Match the column Column I (Molecules/ions) Column II (Properties) (A) ICl 4 ⊖ (P) d z 2 orbital is involved in hybridisation of central atom (B) OClF 3 (Q) Maximum number of atoms in a plane = 5

Question

Match the column

(A) $ICl_{4}^{⊖}$	(P) $d_{z} 2$ orbital is involved in hybridisation of central atom
Column I (Molecules/ions)	Column II (Properties)
(B) $OClF_{3}$	(Q) Maximum number of atoms in a plane $= 5$
(C) $POCl_{3}$	(R) The axial (M-X) bond length is shorter than the equitorial (M-X) bond length
(D) $IF_{7}$	(S) It has atleast one angle less than $12 0^{\circ}$
	(T) It has two lone pair on central atom

TestHub · Accepted Answer

A) ICl 4 ​ ⊖ - P, Q, S, T Hybridisation: sp³d² → involves d(z²) Shape: square planar → all 5 atoms in one plane Has 2 lone pairs - (T) Bond angles = 90° → less than 120° (B) OClF 3 ​ - P, S Hybridisation: sp³d → involves d(z²) Shape: see-saw → has angles < 120° No max planar 5 atoms, no 2 lone pairs (C) POCl 3 ​ - S Hybridisation: sp³ (tetrahedral) Has bond angles slightly less than 120° (due to double bond + electronegativity effects) No d-orbital needed, no lone pair (D) IF 7 ​ - P, R, S Hybridisation: sp³d³ → involves d(z²) Shape: pentagonal bipyramidal Axial bonds shorter than equatorial Has angles < 120° (72° in pentagon)

Chemistry - CHEMICAL BONDING Question with Solution | TestHub

Options:

A) $ICl_{4}^{⊖}$ - P, Q, S, T

(B) $OClF_{3}$ - P, S

(C) $POCl_{3}$ - S

(D) $IF_{7}$ - P, R, S

Doubts & Discussion

Chemistry - CHEMICAL BONDING Question with Solution | TestHub

Options:

A) ICl4​⊖ - P, Q, S, T

(B) OClF3​ - P, S

(C) POCl3​ - S

(D) IF7​ - P, R, S

Doubts & Discussion

A) $ICl_{4}^{⊖}$ - P, Q, S, T

(B) $OClF_{3}$ - P, S

(C) $POCl_{3}$ - S

(D) $IF_{7}$ - P, R, S