Classification of Elements and Periodicity in Properties · Lesson 8 of 10

Valence, Oxidation States, and Second-Period Exceptions

“Use outer electrons to predict combining patterns, then explain where the simplest patterns need refinement.”

Learning Objectives

• Relate common main-group valences to outer electron counts. • Distinguish valence, oxidation state, and coordination or covalency counts. • Predict simple binary formulas while recognising variable valence. • Explain why first members of groups can differ from heavier members. • Describe diagonal relationships and the greater multiple-bond tendency of second-period p elements.

From electron counts to combining patterns

Elements in a group share an outer configuration, so their compounds often share a pattern of formulas. Group 1 metals commonly form singly positive ions, whereas halogens commonly complete their outer arrangement by gaining or sharing one electron. These tendencies connect the microscopic electron arrangement to the composition of visible substances.

Definition

Valence

An element’s combining capacity in a specified chemical context. In simple compounds it can often be described by the number of electrons lost, gained, or shared. It is an unsigned combining count and is not automatically identical to oxidation state.

For many main-group compounds, a useful starting pattern is the number of outer electrons for Groups 1, 2, 13, and 14, and eight minus that number for common simple compounds of Groups 15, 16, and 17. This predicts the familiar pattern 1, 2, 3, 4, 3, 2, 1. It is a guide, not an absolute limit on the chemistry of all members.

Group	1	2	13	14	15	16	17	18
Outer s and p electrons	1	2	3	4	5	6	7	8, except He has 2
Common simple valence	1	2	3	4	3	2	1	Usually 0

Some members show additional combining states: phosphorus can form both three- and five-bond compounds, and sulphur can occur in different oxidation states. Transition and actinoid elements are especially known for variable oxidation states. An outer-electron count should therefore guide the first prediction and then be checked against the specified compound.

Group	Representative hydrides	Representative oxides
1	$L i H$ , $N a H$ , $K H$	$L i_{2} O$ , $N a_{2} O$ , $K_{2} O$
2	$C a H_{2}$	$M g O$ , $C a O$ , $S r O$ , $B a O$
13	$B_{2} H_{6}$ , $A l H_{3}$	$B_{2} O_{3}$ , $A l_{2} O_{3}$ , $G a_{2} O_{3}$ , $I n_{2} O_{3}$
14	$C H_{4}$ , $S i H_{4}$ , $G e H_{4}$ , $S n H_{4}$	$C O_{2}$ , $S i O_{2}$ , $G e O_{2}$ , $S n O_{2}$ , $P b O_{2}$
15	$N H_{3}$ , $P H_{3}$ , $A s H_{3}$	$N_{2} O_{3}$ , $N_{2} O_{5}$ , $P_{4} O_{6}$ , $P_{4} O_{10}$ , $A s_{2} O_{3}$ , $A s_{2} O_{5}$ , $S b_{2} O_{3}$ , $S b_{2} O_{5}$ , $B i_{2} O_{3}$
16	$H_{2} O$ , $H_{2} S$ , $H_{2} S e$ , $H_{2} T e$	$S O_{3}$ , $S e O_{3}$ , $T e O_{3}$
17	$H F$ , $H C l$ , $H B r$ , $H I$	$C l_{2} O_{7}$ is an example of a high oxide

The repeated formulas reveal family relationships, while multiple oxides show that the valence or oxidation pattern need not be unique. The table lists examples rather than every possible compound. Formula ratios also do not always give a molecule directly: boron hydride is written $B_{2} H_{6}$ , not merely the simplest ratio $B H_{3}$ .

Example — Predicting silicon bromide

Problem
Use the simple valences to predict the formula formed by silicon and bromine.

1.Silicon belongs to Group 14 and commonly has valence 4 in such a binary covalent compound.
2.Bromine belongs to Group 17 and commonly has valence 1.
3.One Si therefore combines with four Br atoms, giving $S i B r_{4}$ .
4.The formula is based on a specified combining pattern; it is not obtained by copying group numbers as subscripts.

Example — Predicting aluminium sulphide

Problem
Find the simplest formula for aluminium sulphide.

1.Aluminium commonly forms $A l^{3 +}$ and sulphur forms sulphide $S^{2 -}$ .
2.The total positive and negative charges must cancel. The smallest common total is six.
3.Two aluminium ions contribute $+ 6$ and three sulphide ions contribute $- 6$ .
4.The ratio is 2:3, giving $A l_{2} S_{3}$ . In a solid ionic compound this is a formula-unit ratio, not necessarily a separate molecule.

Example — A fluoride of element 71

Problem
Predict the common fluoride formula of the lanthanoid with atomic number 71.

1.The element is lutetium, Lu, with the characteristic configuration $[X e] 4 f^{14} 5 d^{1} 6 s^{2}$ .
2.Lanthanoids commonly form $3 +$ ions. For Lu, removing the two 6s electrons and the 5d electron leaves a filled 4f subshell.
3.Fluoride is $F^{-}$ , so three fluoride ions balance one $L u^{3 +}$ ion.
4.The common formula is $Lu F_{3}$ . A full chemical prediction still uses known combining behaviour, not merely the last printed orbital label.

Oxidation state is electron bookkeeping

A shared bond does not always involve a complete physical transfer of electrons. Oxidation state uses a consistent bookkeeping convention: assign a bond’s electrons to the more electronegative partner, and calculate the resulting charge for each atom. This gives a useful signed number without claiming that every atom carries that exact measurable charge.

Definition

Oxidation state

The charge assigned to an atom after bond electrons are allocated according to an ionic, electronegativity-based approximation. The sum of oxidation states equals the overall charge of the molecule or ion.

Oxygen illustrates why oxidation state is not fixed for every compound. The electronegativity order is $F > O > N a$ . In $O F_{2}$ , each fluorine is assigned oxidation state $- 1$ , leaving oxygen at $+ 2$ . In $N a_{2} O$ , sodium is assigned $+ 1$ and oxygen $- 2$ . Oxygen still has combining capacity two in these examples, but its oxidation state changes sign.

Example — Opposite oxygen oxidation states

Problem
Determine oxygen’s oxidation state in $O F_{2}$ and $N a_{2} O$ .

1.For neutral $O F_{2}$ , let oxygen’s oxidation state be $x$ . The sum is $x + 2 (- 1) = 0$ .
2.Solving gives $x = + 2$ , because fluorine takes bond-electron ownership.
3.For neutral $N a_{2} O$ , use $2 (+ 1) + x = 0$ , giving $x = - 2$ .
4.This is an electronegativity-based assignment, not proof of a free $O^{2 +}$ ion in $O F_{2}$ .

Example — Counting bonds and assigning charge

Problem
Are aluminium’s oxidation state and the source’s covalency count the same in $[A l C l (H_{2} O)_{5}]^{2 +}$ ?

1.The five water ligands are neutral and the chloride ligand is assigned $- 1$ . A ligand is a group bound to the central atom.
2.If aluminium’s oxidation state is $x$ , then $x - 1 = + 2$ , giving $x = + 3$ .
3.There are six directly attached donor atoms: one from Cl and five from the water molecules. The coordination count is six.
4.The source calls that bond count covalency 6. Whatever bond-count terminology is used, it must not be confused with oxidation state $+ 3$ .

Common mistake

Eight outer electrons do not imply a valence of eight. Noble gases usually have valence zero in ordinary chemistry. Also keep an unsigned combining count, a signed oxidation state, and a coordination count separate.

Why the first members can be unusual

Lithium, beryllium, and the second-period p-block elements are much smaller than their heavier group members. Small size gives strong attraction for electrons and, for their ions, a high charge-to-size ratio. These factors increase the ability to distort a neighbouring electron cloud and help explain stronger covalent character in lithium and beryllium compounds compared with simple heavier-group analogues.

Element pair	Approximate atomic radii / pm	Example ionic radii / pm
Li and Na	152 and 186	$L i^{+}$ : 76; $N a^{+}$ : 102
Be and Mg	111 and 160	$B e^{2 +}$ : 31; $M g^{2 +}$ : 72
B and Al	88 and 143	—

These tabulated ionic values refer to particular radius conventions and environments, so they need not equal a value from another ionic-radius table. Their teaching purpose is the strong size contrast. Charge is also important: $B e^{2 +}$ is much smaller and more strongly polarizing than a similarly located singly charged cation.

Definition

Diagonal relationship

A resemblance between a second-period element and the third-period element one group to its right. Li–Mg and Be–Al are the examples highlighted here. Opposing changes across and down the table can produce comparable sizes and other properties.

Diagonal relationships— The highlighted links are similarities between selected neighbours in successive periods; they are not claims that the paired elements have identical chemistry.

Second-period atoms have only 2s and 2p valence orbitals: one plus three gives four. There are no 2d orbitals because the allowed values for $n = 2$ are only $l = 0$ and $l = 1$ . These atoms do not form ordinary expanded-octet Lewis structures. A familiar contrast is four-coordinate $[B F_{4}]^{-}$ versus six-coordinate $[A l F_{6}]^{3 -}$ . Larger atoms can accommodate more neighbours.

Keep orbital counting and bonding models distinct

The third shell permits 3s, 3p, and 3d orbitals, a total of nine. That quantum-number count is correct, but simply saying “available d orbitals cause expanded bonding” is an oversimplified bonding model. The structural contrast can be understood here through size and coordination without assuming that every extra bond is a localised d-orbital bond.

Second-period p orbitals are also compact enough to overlap effectively side-on, making strong p–p multiple bonding common. Carbon forms $C = C$ and $C \equiv C$ bonds, nitrogen forms $N = N$ and $N \equiv N$ , and combinations such as $C = O$ , $C = N$ , $C \equiv N$ , and $N = O$ are important. This is the pπ–pπ multiple-bond tendency mentioned in the source. Larger, more diffuse orbitals of heavier members often overlap less effectively in this way.

Example — Explaining an exception without discarding a family

Problem
Why can Be compounds resemble Al compounds in some respects more than Mg compounds?

1.Be is the small first member of Group 2; Al lies diagonally below and to its right.
2.Small size and a relatively large charge-to-size ratio favour strong attraction and distortion of neighbouring electron clouds.
3.Be and Al therefore share some covalent or polarizing behaviour despite belonging to different groups.
4.This is a diagonal relationship, not a replacement for their group membership or a claim that all their compounds are alike.

Quick check

Which is the usual simple valence of a Group 16 element in a hydride?

Quick check

What formula follows from $A l^{3 +}$ and $S^{2 -}$ in a neutral ionic compound?

Quick check

What is oxygen’s oxidation state in $O F_{2}$ ?

Quick check

What are Al’s oxidation state and coordination count in $[A l C l (H_{2} O)_{5}]^{2 +}$ ?

Quick check

Which is a diagonal-relationship pair highlighted in this chapter?

Quick check

Why are there no 2d orbitals?

Practice Problems

Explain the common simple-valence pattern across the main groups, including its limitations.
Predict the formulas for lithium oxide, magnesium nitride, aluminium iodide, and silicon oxide using specified common combining states.
Explain why phosphorus and fluorine can form $P F_{3}$ and $P F_{5}$ , so one blindly applied formula rule is insufficient.
Compare oxygen’s oxidation state in $O F_{2}$ and $N a_{2} O$ without claiming that oxidation state equals actual ionic charge in every compound.
Distinguish oxidation state, valence, and coordination count using the aluminium complex.
Explain the anomalous behaviour of Li and Be in terms of size and electron-cloud distortion.
Compare $[B F_{4}]^{-}$ with $[A l F_{6}]^{3 -}$ and explain what the comparison does and does not establish about orbitals.
Give two examples each of multiple bonds between identical second-period atoms and between different second-period atoms.

Key Takeaways

Outer configurations explain recurring combining patterns, but many elements have variable states. Simple formulas follow combining capacities or charge balance, not the direct use of group numbers as subscripts. Oxidation state is a signed electron-bookkeeping assignment and differs from bond or coordination counts. Small size, strong attraction, and compact valence orbitals explain important second-period anomalies. Diagonal relationships and multiple-bond tendencies refine, rather than replace, the group pattern.

Previous · Lesson 7

Electron Gain Enthalpy and Electronegativity

Next · Lesson 9

Periodicity, Reactivity, and Oxides

From electron counts to combining patterns

Oxidation state is electron bookkeeping

Why the first members can be unusual

Quiz

Practice Problems

Key Takeaways