Chemistry - CHEMICAL BONDING Question with Solution | TestHub
List - I (Pair of species) | List - II (Same characteristics in pair) |
|---|---|
(P) ( ) | (1) d-orbital participation in hybridisation of central atom is observed |
(Q) | (2) Tetrahedral electron geometry of central atom |
(R) ( ) | (3) Total number of lone pairs are equal |
(S) ( ) | (4) bond is present |
| (5) Both molecules have same hybridisation on central atom |
Options:
Answer:
Solution:
P: NF₃ (sp³, tetrahedral electron geometry, trigonal pyramidal molecular geometry, 1 lone pair), OF₂ (sp³, tetrahedral electron geometry, bent molecular geometry, 2 lone pairs). Both have tetrahedral electron geometry and sp³ hybridization. So, P -> 2, 5.
Q: PCl₅ (sp³d, trigonal bipyramidal), ICl₃ (sp³d², trigonal bipyramidal electron geometry, T-shaped molecular geometry, 2 lone pairs). PCl₅ involves d-orbital participation (sp³d). ICl₃ also involves d-orbital participation (sp³d). Both have d-orbital participation. So, Q -> 1,5.
R: BeF₂ (sp, linear, 6 lone pairs), SO₃ (sp², trigonal planar, 6 lone pairs). So, R -> 3.
S: CO₂ (sp, linear, pπ-pπ bonds), SO₂ (sp², bent, pπ-pπ and pπ-dπ bonds). Both have pπ-pπ bonds. So, S -> 4.
Therefore, the correct match is P -> 2, 5; Q -> 1; R -> 3; S -> 4.
Final Answer: The final answer is B