Valence bond theory, hybridization (s, p, d), resonance
- Identify the correct answer. Answer: 1. Three canonical forms can be drawn for CO₃²⁻ ion. Key idea: check each statement. mathrmCO₃²⁻ : three equivalent canonical forms.... 1
- Match List I with List II.List I (Molecule): A. ethane B. ethene C. carbon molecule, C_2 D. ethyneList II (Number and types of bond/s between two carbon atoms): I. one σ-bond and two π-bonds II. two π-bonds III. one σ-bond IV. one σ-bond and one π-bo… Answer: 4. A-III, B-IV, C-II, D-I Key idea: count the bond between the two carbons. Ethane H₃C-CH₃ : one σ bond (III). Ethene: one σ + one π... 1
- The number of σ bonds, π bonds and lone pair of electrons in pyridine, respectively are : Answer: 4. 11, 3, 1 Key idea: Count bonds in pyridine, C₅H₅N (a six-membered ring with 3 C=C/C=N double bonds). σ bonds: 6 ring bonds + 5 C-... 1
- BF_3 is planar and electron deficient compound. Hybridization and number of electrons around the central atom, respectively are : Answer: 3. sp² and 6 Key idea: BF₃ has three bond pairs and no lone pair on boron. Three electron domains give sp² hybridisation and trigona... 1
- The hybridizations of atomic orbitals of nitrogen in NO_2^+, NO_3^- and NH_4^+ respectively are Answer: 2. sp , sp² and sp³ Key idea: hybridisation = (bond pairs + lone pairs) on N. NO₂^+ : O=N=O, 2 bond pairs, no lone pair, so sp . NO₃... 1
- Among the following, which one is a wrong statement? Answer: 2. SeF₄ and CH₄ have same shape. Key idea: compare VSEPR shapes. SO₂ has pπ - dπ bonding: correct. SeF₄ is see-saw (4 bond pairs + 1... 1
- Match the compounds given in column I with the hybridisation and shape given in column II and mark the correct option.Column I: (a) XeF_6; (b) XeO_3; (c) XeOF_4; (d) XeF_4Column II: (i) distorted octahedral; (ii) square planar; (iii) pyramidal; (iv)… Answer: 1. (i) (iii) (iv) (ii) Key idea: use VSEPR for each xenon compound. XeF₆ : sp³d³ , 6 bond pairs + 1 lone pair, distorted octahedral... 1