Question

the distribution of d7 electron in high spin octahedral field is
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Answer 1

Answer:

Distribution of Electrons in an Octahedral Complex Metal ions with 4 – 7 electrons in the d orbital can exist as high-spin or low-spin complexes. Weaker ligands tend to give high-spin complexes, whereas stronger ligands tend to give low-spin complexes.

Answer 2

For high spin octahedral field, $d^7$ ion has the configuration of  $t_{2g}^5e_{g}^2$.

Explanation:

In octahedral complex, the splitting of d-orbitals takes place into two sets - $t_{2g}$ and $e_g$.

There will be three orbitals ($d_{xy}$, $d_{yz}$ and $d_{zx}$) in the  $t_{2g}$ set and there will be two orbitals ($d_{x^2-y^2}$, and $d_{z^2}$) in the  $e_{g}$ set.

High spin complex means, the electrons are unpaired to the maximum extent.

First filling starts from  $t_{2g}$ level and then if required  $e_g$ level is filled.

For $d^3$ ion : the configuration will be  $t_{2g}^3$ (For both high or low spin complex)

For $d^4$ ion : the configuration will be  $t_{2g}^3e_{g}^1$ (For high spin) and $t_{2g}^4$ (For low spin complex)

For $d^5$ ion : the configuration will be  $t_{2g}^3e_{g}^2$ (For high spin) and $t_{2g}^5$ (For low spin complex)

For $d^6$ ion : the configuration will be  $t_{2g}^4e_{g}^2$ (For high spin) and $t_{2g}^6$ (For low spin complex)

For $d^7$ ion : the configuration will be  $t_{2g}^5e_{g}^2$ (For high spin) and $t_{2g}^6 e_g^1$ (For low spin complex).

And so on.