Consider the halogenation of ethene, where x is a generic halogen: h2c=ch2(g)+x2(g)→h2xc−ch2x(g) you may want to reference (page 412) section 9.10 while completing this problem. part a use bond energies to determine which halogen produces the most exothermic halogenation reaction with ethene. bond bond energy (\rm kj/mol) h−c 414 c−c 347 c=c 611 c−f 552 c−cl 339 c−br 280 c−i 209 f−f 159 cl−cl 243 br−br 193 i−i 151 use bond energies to determine which halogen produces the most exothermic halogenation reaction with ethene. bond bond energy (\rm kj/mol) 414 347 611 552 339 280 209 159 243 193 151 iodine chlorine bromine fluorine

Consider the halogenation of ethene is as follows:
CH₂=CH₂(g) + X₂(g) → H₂CX-CH₂X(g)
We can expect that this reaction occurring by breaking of a C=C bond and forming of two C-X bonds.
When bond break it is endothermic and when bond is formed it is exothermic.
So we can calculate the overall enthalpy change as a sum of the required bonds in the products:
Part a) 
C=C break = +611 kJ
2 C-F formed = (2 * - 552) = -1104 kJ
Δ H = + 611 - 1104 = - 493 kJ

2C-Cl formed = (2 * -339) = - 678 kJ
ΔH = + 611 - 678 = -67 kJ

2 C-Br formed = (2 * -280) = -560 kJ
ΔH = + 611 - 560 = + 51 kJ

2 C-I Formed = (2 * -209) = -418 kJ
ΔH = + 611 - 418 = + 193 kJ

Part b)
As we can see that the highest exothermic bond formed is C-F bond so from bond energies we can found that addition of fluoride is the most exothermic reaction