Coordination Compounds

The complexes with formula [Mabcd]^n± can have three geometrical isomers. As NO²⁻ and SCN⁻ are ambidentate ligands, therefore 4 × 3 = 12 geometrical isomers will be possible.

The structure of [Co₂(CO)₈] is

From the structure, we can see that there is one CO−CO bond, six terminals CO and two bridging CO.

Homoleptic complexes are those compounds in which all the ligands bound to the central metal are identical. Thus, these types of complexes have only one type of ligands. In the complex [Co(NH₃)₆]Cl₃ , central atom Co has ammonia ligands as all six.

Heteroleptic complexes are those compounds in which central transition metal has more than one type of ligands. So, rest of the options are examples of heteroleptic complexes.

[Cr(H₂O)₆]Cl₃ has the highest primary valency among the given complexes, that is, three, therefore, it will consume three moles of AgNO₃ and precipitate three moles of AgCl.

[Cr(H₂O)₆]Cl₃ + 3AgNo₃ $\to$ 3AgCl + [Cr(H₂O)₆](NO)₃

$\Delta$₀ of complex [Mo(H₂O)₆]²⁺ is greater than that of [Cr(H₂O)₆]²⁺. It is due to the fact that Mo²⁺ belongs to second row transition element while Cr²⁺ is first row transition element.

Magnitude of crystal field splitting energy ($\Delta$₀) depends on the charge on the metal ion. Greater the charge, greater is the crystal field splitting energy. Therefore, [Ti(H₂O)₆]³⁺ > [Ti(H₂O)₆]²⁺.

(A) [Mg (H₂O)₆ ]²⁺ + (EDTA)^4- $\buildrel {excess\,NaOH} \over \longrightarrow $ [Mg (EDTA) ]²⁺ + 6H₂O

The above equation is incorrect because the product formed would be [Mg (EDTA)]²⁻ .

(B) CuSO₄ + KCN $\to$ K₂[Cu (CN)₄] + K₂SO₄

The above equation is incorrect. Thus, the correct equation is

2KCN + CuSO₄ $\to$ K₂SO₄ + Cu + (CN)₂(Cyanogen Gas)

(C) Li₂O + 2KCl $\to$ 2LiCl + K₂O

The above equation is incorrect because K₂O, a stronger base cannot be generated by a weaker base, Li₂O.

(D) [CoCl(NH₃)₅]⁺ + 5H⁺ $\to$ Co²⁺ + $5NH_4^+$ + Cl^-

The above equation is correct because amine complexes decomposes under acidic medium. Thus, the complex [CoCl(NH₃)₅] decomposes under acidic conditions to give ammonium ions.

The IUPAC name for the complex $[\text{Co(NO}_2)(\text{NH}_3)_5]\text{Cl}_2$ is: Pentaammine(nitrito-$\text{N,O}$)cobalt(III) chloride

Here's how to arrive at the name:

The complex is a cobalt complex, and cobalt has a +3 charge in this compound (since it has one negatively charged NO$_2$ ligand and one negatively charged Cl ion, for a total of -2, to balance the +3 charge on cobalt).

The ligands are NH$_3$ (ammonia) and NO$_2$ (nitrito).

Ammonia is a neutral ligand, so its name is just "ammine" and it is designated by the prefix "penta" because there are five of them.

Nitrito is a negatively charged ligand, so its name is "nitrito-$\text{N,O}$" (the "-$\text{N,O}$" indicates that the NO$_2$ ligand is bound through both the nitrogen and oxygen atoms).

Finally, the compound has a chloride ion, so its name is "chloride" and it is designated by the prefix "di" because there are two of them.

Putting it all together, we get the name "pentaammine(nitrito-$\text{N,O}$)cobalt(III) chloride".

The complex compound Co(NH₃)₅Cl₃ dissociates in water to form three moles of ions. This dissociation can be represented as:

$ \text{Co(NH}_3\text{)}_5\text{Cl}_3\,\,\leftrightharpoons\,\,\left[\text{Co(NH}_3\text{)}_5\text{Cl}\right]^{+2} + 2\text{Cl}^- $

Therefore, the structure of the complex is:

$ \left[\text{Co(NH}_3\text{)}_5\text{Cl}\right]\text{Cl}_2 $

When this complex reacts with two moles of AgNO₃ solution, it produces:

$ \left[\text{Co(NH}_3\text{)}_5\text{Cl}\right]\text{Cl}_2 + 2\text{AgNO}_3 \to \left[\text{Co(NH}_3\text{)}_5\text{Cl}\right](\text{NO}_3)_2 + 2\text{AgCl} $

This reaction yields two moles of AgCl, confirming the complex structure as:

$ \left[\text{Co(NH}_3\text{)}_5\text{Cl}\right]\text{Cl}_2 $