Oxidation states

What are they?

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In chem­istry, the de­scrip­tion of dif­fer­ent ox­i­diz­ing-re­duc­ing pro­cess­es is im­pos­si­ble with­out ox­i­da­tion states – spe­cial con­di­tion­al num­bers which are used to de­ter­mine the charge of an atom of any el­e­ment.

If you imag­ine an ox­i­da­tion state (not to be con­fused with va­lence! In some cas­es, they do not co­in­cide) as a note in an ex­er­cise book, we will sim­ply see num­bers with the fig­ure zero (0 in a sim­ple sub­stance), plus (+) or mi­nus (-) above the sub­stance that in­ter­ests us. As it hap­pens, they play an enor­mous role in chem­istry, and the abil­i­ty to de­ter­mine the OS (ox­i­da­tion state) is es­sen­tial in the study of this sub­ject, with­out which fur­ther ac­tions are point­less.

We use the OS for the de­scrip­tion of the chem­i­cal prop­er­ties of a sub­stance (or sep­a­rate el­e­ment), the cor­rect writ­ing of its in­ter­na­tion­al name (un­der­stand­able for any coun­try and na­tion re­gard­less of the lan­guage used) and for­mu­la, and also for clas­si­fi­ca­tion by prop­er­ties.

There can be three types of state: high­est (to de­ter­mine this we must know the group that the el­e­ment is lo­cat­ed in), in­ter­me­di­ary, and low­est (we must sub­tract the num­ber of the group in which the el­e­ment is lo­cat­ed from the num­ber 8; nat­u­ral­ly, we take the num­ber 8 be­cause there are a to­tal of 8 groups in the Pe­ri­od­ic Ta­ble). We will look at how to de­ter­mine the ox­i­da­tion state and to bal­ance it cor­rect­ly in more de­tail be­low.

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How to de­ter­mine the ox­i­da­tion state: con­stant OS

First­ly, the OS may be vari­able or con­stant. De­ter­min­ing a con­stant ox­i­da­tion state is not very dif­fi­cult, so it’s best to start the les­son with this: all we need is to be able to use the pe­ri­od­ic ta­ble. There are a num­ber of cer­tain rules:

  1. Zero state. We men­tioned this above – only sim­ple sub­stances have it: S, O2, Al, K and so on.

  2. If mol­e­cules are neu­tral (in oth­er words, they do not have an elec­tric charge), the sum of their ox­i­da­tion state is equal to zero. How­ev­er, in the case of ions the sum should be equal to the charge of the ion it­self.

  3. Pri­mar­i­ly met­als are lo­cat­ed in groups I, II and III of the Pe­ri­od­ic Ta­ble. El­e­ments of these groups have a pos­i­tive charge, the val­ue of which cor­re­sponds to the num­ber of the group (+1, +2, or +3). An ex­cep­tion is sev­er­al met­als main­ly from side groups – for ex­am­ple, iron (Fe) – its OS can be +2 or +3.

  4. The OS of hy­dro­gen (H) is usu­al­ly +1 (in in­ter­ac­tion with non-met­als: HCl, H₂S), but in some cas­es it is -1 (in the for­ma­tion of hy­drides in com­pounds with met­als: KH, MgH₂).

  5. The OS of oxy­gen (O) is +2. Com­pounds with this el­e­ment form ox­ides (MgO, Na₂O, H₂O – wa­ter). How­ev­er, there are also cas­es when oxy­gen has an ox­i­da­tion state of -1 (in the for­ma­tion of per­ox­ides), or acts as a re­duc­er (in com­pounds with flu­o­rine F, be­cause the ox­i­diz­ing prop­er­ties of oxy­gen are weak­er). On the ba­sis of this in­for­ma­tion, we can bal­ance the ox­i­da­tion states in many com­plex sub­stances, de­scribe ox­i­diz­ing-re­duc­ing re­ac­tions and so on, but more about this lat­er.

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Vari­able OS

Some el­e­ments are dis­tin­guished by the fact that they do not have a sin­gle ox­i­da­tion state and change de­pend­ing on the for­mu­la they are in. Ac­cord­ing to the rules, the sum of all ox­i­da­tion states should also be equal to zero, but to find this, cer­tain cal­cu­la­tions must be made. In writ­ten form, this looks like a sim­ple al­ge­bra­ic equa­tion, but over time we “get the hang of it”, and eas­i­ly learn to per­form the en­tire op­er­a­tion in our heads. It will not be so easy to de­scribe this in words, so let’s move straight to prac­tice:

HNO₃ – de­ter­mine the ox­i­da­tion state of ni­tro­gen (N) in this for­mu­la.

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In chem­istry we also read the names of el­e­ments, and ap­proach the plac­ing of ox­i­da­tion states from the end as well. So, we know that the OS of oxy­gen is 02. We must mul­ti­ply the ox­i­da­tion state by the co­ef­fi­cient on the right (if there is one): -2•3=-6. We then move to hy­dro­gen (H): its CO in the equa­tion will be +1. So for the sum of the OS to be zero, we must add 5. We make sure: +1+5-6=-0.

Ad­di­tion­al ex­er­cis­es can be found at the end, but first we must de­ter­mine which el­e­ments have a vari­able ox­i­da­tion state. In prin­ci­ple, all el­e­ments, not count­ing the first three groups, change their states. The most strik­ing ex­am­ple is the halo­gens (el­e­ments of group VII, not count­ing flu­o­rine F), group IV and no­ble gas­es. Be­low you will see a list of sev­er­al met­als and non-met­als with a vari­able state:

  • H (+1, -1);
  • B (-3, +3);
  • C (-4, -2, +2, +4);
  • N (-3, -2, -1, +1, +2, +3, +4, +5);
  • O (-2, -1);
  • Si (-4, -2, +2, +4);
  • P (-3, +1, +3, +5);
  • S (-2, +2, +4, +6);
  • Cl (-1, +1, +3, +5, +7).

This is just a small se­lec­tion of el­e­ments. To learn to de­ter­mine the OS, you will need study and prac­tice, but this does not mean that you need to learn all the con­stant and vari­able OS by heart: just re­mem­ber that the first and last num­bers are en­coun­tered much more of­ten. An im­por­tant role is of­ten played by the co­ef­fi­cient, and what sub­stance is un­der dis­cus­sion – for ex­am­ple, in sul­fides sul­fur (S) takes a neg­a­tive ox­i­da­tion state, in ox­ides - oxy­gen (O), and in chlo­rides – chlo­rine (Cl). Ac­cord­ing­ly, in these salts a dif­fer­ent el­e­ment takes a pos­i­tive state (and is said to be ox­i­dized in this sit­u­a­tion).

Solv­ing prob­lems to de­ter­mine ox­i­da­tion states

Now we have come to the most im­por­tant part – prac­tice. Try to solve the fol­low­ing prob­lems your­self, and then look at the so­lu­tion and check your an­swers:

  1. K₂Cr₂O₇ –find the ox­i­da­tion state of chromi­um The OS of oxy­gen is -2, for potas­si­um it is +1, while we in­di­cate chromi­um by the un­known vari­able x for the mean­time. Ac­cord­ing­ly, we draw up the equa­tion: +1•2+2•x-2•7=0. Af­ter solv­ing it, we re­ceive the an­swer 6. We check this, and ev­ery­thing match­es up, so the prob­lem has been solved.

  2. H₂­SO₄ – find the OS of sul­fur Ac­cord­ing to the same con­cept we draw up an equa­tion: +1•2+x-2•4=0. Then: 2+x-8=0.x=8-2; x=6. Click here to do en­ter­tain­ing ex­per­i­ments for bet­ter un­der­stand­ing how to de­ter­mine ox­i­da­tion states.

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In con­clu­sion

To learn to de­ter­mine the ox­i­da­tion state in­de­pen­dent­ly, you must not only be able to draw up equa­tions, but also study the prop­er­ties of el­e­ments of dif­fer­ent groups thor­ough­ly, re­mem­ber al­ge­bra lessons, draw­ing up and solv­ing equa­tions with an un­known vari­able. Don’t for­get that rules have ex­cep­tions, and you shouldn’t for­get them: this con­cerns el­e­ments with a vari­able OS. You must also learn to bal­ance co­ef­fi­cients to solve many prob­lems and equa­tions (and know why you are do­ing so).