The correct order of increasing oxidation number of the transition metal ions for the compounds K2Cr2O7, V2O5 and KMnO4 is

Answer Details

The oxidation state or number is a measure of the degree of oxidation of an atom in a compound, which is determined by counting the number of electrons transferred from or to the element when it forms a compound. In this question, we are given three transition metal compounds, K2Cr2O7, V2O5, and KMnO4, and we are asked to arrange them in order of increasing oxidation number of the metal ion. The oxidation number of the metal ion in each compound can be determined by assigning electrons to the metal and oxygen atoms based on their electronegativity values and the overall charge of the compound. Oxygen is generally assigned an oxidation number of -2, and the sum of the oxidation numbers of all the atoms in the compound should equal the overall charge of the compound. For K2Cr2O7, we can assign an oxidation state of +6 to the chromium ion, since each oxygen atom contributes -2 to the total charge, and the two potassium ions each contribute +1. For V2O5, we can assign an oxidation state of +5 to the vanadium ion, since each oxygen atom contributes -2 to the total charge. For KMnO4, we can assign an oxidation state of +7 to the manganese ion, since each oxygen atom contributes -2 to the total charge, and the potassium ion contributes +1. Therefore, the correct order of increasing oxidation number of the metal ions in these compounds is: V2O5 < K2Cr2O7 < KMnO4.