The total number of unpaired electrons present in the complex K3[Cr(oxalate)3] is _____________.
Answer (integer)
3
Solution
In ${K_3}\left[ {Cr\left( {{C_2}{O_4}} \right)} \right]$, oxidation number of Cr :<br><br>$3( + 1) + x + 3( - 2) = 0$<br><br>$\Rightarrow x = + 3$<br><br>$\therefore$ ${}_{24}C{r^{ + 3}} = \left[ {Ar} \right]3{d^3}$<br><br>$\therefore$ Number of unpaired electrons = 3
About this question
Subject: Chemistry · Chapter: Coordination Compounds · Topic: Ligands and Coordination Number
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