Question

The number of unpaired electrons calculated in $[Co(N{H}_3{)}_6{]}^{3+}$ 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)

In both $[Co(N{H}_3{)}_6{]}^{3+}$ and $[Co{F}_6{]}^{3-}Co$ is present as $C{o}^{3+}$ Thus, the electronic configuration of $Co$ is
${}_{27}Co=[Ar]3d^7,4s^2$
${}_{27}C{o}^{3+}=[Ar]3d^6,4s^0$
In case of $[Co(N{H}_3{)}_6{]}^{3+}N{H}_3$ is a strong field ligand, so pairing of electrons in $3d$-orbital takes place.
${}_{27}C{o}^{3+}=[Ar]3d^6,4s^0$

In $[Co{F}_6{]}^{3-},F$ is a weak field ligand, thus does not cause pairing. Hence,
${}_{27}C{o}^{3+}=[Ar]3d^6,4s^0$