Column-I Column-II AlCl 3 + Cl − ⟶ (P) Adjacent bond angle at under line atom is decreases S i F 4 + 2 F − ⟶ (Q) Underline species act as lewis base NH 3 + H + ⟶ (R) Out of given reactions Hybri

Question

Column-I Column-II AlCl 3 ​ + Cl − ⟶ (P) Adjacent bond angle at under line atom is decreases S i F 4 ​ + 2 F − ⟶ (Q) Underline species act as lewis base NH 3 ​ + H + ⟶ (R) Out of given reactions Hybrid orbitals of product have maximum s-character BeCl 2 ​ + Cl − ⟶ (S) d-orbital(s) used in reaction

TestHub · Accepted Answer

(A) AlCl 3 + Cl⁻ → AlCl₄⁻ sp² → sp³ Bond angle decreases → (P) (B) SiF 4 + 2F⁻ → SiF₆²⁻ sp³ → sp³d² Bond angle decreases d-orbitals involved → (P), (S) (C) NH₃ + H⁺ → NH₄⁺ NH₃ donates lone pair Lewis base → (Q) (D) Beryllium chloride + Cl⁻ → BeCl₃⁻ sp → sp² Bond angle decreases Maximum s-character (sp = 50%) → (P), (R)

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Doubts & Discussion

Column-I	Column-II
$AlCl_{3} + Cl^{-} ⟶$	(P) Adjacent bond angle at under line atom is decreases
$S i F_{4} + 2 F^{-} ⟶$	(Q) Underline species act as lewis base
$NH_{3} + H^{+} ⟶$	(R) Out of given reactions Hybrid orbitals of product have maximum s-character
$BeCl_{2} + Cl^{-} ⟶$	(S) d-orbital(s) used in reaction