Chemistry 65.100 A, V Midterm Test #1 October 12, 2001

Part A. Answer each question with a few words or sentences where necessary. (5 Marks each)

A1. What type of hybrid orbitals does the sulphur atom use to make the compound SF5?

The sulphur atom uses sp3d orbitals to bond with the five F atoms.

A2. Write out the complete electronic configuration (i.e. 1s2 2s2....) of chromium (Cr). Then explain why this differs from reality, which is [Ar] 4s13d5.

The predicted configuration is 1s2 2s2 2p6 3s2 3p6 4s2 3d4. However, the atom can become more energetically stable (i.e. go to a lower energy) by moving one of the 4s electrons to the 3d orbitals. This results in the outer electrons being in two half-filled shells, a very stable configuration.

A3. Why does Sc (atomic number 21) have a smaller atomic radius than Ca (atomic number 20)?

Although Sc has one more electron, it also has one more proton than Ca. This extra proton exerts a larger electrostatic pull on the valence electrons, reducing the diameter of the atom.

A4. Which of LiCl or NaCl has the larger lattice energy? Why? Which of LiF or AlF3 has the larger lattice energy? Why?

LiCl has the larger lattice energy because the Li+ ion is smaller than the Na+ ion.

AlF3 has the larger lattice energy because the Al+3 ion has a larger charge than the Li+ ion.

A5. Why does the spectrum of an element contain only certain colors?

This is because light is produced when electrons undergo transitions in the atoms from one energy level to another. These energy levels are quantized (i.e. have discreet values) in the atom, so the difference in energy (which equals the energy of the resultant photons) is also quantized.

A6. Why is the H-O-H bond angle in water less than the ideal tetrahedral angle of 109.5o

The two lone pairs of electrons on the oxygen atom repel the bonding pairs harder than the bonding pairs repel one another, resulting in the H-atoms beind pushed together slightly, reducing the H-O-H bond angle.

Part B. Answer each question. Best two answers will count. (20 marks each)

B1. Using VSEPR theory, predict the shapes of each of the following species. Wrong name of shape = zero marks!

(a) SiCl4

Valence electrons = 4 + (4 x 7) = 32. Making single bonds from the Si to the Cl atoms and completing the octets around the Cl atoms uses all 32 electrons. The molecule is thus of the type AX4, and must therefore be tetrahedral.

(b) PCl5

Valence electrons = 5 + (5 x 7) = 40. Making single bonds from the P to the Cl atoms and completing the octets around the Cl atoms uses all 40 electrons. The molecule is thus of the type AX5, and must therefore be trigonal bipyramidal.

(c) SCl3-

Valence electrons = 6 + (3 x 7) + 1 = 28. Making single bonds from the S to the Cl atoms and completing the octets around the Cl atoms uses 24 electrons, leaving 4 electron to be placed as two lone pairs in the Si atom. The molecule is thus of the type AX3E2, and must therefore be T-shaped.

(d) BF3

Valence electrons = 3 + (3 x 7) = 24. Making single bonds from the B to the F atoms and completing the octets around the F atoms uses all 24 electrons. The molecule is thus of the type AX3, and must therefore be trigonal planar.

 

B2. Lines in the Paschen series of the hydrogen spectrum come from transitions to the m = 3 level. Calculate:
(a) The wavelength (in nm) of the highest energy transition in this series.
(b) The frequency (in s-1) of the lowest energy transition in this series.
(c) The energy of the photons (in kJ/mol) of the transition from n = 6.

(a) The highest energy transition comes from m = ¥ :

Thus,

(b) The lowest energy transition comes from m = 4:

Thus,

But,

(c) For the transition from n = 6,

But,

To find the energy per mole of photons, multiply by Avogadro's number:

 

B3. Calculate the energy of formation of KF(s) from the elements, given the following data:
Energy of sublimation of K(s) 90 kJ mol-1
First Ionization Potential of K(g) 419 kJ mol-1
Energy of dissociation of F2(g) 159 kJ (mol F2) -1
Electron affinity of F -330 kJ mol-1
Lattice Energy of KF(s) 821 kJ mol-1
It is necessary to write out each step as a chemical reaction. Simply adding the above numbers will get you nowhere.

K(s) ® K(g)                             +90 kJ/mol

K(g) ® K+(g) + e-                      +419 kJ/mol

1/2 F2(g) ® F(g)                      +159 / 2 = +79.5 kJ/mol F

F + e- ® F-(g)                             -330 kJ/mol

K+(g) + F-(g) ® KF(s)             - (+821 kJ/mol) = -821 kJ/mol

Adding the above gives the desired reaction. Adding the energies therefore gives the energy of the total reaction:

K(s) + 1/2 F2(g) ® KF(s)              -562.5 kJ/mol