chemistry : class11 Ortho and Para Hydrogen : Properties of hydrogen : Hydrogen MCQ Questions and Answers
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Ortho and Para Hydrogen
MCQ questions and answers.
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MCQ Questions
1. Which combination correctly describes the nuclear-spin states and rotational-level restrictions of molecular hydrogen?
A.
Ortho-hydrogen has antiparallel proton spins and occupies only even rotational levels
B.
Ortho-hydrogen has parallel proton spins and occupies only odd rotational levels
C.
Para-hydrogen has parallel proton spins and occupies only odd rotational levels
D.
Both forms have identical proton-spin alignment and differ only in electronic energy
Correct Answer:
Ortho-hydrogen has parallel proton spins and occupies only odd rotational levels
Explanation:
In ortho-hydrogen, the two proton spins are parallel, giving a symmetric nuclear-spin function. Because the total molecular wavefunction must obey proton-exchange symmetry, ortho-hydrogen is associated with antisymmetric rotational functions, corresponding to odd rotational quantum numbers such as J = 1, 3, and so on.
2. Why is para-hydrogen the more stable modification of molecular hydrogen at very low temperature?
A.
Its molecules contain fewer electrons than ortho-hydrogen
B.
Its lowest allowed rotational level is J = 0, whereas ortho-hydrogen must begin at J = 1
C.
Its protons have parallel spins and therefore zero rotational energy
D.
It has a higher electronic bond energy than ordinary hydrogen
Correct Answer:
Its lowest allowed rotational level is J = 0, whereas ortho-hydrogen must begin at J = 1
Explanation:
Para-hydrogen has antiparallel proton spins and is allowed to occupy the J = 0 rotational ground state. Ortho-hydrogen is restricted to odd rotational levels and therefore has J = 1 as its lowest allowed level, making para-hydrogen thermodynamically favored at very low temperature.
3. At ordinary temperatures, the approximate equilibrium composition of hydrogen is best represented by which ratio?
A.
Ortho : para = 1 : 3
B.
Ortho : para = 1 : 1
C.
Ortho : para = 3 : 1
D.
Ortho : para = 9 : 1
Correct Answer:
Ortho : para = 3 : 1
Explanation:
At ordinary temperatures, the equilibrium mixture contains approximately 75% ortho-hydrogen and 25% para-hydrogen, giving an ortho : para ratio of about 3 : 1. The factor of three arises from the three possible triplet nuclear-spin states of ortho-hydrogen compared with the single singlet state of para-hydrogen.
4. Which statement correctly explains why the low-temperature equilibrium mixture of hydrogen differs from the ordinary-temperature mixture?
A.
The electronic configuration of hydrogen changes from 1s1 to 1s2
B.
Para-hydrogen becomes favored because its J = 0 state has lower rotational energy
C.
Ortho-hydrogen becomes favored because parallel spins reduce the bond length
D.
The proton-to-electron ratio changes upon cooling
Correct Answer:
Para-hydrogen becomes favored because its J = 0 state has lower rotational energy
Explanation:
Cooling reduces the population of higher rotational states. Since para-hydrogen can occupy the rotational ground state J = 0, while ortho-hydrogen cannot occupy that state, the equilibrium composition shifts strongly toward para-hydrogen at sufficiently low temperature.
5. Which nuclear-spin description is correct for para-hydrogen?
A.
A triplet state with total nuclear spin I = 1
B.
A singlet state with total nuclear spin I = 0
C.
A triplet state with total nuclear spin I = 2
D.
A singlet state with total nuclear spin I = 1
Correct Answer:
A singlet state with total nuclear spin I = 0
Explanation:
Para-hydrogen consists of two proton spins arranged antiparallel in an antisymmetric singlet nuclear-spin state. The total nuclear spin is I = 0, and this spin symmetry is coupled with the allowed even rotational levels.
6. Which statement about the magnetic behavior of the two spin isomers is most appropriate?
A.
Both are strongly ferromagnetic because hydrogen contains protons
B.
Ortho-hydrogen is associated with a triplet nuclear-spin state, whereas para-hydrogen is associated with a singlet state
C.
Para-hydrogen is triplet and ortho-hydrogen is singlet
D.
Their magnetic distinction arises from different numbers of electrons in the molecules
Correct Answer:
Ortho-hydrogen is associated with a triplet nuclear-spin state, whereas para-hydrogen is associated with a singlet state
Explanation:
Ortho-hydrogen has parallel proton spins and therefore forms a triplet nuclear-spin state with I = 1. Para-hydrogen has antiparallel proton spins and forms a singlet state with I = 0. Their distinction is therefore based on nuclear-spin coupling, not on different electron counts.
7. A sample of hydrogen is cooled to very low temperature and allowed to reach equilibrium. Which change is expected in its ortho-to-para composition?
A.
The proportion of ortho-hydrogen increases toward 100%
B.
The proportion of para-hydrogen increases substantially because the J = 0 state is accessible only to para-hydrogen
C.
The ortho and para proportions remain fixed at 3 : 1 because they are separate elements
D.
Both forms disappear because molecular hydrogen dissociates completely
Correct Answer:
The proportion of para-hydrogen increases substantially because the J = 0 state is accessible only to para-hydrogen
Explanation:
At low temperature, molecules preferentially occupy the lowest available energy state. Para-hydrogen can occupy J = 0, whereas ortho-hydrogen begins at J = 1, so equilibrium shifts toward para-hydrogen as the temperature decreases.
8. Why does a freshly prepared hydrogen sample generally not attain its equilibrium ortho-para composition instantly?
A.
Conversion requires interconversion of nuclear-spin and rotational states and is slow without suitable conditions
B.
The two forms have different molecular formulas and must undergo combustion first
C.
The conversion is prevented because hydrogen molecules cannot rotate
D.
Only the electronic spin changes, and electronic spin changes are forbidden in all circumstances
Correct Answer:
Conversion requires interconversion of nuclear-spin and rotational states and is slow without suitable conditions
Explanation:
Ortho- and para-hydrogen are spin isomers, and their interconversion involves changes in nuclear-spin symmetry together with the allowed rotational state. The process is slow in ordinary conditions but can be promoted by suitable catalysts or contact with certain surfaces.
9. Which method is most suitable for accelerating the conversion of ortho-hydrogen into para-hydrogen?
A.
Passing hydrogen over a suitable catalyst at low temperature
B.
Increasing the electron count of hydrogen by electrolysis
C.
Exposing hydrogen only to visible light in an evacuated vessel
D.
Compressing hydrogen without changing temperature or providing a conversion surface
Correct Answer:
Passing hydrogen over a suitable catalyst at low temperature
Explanation:
The ortho-to-para conversion is intrinsically slow but can be accelerated by catalysts or suitable paramagnetic surfaces. Low temperature favors the para form thermodynamically, while the catalyst provides a pathway for faster interconversion.
10. Which pairing of rotational quantum numbers with the hydrogen spin isomers is correct?