Question
The coordination number and oxidation state of $$Cr$$ in $${K_3}\left[ {Cr{{\left( {{C_2}{O_4}} \right)}_3}} \right]$$ are respectively
A.
3 and +3
B.
3 and 0
C.
6 and +3
D.
4 and +2
Answer :
6 and +3
Solution :
Coordination number of $$Cr$$ is 6 ( oxalate is bidentate ligand ) and oxidation state of $$Cr$$ in $${K_3}\left[ {Cr{{\left( {{C_2}{O_4}} \right)}_3}} \right]$$ is calculated below.
$$\eqalign{
& 3\left( 1 \right) + x + 3\left( { - 2} \right) = 0 \cr
& \,\,\,\,\,\,\,\,\,\,\,3 + x + \left( { - 6} \right) = 0 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 6 - 3 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = + 3 \cr} $$