Valency

Look at a shelf of chemical formulae — $\text{H}_2\text{O}$ , $\text{NaCl}$ , $\text{CaCl}_2$ , $\text{Al}_2\text{O}_3$ — and a question jumps out: why these particular numbers of atoms, and not some other combination? The answer is a single idea called valency, the combining capacity of an atom, and it can be read straight off the atom's electron-shell diagram. Once you know the valency of two elements, writing the correct formula stops being guesswork and becomes a short, mechanical procedure called the criss-cross method. This article builds valency from the shell diagram upward: how to count it, why a simple rule about "4 or fewer" versus "more than 4" valence electrons works, how the criss-cross method turns two valencies into one formula, and how the same idea extends to charged groups of atoms like $\text{OH}^-$ and $\text{SO}_4^{2-}$ .

The core insight

Valency is the number of electrons an atom's outermost shell has to gain, lose, or share to reach a stable duplet or octet, and criss-crossing two atoms' valencies as each other's subscript gives the chemical formula of the compound they form.

The core idea

Valency is the combining capacity of an element, fixed by how many electrons sit in the outermost (valence) shell of its atom. Everything else in this article follows from three counting rules applied to that one number.

A completely filled outermost shell — 2 electrons for the K shell alone (the duplet rule), or 8 electrons for any outer shell beyond that (the octet rule) — means the atom is already stable. Such an atom has zero valency; it does not need to react to become stable.
If the outermost shell holds 4 or fewer electrons, the atom's easiest path to stability is to lose (or share) exactly those electrons, so valency equals the valence-electron count directly.
If the outermost shell holds more than 4 electrons, the shorter path to a full octet is to gain the missing electrons instead, so valency is the shortfall from 8:

\text{valency} = 8 - (\text{number of valence electrons}), \qquad \text{used only when valence electrons} > 4. \qquad (1)

where: the number of valence electrons is read directly from the outermost shell of the atom's electron distribution, and valency is always reported as a positive whole number (a "how many bonds worth of capacity," not a signed charge). The figure below shows this split — count on one side, look up the rule on the other.

Figure

A flowchart showing valency determined from the count of valence electrons in the outermost shell, branching into zero valency for a full duplet or octet, valency equal to the count when four or fewer, and eight minus the count when more than four

Once the valencies of two combining units are known, the criss-cross method writes the formula: the valency of one element becomes the numerical subscript of the other, and the ratio is simplified to smallest whole numbers. The same method works for a polyatomic ion (a charged group of atoms behaving as one unit, such as $\text{OH}^-$ , $\text{SO}_4^{2-}$ , or $\text{NO}_3^-$ ) — its charge magnitude simply plays the role of valency, and if more than one unit of that ion is needed, its formula is enclosed in brackets before the subscript is written, as in $\text{Ca(OH)}_2$ . At this stage valency is used purely as a number taken from a shell diagram or a standard ion table; why electrons actually transfer or are shared during bonding is a question left for later grades.

Derivation AWhy criss-crossing valencies gives a valid formula

The goal is to show, from nothing more than the definition of valency as "combining capacity," why swapping two elements' valencies to use as each other's subscripts is not an arbitrary trick but the unique smallest whole-number combination in which both elements' combining capacities are exactly used up with none left over and none unsatisfied.

Step 1. Fix the idea of valency as a exchangeable "capacity." If element $A$ has valency $a$ , then one atom of $A$ can satisfy exactly $a$ units of combining capacity from whatever it bonds with. Likewise one atom of element $B$ with valency $b$ can satisfy exactly $b$ units of capacity. A correct formula $A_xB_y$ must use up all the capacity on both sides with nothing left unmatched.

Step 2. Write the balance condition. If a formula has $x$ atoms of $A$ and $y$ atoms of $B$ , the total capacity offered by the $A$ atoms is $x \cdot a$ , and the total capacity offered by the $B$ atoms is $y \cdot b$ . For every unit of capacity from $A$ to be matched by a unit from $B$ with nothing spare, these two totals must be equal:

x \cdot a = y \cdot b. \qquad (2)

where: $x$ and $y$ are the whole-number subscripts of $A$ and $B$ in the formula, and $a, b$ are their respective valencies.

Step 3. Solve for the smallest whole-number $x$ and $y$ satisfying $xa = yb$ . Rearranging gives $x/y = b/a$ , so the smallest whole-number solution is simply $x = b$ and $y = a$ — exactly the criss-cross swap, since it makes $x \cdot a = b \cdot a$ and $y \cdot b = a \cdot b$ , which are equal by definition.

Step 4. Simplify if $a$ and $b$ share a common factor. If $a$ and $b$ have a common factor $k$ , the criss-crossed subscripts $b$ and $a$ can both be divided by $k$ and the balance in Step 2 still holds, since dividing both sides of $xa = yb$ by the same $k$ preserves the equality. This is exactly why $\text{Ca}_2\text{O}_2$ (from valencies 2 and 2) must be simplified to $\text{CaO}$ : the raw criss-cross gives a correct ratio, but not always the smallest one.

Worked example

Question: An atom of magnesium has electron distribution $2, 8, 2$ and an atom of chlorine has electron distribution $2, 8, 7$ . Find each valency, write the formula of magnesium chloride by the criss-cross method, and then use the same two elements' valencies to check what formula magnesium oxide would have (oxygen has electron distribution $2, 6$ ).

Step 1. Magnesium's outermost shell has 2 electrons; since $2 \le 4$ , its valency is $2$ directly.

Step 2. Chlorine's outermost shell has 7 electrons; since $7 > 4$ , its valency is $8 - 7 = 1$ .

Step 3. Criss-cross: magnesium's valency (2) becomes the subscript on chlorine, and chlorine's valency (1) becomes the subscript on magnesium, giving $\text{Mg}_1\text{Cl}_2$ . Since a subscript of 1 is never written, this simplifies to $\text{MgCl}_2$ — the correct, well-known formula for magnesium chloride.

Step 4. For magnesium oxide, first find oxygen's valency: its outermost shell has 6 electrons, and since $6 > 4$ , valency $= 8 - 6 = 2$ . Criss-crossing magnesium (valency 2) with oxygen (valency 2) gives $\text{Mg}_2\text{O}_2$ , which by Step 4 of the derivation must be simplified — dividing both subscripts by their common factor 2 gives $\text{MgO}$ .

Final answer: magnesium chloride is $\text{MgCl}_2$ , and magnesium oxide is $\text{MgO}$ — the same magnesium atom (valency 2) gives a different subscript pattern depending on the valency of its partner, exactly as the balance equation $xa = yb$ predicts.

Common traps

Applying the " $8 -$ valence electrons" rule even when the valence-electron count is already 4 or fewer — that rule fires only when the count exceeds 4; below or at 4, valency equals the count directly.
Forgetting to simplify criss-crossed subscripts to the smallest whole-number ratio, writing $\text{Ca}_2\text{O}_2$ instead of the correct $\text{CaO}$ .
Assuming every outermost shell needs exactly 8 electrons for zero valency — the very first shell (K), when it is the outermost shell, is full and stable at just 2 electrons (the duplet rule).
Treating the charge on a polyatomic ion as something that must be derived from a shell diagram — for ions like $\text{SO}_4^{2-}$ or $\text{NH}_4^+$ , the given numerical charge is used directly in the criss-cross step, exactly like a valency.

Ask the tutor: if you want to practice finding valency from more electron distributions or criss-crossing formulae for compounds with polyatomic ions, ask the AI tutor to work through a few more examples with you.

Why does the rule change at "more than 4" valence electrons instead of some other number

Because 4 is the halfway point of the octet. Losing or gaining electrons both aim for the same goal — a full outer shell — so the atom simply takes whichever path is shorter. With 4 or fewer valence electrons, losing them is the shorter route (fewer electrons to remove than to add), so valency equals the count. With more than 4, gaining the missing electrons up to 8 is shorter, so valency is $8$ minus the count. An atom with exactly 4 valence electrons, like carbon, sits at this crossover and can behave either way in real bonding, but for Class 9 counting purposes it is simply taken as valency 4.

How is valency different from the charge on an ion

For simple ions formed by losing or gaining electrons, valency and ionic charge often have the same numerical value — magnesium loses 2 electrons to form $\text{Mg}^{2+}$ , so both its valency and its charge are "2." But valency itself is always reported as a positive whole number describing combining capacity, whereas charge carries a sign and describes the actual electrical state of the ion. For a polyatomic ion such as $\text{SO}_4^{2-}$ , there is no single shell diagram to read a valency from at all — the charge is simply given, and that number is used exactly where valency would be used in the criss-cross method.

Do I need to know why atoms actually transfer or share electrons to use valency

Not at the Class 9 stage covered here. Valency, as used in this article, is a number you read off an electron-shell diagram or look up for a common ion, and then use mechanically in the criss-cross method to write formulae. The deeper question of the actual mechanism — ionic bonding by electron transfer, covalent bonding by electron sharing, and the energetics behind why one happens rather than the other — belongs to later study of chemical bonding and is not required to answer valency or formula-writing questions at this level.

Valency

The core idea

Worked example

Common traps

Before this

Where this leads