The P Block Elements

Step 1:
Classify the oxides as acidic, basic, or amphoteric: - Basic oxides:
- Acidic oxides:
- Amphoteric oxides:

Step 2:
Count the oxides in each category:
- Amphoteric: 3 ( )
- Basic: 3 ( )
- Acidic: 2 ( ) Thus, the correct answer is option (2).

- Statement I: is a basic oxide. This is correct, as behaves as a basic oxide. - Statement II: is more ionic in nature than . This is true. is more ionic compared to due to the difference in oxidation states of titanium. - Statement III: Boron reacts with dinitrogen at high temperature to form BN. This is correct, as boron nitride (BN) is formed under these conditions. Thus, all statements are correct.

In group 14 elements: - Silicon (Si) has a relatively high melting point. - Tin (Sn) has the lowest melting point. Thus, the correct order is Sn (lowest melting point) and C (highest melting point).

Step 1: Identify Acid 'X':
When an aqueous solution of Borax ( ) is acidified (usually with HCl or ), it yields Orthoboric acid ( ). Reaction: So, X is Boric Acid ( ).
Step 2: Analyze the Statements about Boric Acid:

• Statement I: It is a white crystalline solid (not liquid) with a soapy touch. So, "liquid" makes this statement False.
• Statement II: Boric acid is sparingly soluble in cold water but highly soluble in hot water. True.
• Statement III: It is a very weak monobasic acid (not strong). It does not donate directly; rather, it accepts from water. False.
• Statement IV: It acts as a Lewis acid because the central boron atom has an incomplete octet (electron deficient) and accepts an electron pair from ion of water ( ). True.

Step 3: Conclusion:
Statements II and IV are true.

Step 1: Analyze the Nature of Oxides:

• : Germanium(II) oxide is acidic/amphoteric in nature.
• : Carbon monoxide is a well-known neutral oxide (it reacts with neither acids nor bases to form salts).
• : Germanium(IV) oxide is distinctly acidic (and slightly amphoteric).
• : Carbon dioxide is an acidic oxide (forms carbonic acid with water).

Step 2: Identify Neutral Oxides:
Common neutral oxides are , , and .
Step 3: Conclusion:
is the neutral oxide.

Step 1: Check Set I (Atomic Radius):
Trend: Atomic radius decreases from left to right across a period due to increasing effective nuclear charge. Elements: Li ( ), Be ( ), B ( ), C ( ). Correct Order: (or ). Status: Correct.
Step 2: Check Set II (Ionization Enthalpy - IE):
Trend: Generally increases left to right. Exceptions:

• Be ( )>B ( ) (Penetration/Stability).
• Mg ( )>Al ( ) (Stability of full s-orbital).

Elements: Na ( ), Mg ( ), Al ( ), Si ( ). Correct Order: . (Al comes before Mg due to the exception). Status: Correct.
Step 3: Check Set III (Electronegativity):
Trend: Increases left to right across a period. Elements from Group 13 to 15: Al (1.61), Si (1.90), C (2.55), N (3.04). Although Al and Si are period 3 and C and N are period 2, N is the most electronegative, followed by C. Metals/Metalloids (Al, Si) have lower EN. Correct Order: . Status: Correct.
Step 4: Conclusion:
All three sets are correctly matched.

Industrial production of diborane (B H ) commonly uses the reduction of BF with metal hydrides. Reaction I uses LiAlH and Reaction II uses NaBH . Reaction III is less practical industrially. Therefore, I and II are correct reactions.

1. Atomic radii generally increase down a group due to additional electron shells.
2. For group 13 (Al, Ga, In, Tl), expect Al 3. Approximate radii: Al (143 pm), Ga (135 pm), In (167 pm), Tl (170 pm).
4. Thus, the order is Ga 5. Therefore, the correct option is (2) Ga < Al < In < Tl.

1. In group 14, the nature of oxides changes down the group, with acidity decreasing and basicity increasing.
2. Option (1): SnO and PbO are amphoteric oxides, meaning they can react with both acids and bases. For example, SnO + 2HCl → SnCl + H O (acidic) and SnO + 2NaOH → Na SnO + H O (basic). Similarly for PbO. They are not neutral, so this set is incorrectly matched.
3. Option (2): SnO and PbO are amphoteric, reacting with acids (e.g., SnO + 4HCl → SnCl + 2H O) and bases (e.g., SnO + 2NaOH → Na SnO + H O), correct.
4. Option (3): SiO and GeO are acidic oxides. SiO reacts with bases to form silicates (SiO + 2NaOH → Na SiO + H O), and GeO is similar, correct according to standard classifications.
5. Option (4): CO is acidic (forms carbonic acid with water), and GeO is acidic (behaves like an acidic oxide), correct.
6. Therefore, the set not correctly matched is (1) SnO, PbO - Neutral.

The inert pair effect makes Pb(IV) less stable, so PbI does not exist. Fluorides of Sn and Pb are ionic due to the high electronegativity of F. The trend in statement I does not correctly reflect the effect of inert pair.

• Ionization enthalpy: energy required to remove an electron.
• X = 2080 kJ/mol → very high, corresponds to noble gas configuration after first electron removed, so Na → Ne configuration.
• Y = 496 kJ/mol → Mg (low, easy to remove electron).
• Z = 737 kJ/mol → Al.
• Hence elements X, Y, Z = Na, Mg, Al.

1. Metalloids exhibit properties intermediate between metals and nonmetals, typically along the staircase in the p-block.
2. Standard metalloids: B, Si, Ge, As, Sb, Te, Po, At.
3. From the list: Sb (antimony) – yes; Ge (germanium) – yes; Te (tellurium) – yes.
4. Be (beryllium) – metal; P (phosphorus) – nonmetal; S (sulfur) – nonmetal; Cs (cesium) – metal; Tc (technetium) – metal; I (iodine) – nonmetal.
5. Thus, 3 metalloids.
6. Therefore, the correct option is (1) 3.