chemistry : class11 Orbital Wavefunction : Concept of orbitals : Structure of Atom MCQ Questions and Answers
Practice important
Orbital Wavefunction
MCQ questions and answers.
Page 1
of
3
|
30
MCQ Questions
1. For an electron described by a normalized orbital wavefunction ψ, which expression gives the probability of finding the electron within a small volume element dτ at a point in space?
A.
ψ dτ
B.
|ψ| dτ
C.
|ψ|² dτ
D.
ψ² dτ only when ψ is real
Correct Answer:
|ψ|² dτ
Explanation:
The probability density is given by |ψ|², so the probability of finding the electron in a small volume dτ is |ψ|²dτ. The wavefunction itself is not directly an observable probability.
2. An orbital wavefunction is multiplied everywhere by a constant phase factor e^(iφ), where φ is real. What happens to the electron probability distribution?
A.
It becomes zero for all values of φ
B.
It changes because the wavefunction amplitude changes
C.
It remains unchanged because |e^(iφ)ψ|² = |ψ|²
D.
It becomes negative for φ greater than 90°
Correct Answer:
It remains unchanged because |e^(iφ)ψ|² = |ψ|²
Explanation:
Multiplication by a phase factor changes the phase of ψ but not its modulus squared. Therefore, the observable probability density |ψ|² remains unchanged.
3. Which condition must a physically acceptable orbital wavefunction satisfy so that the total probability of finding the electron somewhere in space is unity?
A.
∫ψ dτ = 0
B.
∫|ψ|² dτ = 1
C.
∫|ψ| dτ = 1
D.
∫ψ² dτ = 0
Correct Answer:
∫|ψ|² dτ = 1
Explanation:
A normalized wavefunction must satisfy the condition that the integral of its probability density over all space equals one: ∫|ψ|²dτ = 1.
4. Two normalized orbital wavefunctions ψ₁ and ψ₂ satisfy ∫ψ₁*ψ₂ dτ = 0. What does this mathematical result indicate?
A.
The two orbitals have identical probability distributions
B.
The two wavefunctions are orthogonal and represent mutually distinct orbitals
C.
Both wavefunctions have zero probability everywhere
D.
The sum of the two wavefunctions is necessarily zero at every point
Correct Answer:
The two wavefunctions are orthogonal and represent mutually distinct orbitals
Explanation:
The integral ∫ψ₁*ψ₂dτ is the overlap or inner product of the two wavefunctions. If it is zero, the normalized wavefunctions are orthogonal, as required for distinct eigenfunctions belonging to suitable orbital states.
5. At a nodal surface of an orbital, which statement about its wavefunction and probability density is correct?
A.
ψ is maximum and |ψ|² is minimum but nonzero
B.
ψ is zero and |ψ|² is also zero
C.
ψ is undefined but |ψ|² remains constant
D.
ψ is negative and |ψ|² must also be negative
Correct Answer:
ψ is zero and |ψ|² is also zero
Explanation:
A node is a region or surface where the wavefunction has zero value. Since probability density is |ψ|², the probability density is also zero at that location.
6. For a real orbital wavefunction, two adjacent regions separated by a nodal surface may have opposite signs. What is the significance of this sign change?
A.
It indicates a change in probability density from positive to negative
B.
It represents a phase change of the wavefunction, while the probability density remains non-negative
C.
It proves that the electron has negative probability in one region
D.
It means the wavefunction is not normalized
Correct Answer:
It represents a phase change of the wavefunction, while the probability density remains non-negative
Explanation:
The sign of a real wavefunction denotes its phase. Although ψ may change sign across a node, |ψ|² remains non-negative and determines the observable probability density.
7. An orbital wavefunction is written as ψ = Nf, where f is a known function and N is a real normalization constant. If ∫f²dτ = 4, what is the positive value of N required for normalization?
A.
4
B.
2
C.
1/2
D.
1/4
Correct Answer:
1/2
Explanation:
Normalization requires ∫ψ²dτ = N²∫f²dτ = 1. Thus N²(4) = 1, giving N = 1/2 when the positive normalization constant is chosen.
8. If an orbital wavefunction has a nonzero value at a point, which conclusion is necessarily valid at that point?
A.
The electron must be found there
B.
The probability density is nonzero, but the exact probability at a mathematical point is not finite in the usual continuous description
C.
The electron has zero kinetic energy there
D.
The wavefunction itself is the probability of finding the electron there
Correct Answer:
The probability density is nonzero, but the exact probability at a mathematical point is not finite in the usual continuous description
Explanation:
A nonzero ψ implies a nonzero probability density |ψ|². Probability is assigned to a finite volume element as |ψ|²dτ; a single mathematical point has zero volume.
9. Which transformation changes the wavefunction but leaves all probability densities unchanged?
A.
ψ becomes 2ψ
B.
ψ becomes ψ + 1 everywhere
C.
ψ becomes −ψ
D.
ψ becomes ψ²
Correct Answer:
ψ becomes −ψ
Explanation:
Replacing ψ by −ψ changes only its overall phase by π. Since |−ψ|² = |ψ|², the probability density is unchanged. Multiplying by 2 would also require renormalization and is not itself an unchanged normalized wavefunction.
10. For two orbitals having wavefunctions ψ₁ and ψ₂, under what condition can they be regarded as non-overlapping in the quantum-mechanical orthogonality sense?
A.
Their wavefunctions have equal magnitudes everywhere
B.
Their probability densities have the same maximum value
C.
The integral of ψ₁*ψ₂ over all space is zero
D.
The integral of |ψ₁|²|ψ₂|² over all space is exactly one
Correct Answer:
The integral of ψ₁*ψ₂ over all space is zero
Explanation:
Orthogonality of orbital wavefunctions is defined by the zero inner product, ∫ψ₁*ψ₂dτ = 0. It means the wavefunctions have zero net overlap in the Hilbert-space sense, even if their spatial regions may partly coincide.