1.

Co-ordination number and oxidation state of Cr in K_(3) [Cr(C_(2)O_(4))_(3)]are, respectively

Answer»

3 and + 3
3 and 0
6 and + 3
4 and + 2

Solution :`C_(2)O_(4)^(2-)`oxalate ion
`implies` BIDENTATE ligand , So C.No = 6
`K_3[ulCr(C_2O_4)_3],+3+x+6=,x=+3`


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