In the compound K[Co(C₂O₄)₂(H₂O)₂] (where C₂O₄²^- = oxalate) the oxidation number and coordination number of cobalt are, respectively A)−1 and 4. B)−1 and 6. C)3 and 4. D)3 and 6. E)1 and 6.

Question

Quizplus · Accepted Answer

The oxidation number and coordination number of cobalt in the given compound can be determined as follows: - Oxalate ion (C₂O₄²^-) has a charge of -2. - Therefore, the overall charge on the complex is: K+ [Co(C₂O₄²^-)₂(H₂O)₂]- = K+ [Co(C₂O₄²^-)₂(H₂O)₂]2- = K+ [Co(C2O4)(en)(H2O)]2- - Here, en is ethylenediamine, which has two nitrogen atoms that can be used for coordination with cobalt. - The charge on the complex is -2, so the sum of the oxidation numbers of all the atoms in the complex must be -2. - The oxidation number of potassium (K) is always +1. - The oxidation number of oxygen (O) in H2O is -2, and in oxalate (C2O4) it is -2 as well. - The oxidation number of nitrogen (N) in ethylenediamine (en) is -2. - Let x be the oxidation number of cobalt (Co). Then: (+1) + 2x + (-2) + (-2) + (-2) + (-2) + (-2) = -2 2x - 10 = -2 2x = 8 x = 4 - Therefore, the oxidation number and coordination number of cobalt in the given compound are +4 and 6, respectively.

In the Compound K[Co(C2O4)2(H2O)2] (Where C2O42- = Oxalate) the Oxidation

In the Compound K[Co(C₂O₄)₂(H₂O)₂] (Where C₂O₄²^- = Oxalate) the Oxidation