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

In which one of the following complexes does the metal ion have a $t_{2g}^3 e_g^2$ configuration?

If the $CFSE$ of $[Ti(H_2O)_6]^{3+}$ is $-96.0 \ kJ / mol$,this complex will absorb maximum at wavelength $........nm$. (nearest integer)
Assume Planck's constant $(h) = 6.4 \times 10^{-34} \ Js$,Speed of light $(c) = 3.0 \times 10^8 \ m / s$ and Avogadro's constant $(N_A) = 6 \times 10^{23} / mol$.

Low spin complex of $d^6$ cation in an octahedral field will have the following energy:
($\Delta_o =$ crystal field splitting energy in an octahedral field,$P =$ Electron pairing energy)

Given below are two statements:
Statement $I$: Among $Zn$, $Mn$, $Sc$ and $Cu$, the energy required to remove the third valence electron is highest for $Zn$ and lowest for $Sc$.
Statement $II$: The correct order of the following complexes in terms of $CFSE$ is $[Co(H_2O)_6]^{2+} < [Co(H_2O)_6]^{3+} < [Co(en)_3]^{3+}$.

Match List-$I$ with List-$II$.
List-$I$ (Coordination entity) List-$II$ (Wavelength of light absorbed in $nm$)
$A$. $[CoCl(NH_3)_5]^{2+}$ $I$. $310$
$B$. $[Co(NH_3)_6]^{3+}$ $II$. $475$
$C$. $[Co(CN)_6]^{3-}$ $III$. $535$
$D$. $[Cu(H_2O)_4]^{2+}$ $IV$. $600$

Choose the correct answer from the options given below:

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