Question

Which of the following complexes are diamagnetic :

• A. $[AuC{l}_4{]}^{{-}^{}}$
• B. $\left[Co\right(H_{2}O{)}_6{]}^{3+}$
• C. $[Co{F}_6{]}^{3-}$
• D. $\left[Co\right(CO{)}_4{]}^{-}$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $[AuC{l}_4{]}^{{-}^{}}$
B $\left[Co\right(H_{2}O{)}_6{]}^{3+}$
D $\left[Co\right(CO{)}_4{]}^{-}$

$\left(a\right)[AuC{l}_4{]}^{-},A{u}^{3+}:[Xe]4f^{14}5d^8$

In a square planar complex, the four ligands are only in the $xy$ plane so any orbital in the $xy$ plane has a higher energy level. The absence of ligands along the z-axis relative to an octahedral field stabilizes the $d_{z^2},d_{xz},and{d}_{yz}$ levels, and leaves the $d_{x^2-y^2}$ level the most destabilized.

The $d_{x^2-y^2}$ level is at such a high level that it remains unoccupied in almost all $d^8$ complexes
As the $4C{l}^{-}$ ions, which are weak-field ligands, approach the $A{u}^{3+}$ ion in preparation for bonding, the
5d electrons pair up since pairing energy is low compared to crystal field splitting energy.
Crystal field spilting energy increase in order,
$3d<4d<5d$
$A{u}^{3+}$ in $[AuC{l}_4{]}^{-}$ :
No unpaired electron is there , so (a) is diamagnetic.

(b) $\left[Co\right(H_{2}O{)}_6{]}^{3+}$
Exception:
$H_{2}O$ even though weak ligand , it will cause pairing when $C{o}^{3+}$ is present.
$Co:\left[Ar\right]4s^{2}3d^7$
$C{o}^{3+}:\left[Ar\right]3d^6$

All electrons are paired. So, diamagentic
(c) $[Co{F}_6{]}^{3-}$
$C{o}^{3+}:\left[Ar\right]3d^6$
No exception with fluorine , so:


Unpaired electrons are there, so paramagnetic.
(d): $\left[Co\right(CO{)}_4{]}^{-}$
Electronic config:
$Co:\left[Ar\right]4s^{2}3d^7$
$C{o}^{-}:\left[Ar\right]4s^{2}3d^8$
CO is a strong ligand which causes pairing of electrons.
$C{o}^{-}:\left[Ar\right]4s^{0}3d^{10}$

No unpaired electron, so diamagnetic.
Correct answers are (a),(b) and (d)