Chemistry - A Molecular Science

lowered by one because the outermost electrons

of the atom are lost in forming the ion.

This loss of the outermost shell makes the cation much smaller than the corresponding atom. In addition, the decrease in the number of electrons decreases the shielding experienced by the remaining electrons and increases Z

. The increased effective nuclear eff

charge contracts the outer shell even further.

Anions are larger than their parent atoms

(Figure 4.3)

. Anions have more electrons

than protons; thus, the valence electrons ar

e more effectively shielded from the positive

charge of the nucleus, which decreases Z

. This decrease in effective nuclear charge eff

causes the outer electron shell to expand. As

the negative charge increases, the effective

nuclear charge decreases and the size of

the anion increases. Consequently, N

3- is larger

than O

2-, which is larger than the F

1-.

Example 4.4

Arrange the following in order of increasing size: S

2-, Ar, and Ca

2+.

Ar, S

2- and Ca

2+ each have 18 electrons, so the screening is the same. Therefore,

increasing the number of protons increases Z

and decreases the size. eff

Ca

2+ (20 protons) < Ar (18 protons) < S

2- (16 protons).

Energy

H

®

1+H

Cl

Cl

®

1-

FF®

1-

Cl

Cl

®

1+

(a) HCl

(b) Cl-F

Figure 4.4 Oxidation states in H-Cl and Cl-F

Note that this order is different from that

of atomic size. Calcium atoms are the largest

because they have the highest n quantum num

ber and the smallest effective nuclear

charge. Sulfur is larger than argon because its

effective nuclear charge is less. Thus, the

order of atomic sizes is Ar < S < Ca.

4.4

OXIDATION STATES

The

oxidation state

(or

oxidation number

) of an atom is the charge it would have if its

bonds were ionic. In ionic compounds, the oxidation state of an ion is the charge on the ion. For example, the oxidation states of sodi

um and chlorine in NaCl are +1 and -1,

respectively. However, oxidation states are also used to account for the electrons in compounds that are not ionic (compounds with no metals). In these cases, the bonds are assumed to be ionic by

assigning

all bonding electrons to the more electronegative atom

(atom with lower energy orbitals). Compare

Figures 4.1 and 4.4a. In both, the electron is

transferred from the high-energy orbital into the low-energy orbital to produce ions. The difference is that the orbital energy of hydroge

n is much lower than that of sodium, so the

electron does not actually transfer in HCl;

we simply assume that it does to determine the

oxidation states

. If the transfer took place, the hydrog

en atom would become a +1 ion, so

(a) The valance orbitals of Cl are

lower than the electron in H, so

the bonding electrons would be assigned to Cl in HCl. Loss of an electron by H produces the H

1+ ion, or a +1 oxidation state. Gain of

an electron by Cl produces the Cl

1-, or a -1 oxidation state.

(b) The valence electrons of Cl are

higher than the valence orbitals

on F, so the bonding electrons would be assigned to F in ClF. Loss of an electron from Cl produces the Cl

1+ ion, or a +1 oxidation state.

Gain of an electron to F produces the F

1- ion, or a -1 oxidation state.

5A

6A

7A

-3

-2

-1

anion

(cage)

atom

(sphere)

NO

F

SC

l

Se

Br

Te

I

Figure 4.3 Relative sizes of nonmetal atoms and their anions The char

ge on the anion is

given at the top of each column.

Chapter 4 The Ionic Bond

© by

North

Carolina

State

University