Which of the following is an inner orbital complex as well as diamagnetic in behaviour
[Atomic numbers Zn = 30, Cr = 24, Co = 27, Ni = 28.]

Options

A. `[Zn(NH_(3))_(6)]^(2+)`
B. `[Cr(NH_(3))_(6)]^(3+)`
C. `[Co(NH_(3))_(6)]^(3+)`
D. `[Ni(NH_(3))_(6)]^(2+)`

Explanation

To solve the question of identifying which of the given complexes is an inner orbital complex and also exhibits diamagnetic behavior, we will analyze each option step by step.

### Step-by-Step Solution:

1. **Identify the Complexes:**
The complexes given are:
- A) [Zn(NH3)6]²⁺
- B) [Cr(NH3)6]³⁺
- C) [Co(NH3)6]³⁺
- D) [Ni(NH3)6]²⁺

2. **Determine the Oxidation State:**
For each complex, we will calculate the oxidation state of the central metal ion.

- **A) [Zn(NH3)6]²⁺:**
- Let oxidation state of Zn be \( x \).
- Amine (NH3) is neutral.
- The equation becomes: \( x + 6(0) = +2 \) → \( x = +2 \)
- Oxidation state of Zn is +2.

- **B) [Cr(NH3)6]³⁺:**
- Let oxidation state of Cr be \( x \).
- The equation becomes: \( x + 6(0) = +3 \) → \( x = +3 \)
- Oxidation state of Cr is +3.

- **C) [Co(NH3)6]³⁺:**
- Let oxidation state of Co be \( x \).
- The equation becomes: \( x + 6(0) = +3 \) → \( x = +3 \)
- Oxidation state of Co is +3.

- **D) [Ni(NH3)6]²⁺:**
- Let oxidation state of Ni be \( x \).
- The equation becomes: \( x + 6(0) = +2 \) → \( x = +2 \)
- Oxidation state of Ni is +2.

3. **Write the Electron Configuration:**
Now we will write the electron configurations for the metal ions in their respective oxidation states.

- **A) Zn²⁺:**
- Atomic number of Zn = 30 → Configuration: [Ar] 4s² 3d¹⁰
- Zn²⁺: [Ar] 3d¹⁰ (4s is empty)

- **B) Cr³⁺:**
- Atomic number of Cr = 24 → Configuration: [Ar] 4s² 3d⁵
- Cr³⁺: [Ar] 3d³ (4s is empty, 2 electrons removed from 4s and 1 from 3d)

- **C) Co³⁺:**
- Atomic number of Co = 27 → Configuration: [Ar] 4s² 3d⁷
- Co³⁺: [Ar] 3d⁶ (4s is empty, 2 electrons removed from 4s and 1 from 3d)

- **D) Ni²⁺:**
- Atomic number of Ni = 28 → Configuration: [Ar] 4s² 3d⁸
- Ni²⁺: [Ar] 3d⁸ (4s is empty, 2 electrons removed from 4s)

4. **Determine Hybridization and Magnetic Properties:**
Next, we will analyze the hybridization and whether the complexes are inner or outer orbital complexes and if they are diamagnetic.

- **A) [Zn(NH3)6]²⁺:**
- Hybridization: sp³d² (outer orbital complex)
- Magnetic behavior: Diamagnetic (all electrons paired)

- **B) [Cr(NH3)6]³⁺:**
- Hybridization: d²sp³ (inner orbital complex)
- Magnetic behavior: Paramagnetic (unpaired electrons)

- **C) [Co(NH3)6]³⁺:**
- Hybridization: d²sp³ (inner orbital complex)
- Magnetic behavior: Diamagnetic (all electrons paired after pairing)

- **D) [Ni(NH3)6]²⁺:**
- Hybridization: sp³d² (outer orbital complex)
- Magnetic behavior: Paramagnetic (unpaired electrons)

5. **Conclusion:**
The only complex that is both an inner orbital complex and diamagnetic is **C) [Co(NH3)6]³⁺**.

### Final Answer:
The correct answer is **C) [Co(NH3)6]³⁺**.