Question

Which of the following trend is/are correct?

• A. $Cl{O}_2>S{O}_2>C{O}_2>P_2{O}_5\text{(acidic nature)}$
• B. $Na>Si>Al>Mg\text{(second I. P.)}$
• C. $S>Se>Te>O\text{(electron affinity)}$
• D. $H_{3}P{O}_2>H_{3}P{O}_3>H_{3}P{O}_4\left(K{a}_1\right)$

Answer 1

Answer: (C), (D)

$H_{3}P{O}_2>H_{3}P{O}_3>H_{3}P{O}_4\left(K{a}_1\right)$

More the electronegativity of the central atom more will be the acidic nature of the compound.
Here $C{O}_2$ will be less acidic than $P_2{O}_5$.
So, option (a) is incorrect.
$S{i}^{+}$ has configuration of $\left[Ne\right]3s^{2}3p^1$
$A{l}^{+}$ has configuration of $\left[Ne\right]3s^2$
$A{l}^{+}$ has fully filled 3s subshell whereas $S{i}^{+}$ has an electron in $3p$ subshell.
So, the second ionisation energy of $S{i}^{+}$ will be less than $A{l}^{+}$.
Hence, option (b) is also incorrect.
E.A. of $O$ is least in its group due to its small size and high electron density.
Option (c) is correct.


Due to cross conjugation in (2); (1) is weaker base than (2). So $H_{3}P{O}_2$ is stronger acid than $H_{3}P{O}_3$. Similarly $H_{3}P{O}_4$ will be less acidic than $H_{3}P{O}_3$.