Element "A" has more IE than its adjacent elements in periodic table ; number of exchange pairs in ℓ = 1 subshells of "A" is 9. Then

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Identify Element 'A': The total exchange pairs in the $l = 1$ ( $p$ -subshell) being 9 corresponds to the configuration $2 p^{6} 3 p^{3}$ ( $6 + 3 = 9$ ), identifying element 'A' as Phosphorus (P), which has a higher ionization energy than its neighbors due to a stable half-filled shell.

Identify Oxy-acid (Q): Dehydration of two molecules of phosphorous acid (H₃PO₃) gives pyrophosphorous acid:

$2 H_{3} P O_{3} - H_{2} O H_{4} P_{2} O_{5} (Q)$

Identify Oxy-acid (R): Adding one oxygen atom to H₄P₂O₅ forms hypophosphoric acid:

$H_{4} P_{2} O_{5} + O \to H_{4} P_{2} O_{6} (R)$

Compare Basicity: H₄P₂O₅ contains two $- O H$ groups (basicity $x = 2$ ), whereas H₄P₂O₆ contains four $- O H$ groups (basicity $y = 4$ ). Thus, $y > x$ .

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