Coordination Compounds
8.0 Magnetic Moment
8.0 Magnetic Moment
Low spin complexes are generally diamagnetic because of pairing of electron, where as high spin complexes are usually paramagnetic because of presence of unpaired electron.
1. Larger the number of unparied electrons stronger will be the paramagnetism
2. However megnetic behaviour of the complex can be confirmed from magnetic moment measurement.
$${\text{Magnetic moment}} = \sqrt {n(n + 2)} \quad \left( {n:{\text{number of unpaired electrons}}} \right)$$
Colour of Co-ordination compounds: Colour of a complex is not because of the light absorbed but it is due to light reflected (complementary colour)
- In coordination complex energy differences ($\vartriangle $) between two sets of d-orbitals is small. Radiations of appropriate frequency absorbed from visible region can cause excitation of $d$-electrons from lower energy orbital to higher energy orbitals. Remaining light is transmitted and compound appears coloured.
- If there are no $d$-electrons present or completely filled $g$-orbitals are present then there is no $d-d$ transition possible and complex appears colourless.
Example: ${\left[ {Ti{{({H_2}O)}_6}} \right]^{ + 3}}$ is coloured whereas ${\left[ {Sc{{({H_2}O)}_6}} \right]^{3 + }}$ is colourless why?
Explanation: In ${\left[ {Ti{{({H_2}O)}_6}} \right]^{ + 3}}$ the titanium ion is present in $T{i^{3 + }}$ from with $3{d^1},4{s^0}$ configuration. The single unpaired electron of $3{d^1}$ orbital makes the compound to show the colour. ${\left[ {Sc{{({H_2}O)}_6}} \right]^{3 + }}$ scandium is present as $S{c^{3 + }}$ state with $3{d^0}, 4{s^0}$ configuration. Since no electron is present in $3d$ and $4s$ orbitals, it remain colourless.
