JEE MAIN - Chemistry (2021 - 18th March Morning Shift - No. 21)
The total number of unpaired electrons present in the complex K3[Cr(oxalate)3] is _____________.
Answer
3
Explanation
In $${K_3}\left[ {Cr\left( {{C_2}{O_4}} \right)} \right]$$, oxidation number of Cr :
$$3( + 1) + x + 3( - 2) = 0$$
$$ \Rightarrow x = + 3$$
$$ \therefore $$ $${}_{24}C{r^{ + 3}} = \left[ {Ar} \right]3{d^3}$$
$$ \therefore $$ Number of unpaired electrons = 3
$$3( + 1) + x + 3( - 2) = 0$$
$$ \Rightarrow x = + 3$$
$$ \therefore $$ $${}_{24}C{r^{ + 3}} = \left[ {Ar} \right]3{d^3}$$
$$ \therefore $$ Number of unpaired electrons = 3
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