Match the following
List I List II
(A) Lewis base (P) NaOH
(B) Arrhenius acid (Q) ${BF}_{3}$
(C) Lewis acid (R) NH₃
(D) Arrhenius base (S) $H_{3} {PO}_{4}$
(A) (B) (C) (D)
(1) Q R P S
(2) S R Q P
(3) P Q R S
(4) R S Q P

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answer is D.

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Detailed Solution

${NH}_{3} -$ Electron pair donor – Lewis base
$H_{3} {PO}_{4} - H^{+}$ donor – Arrhenius Acid
${BF}_{3} -$ Electron pair acceptor – Lewis Acid
$NaOH - {OH}^{-}$ donor – Arrhenius Base

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