To determine which of the given complex ions is diamagnetic in nature, we will follow these steps:
### Step 1: Identify the Complex Ion
Let's take the example of the complex ion Ni(CN)₄²⁻ as mentioned in the video transcript.
### Step 2: Calculate the Oxidation State of the Metal
Assume the oxidation state of nickel (Ni) is \( X \). The charge of the cyanide ion (CN) is -1, and since there are four CN ligands, the total contribution from the ligands is -4. The overall charge of the complex ion is -2.
Setting up the equation:
\[ X + 4(-1) = -2 \]
\[ X - 4 = -2 \]
\[ X = +2 \]
Thus, the oxidation state of nickel in this complex is +2.
### Step 3: Determine the Electronic Configuration
Nickel has an atomic number of 28, and its ground state electronic configuration is:
\[ \text{Ni: } [\text{Ar}] 4s^2 3d^8 \]
Since the oxidation state is +2, we remove two electrons from the 4s orbital:
\[ \text{Ni}^{2+}: [\text{Ar}] 3d^8 \]
### Step 4: Analyze the Electron Configuration
For Ni²⁺, we have 8 electrons in the 3d subshell. The distribution of these electrons in the d-orbitals is as follows:
- The 3d orbitals can hold a maximum of 10 electrons.
- The 3d configuration for 8 electrons will be represented as:
\[ \uparrow\downarrow \, \uparrow\downarrow \, \uparrow\downarrow \, \uparrow \, \uparrow \]
This shows that there are 2 unpaired electrons.
### Step 5: Consider the Ligand Field Strength
Cyanide (CN⁻) is a strong field ligand, which means it will cause pairing of electrons in the d-orbitals. Therefore, the 3d electrons will pair up, leading to the following configuration:
\[ \uparrow\downarrow \, \uparrow\downarrow \, \uparrow\downarrow \, \uparrow\downarrow \]
This results in all electrons being paired.
### Step 6: Determine the Magnetic Nature
Since all electrons are paired in the 3d orbitals, the complex ion Ni(CN)₄²⁻ is **diamagnetic**.
### Conclusion
Thus, the complex ion Ni(CN)₄²⁻ is diamagnetic in nature.
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