Valence bond theory and hybridisation — NEET & JEE Questions

Question 1 · difficulty L1 · recall

In valence bond theory, a covalent bond forms mainly by

• A. overlap of atomic orbitals containing electrons
• B. transfer of all electrons to one atom
• C. nuclear fusion
• D. creation of new protons

Question 2 · difficulty L2 · understanding

The lowest oxidation number of an atom in a compound $\mathrm{A}_2 \mathrm{B}$ is -2 . The number of electrons in its valence shell is _______.

Question 3 · difficulty L2 · understanding

Match the orbital overlap figures shown in List I with the description given in List II and select the correct answer using the code given below the lists.

• A. P-2, Q-1, R-3, S-4
• B. P-4, Q-3, R-1, S-2
• C. P-2, Q-3, R-1, S-4
• D. P-4, Q-1, R-3, S-2

Question 4 · difficulty L2 · understanding

An $sp^3$ hybridised atom has four equivalent hybrid orbitals directed approximately toward

• A. the corners of a tetrahedron
• B. a straight line
• C. a trigonal plane only
• D. two opposite axes only

Question 5 · difficulty L3 · understanding

$$\text { Match List I with List II : }$$ List-I List-II A. $\mathrm{C}_2 \mathrm{H}_4$ I. $3 \sigma$ bonds, $2 \pi$ bonds B. $\mathrm{C}_2 \mathrm{H}_2$ II. $3 \sigma$ bonds, one lone pair C. $\mathrm{CH}_4$ III. $4 \sigma$ bonds D. $\mathrm{NH}_3$ IV. $5 \sigma$ bonds, $1 \pi$ bond Choose the correct answer from the options given below :

• A. A-III, B-IV, C-II, D-I
• B. A-IV, B-I, C-III, D-II
• C. A-I, B-II, C-IV, D-III
• D. A-II, B-III, C-I, D-IV

Question 6 · difficulty L3 · application

The central carbon atom in methane is best described as

• A. $sp^3$ hybridised
• B. $sp^2$ hybridised
• C. sp hybridised
• D. unhybridised p only

Question 7 · difficulty L3 · application

The carbon atoms in ethene are primarily

• A. $sp^2$ hybridised
• B. $sp^3$ hybridised
• C. sp hybridised
• D. dsp2 hybridised

Question 8 · difficulty L3 · application

The carbon atoms in ethyne are primarily

• A. sp hybridised
• B. $sp^2$ hybridised
• C. $sp^3$ hybridised
• D. d2sp3 hybridised

Question 9 · difficulty L4 · application

Which combination correctly matches hybridisation and ideal bond angle?

• A. sp—180°, sp2—120°, sp3—109.5°
• B. sp—109.5°, sp2—180°, sp3—120°
• C. sp—90°, sp2—109.5°, sp3—180°
• D. All—120°

Question 10 · difficulty L4 · application

A carbon atom forms three sigma bonds in one plane and one pi bond perpendicular to that plane. Its hybridisation is

• A. $sp^2$
• B. $sp^3$
• C. sp
• D. none