Q. 46

# <span lang="EN-US

(i)

• In [CoF6]3–, the central metal atom is cobalt Co and oxidation state of Co in hereis: x=-6 times (-1)-3=+3

Hence, Co3+has a d6 configuration (1s22s22p63s23p63d6, while losing 1 3d and 2 4s electrons.) So the orbital energy level diagram is:

And the number of unpaired electrons=4.

Therefore, the magnetic moment of Co= = 4.9 BM. (BM stands for Bohr Magneton)

• For [Co(H2O)6]2+ the central metal atom is also Co and the oxidation state of Co here is x=2+; as aqua molecules are neutral.

Hence Co2+ has a d7 configuration (1s22s22p63s23p63d7,while losing only the 2 4s electrons.) So the orbital energy level diagram is:

And the number of unpaired electrons=3.

Therefore, the magnetic moment of Co= = 3.87 BM. (BM stands for Bohr Magneton).

• For [Co(CN)6]3– the central atom is Co, the oxidation state of Co in here is x=-6 times (-1)-3=+3.

Hence, Co3+ has a d6 configuration (1s2 2s2 2p6 3s2 3p6 3d6, while losing 1 3d and 2 4s electrons.) So the orbital energy level diagram is:

There will be no unpaired electrons as it is diamagnetic,(as CN- is a strong field ligand, the energy gap becomes larger and all the electrons get paired) therefore no value of magnetic moment can be found.

(ii).

• In the case of [FeF6]3– the central metal atom here is Fe and the oxidation state of Fe x=-6 times (-1)-3=+3.

Hence, Fe3+ will have a d5 configuration (1s2 2s2 2p6 3s2 3p6 3d5).So the orbital energy level diagram is:

Therefore it is paramagnetic and number of unpaired electrons are =5

Therefore the value of magnetic moment for Fe3+ = = 5.92 BM.

For [Fe(H2O)6]2+the central metal atom is iron Fe and the oxidation state of Fe in here is x=2+.

Hence, it has a d6 configuration (1s2 2s2 2p6 3s2 3p6 3d6).

So the orbital energy level diagram is:

Therefore, it has 4 numbers of unpaired electrons

Hence, the magnetic moment= = 4.9 BM. (BM stands for Bohr Magneton).

For [Fe(CN)6]4– the central metal atom is Fe and its oxidation state is x=-6 times(-1)-4=+2.

Hence, it has a d6 configuration (1s2 2s2 2p6 3s2 3p6 3d6).

So the orbital energy level diagram is:

It has no unpaired electrons as CN-are strong field ligand and the energy gap becomes larger, therefore [Fe(CN)6]4– is diamagnetic.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses
RELATED QUESTIONS :

<span lang="EN-USChemistry - Exemplar

<span lang="EN-USChemistry - Exemplar

<span lang="EN-USChemistry - Exemplar

<span lang="EN-USChemistry - Exemplar

<span lang="EN-USChemistry - Exemplar