JEE Chemistry Chemical Bonding and Molecular Structure MCQs Set 33

Answer: (c) +2 and +4

Explanation:
1. In the NO group N is +2 (one O at -2 balances it)
2. In the NO2 group N is +4
3. The total of +6 balances the three oxygens, and the two values are different.

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) B and D only

Explanation:
1. \( \ce{P4O10} \) has a \( \ce{P4} \) tetrahedron with an oxygen bridging each of the six edges, so there are six P-O-P bridges, not four
2. The P-O-P angle is bent (about 127 degrees)
3. Each P is sp³ with four oxygens (bridging 160 pm and terminal 143 pm), so (A) and (C) 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) all four

Explanation:
1. (A) is true: the ortho isomer chelates and does not associate between molecules. (B) is true (100, -41, -60 degrees C). (C) is true as HF is hydrogen bonded and the rest follow size (19.5, -35, -67, -85). (D) is true (-17, -33, -55, -88).
2. Only option (a) agrees with this working.

Teacher's Note:
1. Net polarity needs both polar bonds and an asymmetric shape; for ionic versus covalent character apply Fajans' rules (small, highly charged cation, large anion means more covalent).

Answer: (d) all four

Explanation:
1. Each contains O-H groups that can bond to oxygen atoms of neighbouring units: \( \ce{KH2PO4} \) in a 3D network, boric acid in layers, benzoic acid in dimers and methanol in chains.
2. Only option (d) agrees with this working.

Teacher's Note:
1. Net polarity needs both polar bonds and an asymmetric shape; for ionic versus covalent character apply Fajans' rules (small, highly charged cation, large anion means more covalent).

Answer: (b) hydrogen bonding

Explanation:
1. Critical temperature rises with the strength of intermolecular attraction
2. Water molecules are held by strong hydrogen bonds, while \( \ce{O2} \) has only weak dispersion forces.

Teacher's Note:
1. Net polarity needs both polar bonds and an asymmetric shape; for ionic versus covalent character apply Fajans' rules (small, highly charged cation, large anion means more covalent).

Answer: (d) A, B and D only

Explanation:
1. Group 12 metals have filled d subshells, so only s electrons bond and their atomisation enthalpy is lowest
2. Metallic bonding is non-directional
3. Group 2 metals supply two electrons, so their bonding is stronger than group 1, so (C) is false
4. Copper (more bonding electrons) is harder than potassium.

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) A

Explanation:
1. A is neon, a noble gas with a complete octet, so it exists as single atoms.
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) B₄

Explanation:
1. B has five valence electrons and is phosphorus
2. Group 15 elements below nitrogen do not form stable P-P multiple bonds, so phosphorus exists as the tetrahedral \( \ce{P4} \) molecule with six P-P single 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: (c) covalent

Explanation:
1. B is phosphorus and C is chlorine
2. Both are non-metals and their electronegativity difference is small (2.2 and 3.2), so they form covalent compounds such as \( \ce{PCl3} \) and \( \ce{PCl5} \).

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) 6 and 6

Explanation:
1. Each \( \ce{BH4^-} \) has two terminal B-H bonds (2c-2e), so 3 x 2 = 6
2. Each also has two bridging B-H-Al bonds (3c-2e), so 3 x 2 = 6
3. The totals are 6 and 6.

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) All the bond angles in \( \ce{CH2F2} \) are not identical

Explanation:
1. In \( \ce{CH2F2} \) the H-C-H angle (about 112 degrees) differs from the F-C-F angle (about 108), so all angles are not equal
2. The s-character is 25, 33 and 50 per cent in sp³, sp² and sp
3. The O-N-O angle is 180 in \( \ce{NO2^+} \), 120 in \( \ce{NO3^-} \) and about 134 in \( \ce{NO2} \)
4. The C-F bond shortens as more F atoms are on carbon, so \( \ce{CF4} \) is the shortest.

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) all of \( \ce{BF3} \), \( \ce{BeF2} \) and \( \ce{BCl3} \)

Explanation:
1. In each molecule the central atom (B or Be) has an empty p orbital and the halogen has filled p orbitals on the halogen
2. The halogen lone pair back-donates into the empty orbital, forming a partial pi bond.

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) intermolecular hydrogen bonding

Explanation:
1. Intermolecular hydrogen bonds link molecules into associated groups that need extra energy to separate, so the boiling point rises
2. Intramolecular hydrogen bonding lowers the boiling point relative to the isomer that associates.

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) perchloric acid

Explanation:
1. Peroxymonosulphuric acid (\( \ce{H2SO5} \)), peroxo titanic species and pernitric acid (\( \ce{HNO4} \)) all contain an O-O group
2. Perchloric acid \( \ce{HClO4} \) is just chlorine in +7 with four oxygens and no O-O link.

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) \( \ce{AlH3} \) combining with \( \ce{H^-} \) and \( \ce{SiF4} \) forming \( \ce{SiF6^2-} \)

Explanation:
1. \( \ce{NH3} \) is sp³ and stays sp³ in both \( \ce{NH4^+} \) and \( \ce{NH2^-} \)
2. \( \ce{AlH3} \) (sp²) changes to \( \ce{AlH4^-} \) (sp³)
3. \( \ce{SiF4} \) (sp³) changes to \( \ce{SiF6^2-} \) (sp³d²).

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, B and D

Explanation:
1. The bond orders are 1, 1.5, 2 and 2.5, so (A) and (B) hold
2. The unpaired electrons are 2, 1, 1 and 0 for \( \ce{O2} \), \( \ce{O2^+} \), \( \ce{O2^-} \), \( \ce{O2^2-} \), so the strict order in (C) is false
3. The antibonding electrons are 8, 7, 6 and 5, so (D) holds.

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) A and C only

Explanation:
1. Acidity grows from basic oxides (\( \ce{MgO} \), \( \ce{Li2O} \)) through amphoteric \( \ce{ZnO} \) and neutral \( \ce{NO} \) to acidic \( \ce{CO2} \), \( \ce{P2O5} \), \( \ce{SO2} \), \( \ce{SO3} \)
2. Sulphur trioxide is more acidic than \( \ce{P2O5} \), which makes (B) wrong
3. In (D), \( \ce{BeO} \) is amphoteric and \( \ce{NO} \) is neutral, so \( \ce{NO} \) is less acidic than \( \ce{CO2} \).

Teacher's Note:
1. Net polarity needs both polar bonds and an asymmetric shape; for ionic versus covalent character apply Fajans' rules (small, highly charged cation, large anion means more covalent).

Answer: (b) 6

Explanation:
1. Borazine ("inorganic benzene") has an alternating B and N hexagonal ring in which B and N are sp² hybridised and the H atoms lie in the same plane
2. All 12 atoms are coplanar, so X = 12 and X/2 = 6.

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) 5

Explanation:
1. The molecules with 3c-2e bonds are \( \ce{Al2(CH3)6} \) (methyl bridges) and \( \ce{B2H6} \)
2. In \( \ce{Al2Cl6} \) the chlorine bridges are 3c-4e bonds
3. \( \ce{Si2H6} \), \( \ce{C2H6} \), \( \ce{Si2Cl6} \) and \( \ce{B3N3H6} \) are normal electron-precise molecules
4. So the molecules without a 3c-2e bond are (b), (d), (e), (f) and (g): 5.

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) 3

Explanation:
1. Overlap of d orbitals can give sigma (end-on, for example dz2 with dz2), pi (side-on, for example dxz with dxz) and delta (four-lobe face overlap, for example dxy with dxy) bonds
2. So 3 types.

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) regular tetrahedral

Explanation:
1. sp³d gives 5 electron pairs in a trigonal bipyramid
2. With lone pairs the shapes are see-saw (\( \ce{SF4} \)), T-shaped (\( \ce{ClF3} \)) and linear (\( \ce{XeF2} \))
3. A regular tetrahedron has only 4 pairs and needs sp³.

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) d_xy

Explanation:
1. sp³d² uses one s, three p and two d orbitals: d_z2 and d_x2-y2. The d_xy orbital is not part of the set.
2. Only option (c) 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: (a) NP, NP, NP, P

Explanation:
1. Benzene and borazine are symmetric planar rings with zero dipole moment
2. In \( \ce{PCl3F2} \) the three Cl are equatorial and the two F axial, giving a symmetric D3h shape and zero dipole
3. In \( \ce{PCl2F3} \) two F are axial and one F plus two Cl are equatorial, so it is unsymmetrical and polar.

Teacher's Note:
1. Net polarity needs both polar bonds and an asymmetric shape; for ionic versus covalent character apply Fajans' rules (small, highly charged cation, large anion means more covalent).

Answer: (d) 1,2; 1,2,3; 1,2,3,4; 1,2,3,4

Explanation:
1. \( \ce{PCl3F2} \) (84 electrons, non-polar) and \( \ce{PCl2F3} \) (76, polar) share only hybridisation and shape
2. \( \ce{BF3} \) and \( \ce{BCl3} \) share hybridisation, shape and zero dipole, but not electron count
3. \( \ce{CO2} \) and \( \ce{CN2^2-} \) both have 22 electrons and are linear, non-polar and sp
4. \( \ce{C6H6} \) and \( \ce{B3N3H6} \) both have 42 electrons, planar shape, sp² atoms and zero dipole.

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.