Input interpretation
tetraborate anion
Lewis structure
Draw the Lewis structure of tetraborate anion. Start by drawing the overall structure of the molecule: Count the total valence electrons of the boron (n_B, val = 3) and oxygen (n_O, val = 6) atoms, including the net charge: 4 n_B, val + 7 n_O, val - n_charge = 56 Calculate the number of electrons needed to completely fill the valence shells for boron (n_B, full = 6) and oxygen (n_O, full = 8): 4 n_B, full + 7 n_O, full = 80 Subtracting these two numbers shows that 80 - 56 = 24 bonding electrons are needed. Each bond has two electrons, so the above diagram has all the necessary bonds. There are 12 bonds and hence 24 bonding electrons in the diagram. Fill in the remaining unbonded electrons on each atom. In total, there remain 56 - 24 = 32 electrons left to draw. Lastly, fill in the formal charges: Answer: | |
General properties
formula | (B_4O_7)^(2-) net ionic charge | -2 alternate names | tetraborate | tetraborate(2-)
Other properties
ion class | anions | oxoanions | polyatomic ions
Thermodynamic properties
molar free energy of formation Δ_fG° | aqueous | -2605 kJ/mol (kilojoules per mole)