Which of the following statements are correct? I. $\quad \mathrm{P}$ is having least negative electron gain…
Which of the following statements are correct?
I. $\quad \mathrm{P}$ is having least negative electron gain enthalpy among $\mathrm{P}, \mathrm{S}, \mathrm{Cl}$, and $\mathrm{F}$
II. In $\mathrm{Eu}, \mathrm{Yb}$ Lanthanoid contraction is not observed
III. $\mathrm{Ce}(\mathrm{OH})_3$ is most basic among lanthanoid hydroxides
IV. The radii of $\mathrm{Na}$ and $\mathrm{Na}^{+}$are $95 \mathrm{pm}, 186 \mathrm{pm}$ respectively
I, III, IV only
II, IV only
I, III only
I, II, III only
Solution
Europium $(\mathrm{Eu})=[\mathrm{Xe}] 4 \mathrm{f}^7 5 \mathrm{~d}^0 6 \mathrm{~s}^2, 185 \mathrm{pm}$ Ytterbium $(\mathrm{Yb})=[\mathrm{Xe}] 4 \mathrm{f}^{14} 5 \mathrm{~d}^0 6 \mathrm{~s}^2, 170 \mathrm{pm}$ In these two elements, the atomic radii do not show expected decrease due to lanthanoid contraction but instead show an increased value.
This is due to an electron entering into the $4 \mathrm{f}$-orbital(inner shell) and giving rise to very stable half-filled and fullyfilled configurations that results in greater screening of the valence electrons (higher screening constant $\sigma$ ) Thus, Statement II is correct.
The value of electron gain enthalpy is more negative for more electronegative elements.
The electronegativity follows the order :$\mathrm{P} < \mathrm{S} < \mathrm{Cl} < \mathrm{F}$.
Thus, $\mathrm{P}$ has the least negative electron gain enthalpy and therefore Statement I is correct.
$\mathrm{Ce}(\mathrm{OH})_3$ is most basic hydroxide due to largest cation size which results in $\mathrm{Ce}(\mathrm{OH})_3$ being most ionic.
Thus, III is also correct.
The cation formed is always smaller than the parent atom or ion and thus $\mathrm{Na}^{+}$will be smaller than $\mathrm{Na}$.
Thus, IV is incorrect.