The deep red colour of `Fe(SCN)_3` and `Fe(SCN)_4^(ɵ)` is destroyed by addition of :

Options

A. `F^(ɵ)`
B. `CN^(ɵ)`
C. `SCN^(ɵ)`
D. `Fe^(ɵ)`

Explanation

To solve the question regarding the destruction of the deep red color of `Fe(SCN)3` and `Fe(SCN)4^0`, we can follow these steps:

### Step 1: Understand the Complexes
`Fe(SCN)3` and `Fe(SCN)4^0` are coordination complexes where iron (Fe) is coordinated with thiocyanate ions (SCN^-). The deep red color is due to the presence of these complexes.

**Hint:** Identify the components of the complexes and their colors.

### Step 2: Identify the Ligands
We need to determine which of the given ligands (F^-, CN^-, SCN^-, and Fe) can replace the SCN^- in the complexes. The ligands are:
- F^- (fluoride ion)
- CN^- (cyanide ion)
- SCN^- (thiocyanate ion)
- Fe (iron)

**Hint:** Consider the properties of the ligands and their ability to form complexes with iron.

### Step 3: Determine the Strength of the Ligands
Among the ligands, fluoride (F^-) is a stronger ligand compared to SCN^-. Stronger ligands can replace weaker ones in coordination complexes.

**Hint:** Compare the ligand strengths based on their electronegativity and ability to form stable complexes.

### Step 4: Predict the Reaction
When F^- is added to `Fe(SCN)3`, it can replace the SCN^- ligands, forming a new complex `FeF3`. This new complex is known to be green in color, which would result in the destruction of the deep red color of the original complex.

**Hint:** Think about the color changes that occur when ligands are replaced in coordination complexes.

### Step 5: Conclusion
The addition of fluoride ions (F^-) will lead to the formation of `FeF3`, which is green, thus destroying the deep red color of `Fe(SCN)3` and `Fe(SCN)4^0`. Therefore, the answer is:

**Final Answer:** F^- (Option A)

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