Chemistry · D And F Block Elements · NEET
There are 3 steps. (1) Roast chromite ore (FeCr2O4) with sodium carbonate in air: 4FeCr2O4 + 8Na2CO3 + 7O2 -> 8Na2CrO4 + 2Fe2O3 + 8CO2. This gives yellow sodium chromate. (2) Acidify the chromate with sulphuric acid to get orange sodium dichromate (Na2Cr2O7). (3) Because sodium dichromate is more soluble than the potassium salt, add potassium chloride: Na2Cr2O7 + 2KCl -> K2Cr2O7 + 2NaCl. The less soluble orange crystals of K2Cr2O7 crystallise out. Remember the order: ore -> chromate -> dichromate -> K salt.
Chromium is +6. In Cr2O7^2-, the total charge is -2. Oxygen is -2 each, so 7 oxygens give -14. If each Cr is x, then 2x + (-14) = -2, so 2x = +12 and x = +6. Important NEET point: chromium is ALSO +6 in the chromate ion CrO4^2-. So the oxidation state of Cr is the SAME (+6) in both chromate and dichromate. NTA loves to test this.
Because chromium is reduced from +6 to +3. Orange Cr2O7^2- (Cr is +6) gains 6 electrons and becomes green Cr3+ (Cr is +3): Cr2O7^2- + 14H+ + 6e- -> 2Cr3+ + 7H2O. The green colour is due to Cr3+ ions (for example green Cr2(SO4)3). So when dichromate oxidises something, IT gets reduced, and the colour tells you it worked: orange -> green means the reaction happened.
The dichromate ion is made of TWO tetrahedra joined at one corner. Each chromium sits at the centre of a CrO4 tetrahedron, and the two tetrahedra share ONE oxygen atom (the bridging oxygen), giving a Cr-O-Cr bridge. The Cr-O-Cr bond angle is 126 degrees. Compare: the chromate ion CrO4^2- is just ONE tetrahedron. Simple picture: chromate = 1 tetrahedron, dichromate = 2 tetrahedra sharing a corner.
Yes, in acidic solution it is a strong oxidising agent (standard potential E = 1.33 V). It oxidises many things while Cr(VI) drops to Cr(III). Common NEET examples: it oxidises Fe2+ to Fe3+, iodide I- to iodine I2, H2S to sulphur S, and SO2 (or sulphite) to sulphate. Half reaction to memorise: Cr2O7^2- + 14H+ + 6e- -> 2Cr3+ + 7H2O. Note it needs H+ (acidic medium) and gains 6 electrons per dichromate ion.
K2Cr2O7 is used as a primary standard in volumetric analysis (titrations) because it is stable, pure, and does not absorb moisture. Industrially it is used in the leather (tanning) industry and to prepare azo dyes. Sodium dichromate is used more in organic chemistry because it is more soluble. For NEET, remember 'primary standard' as the key lab use of the potassium salt.
Which one of the following statements is correct when SO2 is passed through acidified K2Cr2O7 solution?
Identify the incorrect statement.
Which one of the following ions exhibits d-d transition and paramagnetism as well?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Chromium in dichromate is +6, which is d0, so there are no d electrons to make d-d transitions. Instead, the orange colour comes from a ligand-to-metal charge transfer (LMCT), where electrons jump from oxygen to chromium. That is also why dichromate is diamagnetic (no unpaired electrons).
Chromate (CrO4^2-) is yellow and exists in basic (alkaline) solution; dichromate (Cr2O7^2-) is orange and exists in acidic solution. They interconvert with pH: 2CrO4^2- + 2H+ -> Cr2O7^2- + H2O in acid, and Cr2O7^2- + 2OH- -> 2CrO4^2- + H2O in base. Chromium is +6 in both.
Six electrons. The half reaction is Cr2O7^2- + 14H+ + 6e- -> 2Cr3+ + 7H2O. Each of the two chromium atoms goes from +6 to +3, a drop of 3 each, so 2 x 3 = 6 electrons in total. This is why K2Cr2O7 has an n-factor of 6 in acidic titrations.
K2Cr2O7 can be obtained very pure, is stable in air, does not absorb water (non-hygroscopic), and its solution does not decompose on standing. These properties make it a reliable primary standard. KMnO4 cannot be used as a primary standard because it always contains some MnO2 impurity and its solution slowly decomposes.
