Question

The number of unpaired electrons calculated in $[Co (NH_3)_6]^{3+}$ and $Co (F_6)Z^{3-}$ are

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

Answer 1

Answer: (D)

0 and 4

Answer 2

In both $[Co(NH_3 )_6 ]^{3+}$ and $[CoF_6 ]^{3-} Co$ is present as $Co^{3+}$ Thus, the electronic configuration of $Co$ is
$_{27}Co = [Ar] 3d^7 ,4s^2$
$_{27} Co^{3+} = [Ar] 3d^6 , 4s^0$
In case of $[Co(NH_3 )_6 ]^{3+} NH_ 3$ is a strong field ligand, so pairing of electrons in $3d$-orbital takes place.
$_{27} Co^{3+} = [Ar] 3d^6 , 4s^0$

In $[CoF_6 ]^{3-}, F$ is a weak field ligand, thus does not cause pairing. Hence,
$_{27} Co^{3+} = [Ar] 3d^6 , 4s^0$