Question
Which one of the following arrangements does not give the correct picture of the trends indicated against it?
A.
$${F_2} > C{l_2} > B{r_2} > {I_2}$$ Oxidising power
B.
$${F_2} > C{l_2} > B{r_2} > {I_2}$$ Electron gain enthapy
C.
$${F_2} > C{l_2} > B{r_2} > {I_2}$$ Bond dissociation energy
D.
$${F_2} > C{l_2} > B{r_2} > {I_2}$$ Electronegativity
Answer :
$${F_2} > C{l_2} > B{r_2} > {I_2}$$ Bond dissociation energy
Solution :
Generally as the size of the atom increases, bond dissociation energy decreases, so in halogens $${I_2}$$ have lowest bond dissociation energy, but the bond dissociation energy of chlorine is higher than
that of fluorine because in fluorine there is a greater repulsion between non-bonding electrons $$(2p).$$
Hence, the order of bond dissociation energy is
$$\eqalign{
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,C{l_2} > {F_2}\, > B{r_2}\,\, > \,{I_2} \cr
& {\text{Bond}}\,\,{\text{dissociation}}\,\,\,\,\,243\,\,\,\,\,\,\,159\,\,\,\,\,\,\,\,193\,\,\,\,\,\,\,\,\,151 \cr
& {\text{energy (kJ/mol)}} \cr} $$