• Why is $\ce{Ba(OH)2}$ Barium hydroxide not Barium dihydroxide?

  • Why is Lithium carbonate $\ce{Li2CO3}$ instead of $\ce{LiCO3}$?

Is there a formula or is just the case for these two?


3 Answers 3


There is no other Ba(OH)x so there is no need to note a difference between several forms.

This happens with any metal ion that only attains precisely one positive charge, whether it is +1 or +10.

Lithium ions have +1 charge. Carbonate ions have -2 charge. Therefore to balance the charge in the formula we need 2 lithium to 1 carbonate.

This happens with all formulae, and it is a simple mathematical check to decide what the correct ratios should be.

  • $\begingroup$ In a different method that my teacher used to teach me, Lithium carry valence of one and Carbonate carry valence of two, and when they bond together, they interchange their valence which form Li2CO3 ( since we never wrote a valence of one as the subscript). $\endgroup$
    – user27301
    Feb 28, 2016 at 16:03

This is the general process to write the formula for an ionic name.

  1. Name: Metal [Charge] (Ani)de

Ex. Barium [Charge not needed] Hydroxide

Manganese (IV) Oxide

Lithium [Charge not needed] Carbonate

  1. Charges:

    • if charge is given in parentheses, that is the charge for cation; otherwise, it is always the same. (Ex. $\ce {Ba}^{2+}$, $\ce {Mn}^{4+}$, $\ce {Li}^{1+}$)
    • Charges for anions are always the same (Ex. $\ce {OH}^{1-}$, $\ce {O}^{2-}$, $\ce {CO}_3^{2-}$)
  2. Formula

    • The overall formula is the smallest amount of cations/anions in order to be neutral, i.e. the subscript of the cation is the charge of the anion, the subscript of the anion is the charge of the cation, then reduce (Criss-cross charges/subscripts of anions/cations, then reduce).
    • Ex: $\ce {Ba}_1^{2+}{(OH)}_2^{1-}$
    • Ex2: $\ce {Mn}_1^{4+}{O}_2^{2-}$
    • Ex3: $\ce {Li}_2^{1+}{({CO}_3)_1}^{2-}$
  3. "1"s and "charges" are removed since they are implied.
    • Ex: $\ce {Ba}{(OH)}_2$
    • Ex2: $\ce {Mn}{O}_2$
    • Ex3: $\ce {Li}_2{({CO}_3)}$

The reducing subscripts and straightforward nature (only one right way to do it) are due to how these ionic formulas are "formula units" written to be neutral, as in the actual form, the ions form crystal lattices which can be described as large collections of these ratios.


It all depends on the valency of the atoms involved.

If you are have 1st group elements they all have a valency of +1 , and p-block elements have according to their periods*(it's better to remember groups of p-block elements - these start from the boron group to halogen group (fluorine.)*

Chlorine has valency -1 and sodium has valency +1 so multiply them. The valency of chlorine will be the subscript for $\ce{Na}$ and valency of $\ce{Na}$ will be the subscript for $\ce{Cl}$.

Similarly, Mg has valency 2 and oxygen also 2 $\ce{Mg_2O_2}$ will be the compound since both subscripts are common, we generally don't write them and we write $\ce{MgO}$ as compound.

There are still lot of exceptions.

Hope this helped.

  • $\begingroup$ So basically if both charges cancel out, I don't write any subscript? But if it needs balancing I put the charge on the bottom of the unbalanced element? Thanks for replying! $\endgroup$
    – Ali
    Feb 28, 2016 at 16:32
  • $\begingroup$ yes! that's correct $\endgroup$ Feb 28, 2016 at 16:33

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