Mark the correct statements(s)
(1) Manganeses exhibits `+7` oxidation state
(2) Zinc forms coloured ions
(3) `[CoF_(6)]^(3-)` is diamagnetic
(4) Sc forms `+4` oxidation state
(5) Zn exhibits only `+2` oxidation state

Options

A. 1 and 2
B. 1 and 5
C. 2 and 4
D. 3 and 4

Explanation

To determine which statements are correct regarding the given options, we will analyze each statement one by one.

### Step-by-Step Solution:

1. **Statement (1): Manganese exhibits +7 oxidation state.**
- Manganese (Mn) has an atomic number of 25, with an electronic configuration of [Ar] 3d^5 4s^2.
- It can lose all 7 electrons (2 from 4s and 5 from 3d) to achieve a +7 oxidation state.
- **Conclusion**: This statement is **correct**.

2. **Statement (2): Zinc forms coloured ions.**
- Zinc (Zn) has an atomic number of 30, with an electronic configuration of [Ar] 3d^10 4s^2.
- Since the 3d subshell is completely filled (3d^10), there are no unpaired electrons to allow for the formation of colored ions.
- **Conclusion**: This statement is **incorrect**.

3. **Statement (3): [CoF6]^(3-) is diamagnetic.**
- To find the oxidation state of cobalt in [CoF6]^(3-), we set up the equation: X + 6(-1) = -3, where X is the oxidation state of Co.
- Solving gives X = +3, so cobalt is in the +3 oxidation state.
- The electronic configuration of Co is [Ar] 3d^7 4s^2. For Co^(3+), it loses 2 electrons from 4s and 1 from 3d, resulting in [Ar] 3d^6.
- Fluoride (F^-) is a weak field ligand, meaning it does not cause pairing of electrons. Hence, the 3d^6 configuration will have unpaired electrons.
- **Conclusion**: This statement is **incorrect** (it is paramagnetic).

4. **Statement (4): Scandium forms +4 oxidation state.**
- Scandium (Sc) has an atomic number of 21, with an electronic configuration of [Ar] 3d^1 4s^2.
- The common oxidation states of Sc are +2 and +3, but it does not typically exhibit a +4 oxidation state.
- **Conclusion**: This statement is **incorrect**.

5. **Statement (5): Zn exhibits only +2 oxidation state.**
- As previously stated, Zn has a configuration of 3d^10 4s^2. It typically loses the 2 electrons from the 4s subshell to form Zn^(2+).
- Since the 3d subshell is fully filled, Zn does not exhibit any other oxidation states.
- **Conclusion**: This statement is **correct**.

### Final Answer:
The correct statements are **(1) and (5)**.