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Crystal Field Split in Octahedral Complexes

What is elect...

Question

What is electronic arrangement of metal atom/ ion in octahedral complex with $d^{4}$ configuration, if $△_{0} <$ pairing energy?

t_{2 g}^{4} e_{g}^{0}

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e_{g}^{4} t_{2 g}^{0}

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t_{2 g}^{3} e_{g}^{1}

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e_{g}^{2} t_{2 g}^{2}

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Solution

The correct option is C $t_{2 g}^{3} e_{g}^{1}$
When $Δ_{0} < P$ , the electrons will remain unpaired. Three d-electrons will enter $t_{2 g}$ level and fourth d-electron will enter $e_{g}$ level. Thus, the electronic arrangement of the metal ion will be $t_{2 g}^{3} e_{g}^{1}$ .

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Similar questions

If $Δ_{\circ}$ > P (P = pairing energy; $Δ_{\circ}$ change in energy of octahedral complexes), the correct electronic configuration for $d^{4}$ system will be

Q. If

Δ_{0} < P

, the correct electronic configuration for

d^{4}

system will be -

Q. If

△_{0} < P,

where

P

is pairing energy and

△_{0}

is the crystal field spliting energy, the correct electronic configuration for

d^{4}

system will be:

Q. Assertion :Transition metal ion forming octahedral complexes undergo

s p^{3} d^{2}

d^{2} s p^{3}

hybridisation. Reason: Strong field ligands force the unpaired electrons of central metal ion to pair up causing

d^{2} s p^{3}

hybridisation whereas weak field ligands do not affect electronic configuration of the metal ion undergoes in

s p^{3} d^{2}

hybridisation.

Q. What is meant by crystal field splitting energy ? On the basis of crystal field theory, write the electronic configuration of

d^{4}

in terms of

t_{2 g}

and

e_{g}

in an octahedral field when
(i)

Δ_{0} > P

(ii)

Δ_{0} < P

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What is electronic arrangement of metal atom/ ion in octahedral complex with d4 configuration, if △0< pairing energy?

The correct option is C t32ge1gWhen Δ0<P, the electrons will remain unpaired. Three d-electrons will enter t2g level and fourth d-electron will enter eg level. Thus, the electronic arrangement of the metal ion will be t32ge1g.

What is electronic arrangement of metal atom/ ion in octahedral complex with $d^{4}$ configuration, if $△_{0} <$ pairing energy?

The correct option is C $t_{2 g}^{3} e_{g}^{1}$
When $Δ_{0} < P$ , the electrons will remain unpaired. Three d-electrons will enter $t_{2 g}$ level and fourth d-electron will enter $e_{g}$ level. Thus, the electronic arrangement of the metal ion will be $t_{2 g}^{3} e_{g}^{1}$ .