Question

For each of the following octahedral complexes:

a) determine the number of 'd' electrons on the metal atom.

b) determine the number of unpaired electrons on the metal atom.

c) calculate the magnetic moment (spin only value) in units of Bohr magnetons.

d) calculate the ligand field stabilization energy, LFSE, in units of delta0

1)[Mn(H2O)6]2+

2)[CrCl6]4-

3)[NiF6]3-

4)[Co(CN)6]4-

5)[Mn(CO)6]+

Answer 1

1)[Mn(H2O)6]2+

Mn atomic number is 25. In the given complex , Mn exists as Mn+2

Mn+2 , has the electronic configuration: [Ar]4s⁰ 3d⁵

a)Hence the metal has 5 d-electrons.

b) In this complex the metal has 5 unpaired d-electrons.

c) Magnetic moment = square root of   n (n+2) BM

where n= no.of unpaired electrons present in the metal ion.

Magnetic moment = square root of   n (n+2) BM

                           = square root of 5(5+2) = square root of 35

                           = 5.91BM

d) the ligand field stabilization energy, LFSE

As the ligands are weak field ligands, hence the first 3 electrons occupy the t2g orbitals and the remaining 2 electrons occupy the eg orbitals according to Hund's rule and Pauli's exclusion principle.

LFSE = 0.4t2g - 0.6 eg ₀

          = 0.4 x 3 - 0.6x2 ₀

             ₌ 0.0 ₀

2)[CrCl6]4-

Cr atomic number is 24. In the given complex , Cr exists as Cr+2

Cr+2 , has the electronic configuration: [Ar]4s⁰ 3d⁴

a)Hence the metal has 4 d-electrons.

b) In this complex the metal has 4 unpaired d-electrons.

c) Magnetic moment = square root of   n (n+2) BM

where n= no.of unpaired electrons present in the metal ion.

Magnetic moment = square root of   n (n+2) BM

                           = square root of 4(4+2) = square root of 24

                           = 4.89BM

d) the ligand field stabilization energy, LFSE

As the ligands are weak field ligands, hence the first 3 electrons occupy the t2g orbitals and the remaining 1 electrons occupy the eg orbital according to Hund's rule and Pauli's exclusion principle.

LFSE = 0.4t2g - 0.6 eg ₀

          = 0.4 x 3 - 0.6x1 ₀

             ₌ 0.6 ₀

3) [NiF6]3-

Ni atomic number is 28. In the given complex , Ni exists as Ni+3

Ni+3 , has the electronic configuration: [Ar]4s⁰ 3d⁷

a)Hence the metal has 7 d-electrons.

b) In this complex the metal has 3 unpaired d-electrons.

c) Magnetic moment = square root of   n (n+2) BM

where n= no.of unpaired electrons present in the metal ion.

Magnetic moment = square root of   n (n+2) BM

                           = square root of 3(3+2) = square root of 15

                           = 3.87BM

d) the ligand field stabilization energy, LFSE

As the ligands are weak field ligands, hence the first 5 electrons occupy the t2g orbitals and the remaining 2 electrons occupy the eg orbitals according to Hund's rule and Pauli's exclusion principle.

LFSE = 0.4t2g - 0.6 eg ₀

          = 0.4 x 5 - 0.6x2 ₀

             ₌ 0.8 ₀

4) [Co(CN)6]4-

Co atomic number is 27. In the given complex , Co exists as Co+2

Co+2 , has the electronic configuration: [Ar]4s⁰ 3d⁷

a)Hence the metal has 7 d-electrons.

b) In this complex the metal has 1 unpaired d-electron.

c) Magnetic moment = square root of   n (n+2) BM

where n= no.of unpaired electrons present in the metal ion.

Magnetic moment = square root of   n (n+2) BM

                           = square root of 1(1+2) = square root of 35

                           = 1.732BM

d) the ligand field stabilization energy, LFSE

As the ligands are strong field ligands, hence the first 6 electrons occupy the t2g orbitals and the remaining 1 electrons occupy the eg orbitals according to Hund's rule and Pauli's exclusion principle.

LFSE = 0.4t2g - 0.6 eg ₀

          = 0.4 x 6 - 0.6x1 ₀

             ₌ 1.8 ₀

5) [Mn(CO)6]+

Mn atomic number is 25. In the given complex , Mn exists as Mn+

Mn+ , has the electronic configuration: [Ar]4s¹ 3d⁵

a)Hence the metal has 5 d-electrons.

b) In this complex the metal has 6 unpaired electrons.

c) Magnetic moment = square root of   n (n+2) BM

where n= no.of unpaired electrons present in the metal ion.

Magnetic moment = square root of   n (n+2) BM

                           = square root of 6(6+2) = square root of 48

                           = 6.92BM

d) the ligand field stabilization energy, LFSE

As the ligands are strong field ligands, hence the first 6 electrons occupy the t2g orbitals and no electrons occupy the eg orbitals according to Hund's rule and Pauli's exclusion principle.

LFSE = 0.4t2g - 0.6 eg ₀

          = 0.4 x 6 - 0.6x0 ₀

             ₌ 2.4 ₀