Q.Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing pH on a solution of potassium dichromate?

The key is to track the carbon skeleton through two transformations. The starting molecule is isobutane, $(\mathrm{CH}_3)_3\mathrm{CH}$. The first step uses hot $\mathrm{KMnO}_4$, a strong oxidising agent. The second step uses copper at 573 K, a classic dehydration condition for alcohols.

1. Identify the starting compound and the first reaction.

   Isobutane has a tertiary carbon (the CH group) and three methyl groups. Hot $\mathrm{KMnO}_4$ oxidises alkanes at the most substituted carbon, converting a C–H bond to a C–OH bond. So the product X is tert-butyl alcohol: $(\mathrm{CH}_3)_3\mathrm{C}-\mathrm{OH}$.

   Why this works: The tertiary C–H is the weakest, so it’s attacked first; permanganate inserts oxygen.

2. Determine the second reaction.

   Passing an alcohol over copper at 573 K (about 300 °C) causes dehydration. For a tertiary alcohol like tert-butyl alcohol, elimination is very easy — it loses water to form an alkene. The only possible alkene here is isobutylene (2-methylpropene): $(\mathrm{CH}_3)_2\mathrm{C}=\mathrm{CH}_2$.

   Why this works: Tertiary alcohols dehydrate via an E1 mechanism; the carbocation intermediate is stable, and the most substituted alkene (Saytzeff product) forms.

3. Count the hybridisation in Y (isobutylene).

   Draw the structure:

   $$$$

The question asks for the composition of siderite ore. This is a straightforward recall from metallurgy — the study of how metals are extracted from their ores. Each ore has a distinct chemical identity, and siderite is one of the common iron ores.

The key idea: ores are naturally occurring minerals from which metals are extracted. Iron ores include hematite ($\mathrm{Fe_2O_3}$), magnetite ($\mathrm{Fe_3O_4}$), limonite ($\mathrm{Fe_2O_3 \cdot 3H_2O}$), and siderite. The name "siderite" comes from the Greek word for iron, sideros, which hints at its composition.

Let’s go through the options one by one.

1. Option (A): $\mathrm{CuCO_3 \cdot Cu(OH)_2}$

   This is malachite, a copper ore. The dot here indicates a hydrated or basic carbonate — copper carbonate combined with copper hydroxide. Not siderite.

2. Option (B): $\mathrm{FeCO_3}$

   This is iron(II) carbonate. Siderite is indeed ferrous carbonate. It forms in sedimentary environments and is an important but less common iron ore. The iron is in the +2 oxidation state.

3. Option (C): $\mathrm{CuFeS_2}$ …

The relevant concept here is the trend in oxide formation among alkali metals. As you go down Group 1, the metal cation becomes larger and less polarizing. This stabilizes larger, more complex oxygen anions:

• Small Li forms only the normal oxide (O²⁻).
• Na forms the peroxide (O₂²⁻) under normal conditions.
• K, Rb, and Cs form superoxides (O₂⁻) because the large cation stabilizes the big superoxide ion. Also, excess vs. limited oxygen doesn’t change the product for Cs — it always forms the superoxide.

Let’s work through it step by step:

1. Identify the product for Cs with oxygen

   Cesium is the largest alkali metal. Its cation (Cs⁺) is very large and has low charge density, which stabilizes the large superoxide ion (O₂⁻).

   • With excess O₂: Cs + O₂ → CsO₂ (superoxide).
   • With limited O₂: Cs still forms CsO₂ because the superoxide is the most stable oxide for Cs. So X = superoxide.

2. Identify the product for Na with oxygen

   Sodium is smaller than Cs. Its cation (Na⁺) has a higher charge density, which stabilizes the peroxide ion (O₂²⁻) better than the superoxide.

   • Na + O₂ → Na₂O₂ (peroxide). So Y = peroxide.

3. Match with the options …

The question tests your memory of common ores and their classification by chemical nature — oxide, sulphide, carbonate, or carbonate-hydroxide. In metallurgy, ores are grouped by the anion present: oxides (O²⁻), sulphides (S²⁻), carbonates (CO₃²⁻), and sometimes mixed salts like basic carbonates (carbonate-hydroxide). Getting this right is just a matter of recalling the formula of each ore.

Let’s go through each one.

1. Calamine – This is an ore of zinc. Its chemical formula is ZnCO₃. That’s a pure carbonate — no hydroxide group. So it belongs to List-II: IV (Carbonate).

2. Copper pyrites – Also called chalcopyrite, this is the most common copper ore. Its formula is CuFeS₂. The presence of S²⁻ makes it a sulphide. So it matches List-II: II (Sulphide).

3. Zincite – Another zinc ore, but this one is ZnO. That’s straightforward — an oxide. So it goes with List-II: I (Oxide). …

The oxidation state of an element in a compound represents the hypothetical charge it would have if all bonds were ionic. While there are general rules for assigning oxidation states, it's crucial to understand that these rules have exceptions, especially when dealing with peroxides or superoxides.

For oxygen, the most common oxidation state is -2. However, in peroxides (containing an O-O bond), each oxygen atom has an oxidation state of -1. In superoxides, it's -1/2. When calculating oxidation states, if the standard assumption (e.g., O = -2) leads to an impossibly high or low oxidation state for the central atom, it's a strong indicator that an exception to the general rules is present, and the compound's structure needs to be considered.

Let's determine the oxidation state of Cr in CrO$_5$:

1. Initial Calculation (and why it's incorrect):

   If we assume all five oxygen atoms have an oxidation state of -2 (the most common state for oxygen), then for the neutral compound CrO$_5$:

   Let the oxidation state of Cr be $x$.

   $x + 5 \times (-2) = 0$

   $x - 10 = 0$

   $x = +10$

   This result, +10, is impossible for chromium. Chromium is a d-block element in Group 6, and its maximum possible oxidation state is +6 (corresponding to the total number of valence electrons it can lose). An oxidation state of +10 would imply the loss of more electrons than it possesses, which is chemically unfeasible. This tells us that our initial assumption about all oxygen atoms being -2 is incorrect.

2. Understanding the Structure of CrO$_5$:

   The compound CrO$_5$ has a characteristic "butterfly" structure. This structure reveals the presence of peroxide linkages.

   It consists of:

   • One oxygen atom double-bonded to chromium (a normal oxide oxygen).
   • Two peroxide linkages, each containing two oxygen atoms. This means there are four oxygen atoms involved in O-O bonds. So, out of the five oxygen atoms, one is a normal oxide, and four are peroxide oxygens.

3. Assigning Correct Oxidation States to Oxygen:

   • The oxygen atom that is double-bonded to chromium (Cr=O) has an oxidation state of -2. …

The Chemical Oxygen Demand (COD) test is a crucial measure of water quality, indicating the amount of oxygen required to chemically oxidize all oxidizable organic and inorganic substances present in a water sample. This test provides a rapid estimate of the pollution load in wastewater.

The core idea behind COD determination is to subject the water sample to a powerful chemical oxidation process. A strong oxidizing agent is used to break down the complex organic molecules into simpler, stable inorganic end products like carbon dioxide and water. The amount of oxidant consumed is then measured, which is directly proportional to the oxygen equivalent of the oxidizable matter.

Here's why a specific reagent system is chosen:

1. Need for a Strong Oxidizing Agent: To ensure nearly complete oxidation of a wide range of organic compounds, a very strong oxidizing agent is required.

   • Potassium dichromate ($K_2Cr_2O_7$) is an excellent choice because it is a powerful oxidant, stable, and can oxidize almost all types of organic compounds (except for a few aromatic hydrocarbons and pyridine) to carbon dioxide and water. Its oxidation efficiency is high and consistent.
   • Potassium permanganate ($KMnO_4$), while also an oxidant, is generally less effective than dichromate for COD determination. Its oxidizing power can vary with pH, and it does not oxidize all organic compounds as completely as dichromate, leading to lower and less reliable COD values. It is more commonly used in other specific titrations or for measuring oxygen consumed in milder conditions (e.g., Permanganate Value).

2. Role of Acidification: The oxidizing power of potassium dichromate is significantly enhanced in an acidic environment.

   • Sulphuric acid ($H_2SO_4$) is added to provide the necessary acidic medium. In this acidic environment, the dichromate ion ($Cr_2O_7^{2-}$) is converted to chromic acid, which is a much stronger oxidizing agent. The reaction proceeds as follows: $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$ …