Question

Which among the following is diamagnetic?

• A. [Ni(CN)₄]²⁻
• B. [Co(F₆)]³⁻
• C. [NiCl₄]²⁻
• D. [Fe(CN)₆]³⁻

Hint:

For a complex to be diamagnetic, it must have all paired electrons, which is typically seen with strong field ligands like cyanide (CN⁻).

Answer 1

The Correct Option is A

To determine which of the given complexes is diamagnetic, we must analyze their electronic configurations and possible magnetic properties.

Diamagnetic substances have all their electrons paired, while paramagnetic substances have one or more unpaired electrons.

Let's examine each option:

• [Ni(CN)₄]²⁻:
  Nickel in this complex is in the +2 oxidation state, i.e., Ni²⁺. The electronic configuration of Ni is [Ar] 3d⁸ 4s², so Ni²⁺ is [Ar] 3d⁸. Cyanide (CN⁻) is a strong field ligand, causing pairing of the electrons. Thus, all electrons get paired up, making the complex diamagnetic.
• [Co(F₆)]³⁻:
  Cobalt here is in the +3 oxidation state, i.e., Co³⁺. The electronic configuration of Co is [Ar] 3d⁷ 4s², so Co³⁺ is [Ar] 3d⁶. Fluoride (F⁻) is a weak field ligand, resulting in some unpaired electrons, making the complex paramagnetic.
• [NiCl₄]²⁻:
  Nickel in this complex is again in the +2 oxidation state, similar to the first option. Chloride (Cl⁻) is a weak field ligand, so not all electrons pair up, leaving unpaired electrons and resulting in a paramagnetic complex.
• [Fe(CN)₆]³⁻:
  Iron here is in the +3 oxidation state, i.e., Fe³⁺. The electronic configuration is [Ar] 3d⁵. Cyanide causes pairing, but Fe³⁺ has an odd d electron, which remains unpaired even in strong fields, making it typically paramagnetic.

Thus, the complex [Ni(CN)₄]²⁻ is diamagnetic, as it has all electrons paired.

Answer 2

To determine which of the following complexes is diamagnetic, we need to consider their electron configurations and the possibility of unpaired electrons. Diamagnetic substances have all electrons paired, and thus they do not have any net magnetic moment. 
- In [Ni(CN)₄]²⁻, Ni²⁺ has a \(3d^8\) electron configuration. The cyanide ion (CN⁻) is a strong field ligand, which causes pairing of electrons in the lower energy orbitals. As a result, all electrons are paired in this complex, making it diamagnetic. 
- In [Co(F₆)]³⁻, Co³⁺ has a \(3d^6\) configuration, which can have unpaired electrons depending on the ligands. Since fluoride ions are weak field ligands, they do not cause pairing of electrons. 
Thus, this complex is paramagnetic. 
- In [NiCl₄]²⁻, Ni²⁺ has a \(3d^8\) configuration, but chloride ions are weak field ligands, meaning the electrons remain unpaired, making the complex paramagnetic. - In [Fe(CN)₆]³⁻, Fe³⁺ has a \(3d^5\) configuration, which leads to unpaired electrons, making this complex paramagnetic. 

Thus, the correct answer is [Ni(CN)₄]²⁻, which is diamagnetic.