Free Hybridization MCQs with Answers

1. 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.

2. 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.

3. The hybridisation of the nitrogen atom in ammonia is

• A. sp3, with the lone pair occupying one of the four hybrid orbitals
• B. sp2
• C. sp
• D. unhybridised

Explanation: Nitrogen has four regions of electron density, three bonds and one lone pair, so four sp3 hybrids are needed even though only three of them are used for bonding. This is why the observed angle of 107 degrees is close to the tetrahedral value rather than to 120 degrees. Lone pairs must always be counted when deciding hybridisation.

4. The hybridisation of the carbon atom in carbon dioxide is

• A. sp3
• B. sp2
• C. sp
• D. unhybridised

Explanation: The carbon has only two regions of electron density, since each double bond counts as one region, so two sp hybrid orbitals point in opposite directions and the molecule is linear. The two remaining unhybridised p orbitals form the pi bonds. Counting regions rather than bonds is what makes this quick.

5. In an sp3 hybridised carbon, the angle between the hybrid orbitals is

• A. 90 degrees
• B. 109.5 degrees
• C. 120 degrees
• D. 180 degrees

Explanation: Four equivalent orbitals repel each other to the corners of a tetrahedron, which places them at 109.5 degrees. The sp2 arrangement gives 120 degrees in a plane and sp gives 180 degrees in a line. Each hybridisation is associated with a fixed geometry, which is worth memorising as a set.

6. The hybridisation of the carbon atoms in benzene is

• A. sp3
• B. sp2
• C. sp
• D. a mixture of sp2 and sp3

Explanation: Each carbon forms three sigma bonds, two to neighbouring carbons and one to a hydrogen, using sp2 hybrids at 120 degrees, which makes the ring flat and regular. The remaining p orbital on each carbon overlaps sideways all around the ring to form the delocalised pi system. That delocalisation gives benzene its exceptional stability.

7. Correct arrangement of orbital according to size of hybridized orbitals?

• A. sp > sp2 > sp3
• B. sp3 > sp > sp2
• C. sp2 > sp > sp3
• D. sp3 > sp2 > sp

Explanation: The greater the s character, the closer the orbital is held to the nucleus and the smaller it is, and s character rises from 25 per cent in sp3 through 33 per cent in sp2 to 50 per cent in sp. So sp3 is the largest and sp the smallest. The same trend explains why an sp carbon holds its bonding electrons most tightly, making terminal alkynes weakly acidic.