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Question: What is the correct dipole moment of \(N{{H}_{3}}\) and \(N{{F}_{3}}\) respectively? (A)- \(4.90\t...

What is the correct dipole moment of NH3N{{H}_{3}} and NF3N{{F}_{3}} respectively?
(A)- 4.90×10−304.90\times {{10}^{-30}} C m and 0.80×10−300.80\times {{10}^{-30}} C m
(B)- 0.80×10−300.80\times {{10}^{-30}} C m and 4.90×10−304.90\times {{10}^{-30}} C m
(C)- 4.90×10−304.90\times {{10}^{-30}} C m and 4.90×10−304.90\times {{10}^{-30}} C m
(D)- 0.80×10−300.80\times {{10}^{-30}} C m and 0.80×10−300.80\times {{10}^{-30}} C m

Explanation

Solution

Dipole moment is a measure of polarity of a bond. It is the product of the charges and the distance between partial charges. It is a vector quantity and its direction is always given from less electronegative atom to more electronegative atom.
It is generally expressed in debye (D) and 1 D = 3.33564×10−303.33564\times {{10}^{-30}} C m.
Dipole moment of polar molecules containing lone pairs is the vector sum of dipole of lone pair and net dipole moments of bonds.

Complete answer:
Both NH3N{{H}_{3}} and NF3N{{F}_{3}} have trigonal pyramidal shape.

The dipole moment of lone pairs in NH3N{{H}_{3}} and NF3N{{F}_{3}} is away from nitrogen.
Dipole moment of NH3N{{H}_{3}}
We know that nitrogen is more electronegative than hydrogen. Therefore, the dipole moment of N−HN-Hbond will be form H to N. The net dipole moment of three N−HN-H bond will add up to 1.4 D. As we know that 1 D = 3.33564×10−343.33564\times {{10}^{-34}} C m.
Then, 1.4 D will be equal to 1.4×3.33564×10−301.4\times 3.33564\times {{10}^{-30}} C m, i.e. 4.90×10−304.90\times {{10}^{-30}} C m.

Dipole moment of NF3N{{F}_{3}}
Electronegativity of F is more than that of N, thus the direction of dipole moment of N−FN-F bond will be from F to N. As we can see that the direction of N−FN-F bond is opposite to that of the lone pair on N atoms. So, the net dipole moment of NF3N{{F}_{3}} has been found to be 0.24 D.
Multiplying 0.24 D with 3.33564×10−303.33564\times {{10}^{-30}} C m, we get the dipole moment of 0.80×10−300.80\times {{10}^{-30}}C m.
So, the correct answer is “Option A”.

Note: The dipole moment of lone pairs in NH3N{{H}_{3}} and NF3N{{F}_{3}} is away from nitrogen.
Dipole moment of NH3N{{H}_{3}}
We know that nitrogen is more electronegative than hydrogen. Therefore, the dipole moment of N−HN-Hbond will be form H to N. The net dipole moment of three N−HN-H bond will add up to 1.4 D. As we know that 1 D = 3.33564×10−343.33564\times {{10}^{-34}} C m.
Then, 1.4 D will be equal to 1.4×3.33564×10−301.4\times 3.33564\times {{10}^{-30}} C m, i.e. 4.90×10−304.90\times {{10}^{-30}} C m.

Dipole moment of NF3N{{F}_{3}}
Electronegativity of F is more than that of N, thus the direction of dipole moment of N−FN-F bond will be from F to N. As we can see that the direction of N−FN-F bond is opposite to that of the lone pair on N atoms. So, the net dipole moment of NF3N{{F}_{3}} has been found to be 0.24 D.
Multiplying 0.24 D with 3.33564×10−303.33564\times {{10}^{-30}} C m, we get the dipole moment of 0.80×10−300.80\times {{10}^{-30}}C m.