Question

The formula of tetraammineaquachloridocobalt(III) chloride is

• A. $[\text{Co}(\text{NH}_3)_4(\text{H}_2\text{O})\text{Cl}]\text{Cl}_2$
• B. $[\text{Co}(\text{NH}_3)_4\text{Cl}_2] \times \text{H}_2\text{O}$
• C. $[\text{Co}(\text{NH}_3)_4]\text{Cl}_3 \times \text{H}_2\text{O}$
• D. $[\text{Co}(\text{NH}_3)_4(\text{H}_2\text{O})\text{Cl}]\text{Cl}$

Hint:

When writing structural formulas from IUPAC nomenclature: 1. Identify the central metal and its oxidation state. 2. Sum up the charges of all internal ligands to find the net charge of the bracketed complex. 3. Add a sufficient number of outer counter-ions to make the total overall charge of the formula exactly equal to zero.

Answer 1

The Correct Option is A

Step 1: Break the name into pieces.
The name is tetraammineaquachloridocobalt(III) chloride. Reading it part by part tells us the metal, the ligands inside the bracket, and the counter ion outside. Let us translate each word into a formula.
Step 2: Identify the metal and its oxidation state.
The word cobalt(III) means the central metal is cobalt in the \(+3\) state, that is \(Co^{3+}\).
Step 3: List the ligands inside the sphere.
tetraammine gives four neutral \(NH_3\); aqua gives one neutral \(H_2O\); chlorido gives one \(Cl^-\) bound to the metal. So the bracket is \([Co(NH_3)_4(H_2O)Cl]\).
Step 4: Work out the charge on the complex ion.
Add the charges: cobalt \(+3\), four \(NH_3\) contribute \(0\), one \(H_2O\) contributes \(0\), one \(Cl^-\) contributes \(-1\). \[ (+3)+0+0+(-1)=+2 \] So the complex ion is \([Co(NH_3)_4(H_2O)Cl]^{2+}\).
Step 5: Add the counter ions.
The word chloride at the end means free \(Cl^-\) ions sit outside. To cancel a \(+2\) charge we need two of them, giving \(Cl_2\) outside.
Step 6: Assemble and match.
The complete formula is \([Co(NH_3)_4(H_2O)Cl]Cl_2\), which is option 1.
\[ \boxed{[Co(NH_3)_4(H_2O)Cl]Cl_2} \]