Question

${ CuSO }_{ 4 }$ is paramagnetic while ${ ZnSO }_{ 4 }$ is diamagnetic because:
a.) ${ Cu }^{ 2+ }$ ion has ${ 3d }^{ 9 }$ configuration while ${ Zn }^{ 2+ }$ ion has ${ 3d }^{ 10 }$ configuration.
b.) ${ Cu }^{ 2+ }$ ion has configuration while ${ Zn }^{ 2+ }$ ion has configuration.
c.) ${ Cu }^{ 2+ }$ has half-filled orbitals while ${ Zn }^{ 2+ }$ has fully filled orbitals.
d.) ${ CuSO }_{ 4 }$ is blue in color while ${ ZnSO }_{ 4 }$ is white.

Answer 1

Paramagnetic substances are very weakly attracted by the poles of a magnet, but without retaining any permanent magnetism. Diamagnetic materials are repulsed by an attractive field; an applied attractive field makes an induced field in them the other way, causing a repulsive force.

Complete answer:
A. In ${ CuSO }_{ 4 }$, the electronic configuration of Cu is $\left[ Ar \right] { 3d }^{ 10 }{ 4s }^{ 1 }$
${ Cu }^{ 2+ }$ : The electronic configuration is ${ \left[ Ar \right] 3d }^{ 9 }$
${ ZnSO }_{ 4 }$, the electronic configuration of Zn is $\left[ Ar \right] { 3d }^{ 10 }{ 4s }^{ 2 }$
${ Zn }^{ 2+ }$ : The electronic configuration is $\left[ Ar \right] { 3d }^{ 10 }$
As we know that a compound to be paramagnetic it must have an unpaired electron and in ${ Cu }^{ 2+ }$, an unpaired electron is present while ${ Zn }^{ 2+ }$ has no unpaired electron, so it is diamagnetic. Hence, this statement is correct.

B. ${ Cu }^{ 2+ }$ ion has ${ 3d }^{ 5 }$ configuration while ${ Zn }^{ 2+ }$ ion has ${ 3d }^{ 6 }$ configuration. This statement is incorrect as ${ Cu }^{ 2+ }$ ion has ${ 3d }^{ 9 }$ configuration while ${ Zn }^{ 2+ }$ ion has ${ 3d }^{ 10 }$ configuration.
C. In copper sulfate, the blue color is because light energy is used to promote or excite it combined with different things like the sulfate or carbonate ions and so on. But they will come to low energy levels by emitting light and the blue color is absorbed by the molecule and hence it is colored while in ${ ZnSO }_{ 4 }$ it has no unpaired electrons so energy is not able to excite the electrons. That’s why it is colorless.

So, the correct answer is “Option A”.

Note: The possibility to make a mistake is that you may choose option D. But as you can see that the color of the compound depends on the unpaired electrons which when excited and fall in the region give color. But the paramagnetic or diamagnetic depends on the presence or absence of unpaired electrons.