Question

Which of the following(s) represents the correct order?

• A. $O<S<Se<Te<Po:\text{Metallic character}$
• B. $O<S<Se<Te>Po:\text{Boiling point}$
• C. $O<S<Se<Te>Po:\text{Melting point}$
• D. $O>S>Se>Te>Po:\text{Electron gain enthalpy}$

Answer 1

Answer: (A), (B), (C)

$O<S<Se<Te>Po:\text{Melting point}$

On moving down the group, metallic character increases.
$O<S<Se<Te<Po$

Generally on moving down the group, boiling point of group $16$ elements increases but the boiling point of $Te$ is greater than that of $Po$.
$O<S<Se<Te>Po$

Generally on moving down the group, melting point of group $16$ elements increases but the melting point of $Te$ is greater than that of $Po$.
$O<S<Se<Te>Po$

Generally on moving down the group, electron gain enthalpy decrease due to increase in size and decrease in nuclear charge. But $O$ has less electron gain enthalpy than $S$ due to smaller size and compact nature of oxygen so the incoming electron is not easily accomodated.
The correct order is
$S>Se>Te>O$