JEE MAIN - Chemistry (2003 - No. 51)
In the coordination compound, K4[Ni(CN)4], the oxidation state of nickel is :
0
+1
+2
-1
Explanation
Let the $$O.$$ No of $$Ni$$ in
$$\,\,{K_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right]\,\,$$ $$be$$ $$=x$$ then
$$4\left( { + 1} \right) + x + \left( { - 1} \right) \times 4 = 0$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow 4 + x - 4 = 0$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$x=0$$
$$\,\,{K_4}\left[ {Fe{{\left( {CN} \right)}_6}} \right]\,\,$$ $$be$$ $$=x$$ then
$$4\left( { + 1} \right) + x + \left( { - 1} \right) \times 4 = 0$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow 4 + x - 4 = 0$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $$x=0$$
Comments (0)
