Question

The number of unpaired electrons calculated in ${\left[Co{\left(N{H}_3\right)}_6\right]}^{3+}$ and ${\left[Co{F}_6\right]}^{3-}$ are:

• A. 4 and 4
• B. 0 and 2
• C. 2 and 4
• D. 0 and 4

Answer 1

The correct option is D 0 and 4

In ${\left[Co{\left(N{H}_3\right)}_6\right]}^{3+}$
$C{o}^{+3},d^6$ configuration with low spin complex so zero unpaired electron (as ammonia is strong ligand).
And in ${\left[Co{F}_6\right]}^{3-}$,
$C{o}^{+3},d^6$ configuration with high spin complex so 4 unpaired electron (as fluoride is weak ligand).