Consider 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:

  • A
    $C < D < A < B$
  • B
    $D < A < C < B$
  • C
    $A < C < B < D$
  • D
    $B < C < A < D$

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Similar Questions

Match List-$I$ with List-$II$. Choose the correct answer from the options given below:
| List-$I$ (Complex ion) | List-$II$ (Electronic Configuration) |
| :--- | :--- |
| $A. [Cr(H_2O)_6]^{3+}$ | $I. t_{2g}^2 e_g^0$ |
| $B. [Fe(H_2O)_6]^{3+}$ | $II. t_{2g}^3 e_g^0$ |
| $C. [Ni(H_2O)_6]^{2+}$ | $III. t_{2g}^3 e_g^2$ |
| $D. [V(H_2O)_6]^{3+}$ | $IV. t_{2g}^6 e_g^2$ |

Complete removal of both the axial ligands (along the $z-$ axis) from an octahedral complex leads to which of the following splitting patterns? (relative orbital energies not on scale)

An ion $M^{2+}$ forms the complexes $[M(H_2O)_6]^{2+}$,$[M(en)_3]^{2+}$,and $[MBr_6]^{4-}$. Match the complex with the appropriate colour.

The Crystal Field Stabilisation Energy $(CFSE)$ for $[CoCl_{6}]^{4-}$ is $18000 \; cm^{-1}$. The $CFSE$ for $[CoCl_{4}]^{2-}$ will be $...... \ cm^{-1}$.

In the octahedral crystal field,the correct order of splitting for $I^{-}, H_2O, NH_3$ and $CN^{-}$ ligands is:

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