Question

The complex ion [cobr6]3− is a high-spin complex and absorbs light at a wavelength greater than 7. 70×102 nm. how many unpaired electrons will be present in the complex?

Answer 1

There are four unpaired electrons present in the complex ion  $[CoBr_6]^3^-$  

What is Crystal Field Theory (CFT)?

• CFT explains the breaking of orbital degeneracy in the transition metal complexes because of the presence of the ligands.
• It describes the strength of the metal-ligand bonds
• There are two types of ligands - weak field ligands and strong field ligands
• Weak field ligands form high spin complexes and strong field ligands form low spin complexes
• High spin complexes have more number of unpaired electrons
• Low spin complexes have less or zero unpaired electrons

A complex ion  $[CoBr_6]^3^-$  is an octahedral complex. Co is the central atom to which 6 Br atoms are attached. Br is a weak field ligand and forms a high spin complex.

Electronic configuration of Co -  $[Ar] 3d^7 4s^2$

Electronic configuration of $Co^3^+ \rightarrow [Ar] 3d^6$

Electronic configuration of $[CoBr_6]^3^-$ - $t_2_g^4 e_g^2$

Hence, there are 4 unpaired electrons present in the complex ion  $[CoBr_6]^3^-$ .

Learn more about Crystal Field Theory (CFT):

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