For a $d^{4}$ metal ion in an octahedral field,the correct electronic configuration is:
- A$t_{2g}^{4} e_{g}^{0}$ when $\Delta_{o} < P$
- B$e_{g}^{2} t_{2g}^{2}$ when $\Delta_{o} < P$
- C$t_{2g}^{3} e_{g}^{1}$ when $\Delta_{o} < P$
- D$t_{2g}^{3} e_{g}^{1}$ when $\Delta_{o} > P$
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Similar Questions
If $[Cu(H_{2}O)_{4}]^{2+}$ absorbs light of wavelength $600 \ nm$ for $d-d$ transition,then the value of octahedral crystal field splitting energy for $[Cu(H_{2}O)_{6}]^{2+}$ will be $..... \times 10^{-21} \ J$. (Nearest Integer)
(Given: $h = 6.63 \times 10^{-34} \ Js$ and $c = 3.08 \times 10^{8} \ ms^{-1}$)
(Given: $h = 6.63 \times 10^{-34} \ Js$ and $c = 3.08 \times 10^{8} \ ms^{-1}$)
DifficultJEE MAIN 2022
View SolutionWhich of the following produces high crystal field splitting?
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View SolutionThe homoleptic and octahedral complex of $Co^{2+}$ and $H_2O$ has $.......$ unpaired electron$(s)$ in the $t_{2g}$ set of orbitals.
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View SolutionWhat is crystal field splitting energy? How does the magnitude of $\Delta_{o}$ decide the actual configuration of $d$-orbitals in a coordination entity?
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View SolutionWhich sets of the $d$-orbitals are directly oriented towards the ligands in octahedral coordination compounds?
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