A) \[H-O-\overset{\begin{smallmatrix} O \\ \uparrow \end{smallmatrix}}{\mathop{\underset{\begin{align} & | \\ & O \\ & | \\ & H \\ \end{align}}{\mathop{P}}\,}}\,-O-H\]
B) \[O\leftarrow \overset{\begin{smallmatrix} O \\ | \end{smallmatrix}}{\mathop{\underset{\begin{align} & | \\ & O \\ & | \\ & H \\ \end{align}}{\mathop{P}}\,}}\,-O-H\]
C) \[O\leftarrow \overset{\begin{smallmatrix} H \\ | \end{smallmatrix}}{\mathop{\underset{\begin{smallmatrix} | \\ H \end{smallmatrix}}{\mathop{P}}\,}}\,-O-H\]
D) \[H-OP=O\]
Correct Answer: A
Solution :
Ortho phosphoric acid \[({{H}_{3}}P{{O}_{4}})\] is a tribasic acid. Hence its structure can be represented as \[O\leftarrow P{{(OH)}_{3}}\]. \[H-O-\overset{\begin{smallmatrix} O \\ \uparrow \end{smallmatrix}}{\mathop{\underset{\begin{align} & | \\ & O \\ & | \\ & H \\ \end{align}}{\mathop{P}}\,}}\,-O-H\] \[(lp+\sigma bp=1+3=4)\] Hence, hybridisation of P in \[{{H}_{3}}P{{O}_{4}}\] is \[s{{p}^{3}}\] and thus it is tetrahedral in shape.
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