Question 5 of 34intermediate🔍 AnalyzeShort Answer2 marks

[NiCl4]2- is paramagnetic while [Ni(CO)4] is diamagnetic though both are tetrahedral. Why?

Correct Answer

Both [NiCl4]2- and [Ni(CO)4] have tetrahedral geometry with sp3sp^{3} hybridisation. However, they differ in their magnetic properties due to the difference in oxidation state of nickel and the presence of unpaired electrons.

In [NiCl4]2-, nickel is in +2 oxidation state with electronic configuration 3d83d^{8}. The hybridisation is sp3sp^{3} and the compound is paramagnetic since it contains two unpaired electrons in the 3d orbitals.

In [Ni(CO)4], nickel is in zero oxidation state and contains no unpaired electron. Hence, the compound is diamagnetic despite having the same tetrahedral geometry.

Exercise: Intext Questions | Q: 5.6 | (Page Page 18)
For More Understanding

Explanation

The textbook explains that [NiCl4]2- is a high spin complex with Ni in +2 oxidation state (3d83d^{8} configuration), using sp3sp^{3} hybridisation with two unpaired electrons, making it paramagnetic. In contrast, [Ni(CO)4] has Ni in zero oxidation state with no unpaired electrons, making it diamagnetic despite the same tetrahedral geometry.

Solution Steps

  1. Step 1: Identify the oxidation state of Ni - +2 in [NiCl4]2- and 0 in [Ni(CO)4]

  2. Step 2: Determine the electronic configuration - 3d83d^{8} for Ni2Ni^{2}+ in [NiCl4]2-

  3. Step 3: Identify the hybridisation - sp3sp^{3} for both tetrahedral complexes

  4. Step 4: Count unpaired electrons - 2 unpaired electrons in [NiCl4]2-, none in [Ni(CO)4]

  5. Step 5: Conclude magnetic nature - paramagnetic for [NiCl4]2-, diamagnetic for [Ni(CO)4]

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