Chemical Bonding And Molecular Structure Question 434
Question: Dipole moment of $ H_2O $ is 1.85 D. If the bond angle is $ ~105{}^\circ $ and $ O-H $ bond length is $ ~0.94\overset{o}{\mathop{A}}, $ , what is the magnitude of charge on the oxygen atom in water molecule-
Options:
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 $
Show Answer
Answer:
Correct Answer: C
Solution:
- $ \mu =1.85,D=1.85\times {10^{-18}}esu,cm=q\times d $
$ cos52.5{}^\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 $
