Class 12th

8. A violet compound of manganese (A) decomposes on heating to liberate oxygen and compounds (B) and (C) of manganese are formed. Compound (C) reacts with KOH in the presence of potassium nitrate to give compound (B). On heating compound (C) with conc. H₂SO₄and NaCl, chlorine gas is liberated and a compound (D) of manganese along with other products is formed. Identify compounds A to D and also explain the reactions involved.

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Class 12th General Principles and Isolation of Elements

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8. KMnO₄ K₂MnO₄ + MnO₂ + O₂

(A) (B) (C)

MnO₂ + KOH → 2K₂MnO₄ + 2H₂O

MnO₂ + 4NaCl + 4H₂SO₄→ MnCl₂ + 2NaHSO₄+2H₂O + Cl₂

(D)

(A) = KMnO₄

(B) = K₂MnO₄

(D) = MnCl₂

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24. Zone refining is based on the principle that

(i) Impurities of low boiling metals can be separated by distillation.

(ii) Impurities are more soluble in molten metal than in solid metal.

(iii) Different components of a mixture are differently adsorbed on an adsorbent.

(iv) Vapors of volatile compounds can be decomposed in pure metal.

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24. Option (ii)

Purification of metal crystals by making a thin region of crystal undergo melting is known as Zone refining. The molten crystal is then moved up along the crystal to get pure form of it. This process is used to get pure form of Silicon and Germanium. Basic principle of the zone refining process is that the impurities are more soluble in molten metal than solid metal to get pure form of metal.

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7. (A) Transition metals can act as catalysts because these can change their oxidation state.

How does Fe(III) catalyse the reaction between iodide and persulphate ions?

(B) Mention any three processes where transition metals act as catalysts.

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Class 12th General Principles and Isolation of Elements

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7. (A) 2I^- + S₂O₈^2- →Fe (III) I₂ + 2SO₄^2-

role of fe ions:

2Fe³⁺+ 2I^- → 2Fe²⁺ + I₂

2Fe²⁺ + S₂O₈^2- → 2Fe³⁺ + 2SO₄^2-

(B) 1. Fine powdered state of Nickel is used in the hydrogenation of oils into fats.

2. Iron is used in the formation of ammonia in Haber's process.

3. In contact process in manufacture of H₂SO₄, vanadium is used in its oxide form- V₂O₅.

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23. A number of elements are available in earth's crust but most abundant elements are

(i) Al and Fe

(ii) Al and Cu

(iii) Fe and Cu

(iv) Cu and Ag

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Class 12th General Principles and Isolation of Elements

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23. Option (i)

Explanation: Aluminum is the third most abundant metal in the earth's crust (8.3 percent approximately by weight). It is found in a variety of igneous minerals, including mica and clays. The second most abundant metal in the earth's crust is iron.

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22. Which of the following reactions is an example of autoreduction?

(i) Fe₃O₄+ 4CO→3Fe + 4CO₂

(ii) Cu₂O + C→2Cu + CO

(iii) Cu²⁺(aq) + Fe(s)→Cu(s) + Fe²⁺(aq)

(iv) Cu₂O + $\frac{1}{2}$ Cu₂S → 3Cu + 12SO₂

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22. Option (iv)

Copper (I) oxide is reduced by copper (I) sulfide in this reaction. Because copper is reduced by itself in this process, it is referred to as auto reduction.

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6. Mention the type of compounds formed when small atoms like H get trapped inside the crystal lattice of transition metals. Also give physical and chemical characteristics of these compounds.

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Class 12th General Principles and Isolation of Elements

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6. Interstitial compounds are formed when small atoms like H, C and N get trapped inside the crystal lattice of transition metals.

Characteristics:

They are hard and rigid in nature having high melting points.

Like pure metals only, they are conductors. They can gain chemical inertness.

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21. When copper ore is mixed with silica, in a reverberatory furnace copper matte is produced. The copper matte contains

(i) sulfides of copper (II) and iron (II)

(ii) sulfides of copper (II) and iron (III)

(iii) sulfides of copper (I) and iron (II)

(iv) sulfides of copper (I) and iron (III)

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21. Option (iii)

Copper ore is mixed with silica before heating in a reverberatory furnace. When heated in a reverberatory furnace, the iron oxide slags of iron silicate and copper are produced. This product is produced in the form of copper matte in the furnace. Presence of Cu₂S and FeS is observed in copper matte. In these compounds, we observe the sulfides of Copper (I) Cu₂⁺ and Iron (II) Fe₂⁺.

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5. (A) Answer the following questions:

(i) Which element of the first transition series has the highest second ionization enthalpy?

(ii) Which element of the first transition series has highest third ionization enthalpy?

(iii) Which element of the first transition series has lowest enthalpy of atomization?

(B) Identify the metal and justify your answer:

(i) CarbonylmM(CO)₅

(ii) MO₃F

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5. (A) (i) Out of all the elements of the first transition series copper has the highest second ionisation enthalpy.

Electronic configuration of Copper is: 3d¹⁰4s¹

After the Loss of first electron from the 4s copper acquires 3d¹⁰ configuration which is stable. Therefore, removal of second electron from the field 3-D orbital is very difficult and requires high amount of energy.

(ii) Among the elements of first transition series zinc has the highest third ionisation enthalpy. Electronic configuration of zinc is: 3d¹⁰4s²

After the loss of two electrons from 4s orbital, Z and +2 Ion acquires 3d¹⁰ fully filled configuration which is highly sta

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4. On the basis of lanthanoid contraction, explain the following:

(a) Nature of bonding in La₂O₃and Lu₂O₃.

(b) Trends in the stability of oxo-salts of lanthanoids from La to Lu.

(c) Stability of the complexes of lanthanides.

(d) Radii of 4d- and 5d-block elements.

(e) Trends in acidic character of lanthanide oxides.

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4. (a) As per fajan's rule, smaller the size of ion, greater is its tendency to make covalent bonds as the positively charged cation strongly attract the negatively charged electron cloud.

On moving from La to Lu in the lanthanoid series, the atomic size decreases so the covalent character increases. Hence, La₂O₃is ionic and Lu₂O₃ is covalent.

(b) The stability of Oxo salts is directly proportional to the size of atom, as we move from La to Lu, the size of the atom decreases and hence the stability of oxo salts also decreases.

(c) As we move along the lanthanide series, the atomic size decreases. As a result, the charge/size ratio incr

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4. On the basis of lanthanoid contraction, explain the following:

(a) Nature of bonding in La₂O₃and Lu₂O₃.

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