The actinides showing `+7` oxidation state are:

Options

A. U, Np
B. Pu, Am
C. Np, Pu
D. Am, Cm

Explanation

To determine which actinides show a +7 oxidation state, we need to analyze the electronic configurations of the actinides and their ability to lose electrons. Here’s a step-by-step solution:

### Step 1: Identify the Actinides
The actinides are a series of elements from atomic number 89 (Actinium, Ac) to 103 (Lawrencium, Lr). The relevant actinides for this question are Uranium (U), Neptunium (Np), Plutonium (Pu), Americium (Am), and Curium (Cm).

### Step 2: Write the Electronic Configurations
- **Uranium (U, Z=92)**: [Rn] 5f³ 6d¹ 7s²
- **Neptunium (Np, Z=93)**: [Rn] 5f⁴ 7s²
- **Plutonium (Pu, Z=94)**: [Rn] 5f⁶ 7s²
- **Americium (Am, Z=95)**: [Rn] 5f⁷ 7s²
- **Curium (Cm, Z=96)**: [Rn] 5f⁷ 6d¹ 7s²

### Step 3: Analyze Oxidation States
Now, we will analyze each actinide to see if it can exhibit a +7 oxidation state:
- **Uranium (U)**: Can exhibit oxidation states of +3, +4, +5, +6, but not +7 due to the stability of its electronic configuration.
- **Neptunium (Np)**: Can exhibit oxidation states of +3, +4, +5, +6, and +7. It has the ability to lose 7 electrons (4 from 5f and 2 from 7s).
- **Plutonium (Pu)**: Can exhibit oxidation states of +3, +4, +5, +6, and +7. It can also lose 7 electrons (6 from 5f and 2 from 7s).
- **Americium (Am)**: Generally exhibits +3 and +4 oxidation states. The 5f subshell is half-filled and stable, making +7 less likely.
- **Curium (Cm)**: Similar to Americium, it typically exhibits +3 and +4 oxidation states, and the presence of a filled 5f subshell makes +7 unlikely.

### Step 4: Conclusion
The actinides that can show a +7 oxidation state are:
- **Neptunium (Np)**
- **Plutonium (Pu)**

Thus, the answer to the question is that the actinides showing +7 oxidation states are **Np and Pu**.