The oxidation state of cr in [cr(NH3)4cl2] + is
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3
Oxidation number of Cr in [Cr[NH3]4Cl2]+ is +3 x+(4×0)+(−1×2)=+1 x−2=+1 x= +3.
Answered by
1
Answer:
+3
Oxidation number of Cr in [Cr[NH
3
]
4
Cl
2
]
+
is +3
x+(4×0)+(−1×2)=+1
x−2=+1
x=+3
hope it helps!
by shagun
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