A transition metal A has 'spin-only' magnetic moment value of 1.8 Bm. When it is reacted with dilute sulphuric acid in the presence of air, its compound B is formed. B reacts with compound C to give compound D with the liberation of iodine. Then the metal A and compounds B, C and D are respectively

Options

A. `Ti,TiSO_(4), KI and TiI_(2)`
B. `Zn,ZnSO_(4),KI and Zn_(2)I_(2)`
C. `Cu,CuSO_(4),KI and Cu_(2)I_(2)`
D. `Cu,CuSO_(4), Cu_(2)I_(2) and CuI_(2)`

Answer

To solve the problem step by step, we need to analyze the information given about the transition metal A and its reactions.

### Step 1: Determine the identity of transition metal A
The magnetic moment (μ) is given as 1.8 B.M. The formula for calculating the spin-only magnetic moment is:

\[ \mu = \sqrt{n(n + 2)} \]

where \( n \) is the number of unpaired electrons. Setting up the equation:

\[ 1.8 = \sqrt{n(n + 2)} \]

Squaring both sides gives:

\[ 3.24 = n(n + 2) \]

Rearranging this gives:

\[ n^2 + 2n - 3.24 = 0 \]

Using the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[ n = \frac{-2 \pm \sqrt{2^2 + 4 \cdot 3.24}}{2} \]
\[ n = \frac{-2 \pm \sqrt{4 + 12.96}}{2} \]
\[ n = \frac{-2 \pm \sqrt{16.96}}{2} \]
\[ n = \frac{-2 \pm 4.12}{2} \]

Calculating the two possible values for n:

1. \( n = \frac{2.12}{2} = 1.06 \) (not valid, as n must be an integer)
2. \( n = \frac{-6.12}{2} = -3.06 \) (not valid)

Thus, \( n = 1 \) is the only feasible solution, indicating that metal A has 1 unpaired electron. The transition metals with 1 unpaired electron include Copper (Cu).

### Step 2: Identify compound B formed by the reaction of A with dilute sulfuric acid
When Copper (Cu) reacts with dilute sulfuric acid (H₂SO₄) in the presence of air, it forms Copper(II) sulfate (CuSO₄) and hydrogen gas (H₂). Therefore, compound B is:

\[ \text{Compound B: CuSO}_4 \]

### Step 3: Identify compound C
The problem states that compound B reacts with compound C to form compound D with the liberation of iodine. A common reaction involving CuSO₄ is its reaction with potassium iodide (KI). Thus, we can identify compound C as:

\[ \text{Compound C: KI} \]

### Step 4: Identify compound D formed from the reaction of B and C
When CuSO₄ reacts with KI, the reaction produces Copper(I) iodide (CuI) and potassium sulfate (K₂SO₄), along with the liberation of iodine (I₂):

\[ \text{CuSO}_4 + 2 \text{KI} \rightarrow \text{Cu}_2\text{I}_2 + \text{K}_2\text{SO}_4 + \text{I}_2 \]

Thus, compound D is:

\[ \text{Compound D: Cu}_2\text{I}_2 \]

### Final Summary
- Metal A: Copper (Cu)
- Compound B: Copper(II) sulfate (CuSO₄)
- Compound C: Potassium iodide (KI)
- Compound D: Copper(I) iodide (CuI)