JEE Chemistry Chemical Bonding and Molecular Structure MCQs Set 37

Answer: (a) S is sp³ hybridised and the shape is tetrahedral

Explanation:
1. Sulphur has three sigma bonds (O, Cl, Cl) and one lone pair, so it is sp³ and trigonal pyramidal, not tetrahedral
2. The S=O bond is partly p-d back bonding.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Answer: (d) \( \ce{BF3} \) > \( \ce{BCl3} \) > \( \ce{BBr3} \) and \( \ce{BF3} \) < \( \ce{BCl3} \) < \( \ce{BBr3} \)

Explanation:
1. The 2p-2p overlap between B and F is best, and it falls with the larger halogens
2. The more back bonding there is, the less electron-deficient B is, so the Lewis acid strength increases in the reverse order.

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Answer: (b) \( \ce{HNO3} \)

Explanation:
1. In \( \ce{HNO3} \) the nitrogen forms a dative bond to one oxygen (N+ and O- in the octet structure)
2. \( \ce{BaCl2} \) is ionic and \( \ce{HCl} \) and \( \ce{H2O} \) have only normal covalent bonds.

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Answer: (d) \( \ce{B2H6} \)

Explanation:
1. In \( \ce{B2H6} \) there are not enough electrons for ordinary bonds, so 3c-2e bonds form
2. In \( \ce{Al2Cl6} \) and \( \ce{I2Cl6} \) each bridging Cl donates a lone pair, so every atom has a complete octet and there is no electron deficiency.

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Answer: (a) Boron is in its ground state electronic configuration

Explanation:
1. Boron (2s2 2p1) must be promoted to 2s1 2p2 before hybridising, so it is not in the ground state in the molecule
2. The other three statements are true.

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Answer: (c) iodine monochloride: London dispersion force only

Explanation:
1. ICl is a polar molecule, so it has dipole-dipole attraction in addition to dispersion forces
2. The other three matches are correct.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Answer: (a) boric acid

Explanation:
1. In \( \ce{B(OH)3} \) the hydrogen bonding is intermolecular, linking the molecules into layers
2. The other three form chelate rings through intramolecular bonds.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Answer: (d) \( \ce{SbH3} \) because of its higher molecular mass

Explanation:
1. The boiling points are \( \ce{NH3} \) -33, \( \ce{PH3} \) -88, \( \ce{AsH3} \) -55 and \( \ce{SbH3} \) -17 degrees C. Although \( \ce{NH3} \) hydrogen bonds, the dispersion forces in \( \ce{SbH3} \) are so strong that it boils highest.
2. Only option (d) agrees with this working.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Answer: (b) \( \ce{H2O} \) > \( \ce{H2Te} \) > \( \ce{H2Se} \) > \( \ce{H2S} \)

Explanation:
1. \( \ce{H2O} \) is highest owing to hydrogen bonding
2. The others follow molecular mass: \( \ce{H2Te} \) (-2) > \( \ce{H2Se} \) (-41) > \( \ce{H2S} \) (-60 degrees C).

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

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

Explanation:
1. Back bonding in \( \ce{BF3} \) gives partial double-bond character, so the bond is 130 pm
2. In \( \ce{BF4^-} \) and the amine adducts boron is sp³ with no vacant orbital, and the B-F bonds are longer (about 135 to 138 pm).

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Answer: (a) The nitrogen lone pair occupies a p orbital that overlaps with empty p orbitals on each Si

Explanation:
1. The nitrogen lone pair is in a p orbital perpendicular to the NSi3 plane, but it overlaps with the empty 3d orbitals of silicon (p-pi to d-pi bonding), not empty p orbitals
2. All other statements are true.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Answer: (c) \( \ce{BF4^-} \)

Explanation:
1. \( \ce{BF4^-} \) is sp³ and tetrahedral, and the empty 2p orbital of B is used
2. All four B-F bonds are equal
3. \( \ce{B2H6} \) and \( \ce{Al2Cl6} \) have unequal bridging and terminal bonds, and gaseous \( \ce{BeCl2} \) is linear.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Answer: (a) o-nitrophenol

Explanation:
1. o-Nitrophenol has an intramolecular hydrogen bond between the OH and NO2 groups, so it does not associate with other molecules
2. The other three associate through intermolecular hydrogen bonds.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Answer: (d) T, T, T, F

Explanation:
1. HF hydrogen bonds (S1 true)
2. HI is heavier than HBr (S2 true)
3. \( \ce{N2^+} \) has bond order 2.5 and is longer than \( \ce{N2} \) (S3 true)
4. \( \ce{F2} \) is smaller and boils at -188, below \( \ce{Cl2} \) at -34 degrees C (S4 false).

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Answer: (b) I, III and IV only

Explanation:
1. Hydrogen bonding gives \( \ce{H2SO4} \) a high boiling point and viscosity (I)
2. In \( \ce{SO4^2-} \) all four S-O bonds are equal, so (II) is false
3. S-O has p-pi to d-pi character (III), and S is sp³ as is B in \( \ce{B2H6} \) (IV).

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Answer: (d) 7

Explanation:
1. \( \ce{NH3} \), \( \ce{H2O} \), HF, HCOOH, \( \ce{B(OH)3} \), \( \ce{CH3COOH} \) and \( \ce{HCO3^-} \) contain N-H, O-H or F-H bonds with suitable lone pairs and bond to each other
2. \( \ce{CH4} \) and HI do not
3. So there are 7.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Answer: (a) A and B only

Explanation:
1. \( \ce{AlCl3} \) and \( \ce{BCl3} \) have empty p orbitals and accept lone pairs
2. \( \ce{H2O} \) and \( \ce{NH3} \) have lone pairs and are Lewis bases.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Answer: (c) B and C only

Explanation:
1. \( \ce{B2H6} \) and monomeric \( \ce{AlCl3} \) have fewer than 8 electrons around the central atom (3c-2e bonds in \( \ce{B2H6} \))
2. In \( \ce{NaBH4} \) and \( \ce{LiAlH4} \) the central atom has four bonds and a complete octet.

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Answer: (a) all four

Explanation:
1. Alcohols hydrogen bond but ethers do not (A)
2. The open cage structure of ice makes it less dense than water (B)
3. Glycerol has three OH groups and is much more viscous (C)
4. Ammonia, with hydrogen bonds, liquefies at a higher temperature than \( \ce{HCl} \) (D).

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Answer: (d) A, B and D only

Explanation:
1. The P=O bond has p-pi to d-pi character (A)
2. The phosphate anions have equivalent P-O bonds through resonance (B)
3. \( \ce{H3PO4} \) is tribasic, so (C) is false
4. Its extensive hydrogen bonding makes it viscous (D).

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Answer: (b) none of these

Explanation:
1. Boron is sp³ in \( \ce{B2H6} \) and also sp³ in the adduct \( \ce{BH3.THF} \) (four bonds)
2. Oxygen is sp³ in THF both before and after, and carbon is sp³ throughout
3. No atom changes hybridisation.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Answer: (d) \( \ce{C2H6} \)

Explanation:
1. Ethane has only normal 2c-2e C-C and C-H bonds
2. \( \ce{B2H6} \), the \( \ce{BeH2} \) dimer and polymeric \( \ce{BeH2} \) all have bridging hydrogens in 3c-2e bonds.

Teacher's Note:
1. Draw the Lewis structure first: count valence electrons, join atoms, fill octets, then use multiple bonds. Bond order in resonance hybrids is the total number of bonds divided by the number of equivalent positions.

Answer: (a) N tetrahedral sp³, B tetrahedral sp³

Explanation:
1. In \( \ce{H3N->BCl3} \) the lone pair of N forms a dative bond to B. Each atom then has four bonds and no lone pair, so both are sp³ and tetrahedral.
2. Only option (a) agrees with this working.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Answer: (c) 4

Explanation:
1. Four water molecules are coordinated to \( \ce{Cu^2+} \) in a square plane
2. The fifth is held by hydrogen bonds to sulphate and to the coordinated waters.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Answer: (a) four 2c-2e bonds and two 3c-2e bonds

Explanation:
1. Diborane has four terminal B-H bonds, each 2c-2e, and two bridging B-H-B bonds, each 3c-2e
2. The 12 valence electrons are used by 4 x 2 + 2 x 2.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Answer: (d) \( \ce{MCl2} \) is more ionic than \( \ce{MCl4} \)

Explanation:
1. The cation with the lower charge, M2+, polarises Cl- less than M4+. So \( \ce{MCl2} \) has more ionic character, higher melting point, lower volatility and lower solubility in organic solvents, and it hydrolyses less readily.
2. Only option (d) agrees with this working.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Answer: (b) F-H...F

Explanation:
1. Hydrogen-bond strength increases with the electronegativity and smallness of the donor and acceptor atoms
2. F-H...F, as in \( \ce{[HF2]^-} \), is the strongest known.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Answer: (d) Xe

Explanation:
1. Noble gases are held by London dispersion forces, which grow with the number of electrons and polarisability
2. Xenon is the largest and most polarisable, so it boils highest (165 K).

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Answer: (a) There is extensive intramolecular hydrogen bonding in the condensed phase

Explanation:
1. In liquid water and ice the hydrogen bonding is intermolecular, between different molecules, so the statement of intramolecular bonding is false
2. Water is amphoteric, \( \ce{D2O} \) ice is denser than \( \ce{H2O} \) liquid, and photosynthesis oxidises water to \( \ce{O2} \).

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.

Answer: (c) \( \ce{SiH4} \)

Explanation:
1. \( \ce{SiH4} \) has four bonds, so silicon has a complete octet
2. \( \ce{GaH3} \), \( \ce{AlH3} \) and \( \ce{B2H6} \) are group 13 hydrides with fewer than eight electrons around the central atom.

Teacher's Note:
1. Find the number of bond pairs and lone pairs on the central atom to get the electron geometry and hybridisation, then the shape. Bond energy and magnetism follow from the bond order and unpaired electrons.