chemistry : class11 Probability Density : Concept of orbitals : Structure of Atom MCQ Questions and Answers
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Probability Density
MCQ questions and answers.
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MCQ Questions
1. At two points in an orbital, the wavefunction values are ψ₁ = 2.0 × 10⁻³ nm⁻³ᐟ² and ψ₂ = −5.0 × 10⁻³ nm⁻³ᐟ². What is the ratio of the probability densities ρ₁:ρ₂ at these points?
A.
2:5
B.
4:25
C.
−2:5
D.
25:4
Correct Answer:
4:25
Explanation:
Probability density is ρ = |ψ|². Therefore, ρ₁:ρ₂ = |2.0|²:|−5.0|² = 4:25. The negative sign of ψ₂ has no effect on density.
2. Which mathematical condition correctly expresses the total probability of finding an electron anywhere in space when ψ is a normalized wavefunction?
A.
∫ψ dτ = 0
B.
∫|ψ| dτ = 1
C.
∫|ψ|² dτ = 1
D.
∫ψ² dτ = 0
Correct Answer:
∫|ψ|² dτ = 1
Explanation:
The probability density is |ψ|², and its integral over all space must equal unity for a normalized wavefunction.
3. For a real-valued orbital wavefunction, ψ changes sign on crossing a nodal surface. What happens to the probability density at corresponding points on the two sides of the node?
A.
It must change sign because ψ changes sign
B.
It remains unchanged if the magnitudes of ψ are equal
C.
It becomes negative on one side of the node
D.
It becomes infinite at the nodal surface
Correct Answer:
It remains unchanged if the magnitudes of ψ are equal
Explanation:
Probability density depends on |ψ|², not on the sign of ψ. Thus, equal magnitudes of ψ on opposite sides give equal probability densities, even when the wavefunction changes phase.
4. At an exact node of an orbital, the wavefunction is zero. What is the probability of finding the electron in an infinitesimally small volume element centered exactly at that node?
A.
Exactly zero
B.
Exactly one
C.
Equal to the value of ψ at the node
D.
Maximum because the electron is confined there
Correct Answer:
Exactly zero
Explanation:
At a node, ψ = 0, so the probability density |ψ|² is zero. Consequently, the probability for a volume element centered exactly at the node is zero.
5. For a 2p_z orbital, the angular part of the wavefunction is proportional to cos θ, where θ is measured from the z-axis. At two points having the same radial distance, one at θ = 0° and the other at θ = 60°, what is the ratio of their probability densities?
A.
1:2
B.
1:4
C.
2:1
D.
4:1
Correct Answer:
4:1
Explanation:
At the same radial distance, density varies as |cos θ|². Thus, ρ(0°):ρ(60°) = cos²0°:cos²60° = 1:(1/2)² = 4:1.
6. Which statement best distinguishes probability density from the probability of finding an electron in a finite region?
A.
Probability density is obtained by integrating |ψ|², whereas probability is |ψ|² at one point
B.
Probability density is |ψ|² at a point, whereas probability is the integral of |ψ|² over a volume
C.
Both are always numerically identical and dimensionless
D.
Probability density depends on volume, but probability does not
Correct Answer:
Probability density is |ψ|² at a point, whereas probability is the integral of |ψ|² over a volume
Explanation:
The local probability density is ρ = |ψ|². The probability of finding the electron in a finite region is obtained by integrating ρ dτ over that region.
7. Two wavefunctions ψ and −ψ describe the same spatial orbital. Which statement about their probability densities is correct at every point in space?
A.
The density for −ψ is the negative of that for ψ
B.
The density for −ψ is twice that for ψ
C.
The densities are identical
D.
The density for −ψ is zero wherever ψ is nonzero
Correct Answer:
The densities are identical
Explanation:
Since |−ψ|² = |ψ|², multiplying a wavefunction by −1 changes its phase but leaves the probability density unchanged.
8. A small volume element Δτ is located at a point where the probability density is 3.0 × 10⁻²⁰ m⁻³. If Δτ = 2.0 × 10⁻²⁹ m³ and the density is approximately constant throughout the element, what is the probability of finding the electron there?
A.
6.0 × 10⁻⁴⁹
B.
1.5 × 10⁹
C.
6.0 × 10⁻⁴⁷
D.
6.0 × 10⁻⁵¹
Correct Answer:
6.0 × 10⁻⁴⁹
Explanation:
For a sufficiently small volume, probability is approximately ρΔτ. Hence, P = (3.0 × 10⁻²⁰)(2.0 × 10⁻²⁹) = 6.0 × 10⁻⁴⁹.
9. At the nucleus, which comparison of probability densities is correct for hydrogen-like 1s and 2p orbitals?
A.
Both densities are zero
B.
The 1s density is nonzero, whereas the 2p density is zero
C.
The 1s density is zero, whereas the 2p density is nonzero
D.
Both densities are infinite
Correct Answer:
The 1s density is nonzero, whereas the 2p density is zero
Explanation:
The 1s wavefunction has a finite nonzero value at r = 0, so its probability density is nonzero there. A p-orbital has a radial factor proportional to r near the nucleus, making ψ and therefore |ψ|² zero at r = 0.
10. A normalized orbital has a radial probability density P(r) = 4πr²|R(r)|² for a spherically symmetric state. Which statement is correct about P(r) and the local probability density |R(r)|²?
A.
They are identical at every r because both represent probability
B.
P(r) includes the spherical-shell volume factor 4πr², whereas |R(r)|² is a local density factor
C.
P(r) is always zero at every nonzero r
D.
P(r) is obtained by dividing |R(r)|² by 4πr²
Correct Answer:
P(r) includes the spherical-shell volume factor 4πr², whereas |R(r)|² is a local density factor
Explanation:
The local probability density is determined by |ψ|², while the probability in a spherical shell also depends on the shell volume. Therefore, the radial probability distribution contains the factor 4πr².