The oxidation state of V in V2O7-4 is ________ .

Question

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Correct — option (1), plus five. The species in the stem is printed with a subscript two on the vanadium, a subscript seven on the oxygen and a superscript minus four, so it is the ion containing two vanadium atoms and seven oxygen atoms and carrying an overall charge of minus four. The calculation rests on two standard rules. The first is that oxygen is assigned an oxidation number of minus two in its ordinary compounds, the exceptions being the peroxides, in which it is minus one, the superoxides, and the compounds with fluorine, none of which applies here. The second is that in a polyatomic ion the oxidation numbers of all the atoms must add up to the charge on the ion, just as in a neutral compound they must add up to zero. Let the oxidation state of each vanadium atom be x. There are two vanadium atoms, contributing two x, and seven oxygen atoms, contributing seven times minus two, which is minus fourteen. The sum must equal the charge on the ion, which is minus four. So two x minus fourteen equals minus four; two x equals ten; and x equals plus five. Each vanadium atom is therefore in the plus five oxidation state, and option (1) is the answer. Two checks confirm the result. The first is chemical: plus five is the highest oxidation state available to vanadium, which stands in the fifth group of the transition series and has five electrons available in its outermost s and d orbitals, so a vanadium atom that has given up all five of them reaches plus five and can go no further. That is the state found in vanadium pentoxide, the oxide used as the catalyst in the contact process for manufacturing sulphuric acid, and the ion in this question is one of the many oxo-anions vanadium forms in that state. The second check is structural: the ion is the exact analogue of the dichromate ion and of the pyrophosphate ion, two units joined through a shared oxygen atom, and running the same arithmetic on dichromate, with its two chromium atoms, seven oxygen atoms and a charge of minus two, gives plus six for chromium, which is chromium's own highest state. The parallel is worth noticing, because the family of two-metal, seven-oxygen ions recurs and the arithmetic is identical each time. The essential caution in all such calculations is to remember that the answer is the state of one atom: the equation gives two x equals ten, and it is the ten that must be halved before the answer is written down. Option (1) is the answer.

Why the other options are wrong — (b) The magnitude is right and the sign is wrong, which makes this the option for a candidate who has done the arithmetic carelessly or has confused the charge on the ion with the oxidation state of the metal in it. The ion as a whole carries a negative charge, but that charge is accounted for by the seven oxygen atoms, whose combined contribution of minus fourteen more than offsets the positive contribution of the two vanadium atoms; the metal itself is the positive partner in the arrangement. There is also a chemical reason to reject a negative value on sight. Vanadium is a metal and oxygen is one of the most electronegative elements there is, so in a bond between them the shared electrons are counted to the oxygen, which leaves the vanadium positive. A metal bonded only to oxygen can never carry a negative oxidation state, and noticing that alone eliminates this option before any equation is written.  (c) Plus four is a genuine oxidation state of vanadium, which is what makes this option plausible: it is the state of the blue vanadyl ion and of vanadium dioxide, and vanadium is well known for existing in several states, plus two, plus three, plus four and plus five, each with its own characteristic colour in solution. It is not, however, the state in this particular ion, and the arithmetic settles it: if each vanadium were plus four the total from the two of them would be plus eight, which with minus fourteen from the oxygen would give an overall charge of minus six rather than the minus four the formula specifies. The general warning is that in a question about oxidation state, an option naming a state the element really does adopt elsewhere is not thereby the answer; the state must be derived from the formula given, since the whole purpose of the exercise is to determine which of an element's several possible states it occupies in the species named.  (d) Ten is the value of two x, not of x, and this option is the trap for a candidate who solves the equation correctly and then forgets that there are two vanadium atoms sharing the total. Oxidation state is defined per atom, so once the equation has yielded two x equals ten the remaining step is to divide by two. There is also a limit that rules the value out at once: an element's oxidation state cannot exceed the number of electrons it has available to lose, and vanadium, in the fifth group of the transition series, has five such electrons, so plus five is its ceiling. No element in the fourth period reaches plus ten, and the highest oxidation state met in ordinary chemistry is plus eight, in a very small number of compounds of osmium and ruthenium. Checking a computed oxidation state against the group of the element in the periodic table is a quick way of catching exactly this kind of slip.

Concept: The oxidation number of an atom in a compound is the charge it would carry if every bond it forms were treated as fully ionic, the shared electrons being assigned to the more electronegative partner. It is a bookkeeping device rather than a physical charge, but it is indispensable for balancing redox equations and for naming compounds. The rules used to assign it are few. An element in its free state is zero. A monatomic ion takes the charge of the ion. Fluorine is always minus one; oxygen is minus two except in peroxides, where it is minus one, in superoxides, and in its compounds with fluorine, where it is positive; hydrogen is plus one except in metal hydrides, where it is minus one. Alkali metals are plus one and alkaline earth metals plus two in all their compounds. Finally, the oxidation numbers in a neutral molecule add to zero and in a polyatomic ion add to the charge on the ion, which is the rule that makes an unknown solvable. Transition metals are the elements for which these calculations matter most, because they show variable oxidation states: the energies of the outer s and d electrons are close together, so different numbers of electrons can be removed under different conditions, giving the series of coloured ions that characterises the block. Vanadium is a good illustration, existing in the plus two, plus three, plus four and plus five states with distinct colours in solution; its pentoxide is the catalyst of the contact process, and the metal is used in ferrovanadium alloys to harden steel.

State PCS 2022 Q55 — Science & Technology PYQ with Answer

State PCS 2022 Q55 — Science & Technology PYQ with Answer