A metal complex having the composition Cr `(NH_(3))_(4)Cl_(2)` Br has been isolated in two forms A and B. The form A reacts with `AgNO_(3)` to give a white precipitate readily soluble in dilute aqueous ammonia whereas B gives a pale yellow precipitate soluble in concentrated ammonia.
The formula of the complex B is

Options

A. `[Cr(NH_(3))_(4)Br]Cl_(2)`
B. `[Cr(NH_(3))_(4)ClBr]Cl`
C. `[Cr(NH_(3))_(4)Cl_(2)]Br`
D. `[Cr(NH_(3))_(4)]BrCl_(2)`

Explanation

To determine the formula of complex B from the given information, we can follow these steps:

### Step 1: Understand the Composition of the Complex
The complex is given as Cr(NH₃)₄Cl₂Br. This indicates that the chromium (Cr) is coordinated with four ammonia (NH₃) ligands, two chloride (Cl) ligands, and one bromide (Br) ligand.

### Step 2: Analyze the Reaction of Form A
Form A reacts with AgNO₃ to produce a white precipitate that is soluble in dilute aqueous ammonia. This indicates that form A contains chloride ions (Cl⁻) because AgCl is a white precipitate that dissolves in dilute ammonia.

### Step 3: Analyze the Reaction of Form B
Form B produces a pale yellow precipitate when reacted with AgNO₃. This suggests that form B contains bromide ions (Br⁻) because AgBr is a pale yellow precipitate. Furthermore, the pale yellow precipitate is soluble in concentrated ammonia, which is characteristic of AgBr.

### Step 4: Determine the Composition of Form B
Since form B gives a pale yellow precipitate with AgNO₃, it must have Br⁻ as its only halide ion. Therefore, in form B, the two Cl⁻ ions from the original complex must have been replaced by Br⁻ ions.

### Step 5: Write the Formula for Complex B
Given that form B has one bromide ion and no chloride ions, the formula for complex B can be written as:
\[ \text{Cr(NH}_3\text{)}_4\text{Br}_2 \]

### Conclusion
Thus, the formula of complex B is:
\[ \text{Cr(NH}_3\text{)}_4\text{Br}_2 \]

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