Question

For the compounds \[C{H_3}Cl,C{H_3}Br,C{H_3}I{\text{ }}and{\text{ }}C{H_3}F\], the correct order of increasing C-halogen bond length is:
(A) \[C{H_3}F < C{H_3}Cl < C{H_3}Br < C{H_3}I\]
(B) \[C{H_3}F < C{H_3}Br < C{H_3}Cl < C{H_3}I\]
(C) \[C{H_3}F < C{H_3}I < C{H_3}Br < C{H_3}Cl\]
(D) \[C{H_3}Cl < C{H_3}Br < C{H_3}F < C{H_3}I\]

Answer 1

Hint: Bond length of C-X bond in alkyl halide depends on the electronegativity difference between halogen and carbon. The higher the difference is, the less is the bond length. This makes the C-X bond polarized.

Complete step-by-step solution:
In alkyl halides, carbon atoms have a partial positive charge and halogen has a partial negative charge. Electronegativity is defined as the tendency of an atom to attract a shared pair of electrons in a covalent bond. As we move up the periodic table in the halogen family from iodine to fluorine, electronegativity increases and molecular size decreases. As a result, an increment in bond length is observed.
Electronegativity order: \[I < Br < Cl < F\]
In other words, as molecular size increases, the bonds get longer, thereby the strength of those bonds’ decreases. Of the four halogens, the most electronegative is fluorine and least is iodine. This confers that the electron pair in the carbon-fluorine bond will be shifted most towards the halogen side. The electronegativities of carbon and iodine are almost similar so there occur no separation of charge on the bond.
Bond-length order: \[C{H_3}F < C{H_3}Cl < C{H_3}Br < C{H_3}I\]
Molecular size order: \[I > Br > Cl > F\]
The C-X bond breaking results in positive and negative ions. Since the C-F bond is strongest, it is difficult to break it, but easy to break a C-I bond.

Hence, the correct option is (A).

Note: In the case of halogens, inner electrons are not screened due to presence of many protons and because of the increasing positive charge in the nucleus, electrons are more attracted towards the nucleus. This makes them highly electronegative in periodic tables.