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OH− Lewis Structure

Hydroxide ion · What makes a solution basic. Present in every alkali, from drain cleaner to lime.

Lewis structure of OH- (Hydroxide 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.OH-−

The OH- Lewis structure has 8 valence electrons - the 1− charge adds one, drawn as one bond to hydrogen, with three lone pairs in total, all inside square brackets. The central O 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 OH-
Total valence electrons 8
Bonding electrons 2 (1 shared pairs)
Nonbonding electrons 6 (3 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 17.007 g/mol

O Oxygen 6 valence e− H Hydrogen 1 valence e−

How to draw the OH− Lewis structure

Step through it. The bar shows how much of the 8-electron budget is spent at each stage - it never grows, which is the whole constraint.

  1. OH- Lewis structure, stage 1: SkeletonOH−

    2 of 8 placed

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

  2. OH- Lewis structure, stage 2: Outer octetsOH−

    8 of 8 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.

    6 (O) + 1 (H) + 1 (1- charge) = 8 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, 6 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 6, has 6, 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.

    O: 3 (6 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 O: 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 atoms around the centre are not all the same, so the bond dipoles are different sizes and cannot cancel.

    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 OH− 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
O–H 1.24 polar covalent
Why this ion gets set as a problem: The smallest polyatomic ion, and a useful check that formal charge really does land where it should.

Common questions

How many valence electrons does OH- have?

OH- has 8 valence electrons. The atoms bring 7 between them, and the 1− charge adds one. 2 of them are in bonds and 6 sit in lone pairs.

Why does OH- 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 OH-?

OH- is a linear ion. The central O 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 OH-?

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

Is OH- polar or nonpolar?

Neither, strictly. Polar and nonpolar describe neutral molecules, by asking whether their bond dipoles cancel out. OH- 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 atoms around the centre are not all the same, so the bond dipoles are different sizes and cannot cancel.

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.