(II), Hydrogen Atom and Hyrdrogen Molecular Ion Carl W

University of Connecticut OpenCommons@UConn

Chemistry Education Materials Department of Chemistry

March 2008 Advanced Physical Chemistry Problems (II), Hydrogen Atom and Hyrdrogen Molecular Ion Carl W. David University of Connecticut, [email protected]

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Recommended Citation David, Carl W., "Advanced Physical Chemistry Problems (II), Hydrogen Atom and Hyrdrogen Molecular Ion" (2008). Chemistry Education Materials. 56. https://opencommons.uconn.edu/chem_educ/56 Problems for the Advanced Physical Chemistry Student Part 2, The Hydrogen Atom and the Hydrogen Molecular Ion

C. W. David∗ Department of Chemistry University of Connecticut Storrs, Connecticut 06269-3060 (Dated: March 18, 2008)

I. SYNOPSIS so that ∂2ψ ∂ ∂ψ = ∂x This is a set of problems that were used near the turn of ∂x2 ∂x the century and which will be lost when the web site they were on disappears with my demise. Because these prob- which, mechanically, leads to a large number of lems are being taken from the web and are being edited, terms. Repeating the process two more times, once their statements and the hints/answers offered are sub- for y and once for z, results in still more terms which, when combined, results in enormous simpli- ject to the typical editorial errors that ensue when such 2 work is undertaken in the vacuum of a non-teaching situ- fication as r terms cancel. ation. Therefore, I claim any errors for myself, and hate At the end, one obtains an equation involving E, h, to note that there most likely is no point in contacting π, m (electron mass), and some terms which include me about them for obvious reasons. ‘r’. But this resultant equation can not depend on ‘r’, so the coëfficient of ‘r’ must be zero! From that observation α emerges. It turns out that we’ve been II. THE H ATOM dealing with a 2px orbital. 2. Obtain the value of the wave function: zr−r/2 at 1. For the H atom’s electron (in the infinite nuclear the point P (3, 3, −3). mass approximation), the Schrödinger equation has the form z ¯h2 Ze2 − ∇2ψ − ψ = eψ 2melectron r A possible wave function is

xe−αr

Obtain a value for α which makes this true if it can be made to be true.

y Answer and/or Hint r Defining √ ψ(x, y, z) = xe−α x2+y2+z2 P(3,3,−3) one forms x √ √ ∂ψ x 2 2 2 2 2 2 = −xα e−α x +y +z + e−α x +y +z ∂x r FIG. 1: The Cartesian coördinate frame with a vector of the y,z form ~r = xî + yˆj + zkˆ.

∗Electronic address: [email protected]

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Answer and/or Hint 5. The un-normalized 2s wave function for H is:

−r/2 ψ2s = (2 − r)e Since r = p32 + 32 + (−3)2, we have √ What fraction of the electron is contained in the ψ(3, 3, −1) = −3e− 27/2 solid angle which constitutes the ﬁrst octant?

3. Two possible wave functions for H, ψA and ψB are −r2 ψA = (2 − r)e and Answer and/or Hint −r/3 ψB = (6 − r)e By explicit integration, demonstrate whether or not Another good one. Here we need to evaluate these two functions are orthogonal to each other. Z ∞ 2 Z π/2 Z π/2 numerator = r2dr(2−r)e−r/2 sin ϑ dϑdϕ 0 0 0 and divide it by the integral Answer and/or Hint Z ∞ 2 Z π Z 2π 2 −r/2 denominator = r dr(2− r)e sin ϑ dϑdϕ 0 0 0 We need to form We should get a value of (drum roll please) 1/8. ∞ π 2π:4π Z Z Z r2dr sinϑdϑ dϕ (2 − r)e−r/2 × (6 − r)e−r/3 The important point is the limits of integration! 0 0 0 Notice that the r integration cancels from numera- and see what happens. The rest is left to the reader. tor and denominator. Note that had we used px, for example, against s, we would have found othogonality easily in the 6. Obtain a formula for the current equivalent of a angular integrations, but since these are both s type single electron traversing a Bohr orbit in the n=1 orbitals, such simpliﬁcation eludes us. state. 4. The un-normalized 2s wave function for H is: −r/2 ψ2s = (2 − r)e What fraction of the electron is contained in a Answer and/or Hint sphere of radius 0.5?

This constitutes a discussion of magnetic moments, with the answer to the query embedded somewhere in the middle, where it has some use. Answer and/or Hint For a magnetic ﬁeld (vector) B~ acting on an arm of a current loop, a square current loop and a Bohr This is a non-standard but lovely problem which orbit are similar. The force on each single charge requires us to evaluate the integral (q) travelling in the arm comes from the Lorentz Z 0.5 Z π Z 2π 2 force: numerator = r2dr sin ϑdϑ dϕ (2 − r)e−r/2 0 0 0 F~ = q~v × B~ and divide it by the integral Since ~v is perpendicular to B~ (in our case), the Z ∞ Z π Z 2π 2 cross product simpliﬁes. The current is given by: denominator = r2dr sin ϑdϑ dϕ (2 − r)e−r/2 0 0 0 charge stat coul cm stat coul i = n × q × v × σ cm2 = So, the answer appears to be cm3 charge sec sec ((( R 0.5 2 ((−r/22R π R 2π (r dr(((2(− r)e sinϑdϑ dϕ where σ is the cross sectional area of the wire-loop, (0( (0 0 ((( 2 and the force on each charge is: R ∞ 2 ((( −r/2 R π R 2π 0(r(dr( (2 − r)e 0sin ϑdϑ 0 dϕ ( π q|~v||B~ | = q × v × B × sin = qvB The resultant pair of integrals are elementary. 2 3

Since the number of charges is n × a × σ, the force From Bohr Theory we had (for the radius): on one arm of the loop (of length ‘a’) is n2¯h2 r = F = (n × a × σ) × q × v × B = iaB n 2 Zmee which is reversed on the other (opposite arm) leg and the angular momentum is of the loop. mr2ϑ˙ = n¯h = p The loop is a × b in area (and 2a + 2b in circum- ϑ ference), the moment arm about the pivot point is so, solving for ϑ˙ we obtain b sin α if α is the angle between the loop and the 2 n¯h ﬁeld. ϑ˙ = mr2 and then the current (= stat-coul/sec) = e Axis a charge/transit-time, (= τ ), where (we useτ twice F = i a B normal here, once for the torque, and once for the period- α both usages are common) is the period. i τ=( iaB)( b sinα ) ∆ϑ n¯h 2π τ= iAB sin α = ϑ˙ = = A=ab=area ∆t mr2 τ Axis B so F = i a B b charge e en¯h = = i = − period τ 2πmr2 but, since the area of the loop is πr2, we have (THE F Axis ANSWER) normal en¯h en¯h iA = − πr2 = − e− 2πmr2 2m τ= eAB sinα A=π r2 (and some more just to clean up why we’re doing Axis this) which deﬁnes the Bohr magnetic moment

B e¯h F µ = − B 2m (known as the Bohr magnetron). We then have en¯h FIG. 2: A current loop in a magnetic field τ(orque) − iAB sin α = − B sin α 2m In Figure 2, the length of the horizontal arms are which is finally ‘a’ while the moment arm’s length ‘b/2’, i.e., from the axis to a horizontal arm is b/2 (cm). The torque τ(orque) = nµBB sin α (~τ ) is The energy associated with rotating the current b look (Bohr orbit) about the axis perpendicular to ~τ(orque) = 2 sin α iaB 2 the field ( orienting the look relative to the field) is α, final ang pos but, since ab is the area (A) of the loop, we have Z E = τ(α)dα τ = iAB sin α ref pos i.e., Commonly, this torque is related to a magnetic mo- α ment equivalent, i.e., Z E = τ(x)dx 90 ~τ(orque) = ~µ × B~ which gives in analogy with an electric dipole in an electric field. A current loop is equivalent to a magnetic moment, E = −nµBB cos α a tiny bar magnet. or We assume that the above would hold for a Bohr ~ orbit. E = −n~µB · B 4

7. The hybrid orbital:

s = Px + py + pz Answer and/or Hint in polar coordinates, neglecting the radial part, is:

1 + sin ϑ cos ϕ + sin ϑ sin ϕ + cos ϕ We are asked to evaluate the integral Since this orbital is purported to be part of the Z ∞ Z π Z 2π set, it should point at the corner of a tetrahedron. r2dr sin ϑdϑ dϕ To see if this is true, take the partial derivative 0 0 0 of this function with respect to ϕ, set this equal to 1 1 1 1 1 √ ψ2s + √ ψ2py × √ ψ2s + √ ψ2px + √ ψ2py zero, and ﬁnd out the optimal value of ϕ = ϕcorner. 2 2 3 3 3 Next, take the partial derivative with respect to ϑ and search for its optimal value. When done, you but we know that each contributing atomic orbital should have a reasonable result. (AO) is normalized, i.e., the integral of the three domains to the “square” of any one of them is one (1). Further, we know that the AO’s are orthogonal to each other, so the integral of any two diﬀerent ones over the domain will vanish. Answer and/or Hint The rest is mechanics. We write Z ∞ Z π Z 2π 2 ∗ h2s| | |2pxi = r dr sin ϑdϑ dϕψ2sψ2px We have 0 0 0 f = 1 + sin ϑ cos ϕ + sin ϑ sin ϕ + cos ϑ as a notational device, so the orthogonality integral becomes and form 1 1 1 1 1 1 √ √ h2s| | |2si + √ √ h2s| | |2pxi + √ √ h2s| | |2pyi ∂f 2 3 2 3 2 3 ∂ϕ ϑ 1 1 1 1 1 1 √ √ h2py| | |2si + √ √ h2py| | |2pxi + √ √ h2py| | |2pyi obtaining 2 3 2 3 2 3

0 Most of these integrals are zero, since s is orthogo- fϕ = − sin ϑ sin ϕ + sin ϑ cos ϕ → 0 nal to any p, and each p is orthogonal to any other and p orbital. When ﬁnished, we end up with a non-zero value, 0 fϑ = cos ϑ cos ϕ + cos ϑ sin ϕ − sin ϑ → 0 indicating that these orbitals are not orthogonal. The solution of these two simultaneous equations 9. For the hybrid orbital o yields a value of 45 for ϕextremum (ϕcorner)and o 1 1 1 54.7 for ϑextremum. This latter angle is one-half ψhybrid = √ ψ2s − √ ψ2p − √ ψ2p of the tetrahedral H-C-Hangle 109.471o which, by 3 6 x 6 y the way, also happens to be cos−1 − 1 ! Wow! 3 obtain the values of ϑ and ϕ which makes the or- Why one-half? Where are the other extrema? Is bital maximally positive. there one above and one below some plane, each at 54.7o so that the interorbital angle is the desired tetrahedral one?

8. Given two hybrid orbitals Answer and/or Hint 1 1 ψ1 = √ ψ2s + √ ψ2py 2 2 First, notice that the 2s orbital is irrelevant, since and its spherically symmetric. 1 1 1 Now, a 2px orbital has the form ψ2 = √ ψ2s + √ ψ2p + √ ψ2p 3 3 x 3 y −r/2 ψ2px ∼ xe where the contributing atomic orbitals are pre- and normalized, calculate the numerical value of the −r/2 overlap between them. ψ2py ∼ ye 5

so, converting to spherical polar co¨ordinates, we Answer and/or Hint have

−r/2 ψ2px ∼ r sin ϑ cos ϕe The Schrödinger Equation for the s-states of Hy- drogen is and 2 2 ∂r (∂ψtrial) ! 2 −r/2 ¯h 1 Ze ? ψ2py ∼ r sin ϑ sin ϕe ∂r − 2 − ψtrial = Eψtrial 2me r ∂r r which allows us to take the partial derivatives of ψhybrid with respect to ϑ and ϕ separately, set them or equal to zero, solve the set of equations simultane-  2 2 −βr  ously, and achieve the answer desired! 2 ∂r (r ∂(1+αr)e ) ¯h 1 ∂r −  2  10. To compute the tetrahedral angle in hybridization, 2me r ∂r assume there exists a cube surrounding the origin Ze2 [1], with corners (1,1,1), (1,-1,1), (-1,-1,1), − (1 + αr)e−βr =? E(1 + αr)e−βr and (-1,1,1) above the x-y plane and r (1,1,-1), (1,-1,-1), (-1,-1,-1), and (-1,1,-1) below the which is x-y plane. Choosing the points (1,1,1) and (-1,-1,1), 2 which might be places where H atoms could reside ¯h 2(α − β(1 + αr)) −βr in methane (CH ), say, form the dot product of − − βα − β(α − β(1 + αr)) e 4 2me r the vectors joining the origin to these two points. Ze2 Thus − (1 + αr)e−αrψ =? E(1 + αr)e−αrψ r trial trial ˆ ~r1 = î + ˆj + k Your job, if you should choose to take it, is to gather then 1 terms together, force them to zero choos- ˆ r ~r2 = −î − ˆj + k ing β(α) and then use the remaining, non r terms, to ascertain what the correct value of α must be. Then Good luck.

~r1 · ~r2 = |~r1| × |~r2| cos ϑtetra 12. If one could turn oﬀ the electron-electron repulsion in the Helium atom i.e., so, obtain an equation for cos ϑtetra You will have to ﬁgure out the magnitudes of ~r1 and ~r2. Then e2 → 0 solve for the cos ϑtetra and compute ϑtetra via the r arccos. 12 what would be the energy required to remove both We would have ~r1 · ~r2 = −1 − 1 + 1 = 1 and |~r1| × √ √ electrons from the neutral Helium atom if it ex- |~r2| = 3 3 which allows us to solve for the angle isted in the excited state: 2s3pm =0? (The assump- in question, ϑtetra. ` tion here is that both electrons see the full nuclear charge.)

Answer and/or Hint

Answer and/or Hint 11. Starting with the assumption that

−βr ψtrial = (1 + αr)e Although a silly problem, we obtain

substitute this trial 2s candidate wave function into 22Ry 22Ry − − the Schr¨odinger Equation for the H-atom’s electron 32 22 and see if a choice of α and β can be found which makes the candidate wave function a true eigen- if both electrons see the charge Z(=2). Actually, function. one could argue that the outer electron would see a charge of about +1 rather than +2, especially as it gets further and further from the nucleus (and the other remaining electron). Oh well. 6

energy (positive deﬁnite) of the emerging electron? The reaction is:

Ry + − 00 − He(2s3p)− > He (1s) + e (“free ) 2 (The assumption is that both electrons in neutral 3 Helium see the full nuclear charge, i.e., no shield- ing.) − Ry 22 Answer and/or Hint How about 22Ry 22Ry 22Ry − − + 32 22 12

15. For a px orbital of the form

ψ = xe−r/2 Ry 2px − what is the value of the function at its minimum? 12 Answer and/or Hint FIG. 3: Two electrons in an “excited” state of He

13. The simplest model of the Sodium atom, has its The minimum will occur on the −x axis, at ϕ = valence electron in an s state. The experimen- 180o and ϑ = π/2. Then we just need to know how tal ionization potential of Na is 5.138 eV. What far away from the nucleus we need go to ﬁnd the is the value predicted by the simplest Bohr model minimum in the function for Na? Assume that the electron conﬁguration is −r/2 1s22s22p63s, and that the inner electrons shield the re protons of the nucleus. (Give your answer in e.V.) which means taking its derivative with respect to r

∂re−r/2 r = 1 − e−r/2 ∂r 2 Answer and/or Hint and setting it equal to zero. Once we’ve found the values of ρ, call one of them ρ∗, one needs to back Ry How about − 32 ? subtitute these values into the original wave func- 14. An autoionizing state of a poly-electron atom (or tion to see which is bigger, more negative, or what- molecule) is one in which at least two electrons are ever; locating the minimum requires more than just excited, such that when one electron drops down to mechanics! a lower state, the other can become ionized. If one 16. For a 3s orbital of the form could turn oﬀ the electron-electron repulsion in the Helium atom i.e., 6 − 6ρ + ρ2 e−ρ/3 e2 → 0 what is its value at its minimum? r12 and one created from ground state Helium an autoionizing state 2s3p, what would be the kinetic 7

Answer and/or Hint

Here, we need only take the derivative of this ex- Answer and/or Hint pression with respect to ρ and set it equal to zero, solving for the value of ρ which does the dirty deed. The second choice has the third and fourth columns 17. In the accompanying figure, the non-Spin-Orbit identical, which makes the determinant zero which coupling Zeeman effect is show graphically for the makes the choice invalid. Lyman α line showing why this line is split into a triplet under a magnetic field. How many lines are The third choice uses a 2p orbital, i.e., not a ground expected for the 2p → 3d transition? The selection state at all. rules are ∆m` − ±1 and 0. Hence, the first choice is the right choice.

field + 3d III. H 2 no field n=3 + 3s The co¨ordinate system used for H2 is shown in Figure 3p 5 for future reference. n=2 2s 2p z

n=1 r P(x,y,z) predicted A spectrum

A R/2 λ= constant FIG. 4: A simpliﬁed explanation for the Zeeman eﬀect in hydrogen. r B y

Answer and/or Hint B −R/2

φ There will be 3 lines for each p-substate, thereby generating 9 lines. x 18. The ground state of Berylium

1s(1)α(1) 1s(1)β(1) 2s(1)α(1) 2s(1)β(1) + FIG. 5: Standard Co¨ordinate Scheme for H2 Cation 1s(2)α(2) 1s(2)β(2) 2s(2)α(2) 2s(2)β(2) 1s(3)α(3) 1s(3)β(3) 2s(3)α(3) 2s(3)β(3)

1s(4)α(4) 1s(4)β(4) 2s(4)α(4) 2s(4)β(4) 1. The standard co¨ordinate system for diatomic molecules has the two nucleii on the z axis, one

1s(1)α(1) 1s(1)β(1) 2s(1)α(1) 2s(1)α(1) (say A) at +R/2 and the other (say B) at -R/2

1s(2)α(2) 1s(2)β(2) 2s(2)α(2) 2s(2)α(2) (so that the internuclear distance is R). Remember 1s(3)α(3) 1s(3)β(3) 2s(3)α(3) 2s(3)α(3) that 1s(4)α(4) 1s(4)β(4) 2s(4)α(4) 2s(4)α(4) s 2 2 2 R rA = x + y + z − 2 1s(1)α(1) 1s(1)β(1) 2s(1)α(1) 2p(1)α(1) 1s(2)α(2) 1s(2)β(2) 2s(2)α(2) 2p(2)α(2) and 1s(3)α(3) 1s(3)β(3) 2s(3)α(3) 2p(3)α(3)

1s(4)α(4) 1s(4)β(4) 2s(4)α(4) 2p(4)α(4) s R2 r = x2 + y2 + z + Which of the above is correct? B 2

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+ FIG. 7: Standard Co¨ordinate Scheme for H2 Cation

2. The standard co¨ordinate system for diatomic molecules has the two nucleii on the z axis, one (say A) at +R/2 and the other (say B) at -R/2 (so that the internuclear distance is R). Remember + FIG. 6: Standard Co¨ordinate Scheme for H2 Cation that s R2 r = x2 + y2 + z − so, for a trial LCAO-MO (an approximate wave A 2 function, not an eigenfunction of the system), and −αrA −αrB s ψapp = e + e 2 2 2 R rB = x + y + z + (which is intentionally left in un-normalized form) 2 what is the value of this wave function at the point P(x,y,z)= P(R,-R/2,-R)? Use an α value of 1. so, for a trial LCAO-MO (an approximate wave function, not an eigenfunction of the system) R R ψ = z − e−αrA − z + e−αrB LCAO−pσ 2 2

Answer and/or Hint (which is intentionally left in un-normalized form) what is the value of the wave function at the point P(x,y,z)= P(R,-R/2,-R)? Assume α = 1.

s R2 r = R2 + (−R/2)2 + (−R) − A 2 Answer and/or Hint and

s 2 R −αr R −αr 2 2 R A B r = R + (−R/2) + (−R) + ψLCAO−pσ = −R − e − −R + e B 2 2 2 with which can be substituted into the wave function. q q s So, r = 7R2 ,and r = 3R2 . These values are 2 2 A 2 B 2 2 −R R rA = R + + −R − easily back-substituted into ψapp as requested. 2 2 9

and Answer and/or Hint s −R2 R2 r = R2 + + −R + B 2 2 NO! This answer is obtained by explicitly carrying out the required diﬀerentiation and ﬁnding that no 3. For an LCAO-MO factoring allows it to be phrased as the r.h.s. de- mands, i.e., as a constant times the original wave −αrA −αrB ψ ∗ ∼ xe − xe function (Ansatz). LCAO−πx (which is intentionally left in un-normalized form) is the function returned after operating on this IV. EPILOGUE LCAO-MO with the Hamiltonian:

2 2 2 After editing this material for weeks, and continuously ¯h ZAe ZBe − ∇2 − − ﬁnding errors, some small, some huge, I have to wrap it 2m r r e A B up and send this oﬀ. If, in the years 2008-2010 or so, you with come across an error, and you e-mail me, I will try to have it corrected. ? Hopψ = Eψ But since this material is written in LaTeX there is some doubt whether or not I’ll have access to a Linux an eigenfunction? machine, and access to the digitialcommons site. You can try; we’ll see what happens, if anything. Thanks to all the students over the last 45 years who’ve taught me Physical Chemistry.

[1] There are two literature references you might want to look (1990). at before doing this problem, C. J. Kawa, J. Chem. Ed., 65, 884 (1988), and G. H. Duﬀey, J. Chem. Ed., 67,35