Question: 71. Given below are two statements :
Statement I : Both \left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{+3} and \left[\mathrm{CoF}_{6}\right]^{3-} complexes are octahedral but differ in their magnetic behaviour.
Statement II : \left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+} is diamagnetic whereas \left[\mathrm{CoF}_{6}\right]^{3-} is paramagnetic.
In the light of the above statements, choose the correct answer from the options given below :
(1) Both statement I and Statement II are true
(2) Both Statement I and Statement II are false
(3) Statement I is true but Statement II is false
(4) Statement I is false but Statement II is true
Answer: Option (1)
Explanation:
In both complexes, cobalt is in the +3 oxidation state,
so the electronic configuration of \mathrm{Co}^{3+} is 3d^{6}.
Both \left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+} and
\left[\mathrm{CoF}_{6}\right]^{3-} have coordination number 6,
hence both are octahedral complexes.
\mathrm{NH}_{3} is a strong field ligand,
which causes pairing of electrons in the 3d orbitals,
resulting in a low-spin configuration with no unpaired electrons.
Therefore, \left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}\right]^{3+} is diamagnetic.
\mathrm{F}^{-} is a weak field ligand,
which does not cause pairing of electrons, resulting in a high-spin configuration with unpaired electrons.
Hence, \left[\mathrm{CoF}_{6}\right]^{3-} is paramagnetic.
Thus, both Statement I and Statement II are true.