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sulfate anion

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sulfate anion
sulfate anion

Lewis structure

Draw the Lewis structure of sulfate anion. Start by drawing the overall structure of the molecule, ignoring potential double and triple bonds:  Count the total valence electrons of the oxygen (n_O, val = 6) and sulfur (n_S, val = 6) atoms, including the net charge: 4 n_O, val + n_S, val - n_charge = 32 Calculate the number of electrons needed to completely fill the valence shells for oxygen (n_O, full = 8) and sulfur (n_S, full = 8): 4 n_O, full + n_S, full = 40 Subtracting these two numbers shows that 40 - 32 = 8 bonding electrons are needed. Each bond has two electrons, so we expect that the above diagram has all the necessary bonds. However, to minimize formal charge oxygen wants 2 bonds. Identify the atoms that want additional bonds and the number of electrons remaining on each atom. The net charge has been given to the most electronegative atom, oxygen, in 2 places:  In order to minimize their formal charge, atoms with large electronegativities can force atoms with smaller electronegativities on period 3 or higher to expand their valence shells. The electronegativities of the atoms are 2.58 (sulfur) and 3.44 (oxygen). Because the electronegativity of sulfur is smaller than the electronegativity of oxygen, expand the valence shell of sulfur to 6 bonds (the maximum number of bonds it can accomodate). Therefore we add a total of 2 bonds to the diagram, noting the formal charges of the atoms. Double bonding sulfur to the other highlighted oxygen atoms would result in an equivalent molecule: Answer: |   |
Draw the Lewis structure of sulfate anion. Start by drawing the overall structure of the molecule, ignoring potential double and triple bonds: Count the total valence electrons of the oxygen (n_O, val = 6) and sulfur (n_S, val = 6) atoms, including the net charge: 4 n_O, val + n_S, val - n_charge = 32 Calculate the number of electrons needed to completely fill the valence shells for oxygen (n_O, full = 8) and sulfur (n_S, full = 8): 4 n_O, full + n_S, full = 40 Subtracting these two numbers shows that 40 - 32 = 8 bonding electrons are needed. Each bond has two electrons, so we expect that the above diagram has all the necessary bonds. However, to minimize formal charge oxygen wants 2 bonds. Identify the atoms that want additional bonds and the number of electrons remaining on each atom. The net charge has been given to the most electronegative atom, oxygen, in 2 places: In order to minimize their formal charge, atoms with large electronegativities can force atoms with smaller electronegativities on period 3 or higher to expand their valence shells. The electronegativities of the atoms are 2.58 (sulfur) and 3.44 (oxygen). Because the electronegativity of sulfur is smaller than the electronegativity of oxygen, expand the valence shell of sulfur to 6 bonds (the maximum number of bonds it can accomodate). Therefore we add a total of 2 bonds to the diagram, noting the formal charges of the atoms. Double bonding sulfur to the other highlighted oxygen atoms would result in an equivalent molecule: Answer: | |

General properties

formula | (SO_4)^(2-) net ionic charge | -2 alternate names | tetraoxosulfate | tetraoxidosulfate | sulfate | sulfate(2-)
formula | (SO_4)^(2-) net ionic charge | -2 alternate names | tetraoxosulfate | tetraoxidosulfate | sulfate | sulfate(2-)

Ionic radius

thermochemical radius | 258 pm
thermochemical radius | 258 pm

Units

Other properties

ion class | anions | biomolecule ions | ionic conjugate bases | oxoanions | polyatomic ions | ionic weak bases common sources of ion | zirconium(IV) sulfate hydrate (2 eq) | zinc sulfate heptahydrate (1 eq) | yttrium(III) sulfate octahydrate (3 eq) | ytterbium(III) sulfate (3 eq) | vanadium(IV) oxide sulfate hydrate (1 eq) | thulium(III) sulfate octahydrate (3 eq) | thulium(III) sulfate (3 eq)
ion class | anions | biomolecule ions | ionic conjugate bases | oxoanions | polyatomic ions | ionic weak bases common sources of ion | zirconium(IV) sulfate hydrate (2 eq) | zinc sulfate heptahydrate (1 eq) | yttrium(III) sulfate octahydrate (3 eq) | ytterbium(III) sulfate (3 eq) | vanadium(IV) oxide sulfate hydrate (1 eq) | thulium(III) sulfate octahydrate (3 eq) | thulium(III) sulfate (3 eq)

Thermodynamic properties

molar heat capacity c_p | aqueous | -293 J/(mol K) (joules per mole kelvin difference) molar free energy of formation Δ_fG° | aqueous | -744.5 kJ/mol (kilojoules per mole) molar heat of formation Δ_fH° | aqueous | -909.3 kJ/mol (kilojoules per mole) molar entropy S° | aqueous | 20.1 J/(mol K) (joules per mole kelvin)
molar heat capacity c_p | aqueous | -293 J/(mol K) (joules per mole kelvin difference) molar free energy of formation Δ_fG° | aqueous | -744.5 kJ/mol (kilojoules per mole) molar heat of formation Δ_fH° | aqueous | -909.3 kJ/mol (kilojoules per mole) molar entropy S° | aqueous | 20.1 J/(mol K) (joules per mole kelvin)