Chemistry 65.100 A, V
Midterm Test #1
October 2, 1998

ANSWERS

Part A.

1. Hund’s rule states that when putting electrons into degenerate orbitals, they must be put in one at a time until each orbital contains one. Then only can the spin down ones be added. Another way of saying this is that electrons obey a "maximum un-pairing" rule (like people on a bus).

2. The four quantum numbers are:
n – the principle quantum number. This defines the size and energy of the shell.
l – this is the subshell number, which defines the shape of the orbital
m_l – this is the angular momentum number, defining the orientation of the orbital
m_s – this is the spin of the electron

3. To solve this, just look at the group number of the element in the periodic table:
B - 3, Ti - 4, C - 4, Na - 1, Kr - 8

4. The electronic configuration of Curium (Cm) is [Rn]7s²5f ⁷6d¹ because the 5f orbitals are half-filled, which has a lower overall energy than the configuration predicted by AUFBAU.

5. Ionization potential increases going from left to right across a period because electrons are being added to the same shell, and thus have about the same energies. However, the nuclear charge is increasing. Thus, the net attraction between the electrons and the nucleus increases, and so it requires more energy to remove one of the electrons.

6. Arsenic has the ground state electronic configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p³. It must be written in this order for full marks.

1. (a) Using the Rydberg equation, we have 1/l = R[1/m² – 1/n²]. The highest energy transition will occur from a very high level (say n=¥ ) to m=4. Thus,

1/l = R[1/4² – 1/¥ ²]
= R[1/16]
= 0.01097 nm^-1 [1/16]
= 0.000686 nm^-1

Thus, l = 1/(0.000686 nm^-1) = 1458 nm

(b) This is a longer wavelength than visible light, and so must be in the infrared (IR) part of the spectrum.

(c) The energy of the photons is calculated using Planck’s equation, E = hn. Here, we have only the wavelength l to work with, so we make the substitution n = c/l. Thus,

E = h(c/l)
= 6.63 x 10^-34 J s (3.00 x 10⁸ m s^-1)/(1458 x 10^-9 m)
= 1.36 x 10^-19 J (per photon)

per mole, multiply by Avogadro’s number:
E = 1.36 x 10^-19 J x 6.02 x 10²³ mol^-1 = 82,100 J mol^-1
= 82.1 kJ mol^-1

2. (a) The balanced reaction is WO_3(s) + 3 H_2(g) à W_(s) + 3 H₂O_(g)

4.81 kg WO_3(s) x (1000 g / 1 kg) x (1 mol WO₃ / 231.85 g WO₃) x (1 mol W / 1 mol WO₃) x (183.85 g W / 1 mol W) = 3,814 g W produced.

3. Those in the second period (S, Cl and Ar) should be smaller than those in the third period, since n=2 implies electrons closer to the nucleus than n=3. Within a period, the atomic radius decreases going left to right. Thus, in order of increasing atomic radius are:

Ar < Cl < S < Ca < K