Input interpretation
orthotellurate anion
Lewis structure
Draw the Lewis structure of orthotellurate anion. Start by drawing the overall structure of the molecule: Count the total valence electrons of the oxygen (n_O, val = 6) and tellurium (n_Te, val = 6) atoms, including the net charge: 6 n_O, val + n_Te, val - n_charge = 48 Calculate the number of electrons needed to completely fill the valence shells for oxygen (n_O, full = 8) and tellurium (n_Te, full = 8): 6 n_O, full + n_Te, full = 56 Subtracting these two numbers shows that 56 - 48 = 8 bonding electrons are needed, which are already accounted for in the structure. Note that the valence shell of tellurium has been expanded to 6 bonds. After accounting for the expanded valence, there are 6 bonds and hence 12 bonding electrons in the diagram. Fill in the remaining unbonded electrons on each atom. In total, there remain 48 - 12 = 36 electrons left to draw. Lastly, fill in the formal charges: Answer: | |
General properties
formula | (TeO_6)^(6-) net ionic charge | -6 alternate names | hexaoxotellurate | orthotellurate | orthotellurate(6-)
Other properties
ion class | anions | oxoanions | polyatomic ions