How many of these metals can show `+3` oxidation state also.
`Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Zn`

Answer

To determine how many of the given metals can show a +3 oxidation state, we will analyze each metal based on its electronic configuration and the number of valence electrons it possesses.

### Step-by-Step Solution:

1. **Identify the Metals and Their Electronic Configurations**:
- Scandium (Sc): [Ar] 4s² 3d¹
- Titanium (Ti): [Ar] 4s² 3d²
- Vanadium (V): [Ar] 4s² 3d³
- Chromium (Cr): [Ar] 4s¹ 3d⁵
- Manganese (Mn): [Ar] 4s² 3d⁵
- Iron (Fe): [Ar] 4s² 3d⁶
- Cobalt (Co): [Ar] 4s² 3d⁷
- Nickel (Ni): [Ar] 4s² 3d⁸
- Zinc (Zn): [Ar] 4s² 3d¹⁰

2. **Determine the Number of Electrons in the Valence Shell**:
- For an element to exhibit a +3 oxidation state, it should have at least three electrons that can be removed (from the outermost shell).

3. **Analyze Each Metal**:
- **Scandium (Sc)**: Has 3 valence electrons (2 from 4s and 1 from 3d) → Can show +3 oxidation state.
- **Titanium (Ti)**: Has 3 valence electrons (2 from 4s and 1 from 3d) → Can show +3 oxidation state.
- **Vanadium (V)**: Has 3 valence electrons (2 from 4s and 3 from 3d) → Can show +3 oxidation state.
- **Chromium (Cr)**: Has 6 valence electrons (1 from 4s and 5 from 3d) → Can show +3 oxidation state.
- **Manganese (Mn)**: Has 7 valence electrons (2 from 4s and 5 from 3d) → Can show +3 oxidation state.
- **Iron (Fe)**: Has 8 valence electrons (2 from 4s and 6 from 3d) → Can show +3 oxidation state.
- **Cobalt (Co)**: Has 9 valence electrons (2 from 4s and 7 from 3d) → Can show +3 oxidation state.
- **Nickel (Ni)**: Has 10 valence electrons (2 from 4s and 8 from 3d) → Can show +3 oxidation state.
- **Zinc (Zn)**: Has 12 valence electrons (2 from 4s and 10 from 3d) → Cannot show +3 oxidation state (only shows +2).

4. **Count the Metals That Can Show +3 Oxidation State**:
- The metals that can show a +3 oxidation state are Sc, Ti, V, Cr, Mn, Fe, Co, and Ni. This totals to 8 metals.

### Final Answer:
The number of metals that can show a +3 oxidation state is **8**.