Colour exhibited by divalent Sm in aqueous solutions

Options

A. dark blue
B. yellowish brown
C. blood red
D. greenish yellow

Answer

To determine the color exhibited by divalent samarium (Sm²⁺) in aqueous solutions, we can follow these steps:

### Step 1: Identify the Element
- Samarium (Sm) is a lanthanide with the atomic number 62.

### Step 2: Write the Electronic Configuration
- The electronic configuration of samarium (Sm) is:
\[ [Xe] 6s^2 4f^6 \]

### Step 3: Determine the Configuration of Divalent Samarium
- For the divalent ion (Sm²⁺), two electrons are removed, typically from the outermost shell:
\[ Sm^{2+} : [Xe] 4f^6 \]

### Step 4: Analyze the Electrons in the f Orbital
- In the Sm²⁺ ion, the 4f subshell has 6 electrons. The presence of unpaired electrons in the f orbitals is crucial for determining the color.

### Step 5: Identify the Color of Sm²⁺ in Aqueous Solution
- The presence of unpaired electrons in the f orbitals leads to specific electronic transitions when light is absorbed. For Sm²⁺, the color exhibited in aqueous solution is typically **blood red**.

### Conclusion
- Therefore, the color exhibited by divalent samarium (Sm²⁺) in aqueous solutions is **blood red**.