Give the coordination number (C.N.) and oxidation number (O.N.) of the metal atom in the coordination compound [Cr(NH₃)₂(en)Cl₂]. A) C.N. = 4; O.N. = +2. B) C.N. = 5; O.N. = +3. C) C.N. = 5; O.N. = +2. D) C.N. = 6; O.N. = +3. E) C.N. = 6; O.N. = +2.

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Quizplus · Accepted Answer

The coordination number is determined by counting the number of ligands directly attached to the central metal atom, which is 6 in this case (4 sulfur atoms, 1 nitrogen atom from the imidazole ring, and 1 nitrogen atom from the ethylenediamine ligand). The oxidation number of the central chromium atom can be calculated by considering the charge contributed by the ligands and the overall charge on the complex. The total charge of the ligands is (-1) x 13 = -13. The overall charge of the complex is 0, which means that the sum of the oxidation numbers of all the atoms must be equal to 0. Therefore, the oxidation number of chromium can be calculated as (+2) + (-13) = -11. However, this is not the actual oxidation state of the chromium atom, as it is not feasible for a metal ion to have such a high negative oxidation state. Therefore, we assume that each chloride ion carries a charge of (-1), which would make the oxidation state of chromium to be +2. Hence, the correct answer is E, with a coordination number of 6 and an oxidation number of +2.

Give the Coordination Number (C