Electrovalent bonding

Sodium is a metal so reactive it has to be stored under oil, and chlorine is a greenish gas poisonous enough to have been used as a weapon. Put them together and you get table salt — a stable, edible crystal that dissolves quietly in water. Neither element's separate danger survives the reaction; both atoms are transformed into something new. The transformation is electrovalent bonding: sodium does not share anything with chlorine, it simply hands over one electron, and the two resulting ions — one short an electron, one with a spare — lock together by nothing more than the electrostatic pull of opposite charges. This article, part of ICSE Class 9 Chemistry's "Atomic Structure and Chemical Bonding" chapter, builds on the Octet Rule to explain why that electron transfer happens, works through the standard examples ( $NaCl$ , $MgCl_2$ , $CaO$ ) already covered, and adds one more tool ICSE expects: a quick algebraic rule for writing the formula of any ionic compound straight from the valencies of the two ions involved.

The core insight

An electrovalent (ionic) bond forms when a metal atom transfers electrons to a non-metal atom so both reach a stable octet (or duplet), and the resulting oppositely charged ions — a cation and an anion — are held together by electrostatic attraction in a ratio fixed by balancing their charges.

The core idea

Why the transfer happens at all. Atoms bond because a partially filled outer shell is less stable than a complete one of 8 electrons (2 near helium's configuration) — the Octet Rule. A metal atom, just one or two electrons past a noble-gas configuration, finds it far easier to lose those extra electrons than to gain the many it would need to fill the shell. A non-metal atom, just one to three electrons short of a noble-gas configuration, finds the reverse easier: it gains electrons rather than losing most of its shell. Pair a metal with a non-metal and the arithmetic favours a clean transfer: the metal becomes a positively charged cation, the non-metal a negatively charged anion, and both land on a stable octet.

The bond itself. Once the transfer happens, the cation and anion are simply oppositely charged particles, and opposite charges attract — that electrostatic (Coulombic) attraction is the electrovalent bond, with no shared electron cloud as in covalent bonding. This attraction pulls equally in every direction around an ion, so ionic compounds do not exist as separate two-atom molecules; each ion attracts every oppositely charged neighbour, building an extended, repeating three-dimensional lattice. The chemical formula ( $NaCl$ , $MgCl_2$ , $CaO$ ) is therefore a ratio of ions in that lattice, not a free-floating molecule.

Balancing the charge. The number of electrons a metal atom loses, and the number a non-metal atom gains, are both fixed by how many electrons that atom needs to shed or add to reach a stable octet — and this number is the ion's valency (its charge). For the compound as a whole to be electrically neutral, the total positive charge contributed by all the cations must exactly cancel the total negative charge contributed by all the anions:

(\text{charge on cation}) \times (\text{number of cations}) = (\text{charge on anion}) \times (\text{number of anions})

where: the "charge" of each ion is the number of electrons it lost or gained (its valency), and the "number" of each ion is how many formula units of that ion appear per formula unit of the compound. This single balancing rule is what fixes every ionic formula, including the subscript-2 seen in $MgCl_2$ and the 1:1 ratio seen in $CaO$ .

Figure

Electron transfer diagram showing a sodium atom losing one electron to become a sodium cation and a chlorine atom gaining that electron to become a chloride anion which then attract each other electrostatically

Derivation AWriting an ionic formula from the criss-cross valency rule

The goal is to turn two known valencies — the fixed charge a metal ion and a non-metal ion each carry once they reach a stable octet — into the correct chemical formula of the ionic compound they form, using nothing beyond the charge-balance idea already stated above, and to check the result by counting electrons transferred, the way the sodium chloride and magnesium chloride examples already do.

Step 1. Identify the valency (charge) of each ion from its electron arrangement. Aluminium ( $Z = 13$ : 2, 8, 3) loses 3 electrons to reach 2, 8, so its ion is $Al^{3+}$ , valency $3$ . Oxygen ( $Z = 8$ : 2, 6) gains 2 electrons to reach 2, 8, so its ion is $O^{2-}$ , valency $2$ .

Step 2. Apply the charge-balance equation with $a$ aluminium ions and $b$ oxide ions: $3a = 2b$ . Because the two valencies, 3 and 2, share no common factor, the smallest whole-number solution is found by swapping the valencies across — the "criss-cross" — giving $a = 2$ and $b = 3$ .

Step 3. Write the formula using these smallest whole numbers as subscripts: $Al_2O_3$ , aluminium oxide. Read the subscripts back against Step 1 as a check: 2 aluminium ions at $3+$ each contribute a total charge of $+6$ ; 3 oxide ions at $2-$ each contribute a total charge of $-6$ ; the two totals cancel, confirming the compound is electrically neutral.

Step 4. Confirm by direct electron counting, exactly as done for $NaCl$ and $MgCl_2$ . Two aluminium atoms lose $2 \times 3 = 6$ electrons in total. Three oxygen atoms gain $3 \times 2 = 6$ electrons in total. The electrons lost and gained match exactly, so the transfer is balanced and $Al_2O_3$ is the correct formula — the same criss-cross shortcut, applied to sodium (valency 1) and chlorine (valency 1), would give $Na_1Cl_1$ , written simply as $NaCl$ , since a subscript of 1 is never shown. Verified: charge balance $3(2) = 2(3) = 6$ ; electron count $2\times3 = 3\times2 = 6$ electrons transferred in both directions.

Worked example

Question: Using the criss-cross valency rule, work out the formula of calcium chloride, given that calcium forms $Ca^{2+}$ and chlorine forms $Cl^-$ , and describe the electron transfer that supports it.

Step 1. Read off the valencies. Calcium's ion is $Ca^{2+}$ , valency $2$ . Chlorine's ion is $Cl^-$ , valency $1$ .

Step 2. Apply the criss-cross rule: the valency of one ion becomes the subscript of the other. Calcium's valency of 2 becomes the subscript on chlorine, and chlorine's valency of 1 becomes the (unwritten) subscript on calcium. This gives $Ca_1Cl_2$ , written as $CaCl_2$ .

Step 3. Check by charge balance: one $Ca^{2+}$ contributes $+2$ ; two $Cl^-$ ions contribute $2 \times (-1) = -2$ . The totals cancel, so the compound is neutral.

Step 4. Confirm by electron transfer. Calcium ( $Z=20$ : 2, 8, 8, 2) loses its 2 outer electrons to reach the stable arrangement 2, 8, 8, forming $Ca^{2+}$ . Each chlorine atom (2, 8, 7) needs only 1 electron to complete its octet, so calcium's 2 electrons go to two separate chlorine atoms, each becoming $Cl^-$ (2, 8, 8).

The formula of calcium chloride is $CaCl_2$ , with one calcium ion donating one electron each to two chloride ions — the same pattern already seen for magnesium chloride, since calcium and magnesium both have valency 2.

Competitive stretch

The same rule instantly predicts formulas examiners like to test without atomic-structure diagrams: calcium (valency 2) with oxygen (valency 2) gives $Ca_2O_2$ , which reduces to $CaO$ because both subscripts share the common factor 2 — matching the $CaO$ result already known. Aluminium (valency 3) with chlorine (valency 1) gives $AlCl_3$ directly, with no reduction needed since 3 and 1 share no common factor beyond 1.

Common traps

Forgetting to reduce the criss-crossed subscripts to their smallest whole-number ratio when the two valencies share a common factor — calcium and oxygen (2 and 2) must reduce to $CaO$ , not be left as $Ca_2O_2$ .
Applying the criss-cross rule but then forgetting to verify it with a charge-balance check; an error in reading the valency off an electron arrangement will still produce a formula, just the wrong one, so the check in Step 3 above is not optional.
Treating the criss-cross rule as chemistry rather than as bookkeeping — it works because it enforces the same physical fact already established: total electrons lost by the metal must equal total electrons gained by the non-metal.
Writing a subscript of 1 explicitly, as in $Na_1Cl_1$ — chemical formula convention always omits a subscript of 1, giving $NaCl$ .

Ask the tutor: Want to practise the criss-cross rule on a compound not covered here, such as iron(III) oxide or potassium sulphide, or revisit why the Octet Rule makes metals lose and non-metals gain electrons in the first place? Ask the AI tutor to work through it with you.

Does the criss-cross rule always give the correct formula, or can it fail?

It works reliably whenever you have the correct valency (charge) for each ion, because it is just a shortcut for the charge-balance equation already used to check $NaCl$ , $MgCl_2$ and $CaO$ . Where it can go wrong is if the wrong valency is used to start with — some metals, especially transition metals, can form ions of more than one possible charge, so the rule needs the actual valency for that specific compound, not just guessed from the metal's usual behaviour. For the fixed-valency metals and non-metals in the ICSE Class 9 syllabus, the rule is completely reliable.

If ionic compounds are made of ions, not molecules, why do we still write a formula like NaCl as if it were one unit?

The formula $NaCl$ is called a formula unit, not a molecule, and it represents the simplest whole-number ratio of ions in the giant lattice, not two atoms bonded in isolation. In the actual solid, every $Na^+$ ion is surrounded by several $Cl^-$ neighbours and every $Cl^-$ is surrounded by several $Na^+$ neighbours, extending in three dimensions with no natural pairing-off into individual "molecules." Writing $NaCl$ is a convenient shorthand for that 1:1 ratio, exactly the way $MgCl_2$ shorthand-writes a 1:2 ratio.

How is electrovalent bonding different from covalent bonding in terms of what actually happens to the electrons?

In electrovalent bonding, electrons are completely transferred from the metal atom to the non-metal atom — after bonding, those electrons belong entirely to the anion, and the two resulting ions attract each other purely electrostatically, with no shared electron cloud between them. In covalent bonding, by contrast, electrons are shared between two non-metal atoms, with both nuclei attracting the same pair of electrons rather than one atom giving them up outright. This is why electrovalent compounds form extended ionic lattices held by charge attraction, while covalent compounds form discrete molecules held by shared electron pairs — the two bond types are covered as a matched pair in this chapter precisely so the contrast is clear.

Electrovalent bonding

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads