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Is there a way to remember polyatomic ions (i.e $\ce{PO4^{3-}}$, $\ce{SO4^{2-}}$ etc.). For example, our professor will ask on an exam what the equation for iron (III) carbonate. Obviously the (III) means iron has a 3+ oxidation state. However, if one was not too familiar with the number of oyxgens or charges on specific polyatomic ions, how would you know the symbol for carbonate ($\ce{CO3^{2-}}$)? She gives us a periodic table but I dont think that will help if you dont have carbonate memorized?

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    $\begingroup$ Pfft, it's crystal clear in comparison with traditional names of compounds. $\endgroup$
    – Mithoron
    Commented Sep 18, 2016 at 23:09

2 Answers 2

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Initially the idea was that -ate suffix is reserved for a mononuclear acid with highest oxidation state, while -ite suffix was reserved for a mononuclear acid with second highest oxidation state. The per-...-ate construct was reserved for anion with 'extra' oxygen, typically a peroxocompound and hypo-...-ite construct was reserved for acid extra low on oxygen.

However, since for halogens $\ce{XO4^-}$ anions were obtained rather lately, perhalogenates has this formula and simple halognates are $\ce{XO3^-}$. Mn subgroup (Mn, Tc, Re) inherited this in the sense that highest oxidation state corresponds to per- ... -ate construct

Similarly, ortho- prefix was reserved for mononuclear acids with highest number of hydroxogroups possible, while meta- prefix was reserved for an acid with one water molecule less (typically, polymer/oligomer)

Putting it all together using phosphorous as an example

  • Perphosphate $\ce{PO5^{3-}}$
  • Ortophosphate - $\ce{PO4^{3-}}$
  • Metaphosphate - $\ce{(PO3^-)_n}$
  • Phosphite $\ce{HPO3^{2-}}$ (since $\ce{H3PO3}$ is actually $\ce{H2[HPO3]}$ for some arcane reasons) Note: depending on the chemistry course you are taking (less advanced courses mainly), some will use $\ce{PO3^{3-}}$ as phosphite
  • Hypophosphite - $\ce{H2PO2^-}$ (since $\ce{H3PO2}$ is actually $\ce{H[H2PO2]}$

And for chlorine (see above about perhalogenates)

  • Perchlorate $\ce{ClO4-}$
  • Chlorate $\ce{ClO3-}$
  • Chlorite $\ce{ClO2-}$
  • Hypochlorite $\ce{ClO-}$

Not every degree needs to be present, for example only simple carbonates and percarbonates exist, and sometimes extra stages (usually polynuclear) occur, such as dithionites. This system is widespread in areas heavy on traditions, such as medicine and heavy industry.

There is a gentle pressure from IUPAC to move to much less confusing systematic naming, but the success is limited, especially considering widespread chemicals, which are familiar anyway. It is similar to situation with RJ plugs: everyone recognises phone and Ethernet connectors, but very few care about their full formal name.

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I tried giving my extra-teaching pupils the following guidance:

  1. Determine the element whose name is central.

    • Carbonate → carbon
    • Phosphite → phosphorus
    • Periodate → iodine
    • Hypochlorite → chlorine

    Initially drop all per- or hypo- prefixes; we’ll get back to them later. Also, initially assume -ate, we’ll determine -ite later.

  2. If this element is in the second period:

    1. for an -ate name, add three oxygens.
      • carbonate → $\ce{CO3^n-}$
    2. assume the highest oxidation state of the central element.
      • For carbon, that is $\mathrm{+IV}$ → $\ce{C^{IV}O3^n-}$
  3. If this element is a halide:

    1. for an -ate name, add three oxygens.
      • iodate → $\ce{IO3^n-}$
    2. assume an oxidation state of $\mathrm{+V}$; not the maximum oxidation state!
      • iodate → $\ce{I^{V}O3^n-}$
  4. For all other elements (non-halogens, third and higher periods):

    1. for an -ate name add four oxygens.
      • phosphate → $\ce{PO4^n-}$.
    2. assume the highest oxidation state.
      • for phosphorus, that is $\mathrm{+V}$ → $\ce{P^{V}O4^n-}$
  5. Calculate the ion’s charge by assuming oxygen’s oxidation state to be $\mathrm{-II}$ (the standard).

    • Carbonate: $(+4) + 3 \times (-2) = 4 - 6 = -2$
      Thus, carbonate is $\ce{CO3^2-}$
    • Iodate: $(+5) + 3 \times (-2) = 5 - 6 = -1$
      Thus, iodate is $\ce{IO3-}$
    • Phosphate: $(+5) + 4 \times (-2) = 5 - 8 = -3$
      Thus, phosphate is $\ce{PO4^3-}$

    And because we ignored hypochlorite so far, but we must do the same for chlorate:

    • Chlorate: $(+5) + 3 \times (-2) = 5 - 6 = -1$

    You have successfully determined charge and sum formula of an -ate ion.

  6. If your ion ends in -ite, subtract one oxygen leaving the rest unchanged.

    • Phosphate is $\ce{PO4^3-}$ thus phosphite is $\ce{PO3^3-}$
    • Chlorate is $\ce{ClO3-}$ thus chlorite is $\ce{ClO2-}$
  7. If your ion includes a hypo- prefix, subtract another oxygen from the -ite species, leaving the rest unchanged.

    • Chlorite is $\ce{ClO2-}$ thus hypochlorite is $\ce{ClO-}$
  8. If your ion includes a per- prefix, add another oxygen onto the -ate species leaving the rest unchanged.

    • Iodate is $\ce{IO3-}$ thus periodate is $\ce{IO4-}$
  9. Any additional hydrogen- prefixes are added as $\ce{H+}$ ions, thus they reduce the charge accordingly.

This method reduces the problem to correctly identifying the element in the periodic table. It may be a lot to remember, but it’s basically the extremes of the main groups (top period and left-most non-noble gaseous group) that have three oxygens while all the rest have four. Remember the halogens’ special case with respect to oxidation states and you’re good.

In time, this will happen automatically, i.e. you will see arsenate and automatically read $\ce{AsO4^3-}$.

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