Question 1

How does bond polarity change from C–F to C–I?

Accepted Answer

C–X bond polarity decreases in the order C–F > C–Cl > C–Br > C–I. Carbon is the positive end and the halogen is the negative end, so the bond-dipole arrow points from carbon toward halogen. A molecular dipole is not determined by electronegativity alone, because bond length and charge separation also contribute.

Question 2

Why does the C–X bond become longer and weaker from fluorine to iodine?

Accepted Answer

The halogen atom becomes larger down the group, so the bonding orbitals overlap at a greater distance and the C–X bond length increases. Bond enthalpy therefore decreases: C–F is strongest and C–I is weakest. Typical mean values are 339 kJ/mol for C–Cl, 285 kJ/mol for C–Br and 213 kJ/mol for C–I.

Question 3

Why do the boiling points of methyl halides increase even though their C–X bond polarity decreases?

Accepted Answer

Moving from CH₃F to CH₃I increases molecular size, electron count and polarizability. These changes strengthen London dispersion forces, which become large enough to outweigh the reduction in C–X bond polarity. The boiling points rise to −24.2 °C for CH₃Cl, 3.6 °C for CH₃Br and 42.4 °C for CH₃I.

Question 4

Why is carbon tetrachloride nonpolar while chloroform is polar?

Accepted Answer

Both molecules are tetrahedral around carbon, but CCl₄ has four identical C–Cl bonds. Their equal bond dipoles cancel because of the molecule's symmetry, giving zero molecular dipole. CHCl₃ has three C–Cl bonds and one C–H bond, so the bond dipoles do not cancel completely and the molecule is polar.

Question 5

Why does methyl iodide form a lower layer in water?

Accepted Answer

CH₃I has a liquid density of approximately 2.28 g/cm³, compared with about 1.00 g/cm³ for water. It is only partly soluble, so the two liquids form phases rather than one uniform mixture. The denser methyl iodide-rich phase settles below the water-rich phase.

The C–X Bond: Polarity, Strength and Physical Properties

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