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ClO4− Lewis Structure

Perchlorate ion · A rocket-propellant oxidiser and a persistent groundwater contaminant.

Lewis structure of ClO4- (Perchlorate ion)Each of the four O atoms carries a 1− formal charge and the Cl a 3+; add them up and you get 1−, the charge on the ion. The charge belongs to the ion as a whole, which is what the brackets around the drawing mean.ClOOOO3+----−

The ClO4- Lewis structure has 32 valence electrons - the 1− charge adds one, drawn as four Cl–O bonds, with twelve lone pairs in total, all inside square brackets. The central Cl is sp3 hybridized, which makes it a tetrahedral ion with a bond angle of 109.5 degrees. Each of the four O atoms carries a 1− formal charge and the Cl a 3+; add them up and you get 1−, the charge on the ion. 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−.

ClO4− molecular geometry

ClO4- molecular geometry: tetrahedralOClOOO

Tetrahedral, from 4 electron domains on the central Cl: 4 bonded atoms (AX4). The electron geometry is tetrahedral. Wedges point toward you, hashed wedges away. Every shape in the VSEPR chart →

Calculated properties

Calculated properties of ClO4-
Total valence electrons 32
Bonding electrons 8 (4 shared pairs)
Nonbonding electrons 24 (12 lone pairs)
Lone pairs on Cl 0
Electron domains (steric number) 4
VSEPR notation AX4
Electron geometry tetrahedral
Molecular geometry tetrahedral
Bond angle 109.5°
Hybridization sp3
Charge 1−
Charge distribution localised
Formal charges Cl: +3, O: -1, O: -1, O: -1, O: -1
Resonance structures none
Molar mass 99.446 g/mol

Cl Chlorine 7 valence e− O Oxygen 4 atoms, 6 valence e− each

How to draw the ClO4− Lewis structure

Step through it. The bar shows how much of the 32-electron budget is spent at each stage - it never grows, which is the whole constraint.

  1. ClO4- Lewis structure, stage 1: SkeletonClOOOO−

    8 of 32 placed

    Skeleton. Join every connected pair with one bond. 4 bonds spends 8 electrons.

  2. ClO4- Lewis structure, stage 2: Outer octetsClOOOO−

    32 of 32 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) + 4 x 6 (O) + 1 (1- charge) = 32 valence electrons
  2. Choose the central atom

    Cl is the least electronegative atom here, so it takes the centre. Hydrogen is never central: it can only form one bond.

  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.

    4 bonds x 2 = 8 electrons used, 24 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 24, has 24, 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.

    O1: 3, O2: 3, O3: 3, O4: 3 (24 electrons)
  6. Check the central atom

    Cl ends up with a full octet. Every electron in the budget is now placed.

    Cl: 8 bonding + 0 nonbonding = 8
  7. 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.

    Cl: 7 - 0 - 4 = +3 ; O1: 6 - 6 - 1 = -1 ; O2: 6 - 6 - 1 = -1 ; O3: 6 - 6 - 1 = -1 ; O4: 6 - 6 - 1 = -1
  8. Work out the shape

    Count the electron domains on Cl: 4 bonded groups. 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 tetrahedral arrangement.

    steric number 4 -> sp3 -> tetrahedral, bond angle 109.5
  9. Decide whether it is polar

    The bond dipoles are all the same size and the shape places them symmetrically around the central Cl, so they cancel exactly.

    dipoles cancel -> nonpolar

The expanded-octet version

Courses disagree about ClO4-. The structure at the top of this page obeys the octet rule and leaves formal charges behind. The one below puts more than eight electrons on the central atom and brings every formal charge to zero. Check which convention your course uses.

ClO4- drawn with an expanded octetClOOOO+--−
Every formal charge is zero here, but the central atom holds more than eight electrons.

Where the 1− charge sits

Each of the four O atoms carries a 1− formal charge and the Cl a 3+; add them up and you get 1−, the charge on the ion. The charge belongs to the ion as a whole, which is what the brackets around the drawing mean.

People also ask whether ClO4− 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: The end of the chlorine oxyanion series, and the ion where the expanded-octet convention is hardest to defend.

Common questions

How many valence electrons does ClO4- have?

ClO4- has 32 valence electrons. The atoms bring 31 between them, and the 1− charge adds one. 8 of them are in bonds and 24 sit in lone pairs.

Why does ClO4- have a 1− charge?

Each of the four O atoms carries a 1− formal charge and the Cl a 3+; add them up and you get 1−, the charge on the ion. 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 ClO4-?

ClO4- is a tetrahedral ion. The central Cl has 4 bonded groups, a steric number of 4, which gives tetrahedral electron geometry and a tetrahedral ion.

What is the bond angle in ClO4-?

The bond angle in ClO4- is 109.5 degrees. That is the ideal tetrahedral angle, and nothing distorts it here.

What is the hybridization of ClO4-?

The central Cl in ClO4- is sp3 hybridized. Steric number 4 means 4 orbitals have to be mixed, which is exactly what sp3 gives you.

Is ClO4- polar or nonpolar?

Neither, strictly. Polar and nonpolar describe neutral molecules, by asking whether their bond dipoles cancel out. ClO4- 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, because the tetrahedral arrangement puts every identical bond at the same angle to the others.

Does ClO4- obey the octet rule?

It can be drawn either way, and different courses teach different conventions. The octet structure keeps eight electrons on every atom but leaves formal charges behind. The expanded-octet structure puts more than eight electrons on the central atom and brings every formal charge to zero. Both are shown on this page.

How this page was produced. The formula was parsed, the connectivity resolved (central atom), 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.