A) \[H-O-\underset{H}{\mathop{\overset{O}{\mathop{\overset{\uparrow }{\mathop{\underset{|}{\mathop{\underset{O}{\mathop{\underset{|}{\mathop{P}}\,}}\,}}\,}}\,}}\,}}\,-O-H\]
B) \[O\leftarrow \underset{H}{\mathop{\overset{H}{\mathop{\overset{|}{\mathop{\underset{\begin{smallmatrix} | \\ O \\ | \end{smallmatrix}}{\mathop{P}}\,}}\,}}\,}}\,-O-H\]
C) \[O\leftarrow \underset{H}{\mathop{\overset{H}{\mathop{\overset{|}{\mathop{\underset{|}{\mathop{P}}\,}}\,}}\,}}\,-O-H\]
D) \[H-\overset{O}{\mathop{\overset{|}{\mathop{OP}}\,}}\,=O\]
Correct Answer: A
Solution :
Ortho phosphoric acid \[({{H}_{3}}P{{O}_{4}})\] is a tribasic acid. Hence its: structure can be represented as \[O\leftarrow P{{(OH)}_{3}}\]. \[H-O-\overset{O}{\mathop{\overset{\uparrow }{\mathop{\underset{H}{\mathop{\underset{|}{\mathop{\underset{O}{\mathop{\underset{|}{\mathop{P}}\,}}\,}}\,}}\,}}\,}}\,-O-H\] Hence, hybridisation of P in \[{{H}_{3}}P{{O}_{4}}\] is \[s{{p}^{3}}\] \[s{{p}^{3}}\]and thus it is tetrahedral in shape.
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