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1- Draw the molecular orbital diagram of transition metal ion in high-spin Mn(H2O)4(OH)2 complex, also determine...

1- Draw the molecular orbital diagram of transition metal ion in high-spin Mn(H2O)4(OH)2 complex, also determine the number of unpaired electron.
2- Draw the molecular orbital diagram of transition metal ion in low-spin [Cr(en)2(NH3)2].Cl2 complex, also determine the number of unpaired electron.
3- Draw the molecular orbital diagram of transition metal ion in high-spin K.[Mn(CO)3(OH)3] complex, also determine the number of unpaired electron.

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Answer #1

Mn has the electronic configuration

[Ar] 3d5 4s2

In the complex Mn(H2O)4(OH)2

Mn is in +2 state

So the electronic configuration is [Ar] 3d5 in high spin the metal has the configuration t2g has 3 electrons and eg has 2 electrons.

There are 4+2 = 6 ligands which donate 12 electrons

The MO diagram is as shown the final complex has one unpaired electron

2) In the complex [Cr(en)2(NH3)2].Cl2

Cr is in +2 state

electronic configuration of Cr is [Ar] 3d5 4s1

In +2 state the electronic configuration is [Ar] 3d4

Since it is low spin all the 4 electrons are on t2g and there are no electrons in eg

There are 2 en ligands which are 4 electron donors each and 2 NH3 ligands which are 2 electron donors each so the ligands donate 12 electrons

MO diagram is as shown the final complex has two unpaired electrons

LLJ Metal Orbitals Ligand Orbitals

3)

Mn has the electronic configuration

[Ar] 3d5 4s2

In the complex K.[Mn(CO)3(OH)3]

Mn is in +2 state

So the electronic configuration is [Ar] 3d5 in high spin the metal has the configuration t2g has 3 electrons and eg has 2 electrons.

There are 4+2 = 6 ligands which donate 12 electrons

The MO diagram is as shown the final complex has one unpaired electron

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