Answer

Verified

448.5k+ views

**Hint:**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 = $3.33564\times {{10}^{-30}}$ 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 $N{{H}_{3}}$ and $N{{F}_{3}}$ have trigonal pyramidal shape.

The dipole moment of lone pairs in $N{{H}_{3}}$ and $N{{F}_{3}}$ is away from nitrogen.

Dipole moment of $N{{H}_{3}}$

We know that nitrogen is more electronegative than hydrogen. Therefore, the dipole moment of $N-H$bond will be form H to N. The net dipole moment of three $N-H$ bond will add up to 1.4 D. As we know that 1 D = $3.33564\times {{10}^{-34}}$ C m.

Then, 1.4 D will be equal to $1.4\times 3.33564\times {{10}^{-30}}$ C m, i.e. $4.90\times {{10}^{-30}}$ C m.

Dipole moment of $N{{F}_{3}}$

Electronegativity of F is more than that of N, thus the direction of dipole moment of $N-F$ bond will be from F to N. As we can see that the direction of $N-F$ bond is opposite to that of the lone pair on N atoms. So, the net dipole moment of $N{{F}_{3}}$ has been found to be 0.24 D.

Multiplying 0.24 D with $3.33564\times {{10}^{-30}}$ C m, we get the dipole moment of $0.80\times {{10}^{-30}}$C m.

**So, the correct answer is “Option A”.**

**Note:**The dipole moment of lone pairs in $N{{H}_{3}}$ and $N{{F}_{3}}$ is away from nitrogen.

Dipole moment of $N{{H}_{3}}$

We know that nitrogen is more electronegative than hydrogen. Therefore, the dipole moment of $N-H$bond will be form H to N. The net dipole moment of three $N-H$ bond will add up to 1.4 D. As we know that 1 D = $3.33564\times {{10}^{-34}}$ C m.

Then, 1.4 D will be equal to $1.4\times 3.33564\times {{10}^{-30}}$ C m, i.e. $4.90\times {{10}^{-30}}$ C m.

Dipole moment of $N{{F}_{3}}$

Electronegativity of F is more than that of N, thus the direction of dipole moment of $N-F$ bond will be from F to N. As we can see that the direction of $N-F$ bond is opposite to that of the lone pair on N atoms. So, the net dipole moment of $N{{F}_{3}}$ has been found to be 0.24 D.

Multiplying 0.24 D with $3.33564\times {{10}^{-30}}$ C m, we get the dipole moment of $0.80\times {{10}^{-30}}$C m.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Mark and label the given geoinformation on the outline class 11 social science CBSE

When people say No pun intended what does that mea class 8 english CBSE

Name the states which share their boundary with Indias class 9 social science CBSE

Give an account of the Northern Plains of India class 9 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Trending doubts

Difference Between Plant Cell and Animal Cell

In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE

Which are the Top 10 Largest Countries of the World?

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

How do you graph the function fx 4x class 9 maths CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths