$A$ metal ion $M^{n+}$ having $d^6$ valence electronic configuration combines with three bidentate ligands to form a complex. Assuming crystal field splitting $(\Delta_0) >$ pairing energy,the $d$-orbital electronic configuration would be
- A$t_{2g}^6 e_g^0$
- B$t_{2g}^4 e_g^2$
- C$t_{2g}^3 e_g^3$
- D$t_{2g}^5 e_g^1$
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Similar Questions
The complex ion that will lose its crystal field stabilization energy upon oxidation of its metal to $+3$ state is
DifficultJEE MAIN 2019
View SolutionWrite down the number of $3d$ electrons in each of the following ions:
$Ti^{2+}, V^{2+}, Cr^{3+}, Mn^{2+}, Fe^{2+}, Fe^{3+}, Co^{2+}, Ni^{2+}$ and $Cu^{2+}$
Indicate how you would expect the five $3d$ orbitals to be occupied for these hydrated ions (octahedral).
$Ti^{2+}, V^{2+}, Cr^{3+}, Mn^{2+}, Fe^{2+}, Fe^{3+}, Co^{2+}, Ni^{2+}$ and $Cu^{2+}$
Indicate how you would expect the five $3d$ orbitals to be occupied for these hydrated ions (octahedral).
Difficult
View SolutionConsider the following complexes:
$[CoCl(NH_3)_5]^{2+},$ $[Co(CN)_6]^{3-}$
$(A)$ $(B)$
$[Co(NH_3)_5(H_2O)]^{3+},$ $[Cu(H_2O)_4]^{2+}$
$(C)$ $(D)$
The correct order of $A, B, C,$ and $D$ in terms of the wavenumber of light absorbed is:
$[CoCl(NH_3)_5]^{2+},$ $[Co(CN)_6]^{3-}$
$(A)$ $(B)$
$[Co(NH_3)_5(H_2O)]^{3+},$ $[Cu(H_2O)_4]^{2+}$
$(C)$ $(D)$
The correct order of $A, B, C,$ and $D$ in terms of the wavenumber of light absorbed is:
The configurations of the complexes of manganese $A$ and $B$ respectively are $t_{2g}^3 e_g^1$ and $t_{2g}^4 e_g^0$. Then $A$ and $B$ are:
Which of the following produces high crystal field splitting?
Medium
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