Which of the following are diamagnetic?
(I) `K_(4)[Fe(CN)_(6)]`
(II) `K_(3)[Cr(CN)_(6)]`
(III) `K_(3)[Co(CN)_(6)]`
(IV) `K_(2)[Ni(CN)_(4)]`
Select the correct answer using the codes given below:

Options

A. I, II and IV
B. I,III and IV
C. II and III
D. I and IV

Answer

To determine which of the given coordination compounds are diamagnetic, we need to analyze the electronic configurations of the central metal ions in each complex and check for the presence of unpaired electrons. A compound is diamagnetic if it has no unpaired electrons.

### Step-by-Step Solution:

1. **Identify the oxidation states of the central metal ions:**
- **(I) K₄[Fe(CN)₆]:**
- The charge of the complex is -4 (from K⁺).
- CN⁻ contributes -6, so the oxidation state of Fe (X) is:
\[ X - 6 = -4 \implies X = +2 \]

- **(II) K₃[Cr(CN)₆]:**
- The charge of the complex is -3 (from K⁺).
- CN⁻ contributes -6, so the oxidation state of Cr (X) is:
\[ X - 6 = -3 \implies X = +3 \]

- **(III) K₃[Co(CN)₆]:**
- The charge of the complex is -3 (from K⁺).
- CN⁻ contributes -6, so the oxidation state of Co (X) is:
\[ X - 6 = -3 \implies X = +3 \]

- **(IV) K₂[Ni(CN)₄]:**
- The charge of the complex is -2 (from K⁺).
- CN⁻ contributes -4, so the oxidation state of Ni (X) is:
\[ X - 4 = -2 \implies X = +2 \]

2. **Determine the electronic configurations of the metal ions:**
- **Fe²⁺ (from K₄[Fe(CN)₆]):**
- Atomic number of Fe = 26, configuration: [Ar] 3d⁶ 4s².
- For Fe²⁺: [Ar] 3d⁶ (6 electrons in 3d orbitals).
- In the presence of CN⁻ (a strong field ligand), all electrons pair up, resulting in 0 unpaired electrons. Thus, it is **diamagnetic**.

- **Cr³⁺ (from K₃[Cr(CN)₆]):**
- Atomic number of Cr = 24, configuration: [Ar] 3d⁵ 4s¹.
- For Cr³⁺: [Ar] 3d³ (3 electrons in 3d orbitals).
- With CN⁻, the electrons remain unpaired, resulting in 3 unpaired electrons. Thus, it is **paramagnetic**.

- **Co³⁺ (from K₃[Co(CN)₆]):**
- Atomic number of Co = 27, configuration: [Ar] 3d⁷ 4s².
- For Co³⁺: [Ar] 3d⁶ (6 electrons in 3d orbitals).
- In the presence of CN⁻, all electrons pair up, resulting in 0 unpaired electrons. Thus, it is **diamagnetic**.

- **Ni²⁺ (from K₂[Ni(CN)₄]):**
- Atomic number of Ni = 28, configuration: [Ar] 4s² 3d⁸.
- For Ni²⁺: [Ar] 3d⁸ (8 electrons in 3d orbitals).
- With CN⁻, the electrons pair up, resulting in 0 unpaired electrons. Thus, it is **diamagnetic**.

3. **Conclusion:**
- The diamagnetic complexes are:
- K₄[Fe(CN)₆] (I)
- K₃[Co(CN)₆] (III)
- K₂[Ni(CN)₄] (IV)
- The only paramagnetic complex is:
- K₃[Cr(CN)₆] (II)

### Final Answer:
The diamagnetic complexes are **(I), (III), and (IV)**.