JEE Chemistry Chemical Bonding and Molecular Structure MCQs Set 14

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

Explanation:
1. \( \ce{BF4-} \), \( \ce{SO4^2-} \) and \( \ce{PH4+} \) have four bond pairs and no lone pair, so they are tetrahedral.
2. \( \ce{XeF4} \) has four bond pairs and two lone pairs, so it is square planar.

Teacher's Note:
1. Count bond pairs and lone pairs for each species.
2. Two lone pairs on octahedral arrangement give square planar.

Answer: (c) [\( \ce{NF3} \), \( \ce{H3O+} \)] and [\( \ce{NO3-} \), \( \ce{BF3} \)]

Explanation:
1. \( \ce{NF3} \) and \( \ce{H3O+} \) both have three bond pairs and one lone pair, so both are pyramidal.
2. \( \ce{NO3-} \) and \( \ce{BF3} \) have three bond pairs on the central atom and are trigonal planar.
3. \( \ce{HN3} \) is a bent/linear chain and fits neither pair.

Teacher's Note:
1. Isostructural means same shape and type.
2. Match bond pair and lone pair counts.

Answer: (b) one sigma and two pi bonds

Explanation:
1. The carbide ion is \( \ce{C2^2-} \), which is isoelectronic with \( \ce{N2} \).
2. So the two carbon atoms are joined by a triple bond: one sigma and two pi bonds.

Teacher's Note:
1. Carbide ion is like acetylide: \( \ce{[:C#C:]^2-} \).
2. It reacts with water to give acetylene.

Answer: (c) sp2 for one, sp3 for the other

Explanation:
1. Heterolytic fission gives \( \ce{CH3+} \) and \( \ce{CH3-} \).
2. \( \ce{CH3+} \) has three bond pairs and no lone pair, so it is sp2 (planar).
3. \( \ce{CH3-} \) has three bond pairs and a lone pair, so it is sp3 (pyramidal).

Teacher's Note:
1. A carbocation is sp2 and a carbanion is sp3.
2. Count the electron pairs around carbon.

Answer: (a) \( sp^3d \) and T-shaped

Explanation:
1. Br has 7 valence electrons, with three used for bonds and four forming two lone pairs.
2. Five electron pairs give a trigonal bipyramidal arrangement, \( sp^3d \).
3. The two lone pairs occupy equatorial positions, giving a T-shaped molecule.

Teacher's Note:
1. Lone pairs take equatorial positions in a trigonal bipyramid.
2. Three bond pairs and two lone pairs mean T-shape.

Answer: (c) planar

Explanation:
1. The carbon has three bond pairs and no lone pair.
2. It is sp2 hybridised and trigonal planar.

Teacher's Note:
1. With no lone pair, shape is decided by the bond pairs only.
2. The methyl anion is pyramidal instead.

Answer: (c) \( sp^3d \)

Explanation:
1. Xe has two bond pairs and three lone pairs, i.e. five electron pairs.
2. Five pairs need \( sp^3d \) hybridisation.
3. The three lone pairs sit in the equatorial plane, so the molecule is linear.

Teacher's Note:
1. Steric number 5 means \( sp^3d \).
2. Lone pairs take equatorial sites.

Answer: (d) \( sp^3d^2 \) hybridised state

Explanation:
1. Xe has four bond pairs and two lone pairs, so six electron pairs.
2. Six pairs require \( sp^3d^2 \) hybridisation.
3. The molecule is square planar.

Teacher's Note:
1. Steric number 6 gives \( sp^3d^2 \).
2. With two lone pairs, the shape is square planar.

Answer: (b) 16 sigma, 4 pi

Explanation:
1. Sigma bonds: 6 ring C-C, 4 ring C-H, 1 ring C-C(OOH), 1 C=O, 1 C-O, 1 O-H (acid), 1 C-O (phenol), 1 O-H (phenol) = 16.
2. Pi bonds: 3 in the benzene ring and 1 in the C=O = 4.

Teacher's Note:
1. A double bond is one sigma and one pi bond.
2. Count all single bonds carefully, including C-H and O-H.

Answer: (a) \( sp < sp^2 < sp^3 \)

Explanation:
1. The more p-character a hybrid orbital has, the larger its size.
2. sp is 50 percent p, sp2 is 67 percent and sp3 is 75 percent p.
3. So the size increases as sp < sp2 < sp3.

Teacher's Note:
1. More s-character means a smaller, more tightly held orbital.
2. Electronegativity follows the opposite order to size.

Answer: (a) \( sp > sp^2 > sp^3 \)

Explanation:
1. A hybrid orbital with greater s-character holds electrons nearer to the nucleus.
2. The s-character is 50 percent in sp, 33 percent in sp2 and 25 percent in sp3.
3. So the carbon electronegativity is sp > sp2 > sp3.

Teacher's Note:
1. More s-character gives greater electronegativity.
2. This explains the acidity of terminal alkynes.

Answer: (c) sp and \( sp^2 \) hybridised

Explanation:
1. The structure is \( \ce{(NC)2C=C(CN)2} \).
2. The two carbons of the C=C bond are sp2.
3. The four carbons of the cyanide groups (C#N) are sp.

Teacher's Note:
1. Triple bond carbon is sp.
2. Double bond carbon is sp2.

Answer: (c) I and IV

Explanation:
1. \( \ce{CO2} \) is linear.
2. \( \ce{HgCl2} \) (sp) and \( \ce{C2H2} \) (sp) are also linear.
3. \( \ce{NO2} \) is bent and \( \ce{SnCl4} \) is tetrahedral.

Teacher's Note:
1. Two bond pairs and no lone pair give a linear shape.
2. Check each species for its steric number.

Answer: (c) 4

Explanation:
1. Benzene has 12 sigma bonds (6 C-C and 6 C-H).
2. It has 3 pi bonds (alternate double bonds).
3. Ratio = 12 / 3 = 4.

Teacher's Note:
1. In the Kekule structure there are three double bonds.
2. Count both C-C and C-H sigma bonds.

Answer: (d) \( \ce{ClF3} \)

Explanation:
1. \( \ce{ClF3} \) has three bond pairs and two lone pairs on chlorine (five electron pairs).
2. The lone pairs occupy equatorial positions of a trigonal bipyramid, leaving a T-shape.

Teacher's Note:
1. AB3E2 gives a T-shaped molecule.
2. BrF3 has the same shape.

Answer: (a) (i) and (iii)

Explanation:
1. \( \ce{ICl4-} \) and \( \ce{BrF4-} \) each have four bond pairs and two lone pairs, so they are square planar.
2. \( \ce{XeO4} \) is tetrahedral and \( \ce{I5-} \) is not square planar.

Teacher's Note:
1. AX4E2 is square planar.
2. Count the lone pairs, including those from the negative charge.

Answer: (b) \( \ce{AlH3} \) combines with \( \ce{H-} \) and \( \ce{SiF4} \) forms \( \ce{SiF6^2-} \)

Explanation:
1. \( \ce{AlH3} \) (sp2) becomes \( \ce{AlH4-} \) (sp3), a change.
2. \( \ce{SiF4} \) (sp3) becomes \( \ce{SiF6^2-} \) (\( sp^3d^2 \)), also a change.
3. \( \ce{NH3} \) to \( \ce{NH4+} \) or \( \ce{NH2-} \) stays sp3.

Teacher's Note:
1. Compare steric numbers before and after.
2. Change in hybridisation means change in steric number.

Answer: (a) x-axis

Explanation:
1. \( p_x \) orbitals lie along the x-axis.
2. Head-on overlap, which gives a sigma bond, occurs only along that axis.
3. Along y or z the overlap is sideways (pi) or zero.

Teacher's Note:
1. Sigma means head-on overlap along the internuclear axis.
2. Sideways overlap gives pi.

Answer: (a) Rotation is possible about a sigma bond, but not about a pi bond.

Explanation:
1. A sigma bond is cylindrically symmetrical about the bond axis, so the groups can rotate without breaking it.
2. A pi bond has electron density above and below the axis, and rotation would break it.
3. A p orbital can overlap head-on or sideways, an s orbital cannot form a pi bond, and two atoms share only one sigma bond.

Teacher's Note:
1. Free rotation is the reason ethane has conformations.
2. Pi bonds lock geometry, as in alkenes.

Answer: (b) \( \ce{SF6} \) and \( \ce{SiF6^2-} \)

Explanation:
1. \( \ce{SF6} \) and \( \ce{SiF6^2-} \) each have six bond pairs and no lone pair, so both are octahedral.
2. \( \ce{SF4} \) is see-saw while \( \ce{SiF4} \) is tetrahedral.
3. \( \ce{SeF6^2-} \) and \( \ce{TeF6^2-} \) have a lone pair on the central atom and are distorted.

Teacher's Note:
1. Compare steric number and lone pairs.
2. Species with lone pairs are not isostructural with those without.

Answer: (a) (i) and (ii)

Explanation:
1. \( \ce{NH3} \) has three bond pairs and one lone pair, so it is pyramidal.
2. \( \ce{SeO3^2-} \) and \( \ce{CH3-} \) also have three bond pairs and one lone pair, so they are pyramidal.
3. \( \ce{BH3} \) and \( \ce{CH3+} \) have no lone pair and are trigonal planar.

Teacher's Note:
1. Pyramidal means AX3E.
2. Trigonal planar means AX3 with no lone pair.

Answer: (b) (i), (ii) and (iii) only

Explanation:
1. \( \ce{N2O} \) is N#N-O, \( \ce{N2O3} \) is ON-NO2, and \( \ce{N2O4} \) is O2N-NO2, so all three have N-N bonds.
2. \( \ce{N2O5} \) is O2N-O-NO2, with an N-O-N bridge and no N-N bond.

Teacher's Note:
1. Draw the structures to be sure.
2. N2O5 is the anhydride of nitric acid.

Answer: (d) a dipole moment (and hydrogen bonding)

Explanation:
1. Water is polar and forms strong intermolecular hydrogen bonds.
2. Stronger intermolecular forces mean the gas needs a higher temperature to resist liquefaction.
3. Oxygen is non-polar and has only weak dispersion forces.

Teacher's Note:
1. Critical temperature rises with intermolecular attraction.
2. Polar, hydrogen-bonded liquids have high critical temperatures.

Answer: (c) non-metallic covalent bonding

Explanation:
1. Each carbon in a layer is sp2 hybridised and forms three sigma bonds with neighbours in a hexagonal network.
2. Within a layer the bonds are covalent; the layers are held together by van der Waals forces.

Teacher's Note:
1. Graphite has covalent bonds within layers and weak forces between layers.
2. The delocalised pi electrons make it conduct.

Answer: (c) hydrogen bonding

Explanation:
1. Ethanol has an O-H group and forms intermolecular hydrogen bonds.
2. Dimethyl ether has no O-H and cannot hydrogen bond to itself, so it boils at a much lower temperature.

Teacher's Note:
1. Isomers with O-H or N-H groups boil higher.
2. Hydrogen bonding raises the boiling point.