 sec
