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ClO− Lewis Structure

Hypochlorite ion · The active ingredient in household bleach and in chlorinated pool water.

Lewis structure of ClO- (Hypochlorite ion)The O carries a 1− formal charge, and that is the ion's 1− charge. The charge belongs to the ion as a whole, which is what the brackets around the drawing mean.ClO-−

The ClO- Lewis structure has 14 valence electrons - the 1− charge adds one, drawn as one Cl–O bond, with six lone pairs in total, all inside square brackets. The central Cl is sp3 hybridized and the ion is linear. The O carries a 1− formal charge, and that is the ion's 1− charge. The charge belongs to the ion as a whole, which is what the brackets around the drawing mean.

The brackets are not decoration. An ion's charge belongs to the whole species, not to one atom, so the structure is drawn inside square brackets with the 1− outside. The small circled numbers inside are formal charges, which are a bookkeeping device for choosing between structures - a different thing entirely. They add up to 1−.

Calculated properties

Calculated properties of ClO-
Total valence electrons 14
Bonding electrons 2 (1 shared pairs)
Nonbonding electrons 12 (6 lone pairs)
Electron domains (steric number) 4
VSEPR notation AX1E3
Electron geometry tetrahedral
Molecular geometry linear
Bond angle no angle (only two atoms)
Hybridization sp3
Charge 1−
Charge distribution localised
Formal charges O: -1
Resonance structures none
Molar mass 51.449 g/mol

Cl Chlorine 7 valence e− O Oxygen 6 valence e−

How to draw the ClO− Lewis structure

Step through it. The bar shows how much of the 14-electron budget is spent at each stage - it never grows, which is the whole constraint.

  1. ClO- Lewis structure, stage 1: SkeletonClO−

    2 of 14 placed

    Skeleton. Join every connected pair with one bond. 1 bond spends 2 electrons.

  2. ClO- Lewis structure, stage 2: Outer octetsClO−

    14 of 14 placed

    Outer octets. Spend electrons as lone pairs on the outer atoms. That uses the entire budget, and the central atom has nothing left over.

The reasoning at each step

  1. Count the valence electrons

    Add up the outer-shell electrons every atom brings, then account for the charge. A 1- ion has gained one electron, so it is added to the budget. Skip this and every count that follows will be wrong.

    7 (Cl) + 6 (O) + 1 (1- charge) = 14 valence electrons
  2. Work out what is bonded to what

    Two atoms, one connection. The only question is the bond order.

  3. Join everything with single bonds first

    Every connection starts as one shared pair. Draw them all before worrying about double bonds - the arithmetic in the next step is what tells you where the double bonds have to go.

    1 bond x 2 = 2 electrons used, 12 left
  4. Work out how far short you are

    Add up what every atom still needs to fill its shell - eight electrons for most atoms, two for hydrogen - minus what its single bonds already give it. Compare that with the electrons you have left. The difference decides everything that follows.

    needs 12, has 12, exactly enough
  5. Fill in the lone pairs

    Every electron not in a bond sits as a lone pair on an atom. Each atom takes exactly what it needs to finish its shell - there is no choice left at this point.

    Cl: 3, O: 3 (12 electrons)
  6. Check the formal charges

    Formal charge is valence electrons, minus lone-pair electrons, minus the number of bonds. They are not real charges, and they are not the ion's charge either - but they must add up to it. Adding these up gives 1-, which is the check that the structure is complete.

    O: 6 - 6 - 1 = -1
  7. Work out the shape

    Count the electron domains on Cl: 1 bonded group plus 3 lone pairs. A double or triple bond still counts as one domain, because it points in one direction. That gives tetrahedral electron geometry; ignore the lone pairs and the atoms themselves sit in a linear arrangement. Two atoms can only lie in a straight line, so there is no bond angle to measure.

    steric number 4 -> sp3 -> linear
  8. Decide whether it is polar

    The individual bonds are close to nonpolar, but the lone pair sits on one side of the molecule, so there is still a net dipole.

    dipoles do not cancel -> polar

Where the 1− charge sits

The O carries a 1− formal charge, and that is the ion's 1− charge. The charge belongs to the ion as a whole, which is what the brackets around the drawing mean.

People also ask whether ClO− is polar. Strictly it is neither: polar and nonpolar sort neutral molecules by whether their bond dipoles cancel. A 1− ion is pulled towards any nearby charge whatever its shape, which is why its salts dissolve in water regardless of geometry. The table below is still worth reading - it shows how unevenly each bond holds its electrons, which is what decides whether the charge ends up spread or concentrated.

BondElectronegativity differenceCharacter
Cl–O 0.28 nonpolar covalent
Why this ion gets set as a problem: Two atoms, fourteen electrons, and a formal charge that settles the question of which atom carries the charge.

Common questions

How many valence electrons does ClO- have?

ClO- has 14 valence electrons. The atoms bring 13 between them, and the 1− charge adds one. 2 of them are in bonds and 12 sit in lone pairs.

Why does ClO- have a 1− charge?

The O carries a 1− formal charge, and that is the ion's 1− charge. The charge belongs to the ion as a whole, which is what the brackets around the drawing mean. The brackets around the drawing are part of the answer: they say the charge belongs to the whole ion rather than to whichever atom happens to be nearest.

What is the molecular geometry of ClO-?

ClO- is a linear ion. The central Cl has 1 bonded group and three lone pairs, a steric number of 4, which gives tetrahedral electron geometry and a linear ion.

What is the hybridization of ClO-?

The central Cl in ClO- is sp3 hybridized. Steric number 4 means 4 orbitals have to be mixed, which is exactly what sp3 gives you.

Is ClO- polar or nonpolar?

Neither, strictly. Polar and nonpolar describe neutral molecules, by asking whether their bond dipoles cancel out. ClO- carries a net 1− charge, so it is pulled towards any nearby charge whatever its shape - which is why ionic compounds dissolve in water regardless of geometry. The question the shape does answer is whether the charge is spread evenly: it is not, because the individual bonds are close to nonpolar, but the lone pair sits on one side of the molecule, so there is still a net dipole.

How this page was produced. The formula was parsed, the connectivity resolved (diatomic), and every bond-order arrangement enumerated and scored on octet satisfaction, formal charge and where that charge sits. The structure above is the winner. Bond angles, hybridization and the polarity verdict are read off the resulting geometry, not looked up.