Question

Arrange the following in the order of property indicated against each set:
(i) F₂, I₂, Br₂, Cl₂ (increasing bond dissociation enthalpy)
(ii) NH₃, AsH₃, SbH₃, BiH₃, PH₃ (decreasing base strength)

Answer 1

(i) bond dissociation enthalpy depends on bond length of compound or molecule. in group halogen group { e.g., F, Cl, Br, I } bond length increase , due to this bond dissociation enthalpy decreases . but in case of fluorine , bond dissociation enthalpy is irregular . this is because of its size and intermolecular repulsion between molecules.
so, the perfect increasing order of bond dissociation enthalpy is I₂< F₂ < Br₂ < Cl₂

(ii) we know, basic strength in case of nitrogen family decreases from top to bottom this is because , size of the atom increases. and hence, electron density over the group 15 elements decreases. and hence, basic strength decreases.
so, decreasing order of basic strength is
$NH_3>PH_3>AsH_3>SbH_3>BiH_3$

Answer 2

(i) bond dissociation enthalpy depends on bond length of the concerned molecule. In group halogen group { e.g., F, Cl, Br, I } as we move from F to Cl to Br to I,the atomic radius increaes and hence bond length increases , due to this bond dissociation enthalpy decreases.It can also be explained by electronegetivity.As we move from F to Cl to Br to I,the electronegetivity decreases and hence bond dissociation enthalpy decreases.

so, the perfect increasing order of bond dissociation enthalpy is I₂ < Br₂ < Cl₂< F₂

(ii) we know, in case of nitrogen family , from top to bottom basic strength decreases this is because , size of the atom increases. and hence, electron density over the group 15 elements increases and hence, basic strength increases.

Hence decreasing order of basic strength is

NH₃<PH₃< AsH₃<SbH₃<BiH₃