The S Block Elements

Step 1: Understanding the amalgam.
An amalgam is a type of alloy where mercury is the solvent, and another metal is dissolved in it. When mercury forms an alloy with sodium, it creates a liquid in solid solution, where the mercury (a liquid) dissolves the solid sodium metal.
Step 2: Analyzing the options.
(A) Solid in liquid solution: This is incorrect. In this case, mercury (liquid) is dissolving sodium (solid), not the other way around.
(B) Solid in solid solution: This is incorrect. Sodium is the solid, but it is dissolved in a liquid (mercury), not in another solid.
(C) Liquid in liquid solution: This is incorrect. Sodium is solid, not liquid.
(D) Liquid in solid solution: This is correct. Mercury (liquid) forms an amalgam with sodium (solid), making it a liquid in solid solution.
Step 3: Conclusion.
The correct answer is (D) Liquid in solid solution.

Step 1: Understanding the oxidation number of As.
In H AsO , hydrogen has an oxidation number of +1 and oxygen has an oxidation number of -2. To determine the oxidation number of As, we use the following equation:
Step 2: Conclusion.
The oxidation number of As in H AsO is +3.

Step 1: Understanding hydrides.
Hydrides are compounds that contain hydrogen, but the number of hydrogen atoms varies depending on the metal's ability to bond with hydrogen. Some metal hydrides are deficient in hydrogen, meaning they have fewer hydrogen atoms than expected.
Step 2: Analyzing the options.
(A) LiH: Lithium hydride (LiH) is a simple ionic hydride and has a typical stoichiometry of one hydrogen atom per lithium atom.
(B) KH: Potassium hydride (KH) similarly has one hydrogen atom per potassium atom.
(C) NiH: Nickel hydride (NiH) is deficient in hydrogen compared to the other metal hydrides. This hydride typically has fewer hydrogen atoms than the stoichiometric amount expected.
(D) NaH: Sodium hydride (NaH) has a stoichiometry of one hydrogen atom per sodium atom, similar to LiH and KH.
Step 3: Conclusion.
The correct answer is (C) NiH, as it is deficient in hydrogen.

Step 1: Understanding soft soap production.
Soft soap is typically made by saponifying fats with potassium hydroxide (KOH). Potassium hydroxide produces softer soaps compared to sodium hydroxide, which is used for hard soaps.

Step 2: Analyzing the options.
(A) Caesium: Caesium hydroxide is not used in soap production due to its high reactivity.
(B) Sodium: Sodium hydroxide is used for hard soaps, not soft soap.
(C) Potassium: Correct — Potassium hydroxide is used in the manufacture of soft soap.
(D) Lithium: Lithium hydroxide is not used in soap production.

Step 3: Conclusion.
The correct answer is (C) Potassium, as it is used in the production of soft soap.

Step 1: Identify the composition of each mineral.
Minerals are naturally occurring compounds that contain specific elements. To identify a magnesium mineral, the compound must contain magnesium as a major constituent.
Step 2: Analyze each option.
(A) Limonite: It is an iron ore containing hydrated iron(III) oxide and does not contain magnesium.
(B) Cryolite: It is a mineral of sodium and aluminium with the formula .
(C) Magnesite: Correct, it is magnesium carbonate with the formula .
(D) Magnetite: It is an iron ore with the formula .
Step 3: Conclusion.
Among the given options, magnesite is the mineral that contains magnesium as its main constituent.

Step 1: Understanding octahedral geometry.
An octahedral complex has coordination number 6, meaning six ligand donor atoms are attached to the central metal ion.

Step 2: Checking coordination numbers.
Options (A), (B), and (C) all have six donor atoms around the metal ion, leading to octahedral geometry.

Step 3: Analyzing option (D).
has coordination number 4 and forms a square planar complex, not octahedral.

Step 4: Conclusion.
Hence, is NOT an octahedral complex.

Step 1: Writing electronic configuration of chromium.
Chromium has an anomalous configuration:


Step 2: Counting unpaired electrons.
The five electrons in 3d orbitals are unpaired and one electron in 4s orbital is also unpaired.

Step 3: Total unpaired electrons.


Step 4: Conclusion.
Therefore, chromium has six unpaired electrons in its observed electronic configuration.

Step 1: Recall HVZ reaction condition.
Hell–Volhard–Zelinsky reaction occurs in carboxylic acids having at least one -hydrogen atom.
Step 2: Analyze methanoic acid.
Methanoic acid ( ) has no -carbon and hence no -hydrogen.
Step 3: Analyze other acids.
Ethanoic, propanoic, and butanoic acids contain -hydrogen atoms.
Step 4: Conclusion.
Methanoic acid does not undergo HVZ reaction.

Step 1: Understanding the Raschig method.
In the Raschig method for phenol preparation, a catalyst like calcium phosphate is used to catalyze the reaction between benzene and chlorine to form phenol.
Step 2: Analyzing the options.
(A) Calcium Sulphate: Incorrect. Calcium sulphate is not used in the Raschig method for phenol preparation.
(B) Calcium phosphate: Correct. Calcium phosphate is the catalyst used in the Raschig method for phenol preparation.
(C) Calcium chloride: Incorrect. Calcium chloride is not used in this process.
(D) Calcium carbonate: Incorrect. Calcium carbonate is not involved in this method.
Step 3: Conclusion.
The correct answer is (B) Calcium phosphate.

Step 1: Understanding the first law of thermodynamics.
The change in internal energy ( ) is given by the equation: Where is the heat absorbed by the system and is the work done by the system.
Step 2: Applying the values.
We are given that the heat absorbed is and the work done by the surroundings is . Hence, the change in internal energy is: Step 3: Conclusion.
The correct answer is (C) 4002 kJ.

Step 1: Understand lime water.
Lime water is a dilute aqueous solution of calcium hydroxide .
Step 2: Analyze each reaction.
(A) Produces calcium chloride and ammonia, not lime water.
(B) Forms calcium silicate, a solid compound.
(C) Produces calcium carbonate, which is insoluble in water.
(D) Reaction of quicklime with water produces calcium hydroxide, which on dissolution forms lime water.
Step 3: Conclusion.
Only reaction (D) leads to the formation of lime water.

Step 1: Hydration tendency of alkali metal ions.
Smaller cations have higher hydration energy and a stronger tendency to form hydrates.

Step 2: Analyse lithium ion.
has the smallest size among alkali metal ions and therefore has very high hydration energy.

Step 3: Behaviour of LiCl.
Due to strong hydration of , lithium chloride readily forms hydrated crystals such as .

Step 4: Conclusion.
Lithium chloride is the alkali metal chloride that crystallises in the hydrated form.

Step 1: Understanding imide ion.
Imide ion has the formula . It contains nitrogen bonded to one hydrogen atom with a charge of .

Step 2: Charge balance with lithium.
Lithium forms ions. Two lithium ions are required to balance the charge of the imide ion.

Step 3: Formation of lithium imide.
Combining two ions with one ion gives the formula .

Step 4: Conclusion.
Hence, the correct formula of lithium imide is }.