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Which of these ions has the smallest number of unpaired electrons?

1) Cr2
2) Fe₃⁺
3) Co2
4) V2
5) Ni₂⁺

User Shovalt
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1 Answer

3 votes

Final answer:

Ni²⁺ has the smallest number of unpaired electrons, with a d⁸ configuration leading to two unpaired electrons in an octahedral coordination complex. so, option 5 is the correct answer.

Step-by-step explanation:

To determine which of the given ions has the smallest number of unpaired electrons, we consider the electronic configurations and the effect of their ligands in an octahedral field:

  1. Cr2+ has a d4 configuration; this would typically have several unpaired electrons.
  2. Fe3+ is d5; as a high spin complex with water as a ligand, it has five unpaired electrons.
  3. Co2+ is d7; with water as a weak field ligand, it could have three unpaired electrons (in a high spin configuration).
  4. V2+ is d3; thus, it would have three unpaired electrons.
  5. Ni2+ is d8; with octahedral coordination, it typically has two unpaired electrons.

Among these options, Ni2+ has the smallest number of unpaired electrons with a total of two.

The ion with the smallest number of unpaired electrons is Fe³+. The question provides a list of ions, and we need to determine which one has the fewest unpaired electrons. Since Fe³+ has a d³ configuration, all three of its electrons will be unpaired. Therefore, it has the smallest number of unpaired electrons.

User Govinda
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