Chemistry · Redox Equilibrium · NEET
Stock notation is a naming style where the oxidation state of an element (usually a metal) is written as a Roman numeral inside round brackets, placed right after the element name or symbol. Example: iron in +2 is iron(II) or Fe(II); iron in +3 is iron(III) or Fe(III). It was introduced by Alfred Stock, which is why it is called Stock notation. For NEET, it is a quick way to state 'which oxidation state this atom is in' without writing a full calculation.
Many metals show more than one oxidation state, so the same two elements can form different compounds. Copper forms Cu2O and CuO; iron forms FeCl2 and FeCl3. Just saying 'copper oxide' or 'iron chloride' is not clear enough. Stock notation removes the confusion: Cu2O is copper(I) oxide and CuO is copper(II) oxide. This tells you the oxidation state instantly, which matters in redox chapters where the state decides what gets oxidised or reduced.
No sign is written in the Roman numeral. The Roman numeral shows only the size of the positive oxidation state. So Fe(III) means +3, not -3. Stock notation is normally used for positive oxidation states of metals. You never write Fe(-III) in a normal compound name. If a species is negative, chemists use a different naming style, not Stock notation.
They are closely linked but not the same. The oxidation number is the value you calculate using the rules (for example +3 for iron in Fe2O3). Stock notation is just the way you write that value using a Roman numeral in brackets. So first you find the oxidation number by calculation, then you express it in Stock notation. Fe = +3 (oxidation number) becomes Fe(III) (Stock notation).
Step 1: find the oxidation number of the metal using the rules (oxygen is usually -2, hydrogen usually +1, the total charge of a neutral compound is 0). Step 2: convert that number to a Roman numeral (1=I, 2=II, 3=III, 4=IV, 5=V, 6=VI, 7=VII). Step 3: put it in brackets after the element name. Example: in MnO2, let Mn = x, so x + 2(-2) = 0, x = +4, giving manganese(IV) oxide, or Mn(IV).
In KMnO4, potassium is +1 and each oxygen is -2. Let Mn = x: (+1) + x + 4(-2) = 0, so x = +7. So manganese is in the +7 state, written as Mn(VII), and the compound can be named potassium manganate(VII). This is a classic NEET-favourite state, because Mn(VII) in KMnO4 is a strong oxidising agent.
For zero oxidation state some texts write (0), for example in certain metal carbonyls like Ni(CO)4 the nickel is nickel(0). But Stock notation cannot show fractional or average oxidation states cleanly, because Roman numerals are whole numbers. For species like Fe3O4 (average +8/3) you use the average oxidation number idea, not a single Stock Roman numeral. NEET tests fractional states separately, so do not force them into Stock notation.
Consider the following compounds: KO2, H2O2 and H2SO4. The oxidation states of the underlined elements (K in KO2, O in H2O2, S in H2SO4) in them are, respectively:
The oxidation state of Cr in CrO5 is:
The correct order of N-compounds in their decreasing order of oxidation states is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It is named after the German chemist Alfred Stock, who suggested using Roman numerals in brackets to show oxidation states. That is why NCERT and NEET call it Stock notation.
Yes. It is most useful for metals that have more than one oxidation state, like iron, copper, manganese and chromium. It clears up which compound you mean, for example copper(I) oxide versus copper(II) oxide.
In Cu2O, oxygen is -2 and there are two copper atoms sharing it, so each copper is +1, giving copper(I) oxide. In CuO, one copper balances one oxygen of -2, so copper is +2, giving copper(II) oxide.
Yes, some compounds use (0), such as nickel(0) in Ni(CO)4 or iron(0) in Fe(CO)5. But fractional or average states cannot be written with a single Roman numeral.
NEET often asks you to find or compare oxidation states, and Stock notation is the standard way to state the answer. Getting the Roman numeral right shows you calculated the oxidation number correctly, which is the base skill for the whole redox chapter.
