1. It takes 492 kJ of energy to remove one mole of electrons from the atoms on the surface of solid gold.

a) How much energy does it take to remove a single electron from an atom on the surface of solid gold?

(492 kJ/mol x 10³J/kJ)/(6.022 x 10²³/mol) = 8.17 x 10^-19J

b) What is the maximum wavelength of light capable of doing this?

lambda = (h x c)/E

lambda = (6.626 x 10^-34 J s x 2.998 x 10⁸ m/s)/(8.17 x 10^-19 J)

lambda = 2.43 x 10^-7 m

2. Calculate the longest and shortest wavelengths of light emitted by electrons in the hydrogen atom that begin in the n = 6 state, and then fall to states with smaller values of n.

For longest lambda, n = 6 --> n = 5.

E = -2.178 x 10^-18 (1/36 - 1/25) J = 2.665 x 10^{- 20} J

lambda = (h x c)/E

lambda = (6.626 x 10^-34 J s x 2.998 x 10⁸ m/s)/(2.665 x 10^-20 J)

lambda = 7.454 x 10^-6 m

For shortest lambda, n = 6 --> n = 1.

E = -2.178 x 10^-18 (1/36 - 1/1) J = 2.118 x 10^{- 18} J

lambda = (h x c)/E

lambda = (6.626 x 10^-34 J s x 2.998 x 10⁸ m/s)/(2.118 x 10^-18 J)

lambda = 9.379 x 10^-8 m

3.a) In each of the following sets, which atom or ion has the smallest radius?

i) S, Cl, Kr

Size decreases across a period, and increases down a group. Cl is smallest.

ii) O, P, As

Size decreases across a period, and increases down a group. O is smallest.

iii) O⁺, O, O^-

Cation is smaller than atom; anion is larger than atom. O⁺ is smallest.

b) In each of the following sets, which atom or ion has the smallest ionization energy?

i) Rb, Cs, Ba

I.E. decreases as go down a group, and increases as go across a period. Cs has smallest I.E.

ii) Ne, Cl, Ar

I.E. decreases as go down a group, and increases as go across a period. Cl has smallest I.E.

iii) O^2-, O, O^-

All three species have the same nucleus, but have differing number of electrons. O^2- has smallest I.E.

c) Using data from Chapter 8 of Kotz and Treichel, determine, in kJ/mol, the ionization energy of Cl^- and the electron affinity of Cl⁺.

The ionization energy of Cl^- refers to Cl^-(g) --> Cl(g) + e^-

This process is the reverse of the E.A. for Cl(g).

I.E. here = +349.0 kJ/mol

The electron affinity for Cl⁺ refers to Cl⁺ + e^- --> Cl(g)

Thi process is the reverse of the I.E. for Cl(g).

E.A. here = -1251 kJ/mol

4. For the following molecules and ions:

F₂O, ICl₄⁺, HPO₄^2-, SeF₄, PCl₃, IBr₃, HClO₄, BrO₃^-

a) draw Lewis structures (including resonance forms where appropriate);

b) categorize each species according to the AX_nE_m nomenclature;

c) name the shape of the species;

d) predict whether those species which are neutral molecules will have a dipole moment;

e) for all the species that contain oxygen atoms, calculate the bond order of the central atom to oxygen bonds.

F₂O

F₂O has 20 valence electrons. Its Lewis structure is
 
 

The molecule is AX₂E₂, and has a bent geometry. It has a dipole moment. The two O-F bonds have B.O. = 1.

ICl₄⁺

ICl₄⁺ has 34 valence electrons. Its Lewis structure is
 
 

The ion is AX₄E, and has a seesaw geometry.

HPO₄^2-

HPO₄^2- has 32 valence electrons. Its Lewis structure is
 
 

The ion is AX₄ around P, and has a tetrahedral geometry. B.O. = 1.0 for P-O of P-O-H, and B.O. = 4/3 for P-O with no H.

SeF₄

SeF₄ has 34 valence electrons. Its Lewis structure is
 
 

The molecule is AX₄E, and has a seesaw geometry. It has a dipole moment.

PCl₃

PCl₃ has 26 valence electrons. Its Lewis structure is
 
 

The molecule is AX₃E, and has a triangular pyramidal geometry. It has a dipole moment.

IBr₃

IBr₃ has 28 valence electrons. Its Lewis structure is
 
 

The molecule is AX₃E₂, and has a T-shaped geometry. It has a dipole moment.

HClO₄

HClO₄ has 32 valence electrons. Its Lewis structure is
 
 

The molecule is AX₄ around Cl, and has a tetrahedral geometry around Cl. It has a dipole moment. The Cl-O B.O. for the Cl-O-H bond is 1.0; the other Cl-O B.O.s are 2.0.

BrO₃^-

BrO₃^- has 26 valence electrons. Its Lewis structure is
 
 

The ion is AX₃E, and has a triangular pyramidal geometry. The Br-O B.O. is 5/3.

5. Draw resonance structures and predict bond orders and bond angles for the following:

a) NO₂Cl (N is the central atom);

NO₂Cl has 24 valence electrons. Its Lewis structure is
 
 

N-Cl has a B.O. of 1.0; N-O has a B.O of 1.5; all bond angles are 120^o.

b) HSO₃^- (H is bonded to O);

HSO₃^- has 26 valence electrons. Its Lewis structure is
 
 

S-O B.O for S-O-H is 1.0; S-O B.O. for the other S-O bonds = 1.5. All bond angles are less than 109.5^o.

6. Place the species below in the order of increasing bond length for the N-O bond;

H₂NOH (N is the central atom, and one H is bonded to O), NNO,NO⁺, NO₂^-, NO₃^-.

The Lewis structures are:
 
 

The N-O B.O. for H₂NOH is 1.0.
 
 

The N-O B.O.for NNO is 1.5.
 
 

The N-O B.O. for NO⁺ is 3.0.
 
 

The N-O B.O. for NO₂^- is 1.5.
 
 

The N-O B.O.for NO₃^- is 4/3.

The order of increasing bond lengths is: NO⁺ < NO₂^- = NNO < NO₃^- < H₂NOH.

7. 1.00 g PCl₅ is introduced into a 250 mL flask, the flask is heated to 250^øC, and the dissociation of PCl₅ is allowed to reach equilibrium according to PCl₅(g) = PCl₃(g) + Cl₂(g). The quantity of Cl₂(g) present at equilibrium is found to be 0.250 g. What is the value of the equilibrium constant K_c for this reaction at 250^oC?

Note that initial moles of PCl₅ = 1.00 g/208.2 g/mol = 4.80 x 10^-3 mol

Moles	PCl5	PCl3	Cl2
Initial	4.80 x 10-3	0	0
Change	-x	x	x
Equilibrium	4.80 x 10- 3-x	x	x

Note that problem states that x = 0.25 g Cl₂/70.90 g Cl₂ per mole = 3.92 x 10^-3 mol

K_c = [x/0.250]²/[(4.80 x 10^-3 - x)/0.250)]

K_c = (3.53 x 10^-3)²/(0.250 x 1.27 x 10^-3) = 3.92 x 10^-2 mol/L