Q133P - Chemistry: Molecular Nature Of Matter And Change

Question

Several coordination isomers, with both Co and Cr as$3 +$ ions, have the molecular formula ${\text{CoCrC}}_6{\text{H}}_{18}{\text{N}}_{12}$

(a) Give the name and formula of the isomer in which the Co complex ion has six  ${\text{NH}}_3$ groups. 

(b) Give the name and formula of the isomer in which the Co complex ion has one $CN$ and five $N{H}_3$ groups.

Step-by-Step Solution

Answer

$$\begin{gathered} a)\left(Co{\left(N{H}_3\right)}_6\right)\left(Cr{\left(CN\right)}_6\right)-hexaamminecobalt(III\left)hexacyanochromate\right(III\left) \\ b\right)\left(Co{\left(N{H}_3\right)}_{5}CN\right)\left(CrN{H}_3{\left(CN\right)}_5\right)-pentaamminecyanocobalt\left(III\right) amminepentacyanochromat\left(III\right) \end{gathered}$$

Step 1: Given information

Coordination isomers, with both Co and Cr as $3+$ ions, have the molecular formula $CoCr{C}_6{H}_{18}{N}_{12}$ .

Step 2: Rules for naming coordination compounds

To solve problems like this, mention some rules for naming coordination compounds:

• Cation is known as before anion
• Within the complex ion, name ligands before metal ion, in alphabetical order
• A numerical prefix is that the number of ligands of type
• The number of metal ion is written by roman numerals in parentheses if the metal ion has over one number
• If the complex ion is an anion, add -ate rather than the ending of metal name
  Neutral ligands most frequently have molecule name, but anionic ligands lose the - ide and add -o after the basis name

Step 3: Name and formula of the isomer for

${\text{CoCrC}}_6{\text{H}}_{18}{\text{N}}_{12}$ - molecular formula

a)  complex ion has groups, which means, look at the molecular formula, that Cr complex ion has 6CN groups:  

$\left(Co{\left(N{H}_3\right)}_6\right)\left(Cr{\left(CN\right)}_6\right)-hexaminecobalt\left(III\right)hexacyanochromate\left(III\right)$ 

Step 4: Name and formula of the isomer for part (b)

b) Co complex ion has one CN and five $N{H}_3$ groups, which means, look at the molecular formula, that Cr complex ion has one $N{H}_3$ and five $CN$ groups:

$\left(Co{\left(N{H}_3\right)}_{5}CN\right)\left(CrNH{H}_3{\left(CN\right)}_5\right)-pentaamminecyanocobalt\left(III\right)amminepentacyanochromat\left(III\right)$