Q. 46

# Using crystal field theory, draw energy level diagram, write the electronic configuration of the central metal atom/ion and determine the magnetic moment value in the following:

(i) [CoF_{6}]^{3–}, [Co(H_{2}O)_{6}]^{2+} , [Co(CN)_{6}]^{3–}

(ii) [FeF_{6}]^{3–,} [Fe(H_{2}O)_{6}]^{2+}, [Fe(CN)_{6}]^{4–}

Answer :

(i)

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

Hence, Co^{3+}has a d^{6} configuration (1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{6}, 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(H_{2}O)_{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 Co^{2+} has a d^{7} configuration (1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{7},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, Co^{3+} has a d^{6} configuration (1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{6}, 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 [FeF_{6}]^{3–} the central metal atom here is Fe and the oxidation state of Fe x=-6 times (-1)-3=+3.

Hence, Fe^{3+} will have a d^{5} configuration (1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{5}).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 Fe^{3+} = = 5.92 BM.

• For [Fe(H_{2}O)_{6}]^{2+}the central metal atom is iron Fe and the oxidation state of Fe in here is x=2+.

Hence, it has a d^{6} configuration (1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{6}).

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 d^{6} configuration (1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 3d^{6}).

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.

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Why are different colours observed in octahedral and tetrahedral complexesfor the same metal and same ligands?

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