Question

A metal ion $M^{n+}$ having $d^4$ valence electronic configuration combines with three bidentate ligands to form a complex compound. Assuming $△a>p$:
(i) Write the electronic configuration of $d^4$ ion.
(ii) What type of hybridisation will $M^{n+}$ ion has?
(iii) Name the type of isomerism exhibited by this complex.

Answer 1

$\left(i\right)$ Since $P<{△}_o$so, pairing of electrons takes place and $d^4$ exists as $t_{2g}^{4}e{g}^o$.

$\left(ii\right)$Then hyfridisation is $d^{2}S{p}^3$ due to vaccant two inner $d-$ orbital.

$\left(iii\right)$$\left[M\right(AA{)}_3{]}^{n+}$ shows optical isomerism due to plane to symmetry.