`NiCl_(2) overset(KCN)to "complex"A`
`NiCl_(2) underset("excess") overset(KCl)to "complex"B`
A & B complexes have the co-ordination number 4.
What are the magnetic nature of 'A' & 'B' ?

Options

A. Both diamagnetic
B. A' is diamagnetic & 'B' is paramagnetic with one unpaired electrons.
C. A' is diamagnetic & 'B' is paramagnetic with two unpaired electrons.
D. Both are paragmagnetic

Answer

To determine the magnetic nature of complexes A and B formed from the reaction of NiCl₂ with KCN and KCl respectively, we will follow these steps:

### Step 1: Identify the oxidation state of Nickel in both complexes.
- For Complex A, NiCl₂ reacts with KCN. The oxidation state of Ni can be calculated as follows:
- Let the oxidation state of Ni be \( x \).
- The overall charge of the complex is -2 (since it has 2 CN⁻ ligands).
- The equation can be set up as:
\[ x + 2(-1) = -2 \quad \Rightarrow \quad x - 2 = -2 \quad \Rightarrow \quad x = +2 \]
- Therefore, the oxidation state of Ni in Complex A is +2.

- For Complex B, NiCl₂ reacts with excess KCl. The oxidation state of Ni is the same as in Complex A:
- Using the same method:
\[ x + 2(-1) = 0 \quad \Rightarrow \quad x - 2 = 0 \quad \Rightarrow \quad x = +2 \]
- Therefore, the oxidation state of Ni in Complex B is also +2.

### Step 2: Determine the electronic configuration of Ni²⁺.
- The atomic number of Nickel (Ni) is 28. Its electronic configuration is:
\[ \text{Ni: } [Ar] 3d^8 4s^2 \]
- For Ni²⁺, we remove two electrons (the 4s electrons):
\[ \text{Ni}^{2+}: [Ar] 3d^8 \]

### Step 3: Analyze the ligand field strength and electron pairing.
- **Complex A (Ni(CN)₄²⁻)**:
- CN⁻ is a strong field ligand, which causes pairing of electrons in the 3d orbitals.
- The 3d orbitals will be filled as follows:
\[ 3d: \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \uparrow\downarrow \]
- All electrons are paired, hence Complex A is **diamagnetic**.

- **Complex B (NiCl₄²⁻)**:
- Cl⁻ is a weak field ligand and does not cause pairing of electrons.
- The 3d orbitals will be filled as follows:
\[ 3d: \uparrow \uparrow \uparrow \uparrow \uparrow \quad (8 \text{ electrons}) \]
- There are 2 unpaired electrons, hence Complex B is **paramagnetic**.

### Conclusion:
- Complex A is **diamagnetic** (no unpaired electrons).
- Complex B is **paramagnetic** (2 unpaired electrons).

### Final Answer:
- **Complex A**: Diamagnetic
- **Complex B**: Paramagnetic