Chemistry 1211

Test #5

fall 2009 MW

****Please SHOW ALL WORK

1. What is the maximum number of electrons in an atom that can have the following quantum numbers:[10] {no work, no partial credit}

i) n = 5

For n = 5, the possible l values are 0,1,2,3 and 4. For l = 0 (5s orbital) there is but 1 orbital that houses 2 electrons; for l = 1 (5p orbital) there are 3 orbitals (m= -1,0,+1) that house 6 electrons; for l = 2 (5d orbital) there are 5 orbitals (m = -2,-1,0,1,2) that house 10 electrons; For l = 3 (5f orbital) there is 7 orbitals (m =-3,-2,-1,0,1,2,3) that house 14 electrons.; For l = 4 (5g orbital) there are 9 orbitals (m =-4, -3,-2,-1,0,1,2,3,4) that house 18 electrons. The total is 2 + 6 + 10 +14+18 = 50 electrons.

ii) n = 3 , s = +1/2

For n = 3, the possible l values are 0,1, and 2. For l = 0 (3s orbital) there is but 1 orbital that houses 2 electrons; for l = 1 (3p orbital) there are 3 orbitals (m= -1,0,+1) that house 6 electrons; for l = 2 (3d orbital) there are 5 orbitals (m = -2,-1,0,1,2) that house 10 electrons; The total is 2 + 6 + 10 = 18 electrons, BUT only half of these have a +1/2 spin quantum number so… there are 18/2 or 9 electrons.

iii) n =3, l = 2, s = +1/2

These are valid quantum numbers for electrons in a 3d orbital. There can be 10 electrons in the 3d quantum state, half with spin +1/2, the other half with spin -1/2, so 5 e- are the maximum possible.

iv) n =1, l =1, m =0, s = -1/2

These are NOT valid quantum numbers for ANY electron in ANY orbital. The quantum number n NEVER equals the quantum number l, so NO electrons can reside in such an orbital.

v) n = 3, l = 1, m = 0, s = -1/2

These are valid quantum numbers for an electron in a 3p orbital. Pauli exclusion says that no 2 electrons in the same atom can have the same set of 4 quantum numbers. SO, this represents only 1 electron.

2. Please describe the following using a few appropriate words or terminology: [15]

a) electronic configuration of Cu(I) cation, Cu⁺.

1s² 2s² 2p⁶ 3s² 3p⁶ (4s⁰) 3d¹⁰ (28 electrons), copper loses its 4s electron before its 3d

b) de Broglie hypothesis

de Broglie showed that all matter had a characteristic wavelength, including the electron.

c) Planck’s Quantum theory

Energy is discontinuous and is present only in energy packets having values which are integer multiples of hv.

d) a photon

A photon is a light particle.

e) Heisenberg Uncertainty Principle

There is a fundamental limitation to our ability to make measurements of the electron within the atom. Consequently, we cannot measure simultaneously both the position and velocity of the electron.

3. What is a wavefunction and why is it important in chemistry? [5]

A wavefunction is a solution to the wave equation. It tells us everything that can be known about the electron within the atom. It is made necessary by the Heisenberg Uncertainty Principle since viewing the electron as a particle (deterministically) cannot provide us with the information we need to understand it.

4. Ozone,O₃, absorbs 254 nm radiation and dissociates into O₂ molecules and O atoms.

[O₃(g) + hn ------> O₂(g)+ O(g)]

A 1.00 liter sample of air at 22^oC and 748 mm Hg contains 25 O₃ molecules for every 1x10⁶ gas particles. {R = 0.082 L-Atm/mol-K, h = 6.626 x 10^-34 J-sec, c = 3x10⁸ m/sec}

a) How many gas particles are present in a liter of air at the above conditions?

n = PV/RT = (748/760)(1.0L)/ 0.082 l-atm/mol-K)(295K) = 4.07 x 10^-2 mol

# particles = (4.07 x 10^-2 mol)(6.023 x10²³ gas particles/mol) = 2.45x10²² particles

b) How many O₃ molecules are there in that liter of air?

# ozones = (2.45x10²² particles)(25 O₃/1x10⁶ particles) = 6.12x10¹⁷ O₃

c) How much energy is required to dissociate one O₃ molecule into O₂ and O?

E= hv = hc/wavelength in meters

E= (6.626x10-34 J-sec)(3x108)/254x10-9 m)
E = 7.82x10-19 J

d) How much energy, measured in joules, must be absorbed if all the ozone molecules in the mixture are to completely dissociate? Assume that each photon absorbed causes one O₃ molecule to break apart as shown.

total energy = (7.82x10^-19 J)(6.12x10¹⁷ ozones) = 0.48J

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Chemistry 1211

Test #5

fall 2009 TR

****Please SHOW ALL WORK

1. Please show the maximum number of electrons in an atom that can have the following quantum numbers: [20 points] {no work, no partial credit}

a) n = 4

For n = 4, the possible l values are 0,1,2, and 3. For l = 0 (4s orbital) there is but 1 orbital that houses 2 electrons; for l = 1 (4p orbital) there are 3 orbitals (m= -1,0,+1) that house 6 electrons; for l = 2 (4p orbital) there are 5 orbitals (m = -2,-1,0,1,2) that house 10 electrons; For l = 3 (4f orbital) there is 7 orbitals (m =-3,-2,-1,0,1,2,3) that house 14 electrons. The total is 2 + 6 + 10 +14 = 32 electrons.

b) n = 2 , s = +1/2

For n = 2, the possible l values are 0 and 1. For l = 0 (2s orbital) there is but 1 orbital that houses 2 electrons; for l = 2 (2p orbital) there are 3 orbitals (m= -1,0,+1) that house 6 electrons; only have will have s quantum number s = +1/2, however, so the maximum number of electrons will be (2+6)/2 = 4 electrons

c) n =4, l = 2, s = +1/2

These are valid quantum numbers for electrons in a 4d orbital. There are 5 types of 4d orbital and it can hold a maximum of 10 electrons. We can show that 5 of those electrons will have a spin quantum +1/2, the others spin quantum number -1/2, so (10/2) = 5e are the maximum possible.

d) n =1, l = 0, m =1, s = -1/2

These are NOT valid quantum numbers for ANY electron in ANY orbital. The quantum number m NEVER exceeds the value of the quantum number l, so NO electrons can reside in such an orbital.

e) n = 3, l = 1, m = 0, s = +1/2

These are valid quantum numbers for an electron in a 3p orbital. Pauli exclusion says that no 2 electrons in the same atom can have the same set of 4 quantum numbers. SO, this represents only 1 electron.

2. Please describe the following using appropriate words or terminology: [10]

a) electronic configuration of nitride anion, N^3-

1s² 2s² 2p⁶ (10 electrons)

b) de Broglie hypothesis

de Broglie showed that all matter had a characteristic wavelength, including the electron.

c) Planck’s Quantum theory

Energy is discontinuous and is present only in energy packets having values which are integer multiples of hv.

d) A photon

A photon is a light particle.

e) Penetration

An electron is said to be penetrating if it has a high probability of being found close to the nucleus. Electrons in s orbitals are more penetrating than electrons in p, d or f orbitals.

3. What is a wavefunction and why is it important in chemistry? [5]

A wavefunction is a solution to the wave equation. It tells us everything that can be known about the electron within the atom. It is made necessary by the Heisenberg Uncertainty principle since viewing the electron as a particle (deterministically) cannot provide us with the information we need to understand it.

4. Ozone,O₃, absorbs 254 nm radiation and dissociates into O₂ molecules and O atoms.

[O₃(g) + hn ------> O₂(g)+ O(g)]

A 1.00 liter sample of air at 27^oC and 700 mm Hg contains 75 O₃ molecules for every 1x10⁶ gas particles. {R = 0.082 L-Atm/mol-K, h = 6.626 x 10^-34 J-sec, c = 3x10⁸ m/sec}

a) How many gas particles are present in a liter of air at the above conditions?

n = PV/RT = (700/760)(1.0L)/ 0.082 l-atm/mol-K)(300K) = 3.74 x 10^-2 mol

# particles = (3.74 x 10^-2 mol)(6.023 x10²³ gas particles/mol) = 2.25x10²² particles

b) How many O₃ molecules are there in that liter of air?

# ozones = (2.25x10²² particles)(75 O₃/1x10⁶ particles) = 1.69x10¹⁸ O₃

c) How much energy is required to dissociate one O₃ molecule into O₂ and O?

E= hv = hc/wavelength in meters

E= (6.626x10-34 J-sec)(3x108)/254x10-9 m)
E = 7.82x10-19 J

d) How much energy, measured in joules, must be absorbed if all the ozone molecules in the mixture are to completely dissociate? Assume that each photon absorbed causes one O₃ molecule to break apart as shown.

total energy = (7.82x10^-19 J)(1.69x10¹⁸ ozones) = 1.32J