Category Archives: Inorganic Chemistry

Acid-base equilibria

Acids and bases are easy to deal with when they are strong, as you can assume that any reaction goes to completion. However, weak acids and bases don’t always ionise completely, so you need to look at the equilibria to fully understand them.

Acids dissociate in water to give a proton, (or hydronium ion) and the conjugate base ion.

 HA ⇄ H+ + A

 We can put this into an equilibrium expression, which gives us Ka, the acid ionisation constant:

Ka is an indication of the strength of the acid, the stronger the acid, the larger the Ka.

The same thing applies to bases;

B + H2O ⇄ HB+ + OH

Leading to the Kb expression:

Ka and Kb are the inverse of one another for a given equilibrium.


Most of the stuff for acid-base equilibria is the same as for any other chemical equilibrium problem, however buffers do require a bit more discussion.

A buffer is a solution that can resist changes in pH when small amounts of acid or base are added. They are very important in a broader sense as they regulate the pH of your body and the oceans. Buffers contain a conjugate pair of a weak acid or base, generally in equal parts, for example a carbonate buffer would look like this:

H2CO3 ⇄ H+ + HCO3

where you have added H2CO3 and NaHCO3 to water. The buffer can shift its equilibrium between the acid and conjugate base when small amounts of acid or base are added. The pH will be regulated, because you have such large amounts of H2CO3 and HCO3 to start with, any small change will not affect the overall equilibrium to a large degree.

The Henderson-Hasselbalch equation is used to calculate the pH of a given buffer, and is just a rearrangement of the Ka expression:




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Crystal field theory

We have been discussing metal complexes using the valence bond theory, but while it is a useful method for discussing simple bonds, it cannot account for the colours and magnetic properties of metal complexes. To understand why metal complexes are such beautiful colours, we have to use crystal field theory.

Crystal field theory gives us information on the electronic structure of the metal atom, considering how the d orbitals will be affected by the ligands.

For an octahedral complex, we imagine the ligands are point charges that sit on the Cartesian (x,y,z) axes. The d orbitals are arranged around the nucleus as shown below, with two orbitals pointing directly towards the ligands, and the other three in-between.

The d(x2-y2) and d(z2), which point at the ligands, experience electrostatic repulsion and are therefore at higher energy than the d(xy), d(yz) and d(xz) orbitals. This means the d orbitals are no longer degenerate, and the orbital diagram is split:

The energy difference between the high and low energy orbital is ∆O (the ligand field splitting parameter, if you’re feeling wordy), and it is the size of ∆O that dictates the spin, colour and magnetic properties. The size of the splitting parameter is determined field strength of the ligand. A strong field ligand, such as CN, will give a large splitting parameter and a weak field ligand, I for example, will have a small value for ∆. The ligands are arranged into the spectrochemical series, indicating the strength of the ligand field.

Now that we know about splitting, the next question is how do the d electrons distribute themselves between the non-degenerate orbitals? For example, we have two Fe2+ complexes, [Fe(CN)­6]4- and [Fe(H2O)6]2+. We get two different splitting diagrams:

When we go to fill the orbitals according to Hund’s rule, we should half fill each orbital before going back and pairing the electrons, as it requires energy to put two electrons in an orbital. This is fine, if ∆ is low, as with [Fe(H2O)6]2+, and we get a high spin complex, with a higher number of unpaired electrons. However, if ∆ is large, as with [Fe(CN)­6]4-­, the pairing energy is lower than the energy required to get an electron into the high energy orbitals, so the complex is low spin.

High spin complexes are paramagnetic, as the unpaired electrons give the complex an inherent magnetism. The magnetic moment can be calculated according to the following equation: μs.o. = √{4S(S+1)}. This is the spin-only formula, where S is the number of unpaired electrons.

The colour of transition metal complexes also depends on the magnitude of ∆, as the energy of absorbed light will correspond to the energy difference between the low and high energy orbitals. If ∆ is small, the complex will absorb lower energy light, (red) and appear blue,  the opposite is true for complexes with high ∆.

Some links: (warning – animated gifs)







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Transition metal complexes

Transition metal atoms and ions can act as Lewis acids, accepting electron pairs from molecules or ions with electrons to spare, (Lewis bases).


A complex ion is a metal ion with lewis bases attached through covalent bonds, a metal complex or coordination compound is the same thing, but neutral.

Ligands are the lewis bases attached the metal ion. They may be small molecules or ions

The coordination number is the number of ligands attached to the metal ion.

 Denticity refers to how many bonding sites exist on the ligand – or how many times a single ligand can bond to the central metal ion.


The rules for naming metal complexes are much the same as for other nomenclatures:

  1. Always name the cation before the anion.
  2. The ligands are named first, then the metal atom but all in one word.
  3. The ligands are preceded by a Greek prefix telling you how many of them there are. If you have more than one type of ligand, they are listed in alphabetical order, (ignoring the prefix).
    1. The prefixes are the standard di-, tri-, tetra-… for simple ligands but become bis-, tris-, tetrakis-…  for more complex ligands. As a general rule, if a ligand name already as a prefix in it, (for example ethylenediamine) it is “complex”.
  4. Anionic ligands end in –o, neutral ligands have the same name as the compound, (there are a few exceptions: NH3 = ammine; H2O = aqua).
  5. The metal atom generally has the normal name, but will end with –ate if it is an anion. Some atoms, however, use the old-fashioned names for the anion, (eg. Cu = cuprate; Au = aurate etc). If the symbol on the periodic table doesn’t match the modern name – give it its ye-olde name. You always give the oxidation state of the metal in roman numerals.

Some examples:

[Pt(NH3)Br2]Cl2: tetraamminedichloroplatinum(IV) chloride

Fe(CN)64-: hexacyanoferrate (II)

[Co(en)3]Cl3: tris(ethylenediamine)cobalt (III) chloride


Just like your favourite part of organic chemistry, metal complexes have isomers too! And they are nearly as much fun to name.

There are, as before, two main types of isomers.

Structural isomers deal with differences in the way the atoms are bonded together.

Ionisation isomers: where the ligands and the counterions are exchanged.

[Pt(NH3)4Cl2]Br2 vs [Pt(NH3)4Br2]Cl2

Coordination isomers: are where compounds containing complex anions and cations differ by the distribution of ligands.

[Co(en)3][Cr(CN)­6] vs [Cr(en)3][Co(CN)6]

Linkage isomers: are isomers where a ligand attaches via different atoms.

[Mn(CO)5(SCN)] vs [Mn(CO)5(NCS)]

Stereoisomers have the same bonds, but a different arrangement of them in space.

Geometric isomers: differ in the relative positions of the ligands in space.

If you have a pair of the same ligands, you can have cis- and trans- isomers:

Images courtesy of  Doc Brown (

If you have three, you can get fac- and mer- isomers:

Just to make your life difficult, there are also enantiomers, non-superimposable mirror images:


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Introduction to transition metals

Transition metals are the d-block of the periodic table and therefore have unfilled d-orbitals in common oxidation states. Technically, zinc, cadmium and mercury aren’t transition metals, as they have full d-orbitals, they are generally included anyway.

Trends of the transition metals:

 Electronic configuration:

The electronic configuration of the transition metals is important, as it is the partially filled d-orbitals that are at the root of the interesting chemistry. For the most part, the electronic configuration is predicted by the aufbau principle (which we will cover in greater detail in week 10, but you should know from school. If you’ve forgotten, here’s a worksheet for you.)

Scandium, the first of the transition metals has the configuration [Ar]4s23d1, (remember that the 4s fills before the 3d), and you continue along the period, adding in one electron to the d shell as you go. There are, however, two exceptions. Chromium: [Ar]4s13d5 and copper: [Ar]4s13d10, as the half filled and fully filled d shells are more stable than a full 4s shell.

Chromium - breaking all the rules


Melting and boiling points:

Transition metals all have high melting and boiling points, though some are higher than others.


The melting/boiling point is related to the strength of the metallic bonding, which is determined by the number of unpaired d electrons. The first 5 elements have increasing numbers of unpaired bonds, but they start to pair up after that.


Oxidation states:

The first electrons to be knocked out of a transition metal tend to be the s orbital ones, so the 2+ oxidation state is common across the board, (with a few exceptions). The maximum oxidation state possible is equal to the number of s and d electrons present in the valence shell up to iron, and then decreases after that. Most transition metals have many oxidation states, due to all the different arrangements the d electrons can take.


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Galvanic (or voltaic) cells:

A voltaic cell consists of two half cells, which are connected electrically. A half cell refers to the part of the cell in which one half reaction occurs, and in all text books is depicted as a metal electrode sitting in a beaker of electrolyte solution.

Yeah, you all know this diagram


All voltaic cells share several features, which are essential for operation:

Half cells:

The half cell must contain an electrode, (if the reaction involves reduction to or oxidisation of the metal, the electrode will generally be that metal. If neither species in a half reaction is metal, an inert electrode such as platinum will be used). The solution must contain the relevant ions for the redox reaction to occur.

External circuit:

The half cells need to be connected via a wire, so electrons can be transferred from one half cell to the other. The circuit may be connected via a voltmeter, so the potential of the cell can be measured.

Salt bridge:

If a voltaic cell runs without a salt bridge, charge will quickly build up within the half cells and the cell would stop working. A salt bridge allows ions to flow into each half cell, balancing the charge and allowing extended operation of the cell, but preventing mixing of the two solutions.

Drawing galvanic cells is fairly straightforward – the skeleton two-beaker sketch is a good starting point. We’ll go over how to draw one using an example:

Zn(s) + Cu2+(aq)  →  Cu(s) + Zn2+(aq)

The first thing to do is break it up into the half reactions and identify which is oxidation and which is reduction:

Zn(s) → Zn2+(aq) + 2e                                                                oxidation

Cu2+(aq) + 2e → Cu(s)                                                               reduction

The anode is the half cell in which oxidation occurs, and reduction occurs in the cathode (an ox and a red cat). The anode is always depicted as being on the left, the cathode the right. Label the anode and the cathode, showing what metals and ions are involved:

Show the direction of electron flow (always anode to cathode) and ions in the salt bridge:

This is a fairly clumsy way of depicting cells, and there is a shortcut known as the “line notation”.

Line notation.

As with sketching a cell, the anode is always on the left and the two half cells are separated by a double line, eg:


 Single lines indicate a change in state. The electrodes are always shown on the outer edges, and the reactants are shown in the same order as they would be written in the half reaction. If the half cell includes a gaseous, or two aqueous reactants, an inert electrode is used and any reactants in the same state are separated by a comma, eg:

 Zn(s)|Zn2+(aq)||Fe3+(aq), Fe2+(aq)|Pt

 Cell potentials:

The cell potential tells you how much of a driving force there is behind the cell reaction, and is the difference between the two half-cell potentials.

Cell potential = oxidation potential + reduction potential

Or… because your potentials are usually given as reduction potentials:

Cell potential = reduction potential of cathode – reduction potential of anode

 A note about potentials:

The more positive the standard potential is, the better the species is at oxidising. The more negative, the better the reductant.

Nernst Equation:

The cell potential is easy to work out if you are working in standard conditions, but will change if the concentrations of the components change. Luckily for you, Nernst worked it all out for you and you just need to remember his equation:

Keeping in mind that Q is the reaction quotient and essentially the equilibrium constant (concentration of products/concentration of reactants) when the system isn’t at equilibrium. The reaction is the same for the equilibrium constant, except that at equilibrium the Ecell will be zero.


Some links:


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Oxidation and reduction or, all the things you have forgotten from high school but are expected knowledge at university.

Oxidation/reduction reactions involve the transfer of electrons from one species to another, oxidising one species and reducing the other.

Zn(s) + Cu2+(aq) →  Zn2+(aq) + Cu(s)

In this classic example, the zinc metal gives up two electrons and becomes the zinc (II) ion and the copper (II) ions gain two electrons to become copper metal. The transfer of electrons in redox reactions can be kept track of using oxidation numbers.

Oxidation numbers:

An oxidation number (or state) of a species gives you the hypothetical charge of an atom in a substance, (or actual charge, if it is a monatomic ion) as determined by a few rules. The oxidation numbers will change during an oxidation/reduction reaction, as electrons will be transferred.

The rules for determining the oxidation state are as follows:

  1. The oxidation state of an element (eg. Fe(s), Ca(s), Ar(g)) is always zero.
  2. The oxidation state of an atom in a monatomic ion is the same as the charge on the ion.
  3. The oxidation state of oxygen is always -2, (unless it is a peroxide, in which case it is -1)
  4. The oxidation state of hydrogen is usually +1 (unless it is in a compound such as CaH2, where it is -1)
  5. The oxidation state of fluorine is always -1. Other halogens are usually -1, except when one of the above rules takes precedence.
  6. The sum of the oxidation numbers in a neutral compound is zero. In an ion, the sum of the oxidation numbers is equal to the charge on the ion.

Half reactions:

Fe(s) + Cu2+(aq) → Fe2+(aq) + Cu(s)

Shown above is the overall equation for the oxidation of iron with copper (II), but we can also write this reaction in terms of two half reactions. A half reaction is one part of the oxidation/reduction reaction, showing what happens to each species separately. The reaction shown above can be split accordingly:

Fe(s) → Fe2+(aq) + 2e                            electrons lost by Fe

Cu2+(aq) + 2e → Cu(s)                           electrons gained by Cu

The oxidation reaction is the one in which there is a loss of electrons and increase in the oxidation number, (in this case, the iron). The reduction is the gain of electrons and decrease in oxidation number, (the copper). Confusingly, we refer to each species by what it does, not what happens to it. So the oxidising agent, (or oxidant) is the species that causes oxidation in something else but is itself reduced. The reductant is oxidised, but causes reduction in something else. Don’t get this confused.

You may not always be given the full half equation, you may simply be told the redox couple, (eg, Fe/Fe2+, MnO4/Mn2+) and you will have to balance it yourself. Balancing a half equation is fairly simple, you just need to follow the steps:

  1. Assign oxidation numbers so you know what is oxidised/reduced.
  2. Balance the half equation by:
    1. Balance all the atoms except oxygen and hydrogen
    2. Balance the oxygens by adding the appropriate number of water molecules to the other side of the equation.
    3. Balance the hydrogen atoms, (including the ones you have added in the form of water) by adding H+.
    4. Balance the charge by either i) adding the appropriate number of electrons that have been lost/gained as seen by the change in oxidation number or ii) add electrons where necessary to make sure the equation is neutral overall.
  3. Once you’ve done that for both half equations, add them together. You may need to multiply one or both by a factor to make sure that the electrons have cancelled. If you still have electrons in your overall equation, you’ve done something wrong.

Let’s try an example (adapted from Ebbing General Chemistry):

 Zn(s) + NO3(aq) → Zn2+(aq) + NH4+(aq)

 We can split this one into two half equations:

Zn(s) → Zn2+(aq)­

NO3(aq) → NH4+(aq)

The zinc reaction is easy – the oxidation state starts at zero and ends up at +2, so it is an oxidation requiring two electrons:

Zn(s) → Zn2+(aq)­ + 2e

 The nitrogen half reaction is more complex. First, we need to determine the oxidation states of nitrogen:


The oxygens must each have an oxidation number of -2, and there is an overall -1 charge. Therefore the nitrogen must be +5.


The hydrogens are each +1, and there is an overall +1 charge, so nitrogen must be -3 in this case.

NO3(aq) → NH4+(aq)

Now we need to balance the equation – the number of nitrogens is correct, so we move to the oxygens:

NO3(aq) → NH4+(aq) + 3H2O(l)

Now we need to balance the hydrogens, including those on the ammonium:

NO3(aq) + 10H+(aq) → NH4+(aq) + H2O(l)

Finally, the charge:

NO3(aq) + 10H+(aq) + 8e NH4+(aq) + H2O(l)

We now need to multiply the zinc half reaction by 4, so a total of 8 electrons are being transferred in both reactions, and add them together to give:

4Zn(s) + NO3(aq) + 10H+(aq) → 4Zn2+(aq) + NH4+(aq) + 3H2O(l)

These steps work just fine for acidic reactions, but if your reaction takes place under basic conditions, you need to balance all of the protons with hydroxide ions.

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Filed under Chem 2, Inorganic Chemistry