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Question: Dipole moment of \({H_2}O\) is \(1.85D\). If bond angle is \({1.5^ \circ }\)and \(O - H\) bond lengt...

Dipole moment of H2O{H_2}O is 1.85D1.85D. If bond angle is 1.5∘{1.5^ \circ }and O−HO - H bond length is 0.94A∘0.94\mathop A\limits^ \circ , Determine magnitude of the charge on the oxygen atom in the water molecule.
From the bond angle and vector momentμ=e×l\mu = e \times l .Thus, e (charge)can be determined in esuesu .
(A) 3.23×10−103.23 \times {10^{ - 10}}
(B) 5.43×10−105.43 \times {10^{ - 10}}
(C) 4.43×10−104.43 \times {10^{ - 10}}
(D) None of these

Explanation

Solution

Hint - To solve this problem we should understand about the Dipole moment , bond length by using this information we can easily approach our answer .In this question we have the dipole moment and bond length of water and we have to calculate charge on oxygen atom ,so we should know about the structure of water molecule.

Complete step by step solution:

Dipole moment: A dipole moment is produced in any system in which there is a separation of charge. They can, therefore, occur in ionic bonds as well as in covalent bonds. Dipole moments arise due to the difference in electro -negativity between two chemically bonded atoms. It is a vector quantity having both magnitude and direction.
Bond length: Bond length is defined as the average distance between nuclei of two bonded atoms in a molecule. It is also called bond distance.
Having bond length 0.94A∘0.94\mathop A\limits^ \circ and we know that the angle between the two hydrogen atom is 104.5∘{104.5^ \circ } , so this angle will be 53.5∘{53.5^ \circ } , so from here we know the dipole moment isμ=e×d\mu = e \times d .
So, the displacement distance will be bond length multiply by cos⁡52.5\cos 52.5 so the value comes out asd=lcos⁡52.5d = l\cos 52.5 ,d=0.572×10−8cm0.572 \times {10^{ - 8}}cm so, by using this information we can calculate the charge of atom.
d=l cos⁡52.5∘d = l\,\cos {52.5^ \circ }
=0.94×0.609=0.572A∘= 0.94 \times 0.609 = 0.572\mathop A\limits^ \circ
d=0.572×10−8cmd = 0.572 \times {10^{ - 8}}cm
Also,μ=e×d\mu = e \times d
∴e=μd=1.85×10−18esu cm0.572×10−8cm\therefore e = \dfrac{\mu }{d} = \dfrac{{1.85 \times {{10}^{ - 18}}esu\,cm}}{{0.572 \times {{10}^{ - 8}}cm}}
e=3.23×10−10esue = 3.23 \times {10^{ - 10}}esu

So, the correct answer is option A - 3.23×10−10 esu3.23 \times {10^{ - 10}}\,esu

Note - In water molecules the dipole moment is very important because it allows the molecule to be polar. The oxygen atom is holding the negative charge making the hydrogen atom positively charged, this allows hydrogen bonding of water molecules.