MPSC 2022 Prelims Paper-I · Question 55 of 100

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.

• (2)– 5 — 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.
• (3)+ 4 — 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.
• (4)+ 10 — 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.

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.

Oxidation state questions are among the most dependable items in the chemistry portion of MPSC papers, because they require no memorisation beyond a short list of rules and can be solved in under a minute by writing one equation. The Commission builds them on species that look unfamiliar — pyrophosphates, dichromates, thiosulphates, permanganates, the vanadate in this question — precisely so that a candidate cannot answer from recall and must compute. The option sets are constructed around the three errors the calculation invites, and all three appear here: the sign reversed, the total for all the atoms rather than the value per atom, and a real but wrong oxidation state of the same element. Guarding against them is mechanical. Write the equation rather than attempting it mentally; check the sign against the electronegativity of the partner, since a metal bonded to oxygen must be positive; divide by the number of atoms of the element at the end; and test the result against the element's position in the periodic table, which caps the value it can reach. The ion printed here also rewards a candidate who recognises its family, since the dichromate ion has the identical shape and yields to identical arithmetic.

• In a polyatomic ion the oxidation numbers of all the atoms add up to the charge on the ion; in a neutral compound they add up to zero.
• Oxygen is assigned an oxidation number of minus two in ordinary compounds, the exceptions being peroxides, where it is minus one, superoxides, and its compounds with fluorine.
• For the ion with two vanadium atoms, seven oxygen atoms and a charge of minus four, the equation is two x minus fourteen equals minus four, giving x equal to plus five for each vanadium atom.
• Plus five is the highest oxidation state available to vanadium, which belongs to the fifth group of the transition series and has five electrons in its outer s and d orbitals; it also shows the plus two, plus three and plus four states.
• Vanadium pentoxide, in which vanadium is in the plus five state, is used as the catalyst in the contact process for the manufacture of sulphuric acid.

• Reporting the total for all the atoms of an element rather than the value per atom, which turns plus five into plus ten in this question
• Reversing the sign and making the metal negative, when a metal bonded to oxygen is always the positive partner
• Choosing an oxidation state the element genuinely adopts in some other compound instead of deriving the state required by the formula given
• Applying the value of minus two for oxygen without checking whether the species is a peroxide or a superoxide, where the value is different

MPSC sets oxidation state questions in three shapes. The commonest is the direct calculation asked here, in which an ion or a compound is named and the state of one element in it is wanted. The second asks which element is oxidised or reduced in a given reaction, which requires the same calculation to be done twice and the values compared. The third is descriptive, asking which oxidation states a named transition metal exhibits or which compound contains a metal in its highest state. All three rest on the same short list of rules, so the return on learning those rules thoroughly is high, and the arithmetic should be practised on the standard awkward species — thiosulphate, tetrathionate, dichromate, permanganate — until it is quick and reliable.

No directly related past PYQ was found.

• practice — not a real PYQ

  What is the oxidation state of chromium in the dichromate ion, which contains two chromium atoms, seven oxygen atoms and a charge of minus two ?

  • (a)+ 3
  • (b)+ 6
  • (c)+ 7
  • (d)+ 12

  Answer(b) + 6 — taking oxygen as minus two, the seven oxygen atoms contribute minus fourteen, and the total must equal the charge of minus two, so two x minus fourteen equals minus two, giving two x equal to twelve and x equal to plus six for each chromium atom. Plus twelve is the value of two x and would be the answer of a candidate who forgot to divide by the number of chromium atoms, which is the same slip that the option offering plus ten invites in the vanadium question.

• practice — not a real PYQ

  In which of the following compounds does oxygen have an oxidation number of minus one rather than minus two ?

  • (a)Water
  • (b)Carbon dioxide
  • (c)Hydrogen peroxide
  • (d)Sulphur dioxide

  Answer(c) Hydrogen peroxide — in a peroxide the two oxygen atoms are bonded to each other as well as to hydrogen, and since a bond between two identical atoms contributes nothing to either one's oxidation number, each oxygen ends at minus one rather than minus two. Taking hydrogen as plus one, the two hydrogen atoms give plus two, so the two oxygen atoms must total minus two, that is minus one each. In water, carbon dioxide and sulphur dioxide the usual value of minus two applies.