Which of the following is incorrect match ?

Question

Which of the following is incorrect match ? (A) SiF4: Can act as lewis acid. (B) Benzyne: All C-atoms are sp2 hybridised (C) PBr 3: Non polar (D) CHF = C = CHF: Nodal planes of π -bonds are not lying in same plane

Tardigrade Admin · Accepted Answer

SiF 4 arrow Si have vacant d-orbital so can act as Lewis acid.

→ In benzyne all C-atoms are sp2 hybridised.

→ This is polar molecule.

→ The nodal planes of π - bonds are perpendicular to each other.

Q. Which of the following is incorrect match ?

$S i F_{4}$ : Can act as lewis acid.

Benzyne : All C-atoms are $s p^{2}$ hybridised

$PB r_{3} :$ Non polar

$C H F = C = C H F :$ Nodal planes of $π$ -bonds are not lying in same plane

$S i F_{4} \to$ Si have vacant d-orbital so can act as Lewis acid.
$\to$ In benzyne all C-atoms are $s p^{2}$ hybridised.
$\to$ This is polar molecule.
$\to$ The nodal planes of $π -$ bonds are perpendicular to each other.

Q. Which of the following is incorrect match ?

SiF4​ : Can act as lewis acid.

Benzyne : All C-atoms are sp2 hybridised

PBr3​: Non polar

CHF=C=CHF: Nodal planes of π -bonds are not lying in same plane

SiF4​→ Si have vacant d-orbital so can act as Lewis acid. → In benzyne all C-atoms are sp2 hybridised. → This is polar molecule. → The nodal planes of π− bonds are perpendicular to each other.

$S i F_{4}$ : Can act as lewis acid.

Benzyne : All C-atoms are $s p^{2}$ hybridised

$PB r_{3} :$ Non polar

$C H F = C = C H F :$ Nodal planes of $π$ -bonds are not lying in same plane

$S i F_{4} \to$ Si have vacant d-orbital so can act as Lewis acid.
$\to$ In benzyne all C-atoms are $s p^{2}$ hybridised.
$\to$ This is polar molecule.
$\to$ The nodal planes of $π -$ bonds are perpendicular to each other.