`CoCI_(2)` gives blue colour with `NH_(4)SCN` due to formation of

Options

A. `(NH_(4))_(2)[Co(SCN)_(4)]`
B. `(NH_(4))_(4)[Co(SCN)_(6)]`
C. `(NH_(4))_(3)[Co(SCN)_(6)]`
D. `(NH_(4))[Co(SCN)_(4)]`

Answer

To solve the question regarding the reaction of CoCl₂ with NH₄SCN and the formation of a blue-colored complex, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactants in this case are cobalt(II) chloride (CoCl₂) and ammonium thiocyanate (NH₄SCN).

2. **Understand the Reaction**:
- When CoCl₂ reacts with NH₄SCN, a complex ion is formed. Cobalt(II) ions (Co²⁺) can coordinate with thiocyanate ions (SCN⁻) to form a complex.

3. **Formation of the Complex**:
- The complex formed is known as ammonium tetra-thiocyanocobaltate(II). The chemical formula for this complex is [Co(SCN)₄]²⁻, which can be represented as NH₄[Co(SCN)₄] when combined with ammonium ions.

4. **Color of the Complex**:
- The complex [Co(SCN)₄]²⁻ is known to exhibit a blue color. This is a characteristic property of the cobalt-thiocyanate complex.

5. **Conclusion**:
- Therefore, CoCl₂ gives a blue color with NH₄SCN due to the formation of the complex ammonium tetra-thiocyanocobaltate(II), represented as NH₄[Co(SCN)₄].

### Final Answer:
CoCl₂ gives a blue color with NH₄SCN due to the formation of ammonium tetra-thiocyanocobaltate(II), NH₄[Co(SCN)₄].

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