Q. \( \text{NH}_3 \) bond polarity

Q: Is the molecule \mathrm{NH_3} overall polar?

Yes. \mathrm{NH_3} has a nonzero net dipole due to its trigonal pyramidal shape, caused by a lone pair on nitrogen.

Q: Why doesn’t the \mathrm{N-H} bond polarity cancel out in \mathrm{NH_3} ?

In a symmetric shape (like \mathrm{NH_3} would need a planar arrangement) dipoles could cancel. But trigonal pyramidal geometry makes dipoles add up rather than cancel.

Q: How can VSEPR predict the polarity of \mathrm{NH_3} ?

VSEPR gives \mathrm{AX_3E_1} , a trigonal pyramidal shape. Since it is not symmetric, the net dipole is nonzero, so the molecule is polar.

Q: What comparison shows bond vs molecular polarity in \mathrm{NH_3} ?

Each \mathrm{N-H} bond is polar, but molecular polarity depends on shape. Trigonal pyramidal symmetry is insufficient to cancel bond dipoles, so \mathrm{NH_3} is polar overall.

Answer

NH₃ (ammonia) has three N–H polar bonds because nitrogen is more electronegative than hydrogen. The molecule is trigonal pyramidal and polar overall because the N has a lone pair, so the bond dipoles do not cancel completely.

Final result: N–H bonds are polar and NH₃ is a net dipole (overall polar) molecule.

Detailed Explanation

Step-by-step: Is the \( \text{NH}_3 \) bond polar?

\( \text{NH}_3 \) (ammonia) contains an \( \text{N} \text{–} \text{H} \) bond. The question asks about the bond polarity (whether the individual \( \text{N} \text{–} \text{H} \) bonds are polar).

1) Identify the atoms and their electronegativities

To judge bond polarity, compare electronegativity values (how strongly atoms pull shared electrons).

Typical electronegativities (on the Pauling scale) are:

\( \text{N} \approx 3.0 \)
\( \text{H} \approx 2.2 \)

2) Compare electronegativity difference for the \( \text{N-H} \) bond

The electronegativity difference is:

\[
\Delta \chi = \chi(\text{N}) – \chi(\text{H})
\]
\[
\Delta \chi \approx 3.0 – 2.2 = 0.8
\]

A difference around \(0.4\) to \(1.7\) generally indicates a polar covalent bond (electrons are shared unequally).

3) Determine which side is negative

Because nitrogen is more electronegative than hydrogen, the shared electron density is pulled toward nitrogen.

Therefore, each \( \text{N-H} \) bond has:

\(\text{Nitrogen}\) partially negative, written as \( \delta^- \)
\(\text{Hydrogen}\) partially positive, written as \( \delta^+ \)

Conclusion

Yes. Each \( \text{N-H} \) bond in \( \text{NH}_3 \) is polar covalent because nitrogen is more electronegative than hydrogen.

If you also care about the overall molecule (not required by your prompt): \( \text{NH}_3 \) is overall polar as well due to its bent shape and the lone pair on nitrogen.

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General Chemistry FAQs

Is the \( \mathrm{N-H} \) bond in \( \mathrm{NH_3} \) polar?

Yes. Nitrogen is more electronegative than hydrogen, so electron density shifts toward N. The \( \mathrm{N-H} \) bond has partial charges, making each bond polar.

Is the molecule \( \mathrm{NH_3} \) overall polar?

Yes. \( \mathrm{NH_3} \) has a nonzero net dipole due to its trigonal pyramidal shape, caused by a lone pair on nitrogen.

Why doesn’t the \( \mathrm{N-H} \) bond polarity cancel out in \( \mathrm{NH_3} \)?

In a symmetric shape (like \( \mathrm{NH_3} \) would need a planar arrangement) dipoles could cancel. But trigonal pyramidal geometry makes dipoles add up rather than cancel.

What role does the lone pair on nitrogen play in \( \mathrm{NH_3} \) polarity?

The lone pair repels bonding pairs, bending the geometry into a trigonal pyramid. This creates an asymmetric charge distribution and contributes to the molecular dipole.

What is the molecular geometry and expected dipole direction in \( \mathrm{NH_3} \)?

Geometry is trigonal pyramidal. The dipole points from hydrogen toward nitrogen, because electron density is greater near N.

How can VSEPR predict the polarity of \( \mathrm{NH_3} \)?

VSEPR gives \( \mathrm{AX_3E_1} \), a trigonal pyramidal shape. Since it is not symmetric, the net dipole is nonzero, so the molecule is polar.

What comparison shows bond vs molecular polarity in \( \mathrm{NH_3} \)?

Each \( \mathrm{N-H} \) bond is polar, but molecular polarity depends on shape. Trigonal pyramidal symmetry is insufficient to cancel bond dipoles, so \( \mathrm{NH_3} \) is polar overall.

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