Q.Explain why $BeH_2$ molecule has a zero dipole moment although the Be–H bonds are polar.

Concept & Intuition

A molecule's dipole moment is the vector sum of its bond dipoles. Perfect symmetry can cancel even strong bond dipoles entirely, so both the number of polar bonds and the geometry matter.

Step-by-step reasoning

1. CH$_4$ — tetrahedral, C–H bonds are only weakly polar, and the four bond dipoles cancel by symmetry. Net dipole ≈ 0.

2. CCl$_4$ — tetrahedral with four strongly polar C–Cl bonds, but the perfect tetrahedral symmetry still cancels them. Net dipole ≈ 0.

3. CHCl$_3$ (chloroform) — three C–Cl bonds and one C–H bond; the three C–Cl dipoles don't fully cancel, but their arrangement (109.5° apart) still leaves a comparatively modest net moment: ~1.04 D. …

Concept and Intuition

A molecule has a net dipole moment when the vector sum of all individual bond dipoles is nonzero. Symmetry often cancels dipoles: if a molecule has a center of inversion, a plane of symmetry that mirrors dipoles, or a high-order rotation axis that aligns dipoles oppositely, the net moment is zero. The trick is to visualize the 3D arrangement of polar bonds (like C–Cl or C–Br) and see if they point in opposite directions.

Step-by-Step Reasoning

1. Analyze cis-1,2-dichloroethene (A)

   • Structure: Two chlorine atoms on the same side of the C=C double bond.
   • Each C–Cl bond is polar (Cl more electronegative). In the cis isomer, both C–Cl dipoles point roughly in the same direction (downward if the double bond is horizontal).
   • No symmetry element cancels them; the vector sum is nonzero.
   • Result: Has a net dipole moment.

2. Analyze tetrachloromethane (CCl₄) (B)

   • Tetrahedral geometry with four identical C–Cl bonds.
   • The molecule is highly symmetric: all bond dipoles are equal in magnitude and point toward the corners of a tetrahedron. Their vector sum is exactly zero.
   • Result: No net dipole moment.

3. Analyze o-dichlorobenzene (C)

   • Two chlorine atoms on adjacent carbons of a benzene ring.
   • The C–Cl bond dipoles are not opposite; they are at a 60° angle (in the plane of the ring). Their vector sum is nonzero, pointing roughly between them.
   • The ring itself is symmetric, but the two chlorines break that symmetry.
   • Result: Has a net dipole moment. …

Concept & Intuition

Dipole moment is a vector quantity: it has both magnitude and direction. Each polar substituent on a benzene ring contributes a bond dipole (a vector pointing from the less electronegative atom toward the more electronegative one). The net dipole moment of the molecule is the vector sum of these individual dipoles.

• The nitro group (–NO₂) is strongly electron-withdrawing; its dipole points away from the ring (the nitrogen is positive relative to the oxygens, but the overall group pulls electron density from the ring, so the bond dipole vector points from the ring toward the NO₂ group).
• The amino group (–NH₂) is electron-donating; its dipole points toward the ring (the lone pair on nitrogen donates into the ring, so the bond dipole vector points from the NH₂ toward the ring).

Thus, in a nitroaniline, the two dipoles are in opposite directions relative to the ring. Their relative positions (ortho, meta, para) determine whether they reinforce or cancel each other.

Step-by-step reasoning

1. Identify the bond dipole directions

   For –NO₂: the dipole vector points from the ring carbon outward toward the NO₂ group.

   For –NH₂: the dipole vector points from the NH₂ group inward toward the ring carbon.

   So, if we draw arrows on the ring, the NO₂ arrow points away from the ring, and the NH₂ arrow points toward the ring.

2. Analyze each isomer

   • Option (A): Two NO₂ groups in para positions Both dipoles point outward, directly away from each other. They are equal in magnitude and exactly opposite in direction. Their vector sum is nearly zero (small residual due to ring polarization). So dipole moment is very small.
   • Option (B): ortho-nitroaniline (1,2-) The NO₂ dipole points outward; the NH₂ dipole points inward. The angle between them is 60° (since the ring is a regular hexagon). Vector addition gives partial cancellation. The net dipole is moderate.
   • Option (C): meta-nitroaniline (1,3-) The angle between the two dipoles is 120°. They partially cancel but less than in ortho; the net dipole is larger than ortho but still not maximal. …

Why this approach works

Dipole moment is a vector sum of individual bond polarities. Even if bonds are polar, a symmetric shape (like trigonal planar or tetrahedral with identical substituents) can cancel the vectors to zero. So we must check both molecular geometry and bond polarity for each molecule.

Step-by-step reasoning

1. Analyze NF₃

   • Nitrogen is more electronegative than fluorine, so each N–F bond is polar with the negative end toward F.
   • NF₃ has a trigonal pyramidal shape (lone pair on N), so the three bond dipoles do not cancel.
   • Result: NF₃ has a nonzero dipole moment (≈ 0.24 D).

2. Analyze BF₃

   • Boron is less electronegative than fluorine, so B–F bonds are polar.
   • BF₃ is trigonal planar (120° angles, no lone pair). The three identical bond dipoles sum to zero.
   • Result: BF₃ has zero dipole moment.

3. Analyze CHCl₃ (chloroform)

   • The C–H bond is nearly nonpolar; the three C–Cl bonds are strongly polar (Cl more electronegative).
   • The molecule is tetrahedral, but the three Cl atoms are not symmetric with the H atom — the vector sum of the three C–Cl dipoles does not cancel with the small C–H dipole.
   • Result: CHCl₃ has a nonzero dipole moment (≈ 1.04 D).

4. Analyze CCl₄ (carbon tetrachloride)

   • Four identical C–Cl bonds arranged tetrahedrally.
   • Perfect symmetry: all bond dipoles cancel exactly. …

Evaluating each option:

• (A) Correct. In $\mathrm{NH_3}$ the orbital dipole due to the nitrogen lone pair points in the same direction as the vector sum of the three N–H bond dipoles, so they add ($\mu \approx 1.47\,\mathrm{D}$). (In $\mathrm{NF_3}$ they oppose, giving a small $\mu$.)
• (B) Wrong. $\mathrm{O_2}$ and $\mathrm{H_2}$ are homonuclear molecules; their bonds are non-polar, so no bond dipole exists.
• (C) Wrong. Dipole moment is a vector quantity (it has magnitude and direction). …

Concept: a molecule has zero net dipole moment when its bond dipoles cancel by symmetry.

1. CCl4 - tetrahedral, symmetric; the four C-Cl dipoles cancel -> mu = 0.
2. CO2 - linear O=C=O; the two equal C=O dipoles are opposite and cancel -> mu = 0.
3. NH3 - pyramidal (lone pair); dipoles do not cancel -> mu = 1.47 D.
4. CHCl3 - tetrahedral but one H replaces a Cl, so the dipoles do not cancel -> mu = 1.04 D. …

Concept: A covalent bond between two ATOMS OF THE SAME ELEMENT is non-polar (zero electronegativity difference); between different elements it is polar.

• HCN: H-C and C(triple)N - both polar; no bond between identical atoms. …