In [Cr(O2)(NH3)4H2O]Cl2 , oxidation number of Cr is +3, then oxygen will be in the form

Question

In [Cr(O2)(NH3)4H2O]Cl2 , oxidation number of Cr is +3, then oxygen will be in the form (A) dioxo (B) peroxo (C) superoxo (D) oxo

Tardigrade Admin · Accepted Answer

The complex is [Cr(O2)(NH3)4H2O]Cl2 , let oxidation number of oxygen is x So +3+2x+0× 4+0-2=0 2x=-1 x=-(1/2) . Hence, oxygen is in superoxo form (.O2-.) .

Q. In $[C r (O_{2}) (N H_{3})_{4} H_{2} O] C l_{2}$ , oxidation number of Cr is +3, then oxygen will be in the form

dioxo

peroxo

superoxo

oxo

The complex is $[C r (O_{2}) (N H_{3})_{4} H_{2} O] C l_{2}$ , let oxidation number of oxygen is x

So $+ 3 + 2 x + 0 \times 4 + 0 - 2 = 0$

$2 x = - 1$

$x = - \frac{1}{2}$ .

Hence, oxygen is in superoxo form $(O_{2}^{-})$ .

Q. In [Cr(O2​)(NH3​)4​H2​O]Cl2​ , oxidation number of Cr is +3, then oxygen will be in the form

dioxo

peroxo

superoxo

oxo

The complex is [Cr(O2​)(NH3​)4​H2​O]Cl2​ , let oxidation number of oxygen is x So +3+2x+0×4+0−2=0 2x=−1 x=−21​ . Hence, oxygen is in superoxo form (O2−​) .

Q. In $[C r (O_{2}) (N H_{3})_{4} H_{2} O] C l_{2}$ , oxidation number of Cr is +3, then oxygen will be in the form

The complex is $[C r (O_{2}) (N H_{3})_{4} H_{2} O] C l_{2}$ , let oxidation number of oxygen is x

So $+ 3 + 2 x + 0 \times 4 + 0 - 2 = 0$

$2 x = - 1$

$x = - \frac{1}{2}$ .

Hence, oxygen is in superoxo form $(O_{2}^{-})$ .