HSO4− Lewis Structure
Hydrogen sulfate ion (bisulfate) · Sulfuric acid after it has given away one of its two protons.
The HSO4- Lewis structure has 32 valence electrons - the 1− charge adds one, drawn as four S–O bonds and one bond to hydrogen, with eleven lone pairs in total, all inside square brackets. The central S is sp3 hybridized, which makes it a tetrahedral ion with a bond angle of 109.5 degrees. Each of the three O atoms carries a 1− formal charge and the S a 2+; 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.
- 32 valence e−
- tetrahedral
- 109.5° bond angle
- sp3
- 1− charge
HSO4− molecular geometry
Tetrahedral, from 4 electron domains on the central S: 4 bonded atoms (AX4). The electron geometry is tetrahedral. Wedges point toward you, hashed wedges away. Every shape in the VSEPR chart →
Calculated properties
| Total valence electrons | 32 |
|---|---|
| Bonding electrons | 10 (5 shared pairs) |
| Nonbonding electrons | 22 (11 lone pairs) |
| Lone pairs on S | 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 | S: +2, O: -1, O: -1, O: -1 |
| Resonance structures | none |
| Molar mass | 97.064 g/mol |
H Hydrogen 1 valence e− S Sulfur 6 valence e− O Oxygen 4 atoms, 6 valence e− each
How to draw the HSO4− 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.
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Skeleton. Join every connected pair with one bond. 5 bonds spends 10 electrons.
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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
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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.
1 (H) + 6 (S) + 4 x 6 (O) + 1 (1- charge) = 32 valence electrons -
Work out what is bonded to what
This is an oxyacid. Every oxygen bonds to the central S, and the acidic hydrogens sit on oxygen, not on S.
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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.
5 bonds x 2 = 10 electrons used, 22 left -
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 22, has 22, exactly enough -
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: 2, O2: 3, O3: 3, O4: 3 (22 electrons) -
Check the central atom
S ends up with a full octet. Every electron in the budget is now placed.
S: 8 bonding + 0 nonbonding = 8 -
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.
S: 6 - 0 - 4 = +2 ; O2: 6 - 6 - 1 = -1 ; O3: 6 - 6 - 1 = -1 ; O4: 6 - 6 - 1 = -1 -
Work out the shape
Count the electron domains on S: 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 -
Decide whether it is polar
The O-H bond dipoles do not cancel in this shape, leaving a net dipole.
dipoles do not cancel -> polar
The expanded-octet version
Courses disagree about HSO4-. 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.
Where the 1− charge sits
Each of the three O atoms carries a 1− formal charge and the S a 2+; 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 HSO4− 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.
| Bond | Electronegativity difference | Character |
|---|---|---|
| O–H | 1.24 | polar covalent |
| S–O | 0.86 | polar covalent |
Common questions
How many valence electrons does HSO4- have?
HSO4- has 32 valence electrons. The atoms bring 31 between them, and the 1− charge adds one. 10 of them are in bonds and 22 sit in lone pairs.
Why does HSO4- have a 1− charge?
Each of the three O atoms carries a 1− formal charge and the S a 2+; 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 HSO4-?
HSO4- is a tetrahedral ion. The central S has 4 bonded groups, a steric number of 4, which gives tetrahedral electron geometry and a tetrahedral ion.
What is the bond angle in HSO4-?
The bond angle in HSO4- is 109.5 degrees. That is the ideal tetrahedral angle, and nothing distorts it here.
What is the hybridization of HSO4-?
The central S in HSO4- is sp3 hybridized. Steric number 4 means 4 orbitals have to be mixed, which is exactly what sp3 gives you.
Is HSO4- polar or nonpolar?
Neither, strictly. Polar and nonpolar describe neutral molecules, by asking whether their bond dipoles cancel out. HSO4- 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 O-H bond dipoles do not cancel in this shape, leaving a net dipole.
Does HSO4- 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.