MCQ Questions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure with Answers

Answer: (d) Propyne
Explanation:
The C – C bond length = 1.54 Å, C = C bond length = 1.34 Å and C ≡ C bond length = 1.20 Å.
Since propyne has a triple bond, therefore it has minimum bond length.

Answer: (a) 1
Explanation:
In an antibonding molecular orbital, most of the electron density is located away from the space between the nuclei, as a result of which there is a nodal plane (i.e, a plane at which the electron density is zero) between the nuclei.

Answer: (a) sp²
Explanation:
The hybridisation of sulphur in SO₂ is sp². Sulphur atom has one lone pair of electrons and two bonding domains. Bond angle is <120° and molecular geometry is V-shape, bent or angular

Answer: (c) Acetonitrile
Explanation:
Acetonitrile does not contain sp² hybridized carbon.

Answer: (d) n-Pentane
Explanation:
Boiling point increases with increase in molecular mass. For the compounds with the same molecular mass, boiling point decreases with an increase in branching.
The molecular mass of 2-Methylbutane: 72 g mol^-1
The molecular mass of 2-Methylpropane: 58 g mol^-1
The molecular mass of 2, 2-Dimethylpropane: 72 g mol^-1
The molecular mass of 2-Methylbutane: 72 g mol^-1
2-Methylpropane has the lowest molecular mass among all of the given compounds.
Thus, 2-Methylpropane has the lowest boiling point among the given options.

Answer: (d) AlCl₃
Explanation:
We know that, extent of polarisation ∝ covalent character in ionic bond. Fajans rule states that the polarising power of cation increases, with increase in magnitude of positive charge on the cation Therefore, polarising power ∝ charge of cation.

The polarising power of cation increases with the decrease in the size of a cation. Therefore, polarising (power) ∝ (1)/ (size of cation)
Here the AlCl₃ is satisfying the above two conditions i.e., Al is in +3 oxidation state and also has small size. So it has more covalent character.

Answer: (b) acetonitrile and acetone
Explanation:
Dipole-dipole interactions occur among the polar molecules. Polar molecules have permanent dipoles. The positive pole of one molecule is thus attracted by the negative pole of the other molecule. The magnitude of dipole-dipole forces in different polar molecules is predicted on the basis of the polarity of the molecules, which in turn depends upon the electro negativities of the atoms present in the molecule and the geometry of the molecule (in case of polyatomic molecules, containing more than two atoms in a molecule).

Answer: (a) 3
Explanation:
Be_nAl₂Si₆O₁₈
The oxidation states of each element
Be = +2
Al = +3
Si = +4
O = -2
(2n) + (3 × 2) + (4 + 6) + (−2 × 18) = 0
or 2n + 30 − 36 = 0
or 2n = 6
or n = 3

Answer: (b) sp²
Explanation:
sp² hybrid structures have trigonal planar geometry, which is two dimensional.

Answer: (b) A₂B₃
Explanation:
A has 3 electrons in outermost orbit and B has 6 electrons in its outermost orbits. So A can give three electrons to complete its octet and B needs 2 electrons to complete its octet. So 2 atoms of A will release 6 electrons and 3 atoms of B will need six electrons to complete their octet
So, the formula will be A₂​B₃

Answer: (d) 4
Explanation:
Each water molecule can form a maximum of four hydrogen bonds with neighboring water molecules. The two hydrogens of the water molecule can form hydrogen bonds with other oxygens in ice, and the two lone pair of electrons on oxygen of the water molecule can attract other hydrogens in ice. Hence, 4 possible hydrogen bonds.

Answer: (b) One sigma, two pi
Explanation:
A single bond between two atoms is always considered as sigma bond.
A double bond between two atoms is always considered as one sigma and one pi bond
A triple bond between two atoms is always considered as one sigma bond and two pi bonds.
So according to the given structure CaC₂ (Calcium carbide) has 1 sigma and 2 pi bonds

Answer: (c) NaCl
Explanation:
The highest melting point will be NaCl, it is because, the lattice energy decreases as the size of alkali metal increases so going down the group the melting point decreases, but due to the covalent bonding in LiCl, its melting point is lower than NaCl and so NaCl is expected to have maximum melting point in the alkali chlorides.​

Answer: (c) HCl and He atoms
Explanation:
HCl is polar (μ ≠ 0) and He is non-polar (μ = 0) gives dipole-induced dipole interaction.

Answer: (b) acetonitrile and acetone
Explanation:
Dipole-dipole interactions occur among the polar molecules. Polar molecules have permanent dipoles. The positive pole of one molecule is thus attracted by the negative pole of the other molecule. The magnitude of dipole-dipole forces in different polar molecules is predicted on the basis of the polarity of the molecules, which in turn depends upon the electro negativities of the atoms present in the molecule and the geometry of the molecule (in case of polyatomic molecules, containing more than two atoms in a molecule).

Answer: (d) K⁺ < Ca²⁺ < Mg²⁺ < Be²⁺
Explanation:
High charge and small size of the cations increases polarisation.
As the size of the given cations decreases as
K⁺ > Ca²⁺ > Mg²⁺ > Be²⁺
Hence, polarising power decreases as K⁺ < Ca²⁺ < Mg²⁺ < Be²⁺

Answer: (d) OSF₂
Explanation:
The species having a pyramidal shape according to VSEPR theory is OSF₂. The central S atom has 3 bonding domains (one S = O double bond and two S−F single bonds) and one lone pair of electrons.

The electron pair geometry is tetrahedral and molecular geometry is pyramidal. This is similar to the ammonia molecule.

Answer: (a) Pentagonal bipyramid
Explanation:
IF₇ Hybridization is sp³d³
Structure is Pentagonal bipyramidal.

Answer: (d) 5 sigma (s) and 1 pi (p) bonds
Explanation:
According to valence bond theory, two atoms form a covalent bond through the overlap of individual half-filled valence atomic orbitals, each containing one unpaired electron. In ethene, each hydrogen atom has one unpaired electron and each carbon is sp² hybridized with one electron each sp² orbital. The fourth electron is in the p orbital that will form the pi bond. The bond order for ethene is simply the number of bonds between each atom: the carbon-carbon bond has a bond order of two, and each carbon-hydrogen bond has a bond order of one.

Answer: (b) CO₂
Explanation:
The steric number of central atom of a linear molecule is two. It has two bonded atoms and zero lone pair. All the molecules have two bonded atoms. Thus, we need to work out the number of lone pairs.
In ClO₂, the central atom Cl has 7 valence electrons. Four are used up to form 4 bonds with O atoms. Three are non-bonding electrons. Thus, along with an odd electron, it has a lone pair.
In CO₂, the central C atom has 4 valence electrons. All are used up to form four bonds with O atoms. Thus, it has zero lone pair.
In NO₂, the central N atom has 5 valence electrons. Four are used up to form bonds with oxygen atoms. Thus, one electron is left as an odd electron.
In SO₂, the central S atom has 6 valence electrons. Four are used up to form bonds with oxygen atoms. Two nonbonding electrons form one lone pair.