Quantum Chemistry Questions and Answers

Q: Rigid rotor wavefunctions are given by \(Y_{l,m}(\theta, \phi)\). The wavefunctions \(Y_{1,0}(\theta, \phi)\) and \(Y_{2,0}(\theta, \phi)\) are given below\[ Y_{1,0}(\theta, \phi) = \sqrt{\frac{3}{4\pi}} \cos \theta \]\[ \begin{aligned} Y_{2,0}(\theta, \phi) = &\sqrt{\frac{5}{16\pi}} \\ &\times (3\cos^2 \theta - 1) \end{aligned} \]For a non-polar diatomic molecule, the value of transition dipole moment integral for transition between \(Y_{1,0}(\theta, \phi)\) and \(Y_{2,0}(\theta, \phi)\) is equal to

The integral equals 0.The transition dipole moment for a rotational transition is\[ \begin{aligned} \mu_{fi} = \mu_0 &\int Y_{2,0}^{*} \cos\theta \\ &\times Y_{1,0} \, d\tau \end{aligned} \]where \(\mu_0\) is the molecule's permanent dipole moment. For a non-polar diatomic molecule (H2, N2, O2), \(\mu_0 = 0\), so the integral vanishes whatever the angular factor.This is why homonuclear diatomics have no pure rotational (microwave) spectrum, even though the change \(l = 1 \to 2\) obeys the selection rule \(\Delta l = \pm 1\). The non-zero options would only apply to a polar molecule, and then they would scale with \(\mu_0\).

Q: The correct option with regard to the following statements is(a) Time-independent Schrödinger equation can be exactly solved for Be2+.(b) For a particle confined in a one-dimensional box of length l with infinite potential barriers, the trial variation function \(\phi = \left[\left(\frac{3}{l^3}\right)^{1/2} x\right]\) is not an acceptable trial wavefunction for \(0 \le x \le l\).(c) Wavefunctions for system of Fermions must be anti-symmetric with respect to exchange of any two Fermions in the system.(d) Born-Oppenheimer approximation can be used to separate the vibrational and rotational motion of a molecule.

The correct option is (a) False (b) True (c) True (d) False.(a) False: Be2+ still has two electrons, and the electron-electron repulsion makes the Schrödinger equation impossible to solve exactly; only one-electron species such as H, He+ and Be3+ can be solved exactly.(b) True: a trial function for a box with infinite walls must vanish at both walls. \(\phi \propto x\) is zero at \(x = 0\) but not at \(x = l\), so it is not acceptable.(c) True: this is the Pauli principle in its general form for fermions.(d) False: the Born-Oppenheimer approximation separates electronic from nuclear motion. Separating vibration from rotation is a further, different approximation (the rigid rotor and harmonic oscillator picture). That fixes the pattern F, T, T, F, which only one option has.

Q: The correct option for the average value of kinetic energy and the average value of potential energy of a one-dimensional harmonic oscillator with frequency \(\nu\) in its ground state is

Both averages are \(\frac{h\nu}{4}\), so the answer is \(\frac{h\nu}{4}\) and \(\frac{h\nu}{4}\), respectively.The ground-state energy of the harmonic oscillator is the zero-point energy:\[ \begin{aligned} E_{v} &= \left(v + \tfrac{1}{2}\right)h\nu \\ E_{0} &= \tfrac{1}{2}h\nu \end{aligned} \]For a potential proportional to \(x^{2}\), the virial theorem gives \(2\langle T \rangle = 2\langle V \rangle\), so the kinetic and potential energies are equal on average and each is half the total:\[ \begin{aligned} \langle T \rangle = \langle V \rangle &= \tfrac{1}{2}E_{0} \\ &= \tfrac{1}{4}h\nu \end{aligned} \]The options with \(\frac{h\nu}{2}\) for either term give that term the whole zero-point energy, and the pair \(\frac{h\nu}{2}\), \(\frac{h\nu}{2}\) would make the total \(h\nu\), twice the zero-point energy.

Q: A particle in a one-dimensional box of length L can be excited from n = 1 to n = 2 state with frequency ν. If the box length is doubled, the frequency needed to achieve this transition is :

The frequency becomes ν/4.For a particle in a box the energy levels and the 1 → 2 transition energy are\[ \begin{aligned} E_{n} &= \frac{n^{2}h^{2}}{8mL^{2}} \\ h\nu &= E_{2} - E_{1} = \frac{3h^{2}}{8mL^{2}} \end{aligned} \]The frequency is proportional to 1/L2. Doubling L divides it by 22 = 4, so the new frequency is ν/4.ν/2 assumes the energy varies as 1/L, and 2ν and 4ν have the dependence the wrong way round: a longer box lowers the energy levels and brings them closer together.

Q: If operator \(\hat{D}^{n}\) (n times differentiate with respect to x) operates on eigenfunction eax, the eigenvalue would be :

The eigenvalue is an, printed in the options as αn.Each differentiation of eax brings down one factor of a:\[ \begin{aligned} \hat{D}\,e^{ax} &= a\,e^{ax} \\ \hat{D}^{n}e^{ax} &= a^{n}e^{ax} \end{aligned} \]So eax is an eigenfunction of the n-th derivative operator with eigenvalue an. The options write the constant as α. A product na would come from adding the factor n times, na swaps base and exponent, and n! would arise only for differentiating a power xn, not an exponential.

Q: The eigenfunctions of a rigid rotator are :

The eigenfunctions of a rigid rotator are the spherical harmonics.A rigid rotor has only kinetic energy of rotation, so its Schrodinger equation is the angular part of the Laplacian. Its solutions are the spherical harmonics YJM(θ, φ), products of associated Legendre functions in θ and eiMφ in φ, with energies E = J(J + 1)h2/8π2I.Associated Laguerre polynomials belong to the radial part of the hydrogen atom, and Hermite polynomials to the harmonic oscillator. 'Legendre transformers' is not a family of functions; Legendre transforms are a device of thermodynamics and mechanics.

Q: According to simple HMO theory, which one of the following statements about butadiene is true ?

The true statement is that exciting an electron from the HOMO to the LUMO will not change the π charge densities.Butadiene is an alternant hydrocarbon. By the Coulson-Rushbrooke pairing theorem its bonding and antibonding Huckel orbitals come in pairs with coefficients of equal magnitude at every atom. Since the HOMO and LUMO coefficients have the same squares at each carbon, moving an electron from one to the other leaves every π charge density at 1.The bond orders do change, because the HOMO is bonding and the LUMO antibonding between C1-C2, while the reverse holds for C2-C3, so option B is false. Butadiene's four MOs have energies α ± 1.618β and α ± 0.618β, none non-bonding, and electrophiles attack the terminal carbons, which have the largest HOMO coefficients.

Q: Given below is the wave function of a particle :Which of the following is correct ?

The particle is more likely to be found near A.The plotted quantity is |ψ|2, the probability density. The probability of finding the particle in a small interval dx is |ψ|2dx, the shaded area. The curve has its highest value near A, so the particle is most likely to be found there.At B the curve touches zero, a node, so the particle is least likely to be found near B. The shaded area is a probability, not an energy, and not an amplitude; the amplitude is ψ itself, whose square is plotted, and dx is a length, not an area.

Q: The de Broglie wavelength of an electron travelling with 1% speed of light is about :(mass of electron and speed of light are 9.109 × 10−31 kg and 2.998 × 108 ms−1 respectively) :

The wavelength is about 243 pm.The speed is 1% of c, so v = 2.998 × 106 m s−1, which is slow enough to ignore relativity. Then\[ \begin{aligned} \lambda &= \frac{h}{mv} \\ &= \frac{6.626 \times 10^{-34}}{9.109 \times 10^{-31} \times 2.998 \times 10^{6}} \\ &= 2.43 \times 10^{-10}\ \text{m} = 243\ \text{pm} \end{aligned} \]The other options keep the digits 243 but attach the wrong unit; 243 Å would be 2.43 × 10−8 m, a hundred times too long.

Q: The momentum operator in one-dimension is ...............

The momentum operator is \(-i\hbar \frac{\partial}{\partial x}\).In quantum mechanics the x-component of linear momentum is represented by \(\hat{p}_{x} = -i\hbar\, \partial/\partial x\). Acting on a plane wave \(e^{ikx}\) it returns \(\hbar k\) times the function, the de Broglie momentum, which confirms the form.Without the factor -i the operator would not be Hermitian and its eigenvalues would not be real. Derivatives with respect to time belong to the energy operator, \(i\hbar\, \partial/\partial t\), not to momentum, and the last option adds i to an operator, which has no meaning.

11.

Rigid rotor wavefunctions are given by \(Y_{l,m}(\theta, \phi)\). The wavefunctions \(Y_{1,0}(\theta, \phi)\) and \(Y_{2,0}(\theta, \phi)\) are given below

\[ Y_{1,0}(\theta, \phi) = \sqrt{\frac{3}{4\pi}} \cos \theta \]

\[ \begin{aligned} Y_{2,0}(\theta, \phi) = &\sqrt{\frac{5}{16\pi}} \\ &\times (3\cos^2 \theta - 1) \end{aligned} \]

For a non-polar diatomic molecule, the value of transition dipole moment integral for transition between \(Y_{1,0}(\theta, \phi)\) and \(Y_{2,0}(\theta, \phi)\) is equal to

A.\(1/\sqrt{2\pi}\)
B.0
C.2
D.\(1/\sqrt{4\pi}\)

Exam Year(s)

GATE Chemistry - 2025,

Answer: B

The integral equals 0.

The transition dipole moment for a rotational transition is

\[ \begin{aligned} \mu_{fi} = \mu_0 &\int Y_{2,0}^{*} \cos\theta \\ &\times Y_{1,0} \, d\tau \end{aligned} \]

where \(\mu_0\) is the molecule's permanent dipole moment. For a non-polar diatomic molecule (H₂, N₂, O₂), \(\mu_0 = 0\), so the integral vanishes whatever the angular factor.

This is why homonuclear diatomics have no pure rotational (microwave) spectrum, even though the change \(l = 1 \to 2\) obeys the selection rule \(\Delta l = \pm 1\). The non-zero options would only apply to a polar molecule, and then they would scale with \(\mu_0\).

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12.

The correct option with regard to the following statements is

(a) Time-independent Schrödinger equation can be exactly solved for Be²⁺.
(b) For a particle confined in a one-dimensional box of length l with infinite potential barriers, the trial variation function \(\phi = \left[\left(\frac{3}{l^3}\right)^{1/2} x\right]\) is not an acceptable trial wavefunction for \(0 \le x \le l\).
(c) Wavefunctions for system of Fermions must be anti-symmetric with respect to exchange of any two Fermions in the system.
(d) Born-Oppenheimer approximation can be used to separate the vibrational and rotational motion of a molecule.

A.(a) True (b) False (c) False (d) True
B.(a) True (b) True (c) False (d) False
C.(a) False (b) True (c) True (d) False
D.(a) False (b) True (c) True (d) True

Exam Year(s)

GATE Chemistry - 2025,

Answer: C

The correct option is (a) False (b) True (c) True (d) False.

(a) False: Be²⁺ still has two electrons, and the electron-electron repulsion makes the Schrödinger equation impossible to solve exactly; only one-electron species such as H, He⁺ and Be³⁺ can be solved exactly.

(b) True: a trial function for a box with infinite walls must vanish at both walls. \(\phi \propto x\) is zero at \(x = 0\) but not at \(x = l\), so it is not acceptable.

(d) False: the Born-Oppenheimer approximation separates electronic from nuclear motion. Separating vibration from rotation is a further, different approximation (the rigid rotor and harmonic oscillator picture). That fixes the pattern F, T, T, F, which only one option has.

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13.

The correct option for the average value of kinetic energy and the average value of potential energy of a one-dimensional harmonic oscillator with frequency \(\nu\) in its ground state is

A.\(\frac{h\nu}{2}\) and \(\frac{h\nu}{4}\), respectively
B.\(\frac{h\nu}{2}\) and \(\frac{h\nu}{2}\), respectively
C.\(\frac{h\nu}{4}\) and \(\frac{h\nu}{4}\), respectively
D.\(\frac{h\nu}{4}\) and \(\frac{h\nu}{2}\), respectively

Exam Year(s)

GATE Chemistry - 2026,

Answer: C

Both averages are \(\frac{h\nu}{4}\), so the answer is \(\frac{h\nu}{4}\) and \(\frac{h\nu}{4}\), respectively.

The ground-state energy of the harmonic oscillator is the zero-point energy:

\[ \begin{aligned} E_{v} &= \left(v + \tfrac{1}{2}\right)h\nu \\ E_{0} &= \tfrac{1}{2}h\nu \end{aligned} \]

For a potential proportional to \(x^{2}\), the virial theorem gives \(2\langle T \rangle = 2\langle V \rangle\), so the kinetic and potential energies are equal on average and each is half the total:

\[ \begin{aligned} \langle T \rangle = \langle V \rangle &= \tfrac{1}{2}E_{0} \\ &= \tfrac{1}{4}h\nu \end{aligned} \]

The options with \(\frac{h\nu}{2}\) for either term give that term the whole zero-point energy, and the pair \(\frac{h\nu}{2}\), \(\frac{h\nu}{2}\) would make the total \(h\nu\), twice the zero-point energy.

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14.

A particle in a one-dimensional box of length L can be excited from n = 1 to n = 2 state with frequency ν. If the box length is doubled, the frequency needed to achieve this transition is :

A.\(\nu/4\)
B.\(\nu/2\)
C.\(2\nu\)
D.\(4\nu\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: A

The frequency becomes ν/4.

For a particle in a box the energy levels and the 1 → 2 transition energy are

\[ \begin{aligned} E_{n} &= \frac{n^{2}h^{2}}{8mL^{2}} \\ h\nu &= E_{2} - E_{1} = \frac{3h^{2}}{8mL^{2}} \end{aligned} \]

The frequency is proportional to 1/L². Doubling L divides it by 2² = 4, so the new frequency is ν/4.

ν/2 assumes the energy varies as 1/L, and 2ν and 4ν have the dependence the wrong way round: a longer box lowers the energy levels and brings them closer together.

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15.

If operator \(\hat{D}^{n}\) (n times differentiate with respect to x) operates on eigenfunction e^ax, the eigenvalue would be :

A.\(n\alpha\)
B.\(n^{\alpha}\)
C.\(\alpha^{n}\)
D.\(n!\,\alpha\)

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

The eigenvalue is aⁿ, printed in the options as αⁿ.

Each differentiation of e^ax brings down one factor of a:

\[ \begin{aligned} \hat{D}\,e^{ax} &= a\,e^{ax} \\ \hat{D}^{n}e^{ax} &= a^{n}e^{ax} \end{aligned} \]

So e^ax is an eigenfunction of the n-th derivative operator with eigenvalue aⁿ. The options write the constant as α. A product na would come from adding the factor n times, n^a swaps base and exponent, and n! would arise only for differentiating a power xⁿ, not an exponential.

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16.

The eigenfunctions of a rigid rotator are :

A.Spherical harmonics
B.Laguerre polynomials
C.Hermite polynomials
D.Legendre transformers

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: A

The eigenfunctions of a rigid rotator are the spherical harmonics.

A rigid rotor has only kinetic energy of rotation, so its Schrodinger equation is the angular part of the Laplacian. Its solutions are the spherical harmonics Y_J^M(θ, φ), products of associated Legendre functions in θ and e^iMφ in φ, with energies E = J(J + 1)h²/8π²I.

Associated Laguerre polynomials belong to the radial part of the hydrogen atom, and Hermite polynomials to the harmonic oscillator. 'Legendre transformers' is not a family of functions; Legendre transforms are a device of thermodynamics and mechanics.

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17.

According to simple HMO theory, which one of the following statements about butadiene is true ?

A.It has a non-bonding MO
B.Exciting an electron from the HOMO to the LUMO will not change the π bond orders
C.Exciting an electron from the HOMO to the LUMO will not change the π charge densities
D.It is most likely to undergo electrophilic substitution at one of the two inner carbons

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

The true statement is that exciting an electron from the HOMO to the LUMO will not change the π charge densities.

Butadiene is an alternant hydrocarbon. By the Coulson-Rushbrooke pairing theorem its bonding and antibonding Huckel orbitals come in pairs with coefficients of equal magnitude at every atom. Since the HOMO and LUMO coefficients have the same squares at each carbon, moving an electron from one to the other leaves every π charge density at 1.

The bond orders do change, because the HOMO is bonding and the LUMO antibonding between C1-C2, while the reverse holds for C2-C3, so option B is false. Butadiene's four MOs have energies α ± 1.618β and α ± 0.618β, none non-bonding, and electrophiles attack the terminal carbons, which have the largest HOMO coefficients.

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18.

Given below is the wave function of a particle :

Plot of the square of psi against x: a small hump left of the origin, the tallest peak above point A, a node at point B, a small hump near C, and a thin shaded strip of width dx just left of A.

Which of the following is correct ?

A.The particle is more likely to be found near B
B.The particle is more likely to be found near A
C.The shaded area represents the energy of the particle when it travels through a distance dx
D.The shaded area represents the amplitude of the particle in an area dx

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: B

The particle is more likely to be found near A.

The plotted quantity is |ψ|², the probability density. The probability of finding the particle in a small interval dx is |ψ|²dx, the shaded area. The curve has its highest value near A, so the particle is most likely to be found there.

At B the curve touches zero, a node, so the particle is least likely to be found near B. The shaded area is a probability, not an energy, and not an amplitude; the amplitude is ψ itself, whose square is plotted, and dx is a length, not an area.

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19.

The de Broglie wavelength of an electron travelling with 1% speed of light is about :

(mass of electron and speed of light are 9.109 × 10⁻³¹ kg and 2.998 × 10⁸ ms⁻¹ respectively) :

A.243 m
B.243 cm
C.243 Å
D.243 pm

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

The wavelength is about 243 pm.

The speed is 1% of c, so v = 2.998 × 10⁶ m s⁻¹, which is slow enough to ignore relativity. Then

\[ \begin{aligned} \lambda &= \frac{h}{mv} \\ &= \frac{6.626 \times 10^{-34}}{9.109 \times 10^{-31} \times 2.998 \times 10^{6}} \\ &= 2.43 \times 10^{-10}\ \text{m} = 243\ \text{pm} \end{aligned} \]

The other options keep the digits 243 but attach the wrong unit; 243 Å would be 2.43 × 10⁻⁸ m, a hundred times too long.

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20.

The momentum operator in one-dimension is ...............

A.\(-\hbar \frac{\partial}{\partial x}\)
B.\(-i(\frac{\hbar\partial}{\partial t})-\hbar\frac{\partial}{\partial t}\)
C.\(-i\hbar \frac{\partial}{\partial x}\)
D.\(i - \hbar \frac{\partial}{\partial x}\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: C

The momentum operator is \(-i\hbar \frac{\partial}{\partial x}\).

In quantum mechanics the x-component of linear momentum is represented by \(\hat{p}_{x} = -i\hbar\, \partial/\partial x\). Acting on a plane wave \(e^{ikx}\) it returns \(\hbar k\) times the function, the de Broglie momentum, which confirms the form.

Without the factor -i the operator would not be Hermitian and its eigenvalues would not be real. Derivatives with respect to time belong to the energy operator, \(i\hbar\, \partial/\partial t\), not to momentum, and the last option adds i to an operator, which has no meaning.

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