which of the following statement is correct?

Options

A. The lesser number of oxidation state in `3d`-series in the beginning of the series is due to the presence of too few electrons to loose or share
B. The lesser number of oxidation states in `3d-series` towards the end of the series is due to the presence of too many electrons and thus fewer empty orbital to share electrons with the ligands
C. (A) and (B) both
D. None is correct

Explanation

To solve the question regarding which statement is correct about the oxidation states in the 3d series of transition metals, we will analyze the oxidation states of each element in the series and the reasons behind them.

### Step-by-Step Solution:

1. **Identify the Elements in the 3d Series:**
The 3d series includes the following elements: Scandium (Sc), Titanium (Ti), Vanadium (V), Chromium (Cr), Manganese (Mn), Iron (Fe), Cobalt (Co), Nickel (Ni), Copper (Cu), and Zinc (Zn).

2. **Determine the Electronic Configurations:**
- Sc: [Ar] 4s² 3d¹
- Ti: [Ar] 4s² 3d²
- V: [Ar] 4s² 3d³
- Cr: [Ar] 4s¹ 3d⁵
- Mn: [Ar] 4s² 3d⁵
- Fe: [Ar] 4s² 3d⁶
- Co: [Ar] 4s² 3d⁷
- Ni: [Ar] 4s² 3d⁸
- Cu: [Ar] 4s¹ 3d¹⁰
- Zn: [Ar] 4s² 3d¹⁰

3. **Analyze the Oxidation States:**
- **Scandium (Sc):** Exhibits +3 oxidation state (removes 2 from 4s and 1 from 3d).
- **Titanium (Ti):** Exhibits +3 and +4 oxidation states.
- **Vanadium (V):** Exhibits +2, +3, +4, +5 oxidation states.
- **Chromium (Cr):** Exhibits +1, +2, +3, +6 oxidation states.
- **Manganese (Mn):** Exhibits +2, +3, +4, +6, +7 oxidation states (highest).
- **Iron (Fe):** Exhibits +2 and +3 oxidation states.
- **Cobalt (Co):** Exhibits +2 and +3 oxidation states.
- **Nickel (Ni):** Exhibits +2, +3, and +4 oxidation states.
- **Copper (Cu):** Exhibits +1 and +2 oxidation states.
- **Zinc (Zn):** Exhibits +2 oxidation state (due to fully filled 3d orbital).

4. **Identify Trends in Oxidation States:**
- Early in the series (Scandium to Manganese), there are more oxidation states due to the availability of electrons to lose.
- Towards the end of the series (from Iron to Zinc), the number of oxidation states decreases. This is because there are fewer available empty orbitals for bonding, especially in the case of Zinc, which has a stable full d-orbital configuration.

5. **Evaluate the Statements:**
- Statement A: "The lesser number of oxidation states in the 3d series at the beginning is due to the presence of too few electrons to lose or share." This is correct for Scandium.
- Statement B: "The lesser number of oxidation states in the 3d series towards the end is due to the presence of too many electrons and thus fewer empty orbitals to share electrons with the ligands." This is correct for Zinc.

6. **Conclusion:**
Both statements A and B are correct. Therefore, the correct answer is **C: Both A and B.**