Periodic Table and Valence Electrons

Name _________________________________________ Date _________________________ Period ____

Homework
Chapter 7: Electronic Structure of Atoms
Exercises: Sections 7.4, 7.5 : Ionization Energy and Electron Affinities

1. Write equations that show the process for

(a) The first two ionization energies of gallium;

first: Ga(g) → Ga+(g) + 1 e– second: Ga+(g) → Ga2+(g) + 1 e–

(b) the fourth ionization energy of rhodium.

fourth: Rh3+(g) → Rh4+(g) + 1 e–

2. The difference between the third and fourth ionization energies of scandium is much larger than the difference between the third and fourth ionization energies of titanium. Why?

Scandium = after its 3rd ionization - has an octet of valence electrons in its 3rd energy level, a +3 charge on the cation, and Zeff would have increase substantially – VERY STABLE = requires greater energy to remove additional electrons

→ Sc3+ →

Before: 1s2 2s2 2p6 3s2 3p6 4s2 3d1 1s2 2s2 2p6 3s2 3p6 4s2 3d2

After: 1s2 2s2 2p6 3s2 3p6 1s2 2s2 2p6 3s2 3p6 4s1

new valence e¯ Titanium = after its 3rd ionization - still has 1 unpaired valence electron, so, NO stable octet of valence electrons, does have a +3 charge on the cation, but Zeff has NOT increased – all-in-all, not very stable = requires less energy to remove additional electrons
3. (a) What is the trend in first ionization energies as one moves across the fourth period from K to Kr?

1st ionization energies increase going from left-to-right across a period. (As Z increases, valence electrons are drawn closer to the nucleus, requiring more energy to remove.)

(b) How does this trend compare with the trend in atomic sizes? From left-to-right across a period, as valence electrons are drawn closer to the nucleus the atomic radii decreases …… so as ionization energy ↑, atomic size ↓

4. For each of the following pairs, indicate which element has the larger first ionization energy. In each case provide an explanation in terms of electron configuration and effective nuclear charge.

(i) K, Ca;

Ca b/c its valence electrons are in a pair, , whereas K has

only 1 unpaired valence electron, ; in addition, Zeff for K’s

valence electron is +1, whereas Zeff for Ca’s valence electrons is +2

(b) Si, C;

C both elements are in the same group so both have the same number and configuration of valence

electrons, and , which means each have the same

Zeff, - but carbon’s valence electrons are in period 2 whereas silicon’s are in period 3; ionization energies decrease going down a group b/c the valence electrons get further away from the nucleus

4. Continued:

(c) Mn, Br;
Br
each has 7 valence electrons but in entirely different

configurations and and, it is easier to

remove a lone electron than a paired electron; also, both elements are in the same period so first ionization energies increase from left-to-right because of increasing Zeff

(d) Sn, Xe. Xe Xenon has a stable octet of valence electrons,

and tin does not , ; remember, it is easier to remove

unpaired electrons - also, xenon has a larger Zeff than tin

5. Write the electron configuration of the following ions and determine which has noble-gas configuration.

(a) Si2+ 1s2 2s2 2p6 3s2 = NO

(b) Se2 – 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 = YES (the 4s2 and the 4p6)

6. Identify the element whose ion has the following electron configuration:

(a) a 2+ ion with [Ar]3d9 argon has 18 electrons, this element has 9 more, and b/c it is a 2+ ion add 2 more electrons = 29 total electrons = Copper

(b) How many unpaired electrons does this ion contain? 1 unpaired electron in the 3d subshell

Cu2+: 1s2 2s2 2p6 3s2 3p6 4s0 3d9*

*the Cu2+ ion has lost one 4s electron and one 3d electron

Recall about copper: neutral Cu: predicted = 1s2 2s2 2p6 3s2 3p6 4s2 3d9

neutral Cu: actual = 1s2 2s2 2p6 3s2 3p6 4s1 3d10
7. The most common and stable oxides of gallium, germanium, and arsenic have the formulas Ga2O3, GeO2, and As2O5, respectively. Explain this series of compositions in term of the electronic configurations of the elements.

In order to form these oxides the elements must adopt oxidation numbers of Ga3+, Ga4+, and As5+ respectively. The electron configuration of each neutral atom and ist ion is as follows:

Ga = [Ar] 4s2 3d10 4p1 Ge = [Ar] 4s2 3d10 4p2

Ga2+ = [Ar] 4s0 3d10 4p1 Ge4+ = [Ar] 4s0 3d10 4p0

As = [Ar] 4s2 3d10 4p3

As5+ = [Ar] 4s0 3d10 4p0

8. While the electron affinity of bromine is a negative quantity, it is positive for Kr. Explain this difference in terms of electron configurations of the two elements.

Electron Affinity Lewis Dot Electron Configuration

Br = -325 kJ/ mole [Ar] 4s2 3d10 4p5

Kr > 0 [Ar] 4s2 3d10 4p6

The more negative the electron affinity the greater the attraction for an additional electron. Bromine has 7 valence electrons and would like to acquire 1 more to make a stable octet. Krypton already has a stable octet and has no desire to add another electron since destabilization would be the result.

9. (a) Write an equation for the process that corresponds to the electron affinity of the Mg+ ion.

Mg+ + 1 e– → Mg

(b) Write the electron configurations of the species involved.

Mg+ = [Ne] 3s1 → Mg = [Ne] 3s2

(c) What process does this electron affinity equation correspond to?

the process is the reverse of the first ionization energy for a neutral magnesium atom

(d) What is the magnitude of the energy change in the process? (Hint: The answer is in Figure 7.12 on page 268 of the textbook.)

738

Add an electron to Mg+ ion and the resulting Mg atom will be - 738 “less energetic”

Take away an electron from a neutral Mg atom and the resulting Mg+ ion will be + 738 “more energetic”