chemistry : class11 Hydrogen-Hydrogen Bond : Properties of hydrogen : Hydrogen MCQ Questions and Answers
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Hydrogen-Hydrogen Bond
MCQ questions and answers.
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MCQ Questions
1. According to molecular orbital theory, the bond order of the hydrogen molecule, H2, is:
A.
0
B.
1/2
C.
1
D.
2
Correct Answer:
1
Explanation:
H2 contains two electrons, both occupying the bonding σ1s molecular orbital. Thus, bond order = (number of bonding electrons − number of antibonding electrons) / 2 = (2 − 0) / 2 = 1.
2. Which molecular orbital is formed by the effective head-on overlap of the two 1s orbitals in H2 along the internuclear axis?
A.
σ1s bonding molecular orbital
B.
σ*1s antibonding molecular orbital
C.
π2p bonding molecular orbital
D.
π*2p antibonding molecular orbital
Correct Answer:
σ1s bonding molecular orbital
Explanation:
The two spherical 1s orbitals overlap head-on along the internuclear axis, producing a sigma bonding orbital, written as σ1s. This orbital concentrates electron density between the nuclei and stabilizes the H–H bond.
3. If one electron is removed from H2 to form H2+, the bond order changes from:
A.
1 to 1/2
B.
1 to 0
C.
1/2 to 1
D.
2 to 1
Correct Answer:
1 to 1/2
Explanation:
H2 has two electrons in the bonding σ1s orbital and bond order 1. H2+ has only one bonding electron, so its bond order becomes (1 − 0) / 2 = 1/2.
4. What is expected when H2 absorbs sufficient energy to promote one electron from σ1s to σ*1s?
A.
The bond order becomes zero and the H–H bond is destabilized
B.
The bond order becomes two and the H–H bond shortens
C.
The bond order remains one because both orbitals are bonding
D.
The bond order becomes one-half and the H–H bond becomes stronger
Correct Answer:
The bond order becomes zero and the H–H bond is destabilized
Explanation:
After promotion, H2 would have one electron in σ1s and one in σ*1s. The bond order is therefore (1 − 1) / 2 = 0, so the H–H bond loses its net bonding stabilization.
5. Which statement best explains why H2 is diamagnetic in its ground state?
A.
Its two electrons occupy separate antibonding orbitals with parallel spins
B.
Its two electrons are paired in the bonding σ1s molecular orbital
C.
It contains no electrons in molecular orbitals
D.
Its hydrogen nuclei cancel the magnetic moments of its electrons
Correct Answer:
Its two electrons are paired in the bonding σ1s molecular orbital
Explanation:
The two electrons of H2 occupy the lowest-energy σ1s bonding molecular orbital with opposite spins. Since all electrons are paired, ground-state H2 is diamagnetic.
6. In the formation of H2 from two hydrogen atoms, the potential energy is lowest at the equilibrium internuclear distance because:
A.
Only nucleus–nucleus repulsion operates at that distance
B.
Electron–electron repulsion completely disappears
C.
Attractive and repulsive interactions produce a minimum net energy
D.
The electrons are removed from the molecule at that distance
Correct Answer:
Attractive and repulsive interactions produce a minimum net energy
Explanation:
At the equilibrium H–H separation, attraction between each nucleus and the other atom’s electron balances the repulsive interactions sufficiently to give minimum potential energy. This minimum corresponds to the stable H–H bond length.
7. Which change would generally increase the vibrational frequency of a diatomic hydrogen molecule if the bond force constant remained unchanged?
A.
Replacing H2 by D2
B.
Replacing H2 by T2
C.
Replacing H2 by H2+
D.
Replacing H2 by a molecule with greater reduced mass
Correct Answer:
Replacing H2 by H2+
Explanation:
The vibrational frequency is proportional to the square root of k divided by the reduced mass. H2+ has a smaller reduced mass than H2 because it contains one proton and one electron overall rather than two hydrogen atoms, so its frequency is higher when the force constant is assumed unchanged.
8. The H–H bond in H2 is classified as a sigma bond primarily because:
A.
Its electron density is cylindrically symmetrical about the internuclear axis
B.
Its electrons occupy a pi antibonding orbital
C.
Its bond axis contains a nodal plane
D.
Its formation requires lateral overlap of p orbitals
Correct Answer:
Its electron density is cylindrically symmetrical about the internuclear axis
Explanation:
The H–H bond results from head-on overlap of two 1s orbitals. The resulting electron density is symmetrical around the internuclear axis, which is the defining feature of a sigma bond.
9. Which species has the greatest H–H bond order among H2, H2+, and H2−, assuming the added electron in H2− enters σ*1s?
A.
H2+
B.
H2
C.
H2−
D.
All three have the same bond order
Correct Answer:
H2
Explanation:
The bond orders are H2+ = 1/2, H2 = 1, and H2− = (2 − 1) / 2 = 1/2. Therefore, neutral H2 has the greatest H–H bond order.
10. If the H–H bond dissociation enthalpy is represented by D and the bond is broken homolytically, which process corresponds to D?
A.
H2(g) → H+(g) + H−(g)
B.
H2(g) → 2H·(g)
C.
H2(g) → 2H+(g) + 2e−
D.
H2(g) → H2+(g) + e−
Correct Answer:
H2(g) → 2H·(g)
Explanation:
Homolytic cleavage of the H–H covalent bond gives two neutral hydrogen atoms, each receiving one electron from the shared pair. The enthalpy required for this process is the H–H bond dissociation enthalpy.
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