Match the followingColumn IColumn II(P) Associated Legendre polynomials(I) Harmonic oscillator(Q) Hermite polynomials(II) Particle in a box model(R) Associated Laguerre polynomials(III) Angular part of H atom(S) Trigonometric functions(IV) Radial part of H atom

The correct match is P→III, Q→I, R→IV, S→II.Associated Legendre polynomials appear in the spherical harmonics, the angular part of the hydrogen-atom wavefunctions.Hermite polynomials multiply a Gaussian in the harmonic oscillator wavefunctions.Associated Laguerre polynomials form the radial part of the hydrogen-atom wavefunctions.Trigonometric functions (sines) are the particle-in-a-box wavefunctions.The other options assign Hermite polynomials to the hydrogen atom, swap the radial and angular parts, or give the particle in a box polynomial solutions, all of which contradict the standard solutions.

Quantum Chemistry Questions and Answers

Q: The secular determinant (where \(\alpha\), \(\beta\) and \(E\) have their usual meanings) is given by

The secular determinant is (A), the tridiagonal one with β between neighbouring carbons only.The allyl radical CH2=CH-CH2 has three p orbitals in a chain, C1-C2-C3. In Hückel theory every diagonal element is α - E, an off-diagonal element is β for carbons that are bonded and 0 for carbons that are not. C1 and C2 are bonded, C2 and C3 are bonded, C1 and C3 are not:\[ \begin{vmatrix} \alpha-E & \beta & 0 \\ \beta & \alpha-E & \beta \\ 0 & \beta & \alpha-E \end{vmatrix} = 0 \](B) and (C) each drop one of the two bonds, describing an isolated ethylene plus a lone p orbital. (D) has -β, which flips the sign convention without any physical basis in standard Hückel theory.

Q: The angular part of the wavefunction for the electron in a hydrogen atom is proportional to \(\sin^{2}\theta \cos\theta \, e^{2i\phi}\). The values of the azimuthal quantum number (l) and the magnetic quantum number (m) are, respectively

The quantum numbers are l = 3 and m = 2.The azimuthal factor \(e^{im\phi}\) gives m directly: \(e^{2i\phi}\) means m = +2, not -2.For l, count the powers of the polar functions. A spherical harmonic \(Y_{l}^{m}\) contains \(\sin^{|m|}\theta\) times a polynomial in \(\cos\theta\) of degree \(l - |m|\). Here \(\sin^{2}\theta\) gives |m| = 2 and the single \(\cos\theta\) gives \(l - 2 = 1\), so l = 3. Indeed \(Y_{3}^{2} \propto \sin^{2}\theta\cos\theta\,e^{2i\phi}\), an f orbital.An l = 2 function with m = 2 is \(\sin^{2}\theta\,e^{2i\phi}\) with no cosine factor, so the 2 and 2 option fails, and the -2 options have the wrong sign of the exponent.

Q: The wavelength associated with a particle in one-dimensional box of length L is (n refers to the quantum number)

The wavelength is 2L/n.The wavefunctions of a particle in a box, \(\psi_{n} = \sqrt{2/L}\,\sin(n\pi x/L)\), must vanish at both walls, so they are standing waves with a whole number of half-wavelengths fitting into the box:\[ L = n\frac{\lambda}{2} \Rightarrow \lambda = \frac{2L}{n} \]For the ground state (n = 1) the box holds half a wavelength, so \(\lambda = 2L\). L/n counts whole wavelengths instead of half-wavelengths, L/2n halves the answer again, and nL would make the wavelength grow with the quantum number, when higher states have more nodes and shorter wavelengths.

Q: The INCORRECT statement is

The incorrect statement is The time independent Schrödinger equation for Li2+ cannot be solved exactly.Li2+ has a single electron, so it is a hydrogen-like ion (Z = 3). Its Schrödinger equation separates exactly, like hydrogen's, giving energies \(E_{n} = -Z^{2}R_{H}/n^{2}\) and the familiar orbitals scaled by Z. Exact solution fails only for systems with two or more electrons, such as He or Li+, because of the electron-electron repulsion.The other statements are true. The harmonic oscillator's zero-point energy is \(h\nu/2\). Rigid-rotor levels are \(E_{J} = J(J+1)\hbar^{2}/2I\), inversely proportional to the moment of inertia. The total angular momentum is the (vector) sum of the orbital and spin angular momenta, \(\mathbf{J} = \mathbf{L} + \mathbf{S}\).

Q: The wavefunction for Be3+ in a certain state is given by \(\psi = N\,e^{-\left(\frac{4}{a_{0}}r\right)}\), where N is the normalization constant, r is the distance of electron from the nucleus and \(a_{0}\) is the Bohr radius. The most probable distance of the electron from the nucleus in this state is

The most probable distance is \(a_{0}/4\).The wavefunction \(e^{-4r/a_{0}}\) is the 1s function \(e^{-Zr/a_{0}}\) for \(Z = 4\) (Be3+ is hydrogen-like). The radial probability density is\[ P(r) \propto r^{2}\psi^{2} = r^{2}e^{-8r/a_{0}} \]Setting the derivative to zero:\[ \begin{aligned} 2r - \frac{8r^{2}}{a_{0}} &= 0 \\ r &= \frac{a_{0}}{4} \end{aligned} \]This is the hydrogen result \(a_{0}\) divided by Z. \(a_{0}/8\) uses the exponent of \(\psi^{2}\) as if it were Z, and \(4a_{0}\) and \(8a_{0}\) multiply by Z instead of dividing; a higher nuclear charge pulls the electron closer, not further out.

Q: \(\psi_{1}, \psi_{2}, \psi_{3}\), and \(\psi_{4}\) are four Hückel molecular orbitals of benzene with orbital energies \(E_{1}, E_{2}, E_{3}\), and \(E_{4}\), respectively.\[ \begin{aligned} \psi_{1} &= \tfrac{1}{2}(\phi_{B} + \phi_{C} - \phi_{E} - \phi_{F}) \\ \psi_{2} &= 6^{-1/2}(\phi_{A} - \phi_{B} + \phi_{C} \\ &\qquad - \phi_{D} + \phi_{E} - \phi_{F}) \\ \psi_{3} &= 6^{-1/2}(\phi_{A} + \phi_{B} + \phi_{C} \\ &\qquad + \phi_{D} + \phi_{E} + \phi_{F}) \\ \psi_{4} &= 12^{-1/2}(2\phi_{A} + \phi_{B} - \phi_{C} \\ &\qquad - 2\phi_{D} - \phi_{E} + \phi_{F}) \end{aligned} \]The correct order of the orbital energies is(The six carbon atoms of benzene are denoted by A to F and \(\phi_{J}\) is the \(2p_{z}\) orbital of Jth carbon of benzene.)

The order is \(E_{3} \lt E_{1} = E_{4} \lt E_{2}\).In Hückel theory the energy of a benzene MO is \(\alpha + 2\beta\cos(2\pi k/6)\), with \(\beta \lt 0\), so fewer nodes means lower energy.\(\psi_{3}\) has all coefficients equal, no nodes: \(\alpha + 2\beta\), the lowest.\(\psi_{1}\) (pattern 0, 1, 1, 0, -1, -1) and \(\psi_{4}\) (2, 1, -1, -2, -1, 1) each change sign twice around the ring: they are the degenerate pair at \(\alpha + \beta\).\(\psi_{2}\) alternates in sign at every atom: \(\alpha - 2\beta\), the highest.So \(E_{1} = E_{4}\) lie between \(E_{3}\) and \(E_{2}\). The other orders either break the degeneracy of \(\psi_{1}\) and \(\psi_{4}\) or put the fully alternating orbital too low.

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Hückel molecular orbital theory can be applied to the allene radical

CH₂=CH-ĊH₂

The secular determinant (where \(\alpha\), \(\beta\) and \(E\) have their usual meanings) is given by

A.\[ \begin{vmatrix} \alpha-E & \beta & 0 \\ \beta & \alpha-E & \beta \\ 0 & \beta & \alpha-E \end{vmatrix} \]
B.\[ \begin{vmatrix} \alpha-E & 0 & 0 \\ 0 & \alpha-E & \beta \\ 0 & \beta & \alpha-E \end{vmatrix} \]
C.\[ \begin{vmatrix} \alpha-E & \beta & 0 \\ \beta & \alpha-E & 0 \\ 0 & 0 & \alpha-E \end{vmatrix} \]
D.\[ \begin{vmatrix} \alpha-E & -\beta & 0 \\ -\beta & \alpha-E & -\beta \\ 0 & -\beta & \alpha-E \end{vmatrix} \]

Exam Year(s)

GATE Chemistry - 2012,

Answer: A

The secular determinant is (A), the tridiagonal one with β between neighbouring carbons only.

The allyl radical CH₂=CH-CH₂ has three p orbitals in a chain, C1-C2-C3. In Hückel theory every diagonal element is α - E, an off-diagonal element is β for carbons that are bonded and 0 for carbons that are not. C1 and C2 are bonded, C2 and C3 are bonded, C1 and C3 are not:

\[ \begin{vmatrix} \alpha-E & \beta & 0 \\ \beta & \alpha-E & \beta \\ 0 & \beta & \alpha-E \end{vmatrix} = 0 \]

(B) and (C) each drop one of the two bonds, describing an isolated ethylene plus a lone p orbital. (D) has -β, which flips the sign convention without any physical basis in standard Hückel theory.

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The possible values of \(E\) are

A.\(\alpha+\sqrt{2}\beta, \alpha, \alpha-\sqrt{2}\beta\)
B.\(\alpha+2\sqrt{2}\beta, \alpha, \alpha-2\sqrt{2}\beta\)
C.\(\alpha+\beta, \alpha, \alpha-\beta\)
D.\(\alpha+2\beta, \alpha, \alpha-2\beta\)

Exam Year(s)

GATE Chemistry - 2012,

Answer: A

The energies are \(\alpha + \sqrt{2}\beta,\ \alpha,\ \alpha - \sqrt{2}\beta\).

Dividing each row of the determinant by \(\beta\) and writing \(x = (\alpha - E)/\beta\):

\[ \begin{aligned} \begin{vmatrix} x & 1 & 0 \\ 1 & x & 1 \\ 0 & 1 & x \end{vmatrix} &= x(x^{2}-1) - x \\ &= x^{3} - 2x = 0 \\ x &= 0,\ \pm\sqrt{2} \end{aligned} \]

So \(E = \alpha - x\beta\) gives \(\alpha + \sqrt{2}\beta\) (bonding), \(\alpha\) (non-bonding) and \(\alpha - \sqrt{2}\beta\) (antibonding). The radical's odd electron sits in the non-bonding level.

\(\alpha \pm \beta\) are the ethylene levels, \(\alpha \pm 2\beta\) are the extreme levels of benzene, and \(2\sqrt{2}\) doubles the root of \(x^{2} = 2\).

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If Δy and Δp_y are the uncertainties in the y-coordinate and the y component of the momentum of a particle respectively, then, according to uncertainty principle ΔyΔp_y is (\(\hbar = \dfrac{h}{2\pi}\) and h is Planck’s constant)

A.\(\ge \hbar\)
B.\(\gt \hbar/2\)
C.\(\gt \hbar\)
D.\(\ge \hbar/2\)

Exam Year(s)

GATE Chemistry - 2012,

Answer: D

The uncertainty principle states \(\Delta y\,\Delta p_{y} \ge \hbar/2\).

Heisenberg's relation, in the form derived from the commutator of position and momentum, puts a lower bound on the product of the two standard deviations:

\[ \Delta y\,\Delta p_{y} \ge \frac{\hbar}{2} = \frac{h}{4\pi} \]

The sign must be 'greater than or equal to', because a minimum-uncertainty state (a Gaussian wave packet, or the ground state of a harmonic oscillator) reaches the bound exactly. That rules out both strict inequalities, \(\gt \hbar\) and \(\gt \hbar/2\). The bound is \(\hbar/2\), not \(\hbar\); \(\ge \hbar\) is a looser statement that the oscillator ground state, with a product of exactly \(\hbar/2\), would violate.

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The angular part of the wavefunction for the electron in a hydrogen atom is proportional to \(\sin^{2}\theta \cos\theta \, e^{2i\phi}\). The values of the azimuthal quantum number (l) and the magnetic quantum number (m) are, respectively

A.2 and 2
B.2 and -2
C.3 and 2
D.3 and -2

Exam Year(s)

GATE Chemistry - 2012,

Answer: C

The quantum numbers are l = 3 and m = 2.

The azimuthal factor \(e^{im\phi}\) gives m directly: \(e^{2i\phi}\) means m = +2, not -2.

For l, count the powers of the polar functions. A spherical harmonic \(Y_{l}^{m}\) contains \(\sin^{|m|}\theta\) times a polynomial in \(\cos\theta\) of degree \(l - |m|\). Here \(\sin^{2}\theta\) gives |m| = 2 and the single \(\cos\theta\) gives \(l - 2 = 1\), so l = 3. Indeed \(Y_{3}^{2} \propto \sin^{2}\theta\cos\theta\,e^{2i\phi}\), an f orbital.

An l = 2 function with m = 2 is \(\sin^{2}\theta\,e^{2i\phi}\) with no cosine factor, so the 2 and 2 option fails, and the -2 options have the wrong sign of the exponent.

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The wavefunction of a 1-D harmonic oscillator between \(x = +\infty\) and \(x = -\infty\) is given by \(\psi(x) = N(2x^{2}-1)e^{-x^{2}/2}\). The value of N that normalizes the function \(\psi(x)\) is

(Given: \[ \begin{aligned} &\int_{-\infty}^{+\infty} x^{2n}e^{-x^{2}}dx \\ &= \frac{1\cdot 3\cdot 5\cdots(2n-1)}{2^{n}}\sqrt{\pi} \end{aligned} \])

A.\(\left(\dfrac{1}{8\sqrt{\pi}}\right)^{\frac{1}{2}}\)
B.\(\left(\dfrac{1}{3\sqrt{\pi}}\right)^{\frac{1}{2}}\)
C.\(\left(\dfrac{1}{2\sqrt{\pi}}\right)^{\frac{1}{2}}\)
D.\(\left(\dfrac{1}{4\sqrt{\pi}}\right)^{\frac{1}{2}}\)

Exam Year(s)

GATE Chemistry - 2012,

Answer: C

The normalisation constant is \(N = \left(\dfrac{1}{2\sqrt{\pi}}\right)^{1/2}\).

Normalisation requires \(\int \psi^{2}\,dx = 1\). Squaring the function gives \(N^{2}(4x^{4} - 4x^{2} + 1)e^{-x^{2}}\). Using the given integral with n = 2, 1 and 0:

\[ \begin{aligned} \int x^{4}e^{-x^{2}}dx &= \tfrac{3}{4}\sqrt{\pi} \\ \int x^{2}e^{-x^{2}}dx &= \tfrac{1}{2}\sqrt{\pi} \\ \int e^{-x^{2}}dx &= \sqrt{\pi} \end{aligned} \]

\[ \begin{aligned} N^{2}\sqrt{\pi}\,(4 \cdot \tfrac{3}{4} - 4 \cdot \tfrac{1}{2} + 1) &= 1 \\ 2\sqrt{\pi}\,N^{2} &= 1 \\ N &= (2\sqrt{\pi})^{-1/2} \end{aligned} \]

The 8 in the first option is what the Hermite form 4x² - 2, twice this polynomial, would give. 4 is what dropping the cross term -4x² gives (3 + 1), and 3 is the x⁴ term alone.

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The wavelength associated with a particle in one-dimensional box of length L is (n refers to the quantum number)

A.2L/n
B.L/n
C.nL
D.L/2n

Exam Year(s)

GATE Chemistry - 2017,

Answer: A

The wavelength is 2L/n.

The wavefunctions of a particle in a box, \(\psi_{n} = \sqrt{2/L}\,\sin(n\pi x/L)\), must vanish at both walls, so they are standing waves with a whole number of half-wavelengths fitting into the box:

\[ L = n\frac{\lambda}{2} \Rightarrow \lambda = \frac{2L}{n} \]

For the ground state (n = 1) the box holds half a wavelength, so \(\lambda = 2L\). L/n counts whole wavelengths instead of half-wavelengths, L/2n halves the answer again, and nL would make the wavelength grow with the quantum number, when higher states have more nodes and shorter wavelengths.

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The INCORRECT statement is

A.Zero-point energy of a quantum mechanical harmonic oscillator of frequency \(\nu\) is \(\frac{h\nu}{2}\)
B.Energy level of a quantum mechanical rigid rotor is inversely proportional to its moment of inertia
C.The time independent Schrödinger equation for \(\mathrm{Li^{2+}}\) cannot be solved exactly
D.Total angular momentum of an atomic system is equal to the sum of orbital angular momentum and spin angular momentum

Exam Year(s)

GATE Chemistry - 2023,

Answer: C

The incorrect statement is The time independent Schrödinger equation for Li²⁺ cannot be solved exactly.

Li²⁺ has a single electron, so it is a hydrogen-like ion (Z = 3). Its Schrödinger equation separates exactly, like hydrogen's, giving energies \(E_{n} = -Z^{2}R_{H}/n^{2}\) and the familiar orbitals scaled by Z. Exact solution fails only for systems with two or more electrons, such as He or Li⁺, because of the electron-electron repulsion.

The other statements are true. The harmonic oscillator's zero-point energy is \(h\nu/2\). Rigid-rotor levels are \(E_{J} = J(J+1)\hbar^{2}/2I\), inversely proportional to the moment of inertia. The total angular momentum is the (vector) sum of the orbital and spin angular momenta, \(\mathbf{J} = \mathbf{L} + \mathbf{S}\).

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The wavefunction for Be³⁺ in a certain state is given by \(\psi = N\,e^{-\left(\frac{4}{a_{0}}r\right)}\), where N is the normalization constant, r is the distance of electron from the nucleus and \(a_{0}\) is the Bohr radius. The most probable distance of the electron from the nucleus in this state is

A.\(4a_{0}\)
B.\(\frac{a_{0}}{4}\)
C.\(8a_{0}\)
D.\(\frac{a_{0}}{8}\)

Exam Year(s)

GATE Chemistry - 2023,

Answer: B

The most probable distance is \(a_{0}/4\).

The wavefunction \(e^{-4r/a_{0}}\) is the 1s function \(e^{-Zr/a_{0}}\) for \(Z = 4\) (Be³⁺ is hydrogen-like). The radial probability density is

\[ P(r) \propto r^{2}\psi^{2} = r^{2}e^{-8r/a_{0}} \]

Setting the derivative to zero:

\[ \begin{aligned} 2r - \frac{8r^{2}}{a_{0}} &= 0 \\ r &= \frac{a_{0}}{4} \end{aligned} \]

This is the hydrogen result \(a_{0}\) divided by Z. \(a_{0}/8\) uses the exponent of \(\psi^{2}\) as if it were Z, and \(4a_{0}\) and \(8a_{0}\) multiply by Z instead of dividing; a higher nuclear charge pulls the electron closer, not further out.

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Match the following

Column I	Column II
(P) Associated Legendre polynomials	(I) Harmonic oscillator
(Q) Hermite polynomials	(II) Particle in a box model
(R) Associated Laguerre polynomials	(III) Angular part of H atom
(S) Trigonometric functions	(IV) Radial part of H atom

A.P→III, Q→I, R→IV, S→II
B.P→III, Q→IV, R→II, S→I
C.P→IV, Q→I, R→III, S→II
D.P→II, Q→III, R→IV, S→I

Exam Year(s)

GATE Chemistry - 2023,

Answer: A

The correct match is P→III, Q→I, R→IV, S→II.

Associated Legendre polynomials appear in the spherical harmonics, the angular part of the hydrogen-atom wavefunctions.
Hermite polynomials multiply a Gaussian in the harmonic oscillator wavefunctions.
Associated Laguerre polynomials form the radial part of the hydrogen-atom wavefunctions.
Trigonometric functions (sines) are the particle-in-a-box wavefunctions.

The other options assign Hermite polynomials to the hydrogen atom, swap the radial and angular parts, or give the particle in a box polynomial solutions, all of which contradict the standard solutions.

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10.

\(\psi_{1}, \psi_{2}, \psi_{3}\), and \(\psi_{4}\) are four Hückel molecular orbitals of benzene with orbital energies \(E_{1}, E_{2}, E_{3}\), and \(E_{4}\), respectively.

\[ \begin{aligned} \psi_{1} &= \tfrac{1}{2}(\phi_{B} + \phi_{C} - \phi_{E} - \phi_{F}) \\ \psi_{2} &= 6^{-1/2}(\phi_{A} - \phi_{B} + \phi_{C} \\ &\qquad - \phi_{D} + \phi_{E} - \phi_{F}) \\ \psi_{3} &= 6^{-1/2}(\phi_{A} + \phi_{B} + \phi_{C} \\ &\qquad + \phi_{D} + \phi_{E} + \phi_{F}) \\ \psi_{4} &= 12^{-1/2}(2\phi_{A} + \phi_{B} - \phi_{C} \\ &\qquad - 2\phi_{D} - \phi_{E} + \phi_{F}) \end{aligned} \]

The correct order of the orbital energies is

(The six carbon atoms of benzene are denoted by A to F and \(\phi_{J}\) is the \(2p_{z}\) orbital of J^th carbon of benzene.)

A.\(E_{1} \lt E_{2} = E_{3} \lt E_{4}\)
B.\(E_{4} \lt E_{1} = E_{3} \lt E_{2}\)
C.\(E_{3} \lt E_{1} = E_{4} \lt E_{2}\)
D.\(E_{3} \lt E_{2} \lt E_{1} = E_{4}\)

Exam Year(s)

GATE Chemistry - 2024,

Answer: C

The order is \(E_{3} \lt E_{1} = E_{4} \lt E_{2}\).

In Hückel theory the energy of a benzene MO is \(\alpha + 2\beta\cos(2\pi k/6)\), with \(\beta \lt 0\), so fewer nodes means lower energy.

\(\psi_{3}\) has all coefficients equal, no nodes: \(\alpha + 2\beta\), the lowest.
\(\psi_{1}\) (pattern 0, 1, 1, 0, -1, -1) and \(\psi_{4}\) (2, 1, -1, -2, -1, 1) each change sign twice around the ring: they are the degenerate pair at \(\alpha + \beta\).
\(\psi_{2}\) alternates in sign at every atom: \(\alpha - 2\beta\), the highest.

So \(E_{1} = E_{4}\) lie between \(E_{3}\) and \(E_{2}\). The other orders either break the degeneracy of \(\psi_{1}\) and \(\psi_{4}\) or put the fully alternating orbital too low.

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