JEE Chemistry Chemical Bonding and Molecular Structure MCQs Set 04

Answer: (c) s-p > p-p > s-s

Explanation:
1. Overlap extent: s-p (directional and larger) > p-p > s-s.

Teacher's Note:
1. Greater overlap means a stronger sigma bond.
2. Directional orbitals overlap better than spherical ones.

Answer: (c) First excited state

Explanation:
1. Cl ground state 3s2 3p5 has one unpaired electron.
2. First excited: 3s2 3p4 3d1 has three unpaired electrons, enough for ClF3.

Teacher's Note:
1. One unpaired electron gives valence 1, three give valence 3, five give valence 5, seven give valence 7.

Answer: (c) s-s, s-p or p-p orbitals along the internuclear axis

Explanation:
1. Head-on (axial) overlap of s-s, s-p or p-p orbitals gives a sigma bond.
2. Sideways overlap gives a pi bond.

Teacher's Note:
1. Sigma bonds allow free rotation.

Answer: (b) It is formed from hybrid orbitals

Explanation:
1. Pi bonds are formed from unhybridised p (or d) orbitals by sideways overlap, never from hybrid orbitals.

Teacher's Note:
1. Hybrid orbitals form sigma bonds and hold lone pairs.

Answer: (d) Chlorine

Explanation:
1. Cl2 is formed by axial overlap of two 3p orbitals.
2. H2 (s-s) and HCl, HBr (s-p) differ.

Teacher's Note:
1. Identify the orbitals that carry the unpaired electron.

Answer: (c) \( \ce{XeF5} \)

Explanation:
1. Xenon has 8 valence electrons; fluorine atoms use up an even number of them (2, 4 or 6), leaving paired electrons.
2. XeF5 would leave an unpaired electron, so it cannot exist as a neutral molecule.

Teacher's Note:
1. Noble gas fluorides have even numbers of F atoms.

Answer: (d) Both form stable p-pi multiple bonds with themselves

Explanation:
1. F forms only single bonds (F-F), while O forms O=O. The fluorine atom cannot form pi-bonded multiples with itself.

Teacher's Note:
1. Small, electronegative atoms with lone pairs repel strongly.

Answer: (b) Shape of the molecule

Explanation:
1. O is sp3 in both. H2O is V-shaped (2 lone pairs); H3O+ is pyramidal (1 lone pair).
2. So the hybridisation is unchanged but the shape changes.

Teacher's Note:
1. Count lone pairs and bond pairs.

Answer: (d) sp2 - sp3

Explanation:
1. C2 is part of the double bond (sp2) and C3 is a saturated CH2 (sp3).

Teacher's Note:
1. Hybridisation depends on the number of sigma bonds plus lone pairs.

Answer: (c) \( \ce{NF3} \) and \( \ce{H3O+} \); \( \ce{NO3^-} \) and \( \ce{BF3} \)

Explanation:
1. NF3 and H3O+ are both pyramidal (sp3, one lone pair).
2. NO3- and BF3 are both trigonal planar (sp2).

Teacher's Note:
1. Count valence electron pairs to find shape.

Answer: (d) B and C

Explanation:
1. Second-period elements (B, C) have no d-orbitals, so they cannot show sp3d. P and Cl (period 3) can.

Teacher's Note:
1. d-orbital participation requires n = 3 or higher.

Answer: (a) b and c

Explanation:
1. NH3 and PCl3 are pyramidal.
2. NH2+ (bent, sp2) and CH3+ (trigonal planar, sp2) are planar.

Teacher's Note:
1. sp2 and sp species with no lone pair on top are planar.

Answer: (c) \( \ce{NO2+} \), \( \ce{CO2} \), \( \ce{XeF2} \)

Explanation:
1. NO2+ and CO2 are sp (180 degrees); XeF2 has 3 lone pairs on the equator and is linear.

Teacher's Note:
1. Linear shapes arise from sp or sp3d with three lone pairs.

Answer: (c) \( \ce{Br2O} \) > \( \ce{Cl2O} \) > \( \ce{OF2} \)

Explanation:
1. Br2O (about 116 degrees) > Cl2O (111) > OF2 (103): the angle falls as the terminal atom's electronegativity rises.
2. For PX3 the angle increases from Cl to I.

Teacher's Note:
1. In central atom lone pair compounds, bond pairs are drawn away from the centre when the ligand is electronegative.

Answer: (a) Square planar

Explanation:
1. 6 electron pairs give an octahedral arrangement; with 2 lone pairs opposite each other the shape is square planar (XeF4).

Teacher's Note:
1. Lone pairs sit trans to each other to reduce repulsion.

Answer: (d) Distorted tetrahedral (see-saw)

Explanation:
1. I has 7 - 1 = 6 valence electrons; with 4 bond pairs and 1 lone pair (5 pairs) the structure is sp3d, see-saw.

Teacher's Note:
1. Same as SF4.

Answer: (b) \( \ce{H3O+} \)

Explanation:
1. H3O+ has 3 bond pairs and one lone pair on O (sp3): pyramidal.
2. BF3, NO3- and CO3(2-) are planar.

Teacher's Note:
1. Lone pair on the central atom lifts the plane.

Answer: (b) 2

Explanation:
1. T-shape arises from 5 electron pairs (sp3d trigonal bipyramid) with 3 bond pairs and 2 lone pairs in the equatorial plane (ClF3).

Teacher's Note:
1. Five electron pairs with two lone pairs gives T-shape.

Answer: (c) Only sp3

Explanation:
1. Each ion has 4 electron pairs around Cl (bond pairs plus lone pairs), so all are sp3 hybridised (sigma-bond count approach).

Teacher's Note:
1. Treat Cl-O pi bonds as not counted for hybridisation.

Answer: (b) 180 degrees

Explanation:
1. sp hybrid orbitals are linear: 180 degrees apart.

Teacher's Note:
1. sp: 180, sp2: 120, sp3: 109.5.

Answer: (c) \( \ce{XeF4} \)

Explanation:
1. XeF4 has 4 bond pairs and 2 lone pairs: sp3d2, square planar.

Teacher's Note:
1. Only XeF4 has lone pairs among these.

Answer: (a) sp3 - 2p

Explanation:
1. O is sp3 hybridised; each O-F sigma bond is sp3 (O) with 2p (F).

Teacher's Note:
1. Hybridised central atom overlaps with ligand p orbitals.

Answer: (b) Carbon tetrachloride

Explanation:
1. Only a symmetrical tetrahedral molecule with identical substituents (CCl4) has all angles 109.5 degrees.

Teacher's Note:
1. Mixed substituents distort the angles.

Answer: (b) px and py orbitals

Explanation:
1. sp3: 109.5; px-py: 90; H-O-H: 104.5; sp: 180. The minimum is 90 degrees.

Teacher's Note:
1. Unhybridised p orbitals are mutually perpendicular.

Answer: (d) Tetrahedral, sp3

Explanation:
1. Si has four sigma bonds and no lone pairs: sp3, tetrahedral.

Teacher's Note:
1. Four electron pairs mean tetrahedral.