Question

Choose the correct option(s).

• A. Oxygen has the highest electron gain enthalpy among the chalcogens.
• B. $B{e}^{+2}<L{i}^{+}<N{a}^{+}<K^{+}\text{radii}$.
• C. $F$ is the 2nd most electronegative element.
• D. $Cl$ has the highest negative electron gain enthalpy.

Answer 1

Answer: (B), (D)

The correct options are
B $B{e}^{+2}<L{i}^{+}<N{a}^{+}<K^{+}\text{radii}$.
D $Cl$ has the highest negative electron gain enthalpy.

Electron gain enthalpy decreases down the group and therefore oxygen is expected to have the highest electron gain enthalpy among chalcogens. But due to smaller size, on introducing an electron instability is caused by the inter-electronic repulsions. As a result, some of the energy released is consumed by the instability caused by inter-electronic repulsions.

$\text{Ionic radii}\propto \text{effective nuclear charge (except for transition metals)}$
Thus ionic radii of the given ions increases in the following order:
$B{e}^{+2}<L{i}^{+}<N{a}^{+}<K^{+}$.

Fluorine is the first member of the halogen family and has the smallest size among the halogens. The halogens are the most electronegative elements in their respective periods. This is because they contain 5 electrons in their valence $2p$ subshell and adding one more electron will make them attain the nearest noble gas configuration. Therefore their tendency to abstract the shared pair of electrons in a compound towards themselves is very high. The electrons in fluorine are strongly attracted towards the nucleus since the distance between the nucleus and the electrons in the valence shell in fluorine is the smallest among halogens. Thus, fluorine has the highest electronegativity.

Chlorine has the highest electron gain enthalpy and this is because, in fluorine, the inter-electronic repulsions are high in the $2p$ subshell due to its smaller size. Comparitively, the inter-electronic repulsions in $3p$ subshell of $Cl$ is less due its larger size