question_answer

Dipole moment of\[{{H}_{2}}O\] is 1.85 D. If the bond angle is \[~105{}^\circ \] and\[O-H\] bond length is\[~0.94\overset{\text{o}}{\mathop{\text{A}}}\,\], what is the magnitude of charge on the oxygen atom in water molecule?

A) \[2\times {{10}^{-10}}esu\]      

B) \[4.28\times {{10}^{-10}}esu\]

C) \[3.22\times {{10}^{-10}}esu\]

D) \[1.602\times {{10}^{-19}}C\]

Correct Answer: C

Solution :

[c] \[\mu =1.85\,D=1.85\times {{10}^{-18}}esu\,cm=q\times d\] \[cos\text{ }52.5{}^\circ =\frac{d}{0.94\overset{{}^\circ }{\mathop{A}}\,}\Rightarrow d=0.609\times 0.94\overset{{}^\circ }{\mathop{A}}\,\] \[=0.572\overset{{}^\circ }{\mathop{A}}\,\] \[\therefore \,\mu =q\times d\] \[{{q}_{1}}=\frac{\mu }{d}=\frac{1.85D}{0.572\overset{{}^\circ }{\mathop{A}}\,}\] \[=\frac{1.85\times {{10}^{-18}}esu\,cm}{0.572\times {{10}^{-8}}cm}\] \[=3.2\times {{10}^{-10}}esu\]