The octahedral diamagnetic low spin complex among the following is

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  1. [NiCl4]2-
  2. [CoCl6]3-
  3. [CoF6]3-
  4.  [Co(NH3)6)3+

Answer (Detailed Solution Below)

Option 4 :  [Co(NH3)6)3+
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CONCEPT:

Diamagnetism and Low Spin Complexes

  • Diamagnetic complexes have all paired electrons, resulting in no net magnetic moment.
  • In octahedral complexes, the electronic configuration can be influenced by the nature of ligands, leading to high spin or low spin configurations.
  • Strong field ligands such as CN- and NH3 cause pairing of electrons, resulting in low spin complexes.

EXPLANATION:

  • [NiCl4]2-:
    • Nickel (Ni2+) has a d8 configuration.
    • Chloride (Cl-) is a weak field ligand, leading to a high spin complex.
    • High spin d8 configuration in a tetrahedral field does not result in a diamagnetic complex.
  • [CoCl6]3-:
    • Cobalt (Co3+) has a d6 configuration.
    • Chloride (Cl-) is a weak field ligand, leading to a high spin complex.
    • High spin d6 configuration in an octahedral field does not result in a diamagnetic complex.
  • [CoF6]3-:
    • Cobalt (Co3+) has a d6 configuration.
    • Fluoride (F-) is a weak field ligand, leading to a high spin complex.
    • High spin d6 configuration in an octahedral field does not result in a diamagnetic complex.
  • [Co(NH3)6]3+:

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    • Cobalt (Co3+) has a d6 configuration.
    • Ammonia (NH3) is a strong field ligand, leading to a low spin complex.
    • Low spin d6 configuration in an octahedral field results in all paired electrons, making it diamagnetic.

Therefore, the octahedral diamagnetic low spin complex among the given options is [Co(NH3)6]3+.

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