Periodic Classification Of Elements

Step 1: Structure of diamond.
In diamond, each carbon atom is sp hybridized and forms four strong covalent bonds with four neighbouring carbon atoms.
Step 2: Formation of rigid 3D network.
These covalent bonds extend throughout the crystal forming a rigid three-dimensional tetrahedral lattice.
Step 3: Reason for hardness.
Since every carbon is strongly bonded in all directions, it is extremely difficult to break the structure, hence diamond is very hard.
Step 4: Conclusion.
Thus hardness is due to strong covalent bonding of all four valence electrons.
Final Answer:

Step 1: Oxidation state of oxygen in KO .
In KO , potassium has oxidation state .
Let oxidation state of oxygen be .

Step 2: Oxidation state of oxygen in Na O .
Na O is a peroxide.
In peroxides, oxygen has oxidation state .
Step 3: Final conclusion.
So oxidation numbers are and .
Final Answer:

Step 1: Check Assertion.
, , and form superoxides such as .
So Assertion (A) is true.
Step 2: Check Reason.
Stability of superoxides increases down the group because larger cations stabilize the larger anion due to lower polarizing power, not due to decrease in lattice energy.
So Reason (R) is not correct.
Step 3: Final decision.
Assertion true, Reason false.
Final Answer:

Step 1: Identify chemical formula of each mineral.
Felspar is a potassium aluminosilicate:

Asbestos is a silicate mineral like hornblende group:

Pyrargyrite is silver antimony sulphide:

Diaspore is hydrated alumina:

Step 2: Write final matching.

Final Answer:

Step 1: General trend.
Ionisation energy generally increases across a period due to increasing nuclear charge.
Step 2: Apply known exceptions.
- has lower IE than because loses a electron (easier to remove).
- has lower IE than because has paired electrons in causing extra repulsion.
Step 3: Arrange with exceptions.

Final Answer:

The electron gain enthalpy of sodium ( ) is the negative of its ionisation energy, as it involves gaining an electron. Thus, the electron gain enthalpy is , equal to the ionisation energy value for sodium.

Step 2: Conclusion.
The correct answer is , corresponding to option (a).

The atomic radii of elements generally increase as you move down a group and decrease as you move across a period from left to right. Based on this trend, the correct order of increasing atomic radii is .

Step 2: Conclusion.
The correct order of increasing atomic radii is , corresponding to option (b).

We match the compositions of substances in List-I with the corresponding compositions in List-II. By analyzing the chemical formulas and the substances, the correct matching is found.

Step 2: Conclusion.
The correct matching is A - C, B - D, C - A, D - B, corresponding to option (b).

The order of ionic radii is incorrect because has a smaller ionic radius than , making the sequence incorrect.

Step 1: Understanding Electron Gain Enthalpy.
Group 15 elements (such as nitrogen) have a half-filled p-orbital configuration, which is more stable. This results in a higher electron gain enthalpy compared to group 16 elements, which do not have this half-filled stability.
Step 2: Conclusion.
The correct answer is (C), Half-filled stability of group 15 elements.

Step 1: Reaction of acetic acid with thionyl chloride.
Acetic acid reacts with thionyl chloride (SOCl ) to form acetyl chloride.

Step 2: Friedel-Crafts acylation.
Acetyl chloride reacts with benzene in the presence of AlCl to form acetophenone.

Step 3: Hydrolysis.
The next reaction involves hydrolysis of the nitrile group to form a carboxylic acid, yielding the final product as (benzoic acid).

The standard e.m.f of the cell is given by: Current flows from anode to cathode, and the anode is where oxidation occurs, so electrode A (Cr) will act as the anode, and electrode B (Fe) will act as the cathode. Therefore, the given statements are true, making option (4) correct.

Step 1: Hydrolysis of nitrile.
The hydrolysis of a nitrile (CH CH CN) leads to the formation of an amide (CH CH CONH ).

Step 2: Reaction with ammonia.
When the amide reacts with ammonia, it undergoes reduction to form an amine.

Step 3: Conclusion.
The final product is the primary amine .