This set of Physical Chemistry Multiple Choice Questions & Answers (MCQs) focuses on “Pauli Exclusion Principle”.

1. How many spin functions does two electrons in an orbital have?

a) One

b) Two

c) Three

d) Four

View Answer

Explanation: Each electron can have either spin function α or β. Therefore, the total number of spin functions for two electrons can be written as: α(1), α(2), β(1)β(2), α(1)β(2), and α(2)β(1). This is a combination of each individual electron and electron function.

2. Is it possible to distinguish between two electrons?

a) Yes, because they have different wavefunctions

b) No, because they have different wavefunctions

c) Yes, because although the electrons are different, they have the same wavefunction

d) No, because electrons are identical to one another regardless of the wavefunction

View Answer

Explanation: It is not possible to distinguish between two electrons. The wavefunction must reflect this, therefore the last two spin functions are 2

^{-1/2}[α(1)β(2) + α(2)β(1)] & 2\(^{-\frac{1}{2}}\)(α(1)β(2) – α(2)β(1)) rather than α(1)β(2), and α(2)β(1). One of the reasons for this is that electrons are clouds around atoms rather than a single particle(s).

3. What type of wavefunctions must be antisymmetric with respect to the interchange of two any electrons?

a) Spatial

b) Spatial and spin

c) Spin

d) Neither spatial nor spin

View Answer

Explanation: It has been experimentally proven that spatial and spin wavefunctions must be antisymmetric with respect to the interchange of any two electrons. This fact is known as the Pauli exclusion principle and is the reason why electrons in an orbital must have opposite spin. No two electrons can have all the same quantum numbers.

4. What is the implication of Pauli’s exclusion principle?

a) Electrons must have at least one different quantum number to reside in the same orbital

b) Electrons must have the same quantum numbers to reside in the same orbital

c) Electrons are particles rather than clouds and their location is deterministic

d) Different electrons have the same spin and spatial wavefunction as they are indistinguishable

View Answer

Explanation: Pauli’s exclusion principle states that spatial and spin wavefunctions must be antisymmetric with respect to the interchange of any two electrons. In layman terms, this means that electrons in the same orbital cannot have the same spin, hence having different magnetic quantum number. In non-relativistic quantum mechanics, Pauli’s principle is an additional postulate.

5. What is Pauli’s exclusion principle?

a) Electrons are distinguishable physically and mathematically

b) Electrons in the same orbital must have the same quantum numbers

c) The wavefunction of any system of electrons must be antisymmetric with respect to the interchange of any two electrons

d) The wavefunction of any system of electrons must be symmetric with respect to the interchange of any two electrons

View Answer

Explanation: Pauli’s exclusion principle says that two electrons must be in different eigenstate and cannot have every single quantum number as identical. For a wavefunction to be antisymmetric in this way, the electrons must have different quantum numbers.

For the wavefunction to be antisymmetric, the electrons must have different quantum numbers.

6. What is the ground state electron symmetric description of Helium?

a) 1s(2) 2s(2)

b) 1s(1) 1s(2)

c) 2s(1) 2s(2)

d) 1s(1) 2s(1)

View Answer

Explanation: Helium has two electrons and is right after hydrogen on the periodic table. It fills the first orbital described in quantum mechanics – 1s. Since 1s can hold two electrons, the electronic configuration of Helium is 1s(1) 1s(2). Higher orbitals cannot be filled without filling these lower orbitals.

7. What is the ground state antisymmetric total wavefunction of the Heliium atom?

a) φ = (1s(1)1s(2))2\(^{-\frac{1}{2}}\)[α(1)β(2) – α(2)β(1)]

b) φ = (1s(1)1s(2))2\(^{-\frac{1}{2}}\)

c) φ = 1s(1)1s(2)

d) φ = [alpha;(1)β(2) – α(2)β(1)]

View Answer

Explanation: 1s(1) 1s(2) is the symmetric spin function of the helium atom. Since Pauli’s exclusion principle says that functions must be antisymmetric with respect to one another, the wavefunction needs to be multiplied by an antisymmetric function to be overall antisymmetric. Thus, φ = (1s(1)1s(2))2\(^{-\frac{1}{2}}\)[α(1)β(2)-α(2)β(1)] is the overall antisymmetric wavefunction.

8. What is the Slater determinant?

a) Mathematical expressions for electron probability density

b) Approximate wavefunctions satisfying the symmetric requirements of a wavefunction

c) Determinants of electron wavefunction coefficients

d) Approximate wavefunctions satisfying the antisymmetric requirements of a wavefunction

View Answer

Explanation: Slater developed a mathematical method for approximate wavefunctions satisfying the Pauli exclusion principle in 1929. Elements in a column involve the same spin orbital whereas elements in a row involve the same electron. Electrons not having the same quantum number is automatically satisfied by the Slater determinant.

10. How does the Hamiltonian change when considering spin calculations?

a) An additional magnetic spin energy must be included

b) It does not affect the Hamiltonian

c) An additional spin quantum energy must be included

d) The potential energy term is automatically neglected as it cancels spin

View Answer

Explanation: Spin calculations do not alter energy considerations because the Hamiltonian does not and cannot include spin. Spin is important for excited states of some atoms because of the Pauli exclusion principle. This is seen explicitly in excited states of lithium and helium.

11. What is the general category of particles that requires antisymmetric wavefunctions?

a) Fermions

b) Protons

c) Electrons

d) Bosons

View Answer

Explanation: Fermions are particles that require antisymmetric wavefunctions. They follow a type of statistics called Fermi-Dirac statistics. These particles have half integral spin (s = 1/2, 3/2, ..)

12. What is the general category of particles that requires symmetric wavefunctions?

a) Fermions

b) Protons

c) Electrons

d) Bosons

View Answer

Explanation: Bosons are particles that require symmetric wavefunctions. They follow a type of statistics called Bose-Einstein statistics. These particles have integral spin (s = 0,1,2, ..)

13. What is the Slater determinant for a helium atom?

a) φ = 1s(1)1s(2)

b) φ = 2s(1)2s(2)

c) \(\varphi = \frac{1}{\sqrt 2}\begin{vmatrix}1s(1)\alpha(1) & 1s(1)\beta(1) \\1s(2)\alpha(2) & 1s(2)\beta(2) \end {vmatrix}\)

d) \(\varphi = \begin{vmatrix}1s(1)\alpha(1) & 1s(1)\beta(1) \\1s(2)\alpha(2) & 1s(2)\beta(2) \end {vmatrix}\)

View Answer

Explanation: The Slater determinant must be written in determinant form, or a correct equivalent algebraic expression. Elements in a column involve the same spin orbital whereas elements in a row involve the same electron. For the Helium atom with two electrons, this is written as a superposition of each of the four electron functions present.

14. What is the equivalent algebraic expression for the helium atom Slater determinant?

a) φ = 1s(1)2s(1)

b) \(\varphi = \frac{1}{\sqrt 2}\begin{vmatrix}1s(1)\alpha(1) & 1s(1)\beta(1) \\1s(2)\alpha(2) & 1s(2)\beta(2) \end {vmatrix}\)

c) φ = 1s(1)1s(2)

d) φ = \(\frac{1}{2}\)[(1s(1)α(1))(1s(2)β(2)) – (1s(1)β(1))(1s(2)α(2))]

View Answer

Explanation: Although \(\varphi = \frac{1}{\sqrt 2}\begin{vmatrix}1s(1)\alpha(1) & 1s(1)\beta(1) \\1s(2)\alpha(2) & 1s(2)\beta(2) \end {vmatrix}\) is the Slater determinant for a helium atom, this can be expanded algebraically. Multiplying the diagonal and off-diagonal gives φ = \(\frac{1}{2}\)[(1s(1)α(1))(1s(2)β(2)) – (1s(1)β(1))(1s(2)α(2))].

15. What is the degeneracy of the first excited state of helium?

a) 1

b) 0

c) 2

d) 3

View Answer

Explanation: The actual wavefunction for helium can be written as φ

_{a.b}= \(\frac{1}{\sqrt 2}\)[1s(1)2s(2) ± 1s(2)2s(1)]. Incorporating spin into this wavefunction and multiplying these by antisymmetric functions to give overall antisymmetric wavefunctions yields three possible wavefunctions: φ

_{1}= \(\frac{1}{\sqrt 2}\)[1s(1)2s(2) – 1s(2)2s(1)]α(1)α(2), φ

_{2}= \(\frac{1}{\sqrt 2}\)[1s(1)2s(2) – 1s(2)2s(1)](α(1)β(2) + α(2)β(1)), φ

_{3}= \(\frac{1}{\sqrt 2}\)[1s(1)2s(2) – 1s(2)2s(1)]β(1)β(2). Since there are three possibilities, the degeneracy of this state is 3.

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