Predict the relative strengths of the carbon-halogen bonds in CH3FCH3F, CH3ClCH3Cl, and CH3BrCH3Br. Rank from strongest to weakest. To rank items as equivalent, overlap them.

Question

Parker Clements

Chemistry

31 December, 03:33

Predict the relative strengths of the carbon-halogen bonds in CH3FCH3F, CH3ClCH3Cl, and CH3BrCH3Br. Rank from strongest to weakest. To rank items as equivalent, overlap them.

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Answers (1)

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Isaac Robertson · Answer 1

CH2FCH2F > CH2ClCH2Cl > CH2BrCH2Br

Explanation:

As above, we are comparing:

1, 2 - Difluoloroethane

1, 2 - Dicholoroethane

1, 2 - Dibromoroethane

The three compounds are both in the homologous series with the general formula CnH2n+2X

Where X is the halogen element.

As we go down the periodic table ie fluorine, chlorine, bromine; one electron shell is continuously added to the molecule so therefore bond length between C-X increases. Note that bond length is inversely proportional to bond strength because the shorter the bond length, the stronger the bond.

Also, as more electron shells are added to the halogens down the group, the more further away from the nucleus which gives rise to bond order; the more electrons participate in a bond, the shorter the bond.

In conclusion, the C-F is has the shortest bond and the strongest bond strength.

C-F > C-Cl > C-Br (decreasing bond strength or bond dissociation energy)

CH2FCH2F > CH2ClCH2Cl > CH2BrCH2Br