A metal ion M^{n+} having d^6 valence electro | vedclass

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Question

(A) For an octahedral complex,when the crystal field splitting energy $(\Delta_0)$ is greater than the pairing energy $(P)$,the complex is a low-spin

Vedclass · Accepted Answer

$t_{2g}^6 e_g^0$

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(A) For an octahedral complex,when the crystal field splitting energy $(\Delta_0)$ is greater than the pairing energy $(P)$,the complex is a low-spin complex.In a low-spin $d^6$ configuration,the electrons will first fill the lower energy $t_{2g}$ orbitals completely before occupying the higher energy $e_g$ orbitals.Therefore,all $6$ electrons will occupy the $t_{2g}$ orbitals,resulting in the configuration $t_{2g}^6 e_g^0$.

$A$ 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

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