p-Block Elements

When lead sulphide (PbS) reacts with dilute nitric acid, the reaction is a redox process where lead sulphide is oxidized and nitric acid is reduced.

The overall reaction involves the transformation of PbS and HNO3 to form lead nitrate (Pb(NO3)2), sulphur, and nitrogen oxides, along with water.

The specific nitrogen oxide formed depends on the reaction conditions, including the concentration of nitric acid and the temperature.

Typically, with dilute nitric acid, nitric oxide (NO) and nitrogen dioxide (NO2) are more common products than nitrous oxide (N2O).

The general reaction can be represented as:

Or, under some conditions, sulphur (S) may also be one of the products instead of or in addition to SO2:

From the reaction above, you can see: Sulphur (Option A) can be formed depending on the specific conditions of the reaction, making it a potentially correct product.

Nitric oxide (Option B) is typically produced when dilute nitric acid reacts with PbS, making it a correct product.

Lead nitrate (Option C) is definitely formed in the reaction, making it a correct product.

However, Nitrous oxide (Option D) is not a typical product of the reaction of lead sulphide with dilute nitric acid.

It could be formed under special circumstances in reactions involving nitrogen compounds, but it's not a standard outcome for the conditions described.

Therefore, the correct answer to the question is Option D: Nitrous oxide.

It is not a typical product of the reaction of lead sulphide with dilute nitric acid under normal conditions.

Let's analyze each statement provided about p-block elements and determine their correctness: (A) Non metals have higher electronegativity than metals.

This statement is correct.

Non-metals, especially those in the upper right of the periodic table like fluorine, oxygen, and nitrogen, have higher electronegativity values compared to metals.

(B) Non metals have lower ionisation enthalpy than metals.

This statement is incorrect.

Non-metals generally have higher ionization enthalpy compared to metals because they hold onto their electrons more tightly due to higher nuclear charge and smaller atomic radii.

(C) Compounds formed between highly reactive nonmetals and highly reactive metals are generally ionic.

This statement is correct.

Highly reactive nonmetals (like halogens) tend to gain electrons while highly reactive metals (like alkali metals and alkaline earth metals) tend to lose electrons, resulting in the formation of ionic compounds.

(D) The non-metal oxides are generally basic in nature.

This statement is incorrect.

Non-metal oxides are generally acidic in nature.

They react with bases to form salts and water.

(E) The metal oxides are generally acidic or neutral in nature.

This statement is incorrect.

Metal oxides are usually basic in nature, although some metal oxides can be amphoteric (both acidic and basic).

Most metal oxides react with acids to form salts and water.

After evaluating all the statements, the correct statements are (A) and (C).

Therefore, the correct answer is: Option D: (A) and (C) only

Let's analyze each statement step by step to identify the incorrect ones: (A) Dinitrogen is a diatomic gas which acts like an inert gas at room temperature.

This statement is generally correct.

Dinitrogen (

) is indeed a diatomic gas and is very inert at room temperature due to the strong triple bond between the nitrogen atoms.

(B) The common oxidation states of these elements are

and +5. This statement is also correct. The group 15 elements commonly exhibit oxidation states of

(in nitrides),

(as in

), and +5 (as in

). (C) Nitrogen has unique ability to form

multiple bonds. This statement is correct. Nitrogen has a unique ability to form

multiple bonds, which is why it can form a triple bond (

).

(D) The stability of +5 oxidation states increases down the group.

This statement is incorrect.

The stability of +5 oxidation states actually decreases down the group due to the effect of inert pair effect.

For example, bismuth in +5 state is less stable than nitrogen in +5 state.

(E) Nitrogen shows a maximum covalency of 6.

This statement is incorrect.

Nitrogen can exhibit a maximum covalency of 4 (as in

).

It cannot expand its octet to show a covalency of 6.

Based on the above analysis, the incorrect statements are (D) and (E).

Therefore, the correct answer is: Option B: (D) and (E) only

For test of phosphorus:

To evaluate the statements provided, let's first understand the concepts involved: Assertion A: The stability order of the +1 oxidation state of

, In, and

is Ga < In < Tl. Reason R: The inert pair effect stabilizes the lower oxidation state down the group. The elements

(Gallium),

(Indium), and

(Thallium) belong to Group 13 of the periodic table.

As we move down this group, the +3 oxidation state becomes less stable while the +1 oxidation state becomes more stable.

This trend is attributed to the inert pair effect.

The inert pair effect refers to the reluctance of the s-electrons to participate in bonding as we move down the group in the periodic table.

For Group 13 elements, this effect causes the lower +1 oxidation state to be more stable compared to the +3 oxidation state down the group.

Given this, the Reason R correctly explains why the +1 oxidation state is more stable for

compared to

and

.

Therefore, the stability order of the +1 oxidation state should indeed be Ga < In < Tl.

Conclusion: Both Assertion A and Reason R are true, and Reason R is the correct explanation for Assertion A.

Therefore, the correct answer is: Option D: Both A and R are true and R is the correct explanation of A.

To determine which halogens among

, and

can undergo disproportionation reactions, we need to understand what a disproportionation reaction is.

In a disproportionation reaction, a single substance is simultaneously oxidized and reduced, forming two different products.

The halogen X2 disproportionates in water as follows:

Here, X represents a halogen. Let's analyze each halogen to see if disproportionation is possible: Fluorine (

): Fluorine is the most electronegative element and has a very high oxidation potential.

It does not undergo disproportionation because it prefers to remain in the

oxidation state and does not form higher oxidation states easily. Therefore,

cannot undergo a disproportionation reaction. Chlorine (

): Chlorine can undergo disproportionation. In water, chlorine disproportionates as follows:

Here, chlorine is both reduced to

(chloride, -1 oxidation state) and oxidized to

(hypochlorite, +1 oxidation state). Bromine (

): Bromine can also undergo disproportionation. In water, bromine disproportionates as follows:

Here, bromine is both reduced to

(bromide, -1 oxidation state) and oxidized to

(hypobromite, +1 oxidation state). Iodine (

): Iodine can also undergo disproportionation. In alkaline solutions, iodine disproportionates as follows:

Here, iodine is both reduced to

(iodide, -1 oxidation state) and oxidized to

(hypoiodite, +1 oxidation state). Based on the above analysis, all halogens except

can undergo disproportionation. Therefore, the correct answer is: Option B

and

The most abundant isotope of Boron is

. No. of neutrons in it

Oxidation state of boron in

So,

The significant difference in the melting and boiling points of oxygen and sulfur can be explained by considering atomicity.

Oxygen exists as $\mathrm{O}_2$ (Atomicity = 2).

Sulfur exists as $\mathrm{S}_8$ (Atomicity = 8).

Due to sulfur's higher atomicity (forming S8 molecules), its melting and boiling points are considerably higher than those of oxygen.

(A) Group-13 size :

\mathrm{EN}: \mathrm{B}>\mathrm{TI}>\mathrm{In}>\mathrm{Ga}>\mathrm{Al}

\begin{aligned} \text{(C)} \quad & 2 \mathrm{Al}+6 \mathrm{HCl} \rightarrow 2 \mathrm{AlCl}_3+3 \mathrm{H}_2 \\ & \mathrm{Al}+\mathrm{HNO}_3 \rightarrow \text { No reaction (Due to protective oxide layer)} \end{aligned}

\mathrm{TI}

\mathrm{B}

\mathrm{Al}

\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+} \rightarrow

s p^3 d^2$$

(A) Covalent radius of group 14 elements increases from

to

. (B) Electronegativity decreases from

to

and thereafter it remains more or less constant. (C) Maximum covalency of

is 4 as the outermost shell of

is

shell and it cannot expand its octet.

But other elements can expand their covalency due to the availability of vacant d-orbitals.

(D) Heavier elements do not form

bonds due to their larger atomic size.

(E) Carbon can show negative oxidation states when it is covalently bonded to less electronegative elements like

.