CUET UG Chemistry Booster Test – 2 Structural Concepts

When transition metals are heated directly with oxygen, they form metal oxides. This process is an oxidation reaction because the metal combines with oxygen and its oxidation state increases. In NCERT terminology, oxidation involves addition of oxygen or increase in oxidation number. Transition metals form several oxides because they show variable oxidation states due to the involvement of 3d and 4s electrons. For example, manganese forms oxides such as MnO, MnO₂ and Mn₂O₇ in different oxidation states. Therefore, the formation of stable transition metal oxides by heating metals with oxygen is oxidation.

To arrange the oxides in decreasing order of oxidation state, determine the oxidation number of the central metal in each compound by taking oxygen as –2. In Mn₂O₇, the total oxidation number of oxygen is –14, so the two Mn atoms together contribute +14, giving each Mn an oxidation state of +7. In CrO₃, oxygen contributes –6, so chromium is +6. In V₂O₅, oxygen contributes –10, so the two vanadium atoms together contribute +10, making each V +5. In TiO₂, oxygen contributes –4, so titanium is +4. Since the statements have been rearranged, the correct decreasing order based on the new numbering is 3 → 4 → 2 → 1.

In transition metal oxides, the nature of the oxide changes with the oxidation state of the metal. NCERT explains that oxides in lower oxidation states are generally basic and more ionic, while oxides in higher oxidation states become more acidic and more covalent. As the oxidation number of the metal increases, the metal ion has greater polarising power. This causes greater distortion of the oxide ion electron cloud, increasing covalent character and decreasing ionic character. Therefore, with increase in oxidation number, the ionic character of the resulting transition metal oxide generally decreases.

The nature of vanadium oxides follows the general NCERT trend for transition metal oxides. As the oxidation state of the metal increases, the oxide becomes less basic and more acidic. V₂O₃ contains vanadium in the +3 oxidation state and is basic. V₂O₄ contains vanadium in the +4 oxidation state and is less basic than V₂O₃. V₂O₅ contains vanadium in the +5 oxidation state and is amphoteric, although it is mainly acidic. Therefore, statements 1, 2 and 3 are correct. Statement 4 is wrong because V₂O₅ is not purely basic.

The yellow-coloured chromate ion has the formula CrO₄²⁻. To find the oxidation state of chromium, let it be x. Oxygen usually has oxidation state -2. Therefore, x + 4(-2) = -2. This gives x - 8 = -2, so x = +6. Hence chromium is present in the +6 oxidation state in chromate ion. In IUPAC notation, this is represented as chromium(VI). Chromate and dichromate ions are important oxoanions of chromium in which chromium remains in the +6 oxidation state.

The dichromate ion, Cr₂O₇²⁻, is formed by two tetrahedral CrO₄ units sharing one corner oxygen atom. Each chromium atom is surrounded by four oxygen atoms in a tetrahedral arrangement. One oxygen atom acts as a bridge between the two chromium centres, giving the Cr–O–Cr linkage. NCERT describes the structure of dichromate ion as two tetrahedra sharing one corner. The Cr–O–Cr bond angle is about 126°. Therefore, the structural geometry of dichromate is correctly described as two tetrahedra sharing one corner.

Potassium dichromate is prepared from chromite ore, FeCr₂O₄. During the preparation, chromite ore is fused with sodium carbonate in the presence of air to form sodium chromate, Na₂CrO₄, which gives a yellow aqueous solution. On acidification, sodium chromate is converted into sodium dichromate, Na₂Cr₂O₇·2H₂O, which is a more soluble orange salt. Potassium chloride is then added to form potassium dichromate, K₂Cr₂O₇, which separates as less soluble orange crystals. Therefore, the correct matching is 1-c, 2-a, 3-d and 4-b.

Dichromates are important compounds of chromium and are widely used because of their oxidising nature. Sodium dichromate is more soluble in water than potassium dichromate and is extensively used as an oxidising agent in organic chemistry. Potassium dichromate is used as a primary standard in volumetric analysis because it is stable, pure and suitable for preparing standard solutions. It is also used in the leather industry. However, sodium dichromate is not primarily a reducing catalyst; it is an oxidising agent. Therefore, statements 1, 2 and 3 are correct.

In acidic medium, potassium dichromate acts as a strong oxidising agent. The dichromate ion, Cr₂O₇²⁻, is reduced to Cr³⁺. The half-reaction is Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O. When acidified potassium dichromate oxidises hydrogen sulfide, H₂S is oxidised to sulphur, while dichromate itself is reduced to chromium(III) ion. This is a standard NCERT example of the oxidising action of potassium dichromate in acidic medium. Therefore, the reduction product of dichromate is Cr³⁺.

Chromate ions and dichromate ions are interconvertible in aqueous solution depending upon the pH of the solution. In basic medium, chromate ion, CrO₄²⁻, is favoured and the solution is yellow. In acidic medium, chromate ions are converted into dichromate ions, Cr₂O₇²⁻, and the solution becomes orange. The oxidation state of chromium remains +6 in both chromate and dichromate ions. Therefore, the process is not a redox reaction but an acid-base controlled equilibrium. Hence, conversion of chromate to dichromate is fundamentally a pH-dependent equilibrium reaction.

The passage clearly states that in the laboratory, a manganese(II) ion salt is oxidised by peroxodisulphate to permanganate. This means peroxodisulphate acts as the oxidising agent. In this reaction, manganese is oxidised from the +2 oxidation state in Mn²⁺ to the +7 oxidation state in permanganate ion, MnO₄⁻. NCERT mentions peroxodisulphate as the oxidising agent used in the laboratory preparation of potassium permanganate. Therefore, the correct answer is peroxodisulphate.

The passage states that potassium permanganate is prepared by fusion of MnO₂ with an alkali metal hydroxide and an oxidising agent such as KNO₃. This initial fusion produces dark green K₂MnO₄, which is potassium manganate. Later, manganate(VI) is oxidised or disproportionates to give permanganate. The green colour is characteristic of manganate ion, MnO₄²⁻, while the purple colour is characteristic of permanganate ion, MnO₄⁻. Therefore, the dark green intermediate formed during the initial fusion of MnO₂ is K₂MnO₄.

In acidic medium, permanganate ion, MnO₄⁻, is reduced to Mn²⁺. The relevant half-reaction is MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. The standard electrode potential for this reduction is E° = +1.52 V. This high positive value explains why acidified potassium permanganate is a powerful oxidising agent. Unit verification: electrode potential is measured in volts, represented by V. Therefore, the correct value for the reduction of MnO₄⁻ to Mn²⁺ in acidic medium is +1.52 V.

The products of permanganate reduction differ according to the medium. In acidic medium, MnO₄⁻ is reduced to Mn²⁺, where manganese has oxidation state +2. In neutral or faintly alkaline medium, permanganate is reduced to MnO₂, where manganese is +4. In strongly alkaline medium or one-electron reduction, permanganate may form manganate ion, MnO₄²⁻, where manganese is +6. Therefore, in increasing order of oxidation state of manganese, the order is Mn²⁺, MnO₂ and MnO₄²⁻, corresponding to 1, 2 and 3.

Interstitial compounds are formed when small atoms occupy the empty spaces, or interstitial sites, in the crystal lattices of transition metals. NCERT mentions that small atoms such as hydrogen, carbon and nitrogen can be trapped inside these lattices. Boron is also commonly involved. These compounds are often non-stoichiometric and retain metallic properties such as conductivity. They are generally very hard and have high melting points. Since H, C and N are small enough to fit into interstitial spaces, they are typically involved in the formation of interstitial compounds.

Interstitial compounds of transition metals show characteristic properties. Their melting points are generally higher than those of the pure metals. They are very hard, and NCERT specifically notes that some borides approach diamond in hardness. They retain metallic conductivity because the metallic lattice is still largely preserved. They are also chemically inert. These properties arise because small atoms such as H, C, N or B occupy interstitial spaces in the metal lattice without completely destroying metallic bonding. Therefore, the correct matching is 1-d, 2-c, 3-b and 4-a.

Transition metals readily form alloys because their atomic radii are similar, often within about 15 percent of one another. This allows atoms of one metal to be randomly distributed among the atoms of another metal in the crystal lattice. The resulting alloys are generally hard and often have high melting points, making them industrially useful. NCERT mentions many ferrous alloys involving transition metals such as chromium, manganese, vanadium and tungsten. Statement 4 is incorrect because alloys are not described as only strictly heterogeneous crystal structures; many are substitutional solid solutions. Therefore, statements 1, 2 and 3 are correct.

Mischmetall is a well-known alloy associated with lanthanoids. According to NCERT, it contains about 95% lanthanoid metal, about 5% iron and traces of elements such as sulphur, carbon, calcium and aluminium. It is used in magnesium-based alloys to produce bullets, shell and lighter flint. The name "mischmetall" means mixed metal, reflecting its composition as a mixture of lanthanoid metals. Therefore, the alloy described in the question is mischmetall.

In the production of steels, several transition metals are added to iron to improve its properties. Chromium is added to improve corrosion resistance and hardness, especially in stainless steel. Manganese improves strength and removes impurities such as sulphur. Nickel improves toughness and corrosion resistance. These elements are commonly used in different steel alloys. Mercury, however, is a liquid metal at room temperature and is not typically added as an alloying element in mainstream steel production. Therefore, mercury is the element that is not usually used as a steel alloying metal.

Silver bromide, AgBr, is a light-sensitive compound and forms the basis of traditional photographic films. When AgBr is exposed to light, it undergoes photochemical decomposition to form metallic silver. The formation of finely divided silver produces the dark image on photographic film after development. NCERT mentions the importance of silver compounds such as AgBr in the photographic industry due to their light-sensitive nature. Other transition metal compounds may have industrial uses, but AgBr is specifically associated with photography. Therefore, AgBr is the correct compound.