Free Chemical Bonding MCQs with Answers

1. What type of chemical bond is present in sodium chloride?

• A. Covalent bond
• B. Ionic bond
• C. Metallic bond
• D. Hydrogen bond

Explanation: Sodium transfers its single valence electron to chlorine, producing Na⁺ and Cl⁻ ions held together by electrostatic attraction — an ionic bond. The large electronegativity difference between the two elements is what makes the transfer, rather than sharing, favourable.

2. According to VSEPR theory, the shape of a molecule is decided by

• A. the mutual repulsion of the electron pairs around the central atom, which arrange themselves as far apart as possible
• B. the masses of the atoms attached
• C. the number of neutrons in the central atom
• D. the temperature at which the molecule is examined

Explanation: Bonding and lone pairs are regions of negative charge, so they repel and settle into the arrangement of minimum repulsion, which fixes the geometry. Repulsion is strongest between two lone pairs, weaker between a lone pair and a bond pair and weakest between two bond pairs, and this order explains why bond angles are squeezed when lone pairs are present. Atomic mass plays no part.

3. The shape of the ammonia molecule, NH3, is

• A. trigonal planar
• B. pyramidal, with a bond angle of about 107 degrees
• C. tetrahedral, with an angle of 109.5 degrees
• D. linear

Explanation: Nitrogen has three bond pairs and one lone pair, so the four electron regions point tetrahedrally but only the three bonded atoms are visible, giving a pyramid. The lone pair repels the bond pairs more strongly than they repel each other, so the angle is compressed from 109.5 to about 107 degrees. Describing a shape means naming the arrangement of the atoms, not of the electron pairs.

4. The bond angle in a water molecule is about 104.5 degrees rather than 109.5 degrees because

• A. oxygen is more electronegative than hydrogen
• B. water molecules are hydrogen bonded
• C. the two lone pairs on oxygen repel the bond pairs more strongly than bond pairs repel each other
• D. the molecule is linear

Explanation: Oxygen has four electron regions, two bonding and two lone pairs, so the basic arrangement is tetrahedral, but the two lone pairs squeeze the two O-H bonds closer together than in methane. Ammonia, with only one lone pair, is compressed less, which is why its angle of 107 degrees lies between the two. The progression from methane through ammonia to water is the standard illustration of the repulsion order.

5. The shape of the boron trifluoride molecule, BF3, is

• A. pyramidal
• B. bent
• C. tetrahedral
• D. trigonal planar with bond angles of 120 degrees

Explanation: Boron forms three bonds and has no lone pair, so the three electron regions lie flat at 120 degrees, and the molecule is an exception to the octet rule with only six electrons round the central atom. That electron deficiency is why boron trifluoride readily accepts a lone pair and acts as a Lewis acid. Ammonia looks similar on paper but is pyramidal because of its lone pair.

6. A sigma bond is formed by

• A. sideways overlap of two p orbitals
• B. head on overlap of two orbitals along the axis joining the nuclei
• C. the transfer of electrons from one atom to another
• D. attraction between two ions

Explanation: End on overlap concentrates electron density directly between the nuclei, which makes a sigma bond stronger than a pi bond and allows free rotation about it. Sideways overlap of parallel p orbitals gives a pi bond, with density above and below the internuclear axis. Every single bond is a sigma bond, and every multiple bond contains exactly one.

7. The number of sigma and pi bonds in a molecule of ethyne, C2H2, is

• A. three sigma and two pi
• B. two sigma and three pi
• C. five sigma and no pi
• D. four sigma and one pi

Explanation: The two carbon to hydrogen bonds are sigma, and the carbon to carbon triple bond consists of one sigma plus two pi bonds, giving three sigma and two pi in total. The rule to apply is that a single bond is one sigma, a double bond one sigma and one pi, and a triple bond one sigma and two pi. Counting them correctly is what makes the hybridisation obvious as well.

8. Rotation about a carbon to carbon double bond is restricted because

• A. the sigma bond is too strong
• B. the atoms are too heavy
• C. turning one carbon would break the sideways overlap of the pi bond
• D. double bonds are ionic

Explanation: The pi bond depends on two parallel p orbitals overlapping sideways, and rotating one carbon by ninety degrees would destroy that overlap, which costs far more energy than is available at room temperature. This rigidity is what makes cis and trans isomerism possible in alkenes. Single sigma bonds rotate freely because their overlap is symmetrical about the axis.

9. The hybridisation of the carbon atom in methane is

• A. sp
• B. sp2
• C. sp3
• D. sp3d

Explanation: One 2s and three 2p orbitals mix to give four equivalent sp3 hybrids directed to the corners of a tetrahedron, which explains why all four C-H bonds are identical and the angle is 109.5 degrees. Without hybridisation the s and p orbitals would give bonds of different strengths and 90 degree angles, which contradicts experiment. Counting four electron regions is the quickest route to sp3.

10. The percentage of s character in an sp2 hybrid orbital is

• A. 25 per cent
• B. 33.3 per cent
• C. 50 per cent
• D. 100 per cent

Explanation: An sp2 hybrid is formed from one s and two p orbitals, so the s contribution is one in three. The figures for sp3 and sp are 25 and 50 per cent respectively, and greater s character holds electrons closer to the nucleus, which is why sp hybridised carbon in ethyne makes the attached hydrogen weakly acidic. Bond length also shortens as s character rises.