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Determine the oxidation number of Ni in the complex [Ni(C2O4)2]2− using the equation x + 3×(−2) = −4.

1. The oxidation number of Ni in [Ni(C2O4)2]2− is calculated as x + 3×(−2) = −4, leading to x = −4 + 6 = 2.
2. The oxidation number of Ni in [Ni(C2O4)2]2− is determined by solving x + 3×(−2) = −4, which gives x = −4 + 6 = 2.
3. In the complex [Ni(C2O4)2]2−, the oxidation number of Ni is found using x + 3×(−2) = −4, resulting in x = −4 + 6 = 2.
4. To find the oxidation number of Ni in [Ni(C2O4)2]2−, solve x + 3×(−2) = −4, yielding x = −4 + 6 = 2.

Correct answer: The oxidation number of Ni in [Ni(C2O4)2]2− is calculated as x + 3×(−2) = −4, leading to x = −4 + 6 = 2.

Solution

The oxidation number (O.N.) of Ni in [Ni(C2O4)2]2− is calculated correctly as follows: Oxalate (C2O4)2− is a bidentate ligand with a charge of -2. Since there are two oxalate ligands, their total charge is -4. The overall charge of the complex is -2. Thus, x + (-4) = -2, giving x = +2.

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