The maximum oxidation state shown by `V(Z=23),Cr(Z=24),Co(Z=27),Sc(Z=21)`, are respectively

Options

A. `+5,+6,+3,+3`
B. `+3,+4,+5,+2`
C. `+3,+4,+5,+2`
D. `+4`, in each case

Answer

To determine the maximum oxidation states of vanadium (V), chromium (Cr), cobalt (Co), and scandium (Sc), we need to analyze their electronic configurations and how many electrons they can lose.

### Step-by-Step Solution:

1. **Identify the Atomic Numbers and Electronic Configurations:**
- Vanadium (V) has atomic number 23:
- Electronic configuration: \( [Ar] 3d^3 4s^2 \)
- Chromium (Cr) has atomic number 24:
- Electronic configuration: \( [Ar] 3d^5 4s^1 \)
- Cobalt (Co) has atomic number 27:
- Electronic configuration: \( [Ar] 3d^7 4s^2 \)
- Scandium (Sc) has atomic number 21:
- Electronic configuration: \( [Ar] 3d^1 4s^2 \)

2. **Determine the Maximum Oxidation State for Each Element:**
- **Vanadium (V):**
- Valence electrons: 5 (3d^3 + 4s^2)
- Maximum oxidation state: +5 (by losing all 5 valence electrons)

- **Chromium (Cr):**
- Valence electrons: 6 (3d^5 + 4s^1)
- Maximum oxidation state: +6 (by losing all 6 valence electrons)

- **Cobalt (Co):**
- Valence electrons: 9 (3d^7 + 4s^2)
- Maximum oxidation state: +3 (can lose 2 from 4s and 1 from 3d, resulting in a stable t2g configuration)

- **Scandium (Sc):**
- Valence electrons: 3 (3d^1 + 4s^2)
- Maximum oxidation state: +3 (by losing all 3 valence electrons)

3. **Summarize the Maximum Oxidation States:**
- Vanadium (V): +5
- Chromium (Cr): +6
- Cobalt (Co): +3
- Scandium (Sc): +3

### Final Answer:
The maximum oxidation states shown by V, Cr, Co, and Sc are respectively: **+5, +6, +3, +3.**