Question

The compound ${ CuSO }_{ 4 }{ .5H }_{ 2 }{ O }$ is blue in color while ${ CuSO }_{ 4 }$ is colorless due to:

A) Presence of strong field ligand in ${ CuSO }_{ 4 }{ .5H }_{ 2 }{ O }$.
B) Due to the absence of water (ligand), d-d transitions are not possible in ${ CuSO }_{ 4 }$.
C) Anhydrous undergoes d-d transitions due to crystal field splitting.
D) Color is lost due to unpaired electrons.

Answer 1

Hint: The crystal field stabilization energy (CFSE) is the energy by which the complex is stabilized relative to the free metal atom where there is no splitting of d orbitals.

Complete answer:
d-orbital split into two sets of orbital of different energy(according to crystal field theory ) ${ d }_{ xy }{ ,d }_{ tz }{ ,d }_{ zx }{ ,d }_{ x^{ 2 }{ -y }^{ 2 } }{ ,d }_{ z^{ 2 } }$, so the movement of electron of lower energy set of same energy orbital to higher energy orbital called d-d transition in the case of tetrahedral compounds the trend of energy is reversed.

${ CuSO }_{ 4 }{ .5H }_{ 2 }{ O }$ form the coordinate compound. The geometry of the complex is known to be distorted octahedral. According to crystal field theory, the splitting of d -orbital takes place in the presence of ligands (here water). the two orbitals raise in energy as compared to the other 3d orbitals. The high energy orbitals are ${ e }_{ g }$ and low energy orbitals ${ t }_{ 2 }{ g }$.
The Cu is in the ${ +2 }$ oxidation state. It has ${ 9 }$ electrons in d orbitals. Due to crystal field splitting there is a little energy gap between ${ e }_{ g }$ and ${ t }_{ 2 }{ g }$ orbitals. so the electrons absorb the visible light and excite to eg level and emit blue when comes back to the ground state. That's why ${ CuSO }_{ 4 }{ .5H }_{ 2 }{ O }$ appears blue in color.
In ${ CuSO }_{ 4 }$, no water molecule is present which acts as a ligand, hence no d-d transitions take place and it is colorless.

So, the correct option is B i.e, Due to the absence of water (ligand), d-d transitions are not possible in ${ CuSO }_{ 4 }$.

Note: The possibility to make a mistake is that you may choose option C. But in ${ CuSO }_{ 4 }$, the d-d transitions are not possible as there is no crystal field splitting occurs.