Question

In \[{\left[ {Co{F_6}} \right]^{ - 3}}\], \[C{o^{3 + }}\;\]uses outer d orbits (4d) in sp3d2 hybridization. The number of unpaired electrons present in complex ion is:
A: 0
B: 4
C: 2
D: 3

Answer 1

Hint: Hint: Orbital hybridisation or simply hybridization is the concept in chemistry which states that mixing of the atomic orbitals into the new hybrid orbitals (which possess different shapes, energies, etc., in comparison to component atomic orbitals) is suitable for the electron pairing in order to form chemical bonds in the valence bond theory. Hybridization is generally used to explain molecular geometry of organic compounds.

Step by step answer: The atomic number of cobalt i.e. \[Co\] is 27. Thus, \[Co\] has valence shell electronic configuration of 3d7 4s2. Whereas in \[{\left[ {Co{F_6}} \right]^{ - 3}}\], \[Co\]possess +3 oxidation state.

Now \[C{o^{3 + }}\;\]possess valence shell electronic configuration of 3d64s0.

As we know, ${F^ - }$belongs to the category of weak field ligands (Ligands causing a transition metal to possess a small crystal field splitting, leading to high spin). Therefore, ${F^ - }$ is having 2 paired electrons and remaining 4 unpaired electrons in the 3d orbitals.
In \[{\left[ {Co{F_6}} \right]^{ - 3}}\], \[C{o^{3 + }}\;\]employs outer d orbitals i.e. 4d in sp3d2 hybridization.
As a result, the number of unpaired electrons present in complex ions is 4.

Hence, the correct answer is Option B i.e. 4.

Note: The sequence of hybrids according to energy level is sp < sp2 < sp3. If p character is higher, it means that the energy is more, thus indicating that the electrophilicity is higher and moreover, its affinity for reaction is also higher. It should also be noted that hybrids (sp,sp2,sp3) lead to the formation of σ bonds and pure-breeds on the other hand, lead to the formation of π bonds.