Question Details

Given below are two statements :
Statement I: Both [Co(NH₃)₆]³⁺ and [CoF₆]^3– complexes are octahedral but differ in their magnetic behaviour.
Statement II: [Co(NH₃)₆]³⁺is diamagnetic whereas [CoF₆]^3–is paramagnetic.
In the light of the above statements, choose the correct answer from the options given below:

Options

Both Statement I and Statement II are false

Both Statement I and Statement II are true

Statement I is true but Statement II is false

Statement I is false but Statement II is true

Show Answer

Correct Answer :

Option B

Both Statement I and Statement II are true

Solution :

The correct option is: Both Statement I and Statement II are true.

Let us analyze both statements step-by-step using Valence Bond Theory (VBT) and Crystal Field Theory (CFT).

Analysis of Statement I:
Both complexes have a coordination number of 6, which means both possess an octahedral geometry.
Let us determine the oxidation state and electronic configuration of Cobalt (Co) in both complexes:
The atomic number of Cobalt (Co) is 27, and its ground-state electronic configuration is [Ar] 3d⁷ 4s².
In [Co(NH₃)₆]³⁺, the cobalt ion is in the +3 oxidation state (Co³⁺). The electronic configuration of Co³⁺ is [Ar] 3d⁶.
In [CoF₆]^3-, the cobalt ion is also in the +3 oxidation state (Co³⁺), with the same electronic configuration: [Ar] 3d⁶.
Although both have octahedral geometry, they differ in their magnetic behavior because of the nature of the ligands attached to the central metal ion. Therefore, Statement I is true.

Analysis of Statement II:
Let us look at the nature of the ligands in each complex:

1. For [Co(NH₃)₆]³⁺:
Ammonia (NH₃) acts as a strong-field ligand. It causes the pairing of the 3d electrons in the Co³⁺ ion.
The six 3d electrons of Co³⁺ pair up completely in the three t_2g orbitals:
$t_{2 g} ^{6} e_{g}^{0}$
Since there are no unpaired electrons (number of unpaired electrons, n = 0), [Co(NH₃)₆]³⁺ is diamagnetic.

2. For [CoF₆]^3-:
Fluoride ion (F^-) is a weak-field ligand. It cannot pair up the electrons in the 3d orbitals.
Therefore, the distribution of the six 3d electrons follows Hund's rule, leaving four unpaired electrons:
$t_{2 g} ^{4} e_{g}^{2}$
Since it has 4 unpaired electrons (n = 4), [CoF₆]^3- is paramagnetic.