Question

total structural and stereo isomers of [Co(NO2)2(Nh3)4]NO3

Answer 1

13

Answer 2

Solution:

We start with the complex

$$[Co(NO_2)_2(NH_3)_4]NO_3.$$

Inside the coordination sphere we have Co(III) in an octahedral geometry with four ammonia (A) and two nitrito groups (which we call B), making it of type MA₄B₂. In an octahedron MA₄B₂ there exist two geometric (structural) isomers: cis and trans. Now, note that the nitrito ligand is ambidentate – it may coordinate either through nitrogen (giving “nitro”) or through oxygen (“nitrito”). For two identical NO₂– ligands, the distinct linkage possibilities are:

• Both coordinated as nitro
• Both coordinated as nitrito
• One nitro and one nitrito

When the two B’s are in trans positions these three cases are each unique (the mixed case is not chiral due to internal symmetry). For the cis isomer the mixed (one nitro, one nitrito) arrangement loses a mirror plane and exists as a pair of enantiomers. Hence for the cis case we get:

• Both nitro: 1
• Both nitrito: 1
• Mixed (chiral): 2 (a pair of enantiomers)

That gives a total for the “direct” complex (without ionization) of:

$$\text{trans}:\; 3 \quad + \quad \text{cis}:\; 1+1+2 = 4 \quad \Rightarrow \quad 3+4=7.$$

Additionally, an ionization isomerism is possible by exchange between a coordinated NO₂– and the nitrate counter‐ion. This gives the ionization isomer

$$[Co(NO_2)(NH_3)_4(NO_3)]NO_2.$$

Now the coordination sphere is of type MA₄BC with two different anionic ligands (NO₂ which is ambidentate and NO₃ which (typically) coordinates only through O). Such an MA₄BC complex can also adopt cis and trans arrangements. In the trans form the NO₂ ligand (which can be either nitro or nitrito) gives 2 isomers. In the cis form the unsymmetrical arrangement is chiral; for a given linkage mode of NO₂ one gets a pair of enantiomers. Thus with the two possible ways of NO₂ coordination the cis ionization isomer gives 2×2 = 4 isomers. Hence the ionization isomer contributes

$$2\; (\text{trans}) + 4\; (\text{cis}) = 6.$$

Adding these:

$$\text{Total isomers} = 7\; (\text{direct}) + 6\; (\text{ionization}) = 13.$$

Explanation (minimal):

1. For [Co(NO₂)₂(NH₃)₄]⁺ (MA₄B₂): cis gives (1 nitro + 1 nitrito + 2 (mixed enantiomers)) = 4, trans gives 3; total = 7.
2. Ionization isomer [Co(NO₂)(NH₃)₄(NO₃)]NO₂ (MA₄BC): trans gives 2 (NO₂ linkage possibilities) and cis gives 4 (2 linkage × 2 enantiomers) = 6.
3. Overall total = 7 + 6 = 13.