Why are different colours observed in octahedral and tetrahedral complexes for the same metal and same ligands?

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(N/A) The colour of coordination complexes arises from $d-d$ transitions,which depend on the magnitude of crystal field splitting $(\Delta)$.
For the same metal ion and the same ligands,the crystal field splitting energy in an octahedral complex $(\Delta_o)$ is significantly larger than in a tetrahedral complex $(\Delta_t)$.
The relationship between them is given by $\Delta_t = \frac{4}{9} \Delta_o$.
Since the energy gap $(\Delta)$ is different,the wavelength of light absorbed $(\Delta E = \frac{hc}{\lambda})$ is different,leading to the observation of different complementary colours for the same metal-ligand system.

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The increasing order of the crystal field splitting power of some common ligands is

In an octahedral crystal field,the $t_{2g}$ orbitals are:

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

$A$ $[M(H_2O)_6]^{2+}$ complex typically absorbs at around $600 \ nm$. It is allowed to react with ammonia to form a new complex $[M(NH_3)_6]^{2+}$ that should have absorption at ....... $nm$.

Which complex shows the highest crystal field stabilisation energy?

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