JEE Chemistry Chemical Bonding and Molecular Structure MCQs Set 30

Multiple Choice Questions (MCQs) for JEE Chemistry: Chemical Bonding and Molecular Structure

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Multiple Choice Questions

Question 1: How many hydrogen bonds does each H2O molecule form in an ice crystal?
(a) 4
(b) 6
(c) 2
(d) 3
Show Answer & Explanation

Answer: (a) 4

Explanation:
1. Each water molecule is tetrahedrally surrounded by four others
2. It donates two hydrogens and accepts two through its lone pairs
3. This gives 4 hydrogen bonds per molecule.

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.

Question 2: The structures of \( \ce{AlCl3} \) and \( \ce{PCl3} \) can be described respectively as
(a) both planar
(b) both pyramidal
(c) pyramidal and planar
(d) planar and pyramidal
Show Answer & Explanation

Answer: (d) planar and pyramidal

Explanation:
1. \( \ce{AlCl3} \) has 3 bond pairs and no lone pair, so it is sp2 and trigonal planar
2. \( \ce{PCl3} \) has 3 bond pairs and 1 lone pair, so it is sp3 and pyramidal
3. Hence planar and pyramidal.

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.

Question 3: The bond orders of \( \ce{NO} \) and \( \ce{NO^+} \) are respectively
(a) 2 and 4
(b) 2.5 and 3
(c) 3.5 and 2.5
(d) 4 and 2
Show Answer & Explanation

Answer: (b) 2.5 and 3

Explanation:
1. \( \ce{NO} \) has 11 valence electrons: bonding 8, antibonding 3 in the pi* orbital, so bond order = (8-3)/2 = 2.5
2. \( \ce{NO^+} \) has 10 electrons, bonding 8, antibonding 2 giving (8-2)/2 = 3
3. Hence 2.5 and 3.

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.

Question 4: Which of the following molecules has a non-zero dipole moment?
(a) \( \ce{BF3} \)
(b) \( \ce{PCl3} \)
(c) \( \ce{SiCl4} \)
(d) \( \ce{XeF4} \)
Show Answer & Explanation

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

Explanation:
1. \( \ce{BF3} \) is trigonal planar, \( \ce{SiCl4} \) tetrahedral and \( \ce{XeF4} \) square planar, so their bond dipoles cancel
2. \( \ce{PCl3} \) is pyramidal with a lone pair, so the dipoles do not cancel
3. It has a net dipole moment.

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).

Question 5: According to VSEPR theory the shape of the \( \ce{SF4} \) molecule is
(a) squashed tetrahedral (see-saw)
(b) square pyramidal
(c) tetrahedral
(d) square planar
Show Answer & Explanation

Answer: (a) squashed tetrahedral (see-saw)

Explanation:
1. Sulphur has 4 bond pairs and 1 lone pair, giving 5 electron pairs and trigonal bipyramidal geometry
2. The lone pair takes an equatorial position
3. The resulting shape is see-saw, also called squashed tetrahedral.

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.

Question 6: In which of the following cases does breaking of covalent bonds take place?
(a) boiling of water
(b) melting of KCN
(c) melting of \( \ce{SiO2} \)
(d) boiling of \( \ce{CF4} \)
Show Answer & Explanation

Answer: (c) melting of \( \ce{SiO2} \)

Explanation:
1. \( \ce{SiO2} \) is a covalent network solid, so melting requires breaking Si-O covalent bonds
2. Boiling of water and of \( \ce{CF4} \) overcomes only intermolecular forces
3. Melting of KCN overcomes ionic attraction
4. Hence melting of \( \ce{SiO2} \).

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.

Question 7: The hybrid orbitals used by chlorine in the \( \ce{ClF3} \) molecule are of the type
(a) sp3
(b) sp2
(c) sp3d
(d) sp2d
Show Answer & Explanation

Answer: (c) sp3d

Explanation:
1. Chlorine in \( \ce{ClF3} \) has 3 bond pairs and 2 lone pairs, so 5 electron pairs
2. Five pairs need sp3d hybridisation (T-shaped molecule).

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.

Question 8: In solid \( \ce{CuSO4.5H2O} \), copper is coordinated to
(a) four water molecules
(b) one water molecule
(c) three water molecules
(d) five water molecules
Show Answer & Explanation

Answer: (a) four water molecules

Explanation:
1. In the crystal, four water molecules lie in a square plane around \( \ce{Cu^2+} \) and two sulphate oxygens occupy the axial positions
2. The fifth water is held by hydrogen bonds
3. Copper is therefore coordinated to four water molecules.

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.

Question 9: Which of the following contains the maximum number of electrons in antibonding molecular orbitals?
(a) \( \ce{O2} \)
(b) \( \ce{O2^-} \)
(c) \( \ce{O2^+} \)
(d) \( \ce{O2^2-} \)
Show Answer & Explanation

Answer: (d) \( \ce{O2^2-} \)

Explanation:
1. Antibonding electrons are 6 in \( \ce{O2} \), 5 in \( \ce{O2^+} \), 7 in \( \ce{O2^-} \) and 8 in \( \ce{O2^2-} \)
2. Adding electrons to \( \ce{O2} \) goes into pi* orbitals
3. So \( \ce{O2^2-} \) has the maximum.

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.

Question 10: Which of the following molecules is non-linear?
(a) \( \ce{CO2} \)
(b) \( \ce{SO2} \)
(c) \( \ce{HCN} \)
(d) \( \ce{C2H2} \)
Show Answer & Explanation

Answer: (b) \( \ce{SO2} \)

Explanation:
1. \( \ce{CO2} \), \( \ce{HCN} \) and \( \ce{C2H2} \) have sp carbon and are linear
2. \( \ce{SO2} \) has 2 bond pairs and 1 lone pair on S (sp2), so it is bent.

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.

Question 11: In the ammonium ion,
(a) the four hydrogens are at the corners of a square
(b) all bonds are ionic
(c) all bonds are coordinate
(d) the nitrogen atom carries a formal charge
Show Answer & Explanation

Answer: (d) the nitrogen atom carries a formal charge

Explanation:
1. \( \ce{NH4^+} \) is tetrahedral and its bonds are covalent
2. Three N-H bonds are normal covalent bonds and one is coordinate, and all four become identical
3. The formal charge of +1 sits on nitrogen
4. Hence the nitrogen atom carries a formal charge.

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.

Question 12: Which of the following does not obey the octet rule?
(a) \( \ce{I3^-} \)
(b) \( \ce{CO3^2-} \)
(c) \( \ce{O3} \)
(d) \( \ce{SO2} \)
Show Answer & Explanation

Answer: (a) \( \ce{I3^-} \)

Explanation:
1. Central iodine in \( \ce{I3^-} \) has 3 lone pairs and 2 bond pairs, so 10 electrons around it
2. In the other three species all atoms have an octet in their Lewis structures.

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.

Question 13: The oxidation states of the two sulphur atoms in thiosulphuric acid \( \ce{H2S2O3} \) are
(a) +II and +II
(b) 0 and +IV
(c) -II and +VI
(d) +I and +III
Show Answer & Explanation

Answer: (c) -II and +VI

Explanation:
1. Thiosulphuric acid is sulphuric acid with one O replaced by S, so the two S atoms are not equivalent
2. The central S bonded to O atoms and to the other S is +6, and the terminal (thio) S behaves like sulphide sulphur at -2
3. The sum is +4, which balances 2(+1) from H and 3(-2) from O, giving -4 in total.

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.

Question 14: The bond between the two boron atoms in diborane is of the type
(a) 2-centre 2-electron
(b) 3-centre 3-electron
(c) 3-centre 2-electron
(d) 4-centre 4-electron
Show Answer & Explanation

Answer: (c) 3-centre 2-electron

Explanation:
1. Diborane has two bridging hydrogens, each held by a B-H-B banana bond
2. Each bridge uses only 2 electrons over 3 atoms
3. These are 3-centre 2-electron 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.

Question 15: Assuming a Lewis structure of \( \ce{SO2} \) in which all atoms obey the octet rule, the formal charge on S is
(a) 0
(b) +2
(c) -2
(d) +1
Show Answer & Explanation

Answer: (d) +1

Explanation:
1. The octet structure is O=S(+)-O(-) with one double bond and one single bond
2. S has 3 bonds and 1 lone pair
3. Formal charge = 6 - 2 - 3 = +1.

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.

Question 16: In which of the following pairs are both molecules polar?
(a) \( \ce{O2} \) and \( \ce{H2O} \)
(b) \( \ce{SO2} \) and \( \ce{SCl2} \)
(c) \( \ce{BF3} \) and \( \ce{PCl3} \)
(d) \( \ce{CS2} \) and \( \ce{NO2} \)
Show Answer & Explanation

Answer: (b) \( \ce{SO2} \) and \( \ce{SCl2} \)

Explanation:
1. \( \ce{SO2} \) and \( \ce{SCl2} \) are both bent with lone pairs, so both are polar
2. \( \ce{O2} \) is non-polar, \( \ce{BF3} \) is non-polar and \( \ce{CS2} \) is non-polar
3. Only the pair \( \ce{SO2} \) and \( \ce{SCl2} \) has both 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).

Question 17: Which of the following has the highest bond order?
(a) \( \ce{O2} \)
(b) \( \ce{O2^+} \)
(c) \( \ce{O2^-} \)
(d) \( \ce{O2^2-} \)
Show Answer & Explanation

Answer: (b) \( \ce{O2^+} \)

Explanation:
1. Bond orders are \( \ce{O2^+} \) 2.5, \( \ce{O2} \) 2, \( \ce{O2^-} \) 1.5 and \( \ce{O2^2-} \) 1
2. So \( \ce{O2^+} \) has the highest bond 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.

Question 18: The hybridisation of sulphur in \( \ce{SO3} \), \( \ce{SO4^2-} \) and \( \ce{SF4} \) respectively is
(a) sp, sp3, sp2
(b) sp, sp2, sp3d
(c) sp2, sp, sp3
(d) sp2, sp3, sp3d
Show Answer & Explanation

Answer: (d) sp2, sp3, sp3d

Explanation:
1. \( \ce{SO3} \) has 3 sigma bonds and no lone pair (sp2)
2. \( \ce{SO4^2-} \) has 4 sigma bonds (sp3)
3. \( \ce{SF4} \) has 4 bond pairs and 1 lone pair (sp3d).

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.

Question 19: The correct sequence of molecular orbitals for the carbide ion \( \ce{C2^2-} \) is
(a) \( \sigma 1s^2\ \sigma^* 1s^2\ \sigma 2s^2\ \sigma^* 2s^2\ \pi 2p^4\ \sigma 2p^2 \)
(b) \( \sigma 1s^2\ \sigma^* 1s^2\ \sigma 2s^2\ \sigma^* 2s^2\ \pi 2p^4 \)
(c) \( \sigma 1s^2\ \sigma^* 1s^2\ \sigma 2s^2\ \sigma^* 2s^2\ \pi 2p^4\ \sigma 2p^2\ \pi^* 2p^2 \)
(d) \( \sigma 1s^2\ \sigma^* 1s^2\ \sigma 2s^2\ \sigma^* 2s^2\ \pi 2p^4\ \sigma 2p^2\ \pi^* 2p^4 \)
Show Answer & Explanation

Answer: (a) \( \sigma 1s^2\ \sigma^* 1s^2\ \sigma 2s^2\ \sigma^* 2s^2\ \pi 2p^4\ \sigma 2p^2 \)

Explanation:
1. \( \ce{C2^2-} \) has 14 electrons, like \( \ce{N2} \)
2. Fill 1s2, sigma*1s2, 2s2, sigma*2s2, pi 2p4 and sigma 2p2 (12 + 2 = 14)
3. The pi* orbitals stay empty, giving a bond order of 3.

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.

Question 20: According to VSEPR theory the shape of the \( \ce{IF5} \) molecule is
(a) tetrahedral
(b) trigonal bipyramidal
(c) square pyramidal
(d) trigonal planar
Show Answer & Explanation

Answer: (c) square pyramidal

Explanation:
1. Iodine has 5 bond pairs and 1 lone pair, so 6 electron pairs (octahedral arrangement)
2. One position is the lone pair, leaving a square pyramidal shape.

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.

Question 21: In the structure C6H5-C-N-O (benzonitrile oxide, drawn as the stable Lewis structure with a C-N triple bond), the formal charges on the C, N and O atoms of the CNO group are respectively
(a) 0, +1, -1
(b) 0, -1, +1
(c) -1, +1, -1
(d) +1, 0, -1
Show Answer & Explanation

Answer: (a) 0, +1, -1

Explanation:
1. The structure is Ph-C triple bond N(+)-O(-)
2. Carbon has 4 bonds and no lone pair, so formal charge 0
3. Nitrogen has 4 bonds, so +1
4. Oxygen has 1 bond and 3 lone pairs, so -1.

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.

Question 22: Which of the following compounds does not have a pi bond?
(a) \( \ce{SO2} \)
(b) \( \ce{O2} \)
(c) \( \ce{SO3} \)
(d) \( \ce{SF6} \)
Show Answer & Explanation

Answer: (d) \( \ce{SF6} \)

Explanation:
1. \( \ce{SF6} \) has only six S-F single (sigma) bonds
2. \( \ce{SO2} \), \( \ce{O2} \) and \( \ce{SO3} \) all contain double bonds, which include a pi bond.

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.

Question 23: The relative basic strengths of \( \ce{NH3} \), \( \ce{CH3NH2} \) and \( \ce{NF3} \) are in the order
(a) \( \ce{NH3} \) > \( \ce{CH3NH2} \) > \( \ce{NF3} \)
(b) \( \ce{CH3NH2} \) > \( \ce{NH3} \) > \( \ce{NF3} \)
(c) \( \ce{NF3} \) > \( \ce{CH3NH2} \) > \( \ce{NH3} \)
(d) \( \ce{CH3NH2} \) > \( \ce{NF3} \) > \( \ce{NH3} \)
Show Answer & Explanation

Answer: (b) \( \ce{CH3NH2} \) > \( \ce{NH3} \) > \( \ce{NF3} \)

Explanation:
1. The methyl group donates electrons and makes the lone pair on N more available in \( \ce{CH3NH2} \)
2. In \( \ce{NF3} \) the highly electronegative fluorines withdraw electron density, so it is almost non-basic
3. Order is \( \ce{CH3NH2} \) > \( \ce{NH3} \) > \( \ce{NF3} \).

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).

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Multiple Choice Questions (MCQs) for JEE Chemistry Chemical Bonding and Molecular Structure

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