Q.Which bond is more polar in the following pairs of molecules:

The polarity of a covalent bond is determined by the difference in electronegativity between the two bonded atoms. A larger electronegativity difference results in a more polar bond, as the electron density is unequally shared, creating partial positive ($\delta^+$) and partial negative ($\delta^-$) charges. Electronegativity generally increases across a period and decreases down a group in the periodic table.

Here is the analysis for each pair:

(a) H₃C–H vs H₃C–Br

1. We compare the C–H and C–Br bonds.
2. Carbon (C) has an electronegativity of $2.55$. Hydrogen (H) has $2.20$. Bromine (Br) has $2.96$.
3. The electronegativity difference for C–H is $|2.55 - 2.20| = 0.35$.
4. The electronegativity difference for C–Br is $|2.55 - 2.96| = 0.41$.
5. Since $0.41 > 0.35$, the C–Br bond is more polar.

(b) H₃C–NH₂ vs H₃C–OH

1. We compare the C–N and C–O bonds.
2. Carbon (C) has $2.55$. Nitrogen (N) has $3.04$. Oxygen (O) has $3.44$.
3. The electronegativity difference for C–N is $|2.55 - 3.04| = 0.49$.
4. The electronegativity difference for C–O is $|2.55 - 3.44| = 0.89$.
5. Since $0.89 > 0.49$, the C–O bond is more polar.

(c) H₃C–OH vs H₃C–SH

1. We compare the C–O and C–S bonds.
2. Carbon (C) has $2.55$. Oxygen (O) has $3.44$. Sulphur (S) has $2.58$.
3. The electronegativity difference for C–O is $|2.55 - 3.44| = 0.89$.
4. The electronegativity difference for C–S is $|2.55 - 2.58| = 0.03$.
5. Since $0.89 > 0.03$, the C–O bond is more polar.

Understanding bond polarity is fundamental in chemistry, as it influences a molecule's physical properties (like boiling point and solubility) and chemical reactivity. A bond is considered polar when there is an unequal sharing of electrons between the two bonded atoms. This unequal sharing arises from a difference in their electronegativity.

Electronegativity is a measure of an atom's ability to attract shared electrons in a chemical bond. When two atoms with different electronegativities form a bond, the more electronegative atom pulls the shared electron pair closer to itself. This creates a partial negative charge ($\delta^-$) on the more electronegative atom and a partial positive charge ($\delta^+$) on the less electronegative atom, resulting in a polar bond.

The greater the difference in electronegativity ($\Delta EN$) between two bonded atoms, the more polar the bond will be.

Let's use the Pauling electronegativity values for the relevant atoms to quantify the difference:

• Hydrogen (H): 2.20
• Carbon (C): 2.55
• Nitrogen (N): 3.04
• Oxygen (O): 3.44
• Sulphur (S): 2.58
• Bromine (Br): 2.96

Now, let's analyze each pair of molecules.

(a) H₃C–H vs. H₃C–Br

1. Identify the bonds: We are comparing the C–H bond in methane (H₃C–H) with the C–Br bond in bromomethane (H₃C–Br).
2. Calculate electronegativity difference for C–H:
   • $EN_C = 2.55$
   • $EN_H = 2.20$
   • $\Delta EN_{C-H} = |EN_C - EN_H| = |2.55 - 2.20| = 0.35$
3. Calculate electronegativity difference for C–Br:
   • $EN_C = 2.55$
   • $EN_{Br} = 2.96$
   • $\Delta EN_{C-Br} = |EN_C - EN_{Br}| = |2.55 - 2.96| = 0.41$
4. Compare the differences:
   • Since $0.41 > 0.35$, the C–Br bond has a larger electronegativity difference than the C–H bond.
5. Conclusion: The C–Br bond is more polar. Bromine is more electronegative than carbon, pulling electron density away from carbon. Carbon and hydrogen have relatively similar electronegativities, making the C–H bond only slightly polar.

(b) H₃C–NH₂ vs. H₃C–OH

1. Identify the bonds: We are comparing the C–N bond in methylamine (H₃C–NH₂) with the C–O bond in methanol (H₃C–OH).
2. Calculate electronegativity difference for C–N:
   • $EN_C = 2.55$
   • $EN_N = 3.04$
   • $\Delta EN_{C-N} = |EN_C - EN_N| = |2.55 - 3.04| = 0.49$
3. Calculate electronegativity difference for C–O:
   • $EN_C = 2.55$
   • $EN_O = 3.44$
   • $\Delta EN_{C-O} = |EN_C - EN_O| = |2.55 - 3.44| = 0.89$
4. Compare the differences:
   • Since $0.89 > 0.49$, the C–O bond has a larger electronegativity difference than the C–N bond.
5. Conclusion: The C–O bond is more polar. Oxygen is more electronegative than nitrogen because it is further to the right in the same period (Period 2) of the periodic table. Both oxygen and nitrogen are significantly more electronegative than carbon, but oxygen's higher electronegativity makes the C–O bond considerably more polar.

(c) H₃C–OH vs. H₃C–SH

1. Identify the bonds: We are comparing the C–O bond in methanol (H₃C–OH) with the C–S bond in methanethiol (H₃C–SH).
2. Calculate electronegativity difference for C–O:
   • $EN_C = 2.55$
   • $EN_O = 3.44$
   • $\Delta EN_{C-O} = |EN_C - EN_O| = |2.55 - 3.44| = 0.89$
3. Calculate electronegativity difference for C–S:
   • $EN_C = 2.55$
   • $EN_S = 2.58$
   • $\Delta EN_{C-S} = |EN_C - EN_S| = |2.55 - 2.58| = 0.03$
4. Compare the differences:
   • Since $0.89 > 0.03$, the C–O bond has a much larger electronegativity difference than the C–S bond.
5. Conclusion: The C–O bond is more polar. Oxygen is significantly more electronegative than sulphur. Although both are in Group 16, oxygen is in Period 2 and sulphur is in Period 3. Electronegativity decreases down a group, so oxygen is much more electronegative than sulphur. In fact, sulphur's electronegativity (2.58) is very close to carbon's (2.55), making the C–S bond almost nonpolar.