The sum of stereoisomers of complex-A, complex-B and complex-C in following reaction is ………..
`[PtCl_(4)]^(2-) underset(-2Cl^(-)) overset(+2("pyridine"))to"[Complex-A]" underset(-Cl^(-))overset(+NH_(3)) to"[Complex-B]" underset("-(Pyridine)")overset(+Br^(-))to "[Complex-C]"`

Explanation

To solve the question regarding the sum of stereoisomers of complexes A, B, and C in the given reaction, we will analyze each complex step by step.

### Step 1: Analyze Complex A
- The starting complex is \([PtCl_4]^{2-}\), which is a square planar complex.
- When one pyridine (a bidentate ligand) replaces one chloride ion, we form Complex A.
- The possible configurations for Complex A are:
- **Cis**: where the two pyridine ligands are adjacent to each other.
- **Trans**: where the two pyridine ligands are opposite each other.
- Therefore, Complex A has **2 stereoisomers** (cis and trans).

### Step 2: Analyze Complex B
- Complex A is then treated with ammonia (\(NH_3\)), replacing one chloride ion.
- The resulting configuration for Complex B can also be either:
- **Cis**: where the pyridine and ammonia are adjacent.
- **Trans**: where the pyridine and ammonia are opposite.
- Thus, Complex B also has **2 stereoisomers** (cis and trans).

### Step 3: Analyze Complex C
- Complex B is then treated with bromide ion (\(Br^-\)), which replaces one pyridine.
- The resulting configuration for Complex C can have:
- **Cis** and **Trans** arrangements for the pyridine and chloride.
- **Cis** and **Trans** arrangements for the pyridine and bromide.
- **Cis** and **Trans** arrangements for the ammonia and bromide.
- Therefore, Complex C can have **3 stereoisomers** based on these arrangements.

### Step 4: Calculate Total Stereoisomers
- Now, we can sum the stereoisomers from all three complexes:
- Complex A: 2 stereoisomers
- Complex B: 2 stereoisomers
- Complex C: 3 stereoisomers
- Total = \(2 + 2 + 3 = 7\)

### Final Answer
The sum of stereoisomers of Complex A, Complex B, and Complex C is **7**.

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