Valency

One atom of oxygen always grabs exactly two atoms of hydrogen to make water. One atom of nitrogen always grabs exactly three to make ammonia. One atom of carbon always grabs four to make methane. Why these particular numbers, and not any other combination? Every element has a fixed "combining capacity" — a kind of appetite for hydrogen atoms — and chemists gave this appetite a name: valency. Once you know an element's valency, you can predict how it will combine with other elements and radicals, check whether a written formula is even possible, and build new formulae from scratch instead of memorising each one. This article defines valency the way CISCE Class VIII expects — through hydrogen-combining capacity — and shows how a simple balancing rule turns two known valencies into a correct chemical formula.

The core insight

Valency is the fixed number of hydrogen atoms (or their equivalent) that one atom of an element or one radical combines with, and balancing the valencies of two combining units — swapping each one's valency in as the other's subscript — gives the correct formula of the compound they form.

The core idea

Valency is the combining capacity of an atom of an element, or of a radical, measured by counting how many hydrogen atoms combine with one atom of it to form a stable compound.

\text{Valency of an element} = \text{number of H atoms that combine with one atom of it}

where: the right-hand side is read directly off a known, stable compound's formula — no calculation, just counting atoms. In hydrogen chloride, $\text{HCl}$ , one chlorine atom combines with one hydrogen atom, so chlorine's valency is 1. In water, $\text{H}_2\text{O}$ , one oxygen atom combines with two hydrogen atoms, so oxygen's valency is 2. In ammonia, $\text{NH}_3$ , nitrogen's valency is 3. In methane, $\text{CH}_4$ , carbon's valency is 4. These four compounds are the anchor set worth memorising, because every other valency you meet is cross-checked against them. Most elements never form a simple hydride at all — sodium and calcium are common examples — so their valency is worked out indirectly, by comparing a known compound of theirs against an element whose valency is already fixed (usually oxygen or chlorine). Radicals, groups of atoms that travel together as a single charged unit (the hydroxide radical $\text{OH}^-$ , the sulphate radical $\text{SO}_4^{2-}$ , the carbonate radical $\text{CO}_3^{2-}$ ), are given valencies the same way and used exactly like single atoms in formula-building.

Element / radical	Symbol	Valency
Hydrogen	H	1
Sodium	Na	1
Chlorine	Cl	1
Hydroxide radical	OH	1
Oxygen	O	2
Calcium	Ca	2
Sulphate radical	SO4	2
Nitrogen	N	3
Carbon	C	4

Some elements are not stuck with one number: variable valency elements show different combining capacities in different compounds — iron is the standard example, forming compounds where it behaves as valency 2 (ferrous) and others where it behaves as valency 3 (ferric). At this stage you only need to recognise that variable valency exists and can read it off a given formula; the electronic reason for it belongs to a later class.

Figure

Bar chart showing how many hydrogen atoms combine with one atom of chlorine oxygen nitrogen and carbon giving valencies one two three and four

Why it works

Valency of two combining units always balances out: the total combining power on one side must equal the total combining power on the other. Turning that balance into a formula is a two-step routine that CISCE calls the criss-cross rule, and it is the one piece of arithmetic reasoning this topic asks for. This derivation shows exactly why swapping the two valencies into each other's subscript position produces a formula that is already balanced, without any trial and error.

Step 1. Write the two combining units side by side with their known valencies marked above them, for example an element A of valency $a$ and a radical or element B of valency $b$ : $\overset{a}{\text{A}}\ \overset{b}{\text{B}}$ . Valency $a$ means one atom of A supplies $a$ combining units; valency $b$ means one atom (or one radical) of B supplies $b$ combining units.

Step 2. For the compound to be stable, the total units contributed by all the A atoms must equal the total units contributed by all the B atoms. If $m$ atoms of A combine with $n$ atoms of B, this balance is the single equation $a \times m = b \times n$ .

Step 3. The simplest whole-number solution to $a \times m = b \times n$ is to swap the valencies straight across as subscripts: take $m = b$ and $n = a$ , so that $a \times b = b \times a$ automatically holds. Writing this out gives the formula $\text{A}_b\text{B}_a$ — each element's valency has become the other element's subscript. If $a$ and $b$ share a common factor, divide both subscripts by it to reach the simplest ratio (a subscript of 1 is never written).

Step 4. Check the swap against a known pair: sodium (valency 1) and oxygen (valency 2) criss-cross to $\text{Na}_2\text{O}_1 = \text{Na}_2\text{O}$ , which is exactly the correct, familiar formula for sodium oxide — confirming that the criss-cross rule reproduces a real compound rather than an arbitrary combination.

Worked example

Question: The valency of calcium is 2 and the valency of the hydroxide radical (OH) is 1. Use the balancing rule to find the formula of calcium hydroxide.

Step 1. Mark the valencies: $\overset{2}{\text{Ca}}\ \overset{1}{(\text{OH})}$ . Calcium contributes 2 combining units per atom; each hydroxide radical contributes 1.

Step 2. Criss-cross the valencies into subscripts: calcium's valency (2) becomes hydroxide's subscript, and hydroxide's valency (1) becomes calcium's subscript (and a subscript of 1 is dropped). This gives $\text{Ca(OH)}_2$ .

Step 3. Verify the balance: 1 calcium atom supplies $2$ units; 2 hydroxide radicals supply $2 \times 1 = 2$ units. The two sides match, $2 = 2$ , so the formula is balanced.

1\ \text{Ca (valency 2)} + 2\ \text{OH (valency 1 each)} \rightarrow \text{Ca(OH)}_2

The formula of calcium hydroxide is $\text{Ca(OH)}_2$ .

Competitive stretch

Aluminium has valency 3 and the sulphate radical $\text{SO}_4$ has valency 2. Criss-crossing gives $\text{Al}_2(\text{SO}_4)_3$ — check: 2 aluminium atoms supply $2 \times 3 = 6$ units, and 3 sulphate radicals supply $3 \times 2 = 6$ units, so $6 = 6$ balances, confirming aluminium sulphate is $\text{Al}_2(\text{SO}_4)_3$ .

Common traps

Treating valency as only an elements-only property — radicals like hydroxide, sulphate and carbonate have their own valencies and are criss-crossed exactly like single atoms.
Confusing valency (a ratio, the combining capacity) with the actual atom count in one specific compound — always trace it back to "how many H atoms combine with one atom of this element."
Forgetting to simplify the criss-crossed subscripts to their lowest ratio when the two valencies share a common factor, and forgetting to drop a subscript of 1.
Assuming every element has one fixed valency — variable-valency elements like iron show different valencies in different compounds, so always check against the given formula.

Ask the tutor: Want to practise the criss-cross rule on a few more pairs, like magnesium with chloride, or sodium with carbonate?

Why does hydrogen get used as the reference for defining valency instead of some other element?

Hydrogen forms very simple, well-known compounds with most elements, and it always contributes exactly one combining unit, which makes it a clean, consistent yardstick. Historically chemists found it easiest to compare every element against hydrogen's compounds first, and even elements that never combine with hydrogen directly (like sodium) get their valency worked out by comparing them to elements whose hydrogen-based valency is already fixed.

If an element does not combine with hydrogen at all, how can it have a valency?

Its valency is found indirectly. You look at a compound the element forms with an element whose valency is already known, such as oxygen (valency 2) or chlorine (valency 1), and work backwards using the same balancing idea used in the criss-cross rule. For example, sodium forms $\text{Na}_2\text{O}$ with oxygen, so balancing $a \times 2 = 2 \times 1$ gives sodium's valency as 1.

Does a radical like sulphate count as one particle when working out valency, or do I need to know what is inside it?

For valency and formula-writing purposes, treat the whole radical as a single unit with its own fixed valency, exactly as you would treat a single atom — you do not need to know the internal structure of $\text{SO}_4^{2-}$ to use it correctly. What is inside the radical (why it has that particular charge) is explained later, once bonding and electron transfer are introduced; for now, memorise the radical and its valency as a package.