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 $



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