Quantum Chemistry Questions and Answers

Q: The result of the following operation\(\widehat{A}\,(e^{ikx})\) = &because; \(\widehat{A} = -i\hbar\,\dfrac{d}{dx}\)is :

The result is \(\hbar k\,e^{ikx}\), option C.Apply the momentum operator to the function:\[ \begin{aligned} \widehat{A}\,e^{ikx} &= -i\hbar\,\frac{d}{dx}e^{ikx} \\ &= -i\hbar\,(ik)\,e^{ikx} \\ &= \hbar k\,e^{ikx} \end{aligned} \]since \(-i \times i = 1\). So eikx is an eigenfunction of the momentum operator with eigenvalue \(\hbar k\). Option A would mean the operator had no effect, B leaves out the factor \(\hbar\), and D changes the exponent to kx2, which differentiation of eikx never produces.

Q: By applying variation method, the ground state energy of the electron in hydrogen atom is (in a.u.) :

The variational ground-state energy of hydrogen is −0.5 a.u.With the trial function e−αr, the energy expectation value in atomic units is a function of α that is minimised as follows:\[ \begin{aligned} E(\alpha) &= \frac{\alpha^{2}}{2} - \alpha \\ \frac{dE}{d\alpha} &= \alpha - 1 = 0 \Rightarrow \alpha = 1 \\ E &= \frac{1}{2} - 1 = -0.5\ \text{hartree} \end{aligned} \]This equals the exact energy, −13.6 eV, because the trial function has the form of the true 1s function. −2 a.u. is the energy of He+, −2.75 a.u. is the simple variational estimate for helium with an unscreened charge, and 0 is the ionisation limit.

Q: The position and momentum operators do not commute because :

The correct pair is the first one: x̂P̂xΨ gives only the derivative term, while P̂xx̂Ψ also gives Ψ.With \(\hat{P}_{x} = \frac{\hbar}{i}\frac{d}{dx}\):\[ \begin{aligned} \hat{x}\hat{P}_{x}\Psi &= x\,\frac{\hbar}{i}\frac{d\Psi}{dx} \\ \hat{P}_{x}\hat{x}\Psi &= \frac{\hbar}{i}\frac{d(x\Psi)}{dx} \\ &= \frac{\hbar}{i}\left(\Psi + x\frac{d\Psi}{dx}\right) \end{aligned} \]The product rule produces the extra \(\Psi\), so the two orders differ and \([\hat{x}, \hat{P}_{x}] = i\hbar\). The second option assigns the two results to the wrong orders, the third claims they are equal, and the fourth drops the factor x.

Q: For the particle in 1-D box, the value of quantum number n = 0 is ruled out because :

n = 0 is ruled out because it implies \(\Psi = 0\) everywhere in the box.The wave functions of a particle in a box of length L are \(\Psi_{n} = \sqrt{2/L}\,\sin(n\pi x/L)\). Putting n = 0 gives \(\sin 0 = 0\) at every x, so \(\Psi = 0\) throughout. Then \(|\Psi|^{2}\) is zero everywhere and the probability of finding the particle anywhere is zero, which means there is no particle at all.The energy \(E_{n} = n^{2}h^{2}/8mL^{2}\) would simply be zero, not divergent, and the wave function would vanish rather than diverge. Nothing about n = 0 makes the particle classical; the lowest allowed state is n = 1, with non-zero zero-point energy.

Q: The term representing electron-nucleus interaction in Hamiltonian operator for many electron atom is proportional to :

The electron-nucleus term is \(-\sum_{i} Ze^{2}/r_{i}\).Each electron i is attracted by the nucleus of charge +Ze at distance ri. The potential energy of that attraction is -Ze2/ri (times 1/4πε0 in SI units), and the Hamiltonian sums it over all electrons. The minus sign shows attraction.Terms in e2/rij are the electron-electron repulsions, summed over pairs, and are positive. Ze/ri lacks one factor of the charge and has the wrong sign, so it is not an energy of interaction.

Q: For a butadiene molecule, the energy values of levels obtained by using HMO theory are :(1) α - 1.6 β(2) α + 1.6 β(3) α + 0.6 β(4) α - 0.6 βWhat is the sequence of levels in the order of increasing energy ?

The order of increasing energy is (2), (3), (4), (1).In Huckel theory the resonance integral β is negative, so adding a multiple of β to α lowers the energy. For butadiene the four levels are α + 1.62β, α + 0.62β, α - 0.62β and α - 1.62β, in that order from lowest to highest.Lowest: α + 1.6β (2), the fully bonding orbital.Then α + 0.6β (3), the HOMO.Then α - 0.6β (4), the LUMO.Highest: α - 1.6β (1).Option A lists them as numbered, and option D treats β as if it were positive, which reverses the order.

31.

The result of the following operation

\(\widehat{A}\,(e^{ikx})\) = ∵ \(\widehat{A} = -i\hbar\,\dfrac{d}{dx}\)

is :

A.\(e^{ikx}\)
B.\(k\,e^{ikx}\)
C.\(k\hbar\,e^{ikx}\)
D.\(k\hbar\,e^{ikx^{2}}\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

The result is \(\hbar k\,e^{ikx}\), option C.

Apply the momentum operator to the function:

\[ \begin{aligned} \widehat{A}\,e^{ikx} &= -i\hbar\,\frac{d}{dx}e^{ikx} \\ &= -i\hbar\,(ik)\,e^{ikx} \\ &= \hbar k\,e^{ikx} \end{aligned} \]

since \(-i \times i = 1\). So e^ikx is an eigenfunction of the momentum operator with eigenvalue \(\hbar k\). Option A would mean the operator had no effect, B leaves out the factor \(\hbar\), and D changes the exponent to kx², which differentiation of e^ikx never produces.

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32.

For a stone weighing 100 g located within 0.1 Å, which of the following statements is incorrect ?

(Given : \(\dfrac{h}{4\pi}\) = 5.27 × 10⁻³⁵ kg. m².s⁻¹).

A.Uncertainty in velocity is negligible
B.Both position and velocity can be determined with reasonable precision
C.Position of stone is precisely known
D.Velocity cannot be measured accurately

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: D

The incorrect statement is D, 'Velocity cannot be measured accurately'.

The position is fixed to \(\Delta x = 0.1\ \text{Å} = 10^{-11}\ \text{m}\). The uncertainty principle gives

\[ \begin{aligned} \Delta p &\ge \frac{h/4\pi}{\Delta x} = \frac{5.27\times 10^{-35}}{10^{-11}} \\ &= 5.27\times 10^{-24}\ \text{kg m s}^{-1} \\ \Delta v &= \frac{\Delta p}{m} = \frac{5.27\times 10^{-24}}{0.1} \\ &\approx 5\times 10^{-23}\ \text{m s}^{-1} \end{aligned} \]

This is negligible, so statement A is true. Both position and velocity can therefore be known to very high precision (B), and a position known to 0.1 Å is precisely known for a stone (C). Only D contradicts the calculation.

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33.

By applying variation method, the ground state energy of the electron in hydrogen atom is (in a.u.) :

A.-2.75
B.-0.5
C.-2
D.0

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: B

The variational ground-state energy of hydrogen is −0.5 a.u.

With the trial function e^−αr, the energy expectation value in atomic units is a function of α that is minimised as follows:

\[ \begin{aligned} E(\alpha) &= \frac{\alpha^{2}}{2} - \alpha \\ \frac{dE}{d\alpha} &= \alpha - 1 = 0 \Rightarrow \alpha = 1 \\ E &= \frac{1}{2} - 1 = -0.5\ \text{hartree} \end{aligned} \]

This equals the exact energy, −13.6 eV, because the trial function has the form of the true 1s function. −2 a.u. is the energy of He⁺, −2.75 a.u. is the simple variational estimate for helium with an unscreened charge, and 0 is the ionisation limit.

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34.

The position and momentum operators do not commute because :

A.\(\hat{x}\hat{P}_{x}\Psi = x * \frac{\hbar}{i}\frac{d\Psi}{dx}\) and \(\hat{P}_{x}\hat{x}\Psi = \frac{\hbar}{i}\left(\Psi + x\frac{d\Psi}{dx}\right)\)
B.\(\hat{x}\hat{P}_{x}\Psi = \frac{\hbar}{i}\left(\Psi + x\frac{d\Psi}{dx}\right)\) and \(\hat{P}_{x}\hat{x}\Psi = \frac{x\hbar}{i}\frac{d\Psi}{dx}\)
C.\(\hat{x}\hat{P}_{x}\Psi = \hat{P}_{x}x\Psi = \frac{x\hbar}{i}\frac{d\Psi}{dx}\)
D.\(\hat{x}\hat{P}_{x}\Psi = \frac{\hbar}{i}\frac{d\Psi}{dx}\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: A

The correct pair is the first one: x̂P̂_xΨ gives only the derivative term, while P̂_xx̂Ψ also gives Ψ.

With \(\hat{P}_{x} = \frac{\hbar}{i}\frac{d}{dx}\):

\[ \begin{aligned} \hat{x}\hat{P}_{x}\Psi &= x\,\frac{\hbar}{i}\frac{d\Psi}{dx} \\ \hat{P}_{x}\hat{x}\Psi &= \frac{\hbar}{i}\frac{d(x\Psi)}{dx} \\ &= \frac{\hbar}{i}\left(\Psi + x\frac{d\Psi}{dx}\right) \end{aligned} \]

The product rule produces the extra \(\Psi\), so the two orders differ and \([\hat{x}, \hat{P}_{x}] = i\hbar\). The second option assigns the two results to the wrong orders, the third claims they are equal, and the fourth drops the factor x.

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35.

For the particle in 1-D box, the value of quantum number n = 0 is ruled out because :

A.The energy of the particle becomes diverging at n = 0
B.The wave function becomes divergent at n = 0
C.Particle starts following classical mechanics principle
D.It implies \(\Psi = 0\), everywhere in the box

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: D

n = 0 is ruled out because it implies \(\Psi = 0\) everywhere in the box.

The wave functions of a particle in a box of length L are \(\Psi_{n} = \sqrt{2/L}\,\sin(n\pi x/L)\). Putting n = 0 gives \(\sin 0 = 0\) at every x, so \(\Psi = 0\) throughout. Then \(|\Psi|^{2}\) is zero everywhere and the probability of finding the particle anywhere is zero, which means there is no particle at all.

The energy \(E_{n} = n^{2}h^{2}/8mL^{2}\) would simply be zero, not divergent, and the wave function would vanish rather than diverge. Nothing about n = 0 makes the particle classical; the lowest allowed state is n = 1, with non-zero zero-point energy.

Discuss in Forum

36.

The term representing electron-nucleus interaction in Hamiltonian operator for many electron atom is proportional to :

A.\(\sum_{i}\frac{e^{2}}{r_{ij}}\)
B.\(-\sum_{i}\frac{Ze^{2}}{r_{i}}\)
C.\(\sum_{i}\frac{Ze}{r_{i}}\)
D.\(-\frac{Ze^{2}}{r_{ij}}\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: B

The electron-nucleus term is \(-\sum_{i} Ze^{2}/r_{i}\).

Each electron i is attracted by the nucleus of charge +Ze at distance r_i. The potential energy of that attraction is -Ze²/r_i (times 1/4πε₀ in SI units), and the Hamiltonian sums it over all electrons. The minus sign shows attraction.

Terms in e²/r_ij are the electron-electron repulsions, summed over pairs, and are positive. Ze/r_i lacks one factor of the charge and has the wrong sign, so it is not an energy of interaction.

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37.

For a butadiene molecule, the energy values of levels obtained by using HMO theory are :

(1) α - 1.6 β
(2) α + 1.6 β
(3) α + 0.6 β
(4) α - 0.6 β

What is the sequence of levels in the order of increasing energy ?

A.(1), (2), (3), (4)
B.(3), (2), (4), (1)
C.(2), (3), (4), (1)
D.(4), (3), (1), (2)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: C

The order of increasing energy is (2), (3), (4), (1).

In Huckel theory the resonance integral β is negative, so adding a multiple of β to α lowers the energy. For butadiene the four levels are α + 1.62β, α + 0.62β, α - 0.62β and α - 1.62β, in that order from lowest to highest.

Lowest: α + 1.6β (2), the fully bonding orbital.
Then α + 0.6β (3), the HOMO.
Then α - 0.6β (4), the LUMO.
Highest: α - 1.6β (1).

Option A lists them as numbered, and option D treats β as if it were positive, which reverses the order.

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