Consider the following reaction and statements:
$$[Co(NH_3)_4Br_2]^+ + Br^- \rightarrow [Co(NH_3)_3Br_3] + NH_3$$
(i) Two isomers are produced if the reactant complex ion is a cis-isomer.
(ii) Two isomers are produced if the reactant complex ion is a trans-isomer.
(iii) Only one isomer is produced if the reactant complex ion is a trans-isomer.
(iv) Only one isomer is produced if the reactant complex ion is a cis-isomer.
The correct statements are:

We first note that both $$[Co(NH_3)_4Br_2]^+$$ and $$[Co(NH_3)_3Br_3]$$ are octahedral Co(III) complexes. In an octahedral field we always imagine six positions situated at the corners of a regular octahedron. Each position is cis ( 90° ) to four others and trans ( 180° ) to the remaining one.

The reactant ion $$[Co(NH_3)_4Br_2]^+$$ contains two bromido ligands and four ammine ligands, so it has the general formula $$MA_4B_2$$ (with $$A = NH_3$$ and $$B = Br^-$$). Such a complex shows two possible geometries:

• cis- isomer - the two $$Br^-$$ ligands occupy adjacent positions (90° apart).
• trans-isomer - the two $$Br^-$$ ligands are opposite each other (180° apart).

The product $$[Co(NH_3)_3Br_3]$$ has the general formula $$MA_3B_3$$. For an octahedral $$MA_3B_3$$ species the textbook tells us that TWO geometrical arrangements are possible:

• fac (facial) - all three $$Br^-$$ ligands are mutually cis (form one face of the octahedron).
• mer (meridional) - two of the $$Br^-$$ ligands are trans and the third is cis to both of them.

We now examine how many of these arrangements can actually be produced when one $$NH_3$$ of the reactant is replaced by the incoming $$Br^-$$ ion.

To see this clearly we draw a labelled octahedron. Place the metal at the centre and mark the six sites as

$$\begin{array}{ccc} \text{Site 1} & \text{Site 2} & \text{Site 3}\\ \text{Site 4} & \text{Site 5} & \text{Site 6} \end{array}$$

where (1,4), (2,5) and (3,6) are the three trans pairs. We now start with the two $$Br^-$$ ligands in positions dictated by the chosen isomer of the reactant and then insert a third $$Br^-$$ into one of the four remaining sites while simultaneously letting one $$NH_3$$ leave.

Case 1 - cis-reactant.

Take the two bromido ligands in, say, sites 1 and 2. These are cis. The four empty sites (3,4,5,6) are indistinguishable before substitution. Now add the new $$Br^-$$ to each possible site and examine the resulting $$MA_3B_3$$ geometry.

• New $$Br^-$$ at site 3 - sites 1,2,3 are all mutually cis, giving the fac product.
• New $$Br^-$$ at site 4 - the new arrangement has site 1 trans to 4, while 2 is cis to both; this corresponds to the mer product.
• New $$Br^-$$ at site 5 - identical to the previous mer situation by simple rotation.
• New $$Br^-$$ at site 6 - again gives the same fac product as the first case after rotation.

Therefore both fac and mer products can be formed. Hence two isomers are produced when the reactant is cis. Statement (i) is true, statement (iv) is false.

Case 2 - trans-reactant.

Now place the two bromido ligands in, say, sites 1 and 4 (a trans pair). The vacant sites are 2,3,5,6.

• New $$Br^-$$ at site 2 - the final complex has bromido ligands in sites 1,2,4. Here 1 is trans to 4 and cis to 2, while 2 is cis to 4. This is the mer arrangement.
• New $$Br^-$$ at site 3 - by rotation the same mer pattern appears.
• New $$Br^-$$ at site 5 - likewise gives the identical mer geometry.
• New $$Br^-$$ at site 6 - again the same mer product.

Because two of the bromido ligands are locked in a trans relation from the start, it is impossible to generate a configuration in which all three are mutually cis; hence the fac form cannot appear. Only the mer isomer is obtainable. So exactly one isomer is formed when the reactant is trans. Statement (iii) is true, statement (ii) is false.

Collecting the valid statements, we have (i) and (iii).

Option C lists statements (i) and (iii).

Hence, the correct answer is Option C.