JEE Chemistry Chemical Bonding and Molecular Structure MCQs Set 19

Answer: (a) S in \( \ce{H2S} \) has two lone pairs (\( sp^3 \), V-shape) and P in \( \ce{PCl3} \) has one lone pair (\( sp^3 \), pyramidal)

Explanation:
1. \( \ce{H2S} \) has two bond pairs and two lone pairs on sulphur, giving a V-shape.
2. \( \ce{PCl3} \) has three bond pairs and one lone pair on phosphorus, giving a pyramid.
3. Lone pairs distort the regular tetrahedral arrangement.

Teacher's Note:
1. Both central atoms are \( sp^3 \) hybridised.
2. The lone pairs decide the shape.

Answer: (a) linear, square planar and see-saw

Explanation:
1. \( \ce{XeF2} \) has three lone pairs (linear).
2. \( \ce{XeF4} \) has two lone pairs (square planar).
3. \( \ce{XeO2F2} \) has four bond pairs and one lone pair, so it is see-saw (the lone pair is equatorial).

Teacher's Note:
1. Lone pairs go equatorial in the trigonal bipyramid.
2. XeO2F2 has the same shape as SF4.

Answer: (a) trigonal bipyramidal and square pyramidal

Explanation:
1. \( \ce{PCl5} \) has five bond pairs and no lone pair, so it is trigonal bipyramidal.
2. \( \ce{BrF5} \) has five bond pairs and one lone pair, so it is square pyramidal.

Teacher's Note:
1. AX5 is trigonal bipyramidal, AX5E is square pyramidal.
2. Add the lone pairs to the bond pairs first.

Answer: (a) \( \ce{O2[AsF6]} < \ce{O2} < \ce{KO2} \)

Explanation:
1. \( \ce{O2[AsF6]} \) contains \( \ce{O2+} \) (bond order 2.5), \( \ce{O2} \) has bond order 2, and \( \ce{KO2} \) contains \( \ce{O2-} \) (bond order 1.5).
2. Bond length rises as bond order falls, so \( \ce{O2+} < \ce{O2} < \ce{O2-} \).

Teacher's Note:
1. Link the three compounds to the ions \( \ce{O2+} \), \( \ce{O2} \), \( \ce{O2-} \).
2. Longer bonds mean lower bond order.

Answer: (a) No; \( \ce{N(CH3)3} \) is pyramidal while \( \ce{N(SiH3)3} \) is planar because of p(pi)-d(pi) back bonding

Explanation:
1. Trimethylamine has a lone pair on nitrogen and is pyramidal (\( sp^3 \)).
2. In trisilylamine the lone pair on N is delocalised into the vacant 3d orbitals of Si.
3. The nitrogen becomes \( sp^2 \), so the molecule is planar.

Teacher's Note:
1. Silicon has empty d orbitals that can accept a lone pair.
2. Carbon has none, so it cannot do this.

Answer: (a) 6

Explanation:
1. Melamine has a triazine ring with three ring nitrogen atoms and three \( \ce{NH2} \) groups.
2. Each of the six nitrogen atoms carries one lone pair.
3. Total = 6.

Teacher's Note:
1. Each N in an amine or an aromatic ring has one lone pair.
2. Carbon and hydrogen have none.

Answer: (a) 8

Explanation:
1. Nitroso oxygen has 2 lone pairs and nitroso nitrogen has 1.
2. The nitro nitrogen has none.
3. The doubly bonded nitro O has 2 and the singly bonded (or coordinate) nitro O has 3.
4. Total = 2 + 1 + 0 + 2 + 3 = 8.

Teacher's Note:
1. Draw the full Lewis structure and count carefully.
2. Each O in a single bond has 3 lone pairs.

Answer: (a) 4

Explanation:
1. Linear species are \( \ce{BeCl2} \), \( \ce{N3-} \), \( \ce{N2O} \), \( \ce{NO2+} \), \( \ce{ICl2-} \), \( \ce{I3-} \) and \( \ce{XeF2} \).
2. Of these, \( \ce{ICl2-} \), \( \ce{I3-} \) and \( \ce{XeF2} \) use \( sp^3d \) hybridisation.
3. The remaining four (sp hybridised) are \( \ce{BeCl2} \), \( \ce{N3-} \), \( \ce{N2O} \) and \( \ce{NO2+} \).

Teacher's Note:
1. sp means two electron pairs, no d orbitals.
2. Five pairs (three lone pairs) means \( sp^3d \).

Answer: (a) 6

Explanation:
1. \( \ce{[TeBr6]^2-} \): Te has 6 + 6 + 2 = 14 electrons, 6 bonds, so 1 lone pair.
2. \( \ce{[BrF2]+} \): 7 - 1 = 6 electrons, 2 bonds, so 2 lone pairs.
3. \( \ce{SNF3} \): S forms 4 bonds (plus S#N), 0 lone pairs.
4. \( \ce{[XeF3]-} \): 8 + 1 = 9 electrons, 3 bonds, so 3 lone pairs.
5. Total = 1 + 2 + 0 + 3 = 6.

Teacher's Note:
1. Lone pairs = (valence electrons - bonding electrons) / 2.
2. Adjust for the charge.

Answer: (a) 6

Explanation:
1. Diamagnetic: \( \ce{H2} \), \( \ce{Li2} \), \( \ce{Be2} \) (as a formal MO configuration), \( \ce{C2} \), \( \ce{N2} \) and \( \ce{F2} \).
2. Paramagnetic: \( \ce{He2+} \) (one unpaired electron), \( \ce{B2} \) (two) and \( \ce{O2-} \) (one).
3. So six are diamagnetic.

Teacher's Note:
1. Fill MOs and look for unpaired electrons.
2. Be2 is counted by its MO configuration, although it is not stable.

Answer: (c) 6

Explanation:
1. The triple bond \( \ce{N#N} \) consists of three shared pairs.
2. So 6 electrons take part in the bonding.
3. The other four are lone-pair electrons (one pair per atom).

Teacher's Note:
1. Only shared electrons are bonding electrons.
2. Each bond pair holds 2.

Answer: (a) \( \ce{X+Y-} \)

Explanation:
1. The electropositive element loses an electron to become \( \ce{X+} \).
2. The electronegative element gains it to become \( \ce{Y-} \).
3. The compound is ionic, \( \ce{X+Y-} \).

Teacher's Note:
1. Big electronegativity difference gives an ionic bond.
2. Example: NaCl.

Answer: (a) \( \ce{CO2} \)

Explanation:
1. \( \ce{CO2} \) has two bond pairs and no lone pair on carbon, so it is linear.
2. \( \ce{NO2} \), \( \ce{SO2} \) and \( \ce{ClO2} \) are bent because of a lone pair or an odd electron on the central atom.

Teacher's Note:
1. Two electron pairs on the central atom give a linear shape.
2. Lone pairs bend the molecule.

Answer: (a) \( \ce{CN-} \)

Explanation:
1. CO has 6 + 8 = 14 electrons.
2. \( \ce{CN-} \) has 6 + 7 + 1 = 14 electrons.
3. \( \ce{O2+} \) has 15, \( \ce{O2-} \) has 17 and \( \ce{N2+} \) has 13 electrons.

Teacher's Note:
1. Count total electrons.
2. \( \ce{CN-} \), CO, \( \ce{N2} \) and \( \ce{NO+} \) are all 14 electron species.

Answer: (b) its regular tetrahedral structure

Explanation:
1. Each C-Cl bond is polar, but the four bond dipoles are arranged symmetrically in a regular tetrahedron.
2. Their vector sum is zero.

Teacher's Note:
1. The shape decides the net dipole moment.
2. Symmetry causes the cancellation.

Answer: (d) HCl

Explanation:
1. Hydrogen bonds need H bonded to F, O or N.
2. Phenol, ammonia and water have such bonds.
3. Chlorine is large and not electronegative enough, so HCl does not form hydrogen bonds.

Teacher's Note:
1. Only N, O and F form strong hydrogen bonds.
2. Cl forms very weak ones, if at all.

Answer: (b) two orbitals at 180 degrees

Explanation:
1. sp hybridisation gives two equivalent orbitals.
2. They point in opposite directions, at 180 degrees to each other.

Teacher's Note:
1. sp is linear.
2. sp2 is trigonal planar and sp3 is tetrahedral.

Answer: (c) trans-1,2-dichloroethylene

Explanation:
1. In the trans isomer the two C-Cl bond dipoles point in opposite directions and cancel.
2. In the cis and 1,1-isomers they add up and give a net dipole.

Teacher's Note:
1. Trans isomers of symmetric alkenes are often non-polar.
2. Cis isomers are polar.

Answer: (c) \( \ce{CH3+} \)

Explanation:
1. \( \ce{CH3+} \) has three bond pairs and no lone pair, so carbon is \( sp^2 \).
2. \( \ce{PH3} \), \( \ce{NH3} \) and \( \ce{SbH3} \) have three bond pairs and one lone pair, so the central atom is \( sp^3 \) (or nearly pure p in the heavier hydrides).

Teacher's Note:
1. A lone pair adds to the count of hybrid orbitals.
2. The carbocation is planar.

Answer: (a) 120 degrees and 109.5 degrees

Explanation:
1. In tetrachloroethene each carbon is sp2 with trigonal planar geometry, so the Cl-C-Cl angle is about 120 degrees.
2. In tetrachloromethane carbon is sp3 and the angle is 109.5 degrees.

Teacher's Note:
1. Match the hybridisation to the angle.
2. Alkene carbon: 120 degrees.

Answer: (a) \( \ce{O2-} \)

Explanation:
1. \( \ce{O2-} \) has 17 electrons, with one unpaired electron in an antibonding pi orbital.
2. \( \ce{CN-} \), CO and \( \ce{NO+} \) all have 14 electrons and are diamagnetic.

Teacher's Note:
1. Isoelectronic with \( \ce{N2} \) means diamagnetic.
2. The superoxide ion is paramagnetic.

Answer: (a) \( \ce{PCl3} \)

Explanation:
1. \( \ce{PCl3} \) has three bond pairs and one lone pair, so it is pyramidal.
2. \( \ce{SO3} \), \( \ce{CO3^2-} \) and \( \ce{NO3-} \) have no lone pair on the central atom and are trigonal planar.

Teacher's Note:
1. AX3E is pyramidal.
2. AX3 is planar.

Answer: (b) \( \ce{BF3} \)

Explanation:
1. \( \ce{BF3} \) is planar triangular and symmetrical, so its three B-F bond dipoles cancel.
2. \( \ce{CH2Cl2} \), \( \ce{NF3} \) and \( \ce{ClO2} \) have unsymmetrical shapes and a net dipole.

Teacher's Note:
1. Symmetric shape with identical bonds means zero dipole.
2. BF3 is the standard example.

Answer: (a) \( sp^3 \)

Explanation:
1. Chlorine in \( \ce{ClO2-} \) has two bond pairs and two lone pairs.
2. Four pairs give \( sp^3 \) hybridisation, and the ion is bent.

Teacher's Note:
1. Chlorite is bent like water.
2. Count lone pairs on Cl.