p-Block Elements

Correct order of

(i) Boiling point: $\mathrm{HF}>\mathrm{HI}>\mathrm{HBr}>\mathrm{HCl}$

As we go from $\mathrm{HCl}$ to $\mathrm{HI}$, the size and molar mass increase, so van der Waals forces increase and the boiling point increases.

But $\mathrm{HF}$ has strong hydrogen bonding because fluorine is highly electronegative, so its boiling point becomes the highest.

(ii) Melting point : $\mathrm{HI}>\mathrm{HF}>\mathrm{HBr}>\mathrm{HCl}$

In general, from $\mathrm{HCl}$ to $\mathrm{HI}$, increasing size and mass increase intermolecular forces, so melting point increases.

However, $\mathrm{HF}$ shows hydrogen bonding, so its melting point is higher than $\mathrm{HCl}$ and $\mathrm{HBr}$.

Overall, $\mathrm{HI}$ has the highest melting point among hydrogen halides due to stronger dispersion forces in the solid state.

Hydrides of group 15: $\mathrm{NH_3,\ PH_3,\ AsH_3,\ SbH_3}$

A. Stability decreases down the group — Correct

Down the group, the $\mathrm{E-H}$ bond strength decreases because atomic size increases, so overlap becomes poorer. Hence hydrides become less stable:

$\mathrm{NH_3 > PH_3 > AsH_3 > SbH_3}$

B. Basicity decreases down the group — Correct

Basicity depends on the availability of the lone pair on $\mathrm{E}$. Down the group, the lone pair becomes more diffused and less available for donation, so basicity decreases:

$\mathrm{NH_3 > PH_3 > AsH_3 > SbH_3}$

C. Reducing character increases down the group — Correct

Reducing nature increases as the $\mathrm{E-H}$ bond becomes weaker down the group, so the hydride can donate hydrogen (or get oxidised) more easily:

$\mathrm{NH_3 < PH_3 < AsH_3 < SbH_3}$

D. Boiling point increases down the group — Not correct as a general statement

Boiling points generally increase down the group due to increasing molar mass, but $\mathrm{NH_3}$ is anomalous because of hydrogen bonding, which raises its boiling point above what is expected.

So the trend is not strictly increasing from $\mathrm{NH_3}$ to $\mathrm{SbH_3}$.

Correct option: Option A (A, B & C only)

The lightest element in group 7 is manganese (Mn).

An oxoanion where Mn is in the +6 oxidation state is the manganate ion, $\text{MnO}_4^{2-}$.

Its potassium salt is potassium manganate, $\text{K}_2\text{MnO}_4$, which is green.

Correct option: D) green

Incorrect statement: Option C

• $^{13}\text{C}$ is stable (non-radioactive).

• The common radioactive isotope of carbon is $^{14}\text{C}$ (also $^{11}\text{C}$ is radioactive, but short-lived).

The other statements (A, B, D) are correct in standard chemical context.

$ \mathrm{PbO}_2+\mathrm{Pb} \rightarrow 2 \mathrm{PbO} $

In this reaction, lead changes from $+4$ (in $\mathrm{PbO}_2$) and $0$ (in $\mathrm{Pb}$) to $+2$ (in $\mathrm{PbO}$).

Since $\mathrm{Pb}^{2+}$ is more stable than $\mathrm{Pb}^{4+}$, the reaction prefers to form $\mathrm{Pb}^{2+}$.

So, this reaction is spontaneous, hence $\Delta \mathrm{G}^{\circ}(1)$ is negative.

$ \mathrm{SnO}_2+\mathrm{Sn} \rightarrow 2 \mathrm{SnO} $

In this reaction, tin changes from $+4$ (in $\mathrm{SnO}_2$) and $0$ (in $\mathrm{Sn}$) to $+2$ (in $\mathrm{SnO}$).

But $\mathrm{Sn}^{2+}$ is less stable than $\mathrm{Sn}^{4+}$, so forming $\mathrm{Sn}^{2+}$ is not preferred.

Therefore, this reaction is non-spontaneous, hence $\Delta \mathrm{G}^{\circ}(2)$ is positive.

(A) $\mathrm{B}^{+3}<\mathrm{Al}^{+3}<\mathrm{Ga}^{+3}<\mathrm{In}^{+3}<\mathrm{Tl}^{+3}$ : ionic size. This order means that the ionic radius increases as we go down the group (from $\mathrm{B}^{+3}$ to $\mathrm{Tl}^{+3}$). So the ionic radii do not decrease down the group. Therefore, statement (A) is incorrect.

(B) $\mathrm{B}>\mathrm{Tl}>\mathrm{In}>\mathrm{Ga}>\mathrm{Al}: \mathrm{EN}$. This is the actual order of electronegativity in group 13. It does not simply decrease from top to bottom; instead, there are irregular variations. So the statement that electronegativity decreases down the group is not correct. Therefore, statement (B) is incorrect.

(C) $\mathrm{B}>\mathrm{Tl}>\mathrm{Ga}>\mathrm{Al}>\mathrm{In}:$ IE. From this order of first ionisation enthalpy, we can see that boron has the highest first ionisation enthalpy among the group 13 elements. So statement (C) is correct.

(D) Trichlorides and triiodides of group $13^{\text {th }}$ elements are covalent in nature. The $\mathrm{MCl_3}$ and $\mathrm{MI_3}$ (where M is a group 13 element) have high polarising power due to the small, highly charged $\mathrm{M^{3+}}$ ion, so they form covalent bonds with halide ions. Hence, statement (D) is also correct.

$ \begin{array}{ll} \mathrm{X}=\mathrm{N} & \mathrm{X}_2 \mathrm{O}_3=\mathrm{N}_2 \mathrm{O}_3(\text { Acidic }) \\ \mathrm{Y}=\mathrm{Sb} & \mathrm{Y}_2 \mathrm{O}_3=\mathrm{Sb}_2 \mathrm{O}_3(\text { Amphoteric }) \end{array} $

Statement-I is true

$ \mathrm{BCl}_3+3 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{~B}(\mathrm{OH})_3+3 \mathrm{HCl} $

Statement-II is false

$\mathrm{HF}<\mathrm{HCl}<\mathrm{HBr}<\mathrm{HI}$ (bond length order)

Bond length in $HX$ increases as we go down the halogen group because the size of the halogen atom increases. So the longest $H-X$ bond is in $\mathrm{HI}$.

$\mathrm{F}<\mathrm{Cl}<\mathrm{Br}<\mathrm{I}$ (radius order)

Covalent radius also increases down the group. So iodine has the largest covalent radius, not the smallest. Hence, the halogen that forms the longest bond with hydrogen (iodine) does not have the smallest covalent radius.

Therefore, Statement I is incorrect.

$\mathrm{PH}_3<\mathrm{AsH}_3<\mathrm{NH}_3<\mathrm{SbH}_3<\mathrm{BiH}_3$ (Boiling point order)

From this order, the hydride with the lowest boiling point is $\mathrm{PH}_3$. So in Statement II, the element $E$ is phosphorus.

Maximum possible covalency of phosphorous is 6

Phosphorus can show a maximum covalency of $6$ (for example, in $\mathrm{PCl}_5$ it shows covalency $5$, and in $\mathrm{PF}_6^-$ it shows covalency $6$). So the part “maximum covalency of $E$ is $4$” is false.

Therefore, Statement II is incorrect.

Element $E$ must be nitrogen $(\mathrm{N})$.

The binary cation $\left(\mathrm{EH}_x\right)^{+}$ formed with hydrogen should match a common stable cation. Nitrogen forms the well-known cation $\mathrm{NH}_4^{+}$ (ammonium ion), so here $E = \mathrm{N}$ and $x = 4$.

Also, $\mathrm{EH}_3$ means $\mathrm{NH}_3$ (ammonia). In alkaline medium, $\mathrm{NH}_3$ reacts with $\mathrm{K}_2\mathrm{HgI}_4$ to give a precipitate of basic mercury(II)amido-iodine, which is the characteristic test for ammonia. This again confirms $E = \mathrm{N}$.

Now we compare first ionisation enthalpy values for the first elements of groups 13, 14, 15, and 16: $\mathrm{B}$ (group 13), $\mathrm{C}$ (group 14), $\mathrm{N}$ (group 15), and $\mathrm{O}$ (group 16).

Across a period, first ionisation enthalpy generally increases due to increasing effective nuclear charge. Also, $\mathrm{N}$ has a stable half-filled configuration $2p^3$, so its first ionisation enthalpy is higher than $\mathrm{O}$.

Therefore, among $\mathrm{B}, \mathrm{C}, \mathrm{N}, \mathrm{O}$, nitrogen has the highest first ionisation enthalpy. So the correct value corresponds to $\mathrm{N}$ (approximately $1402\ \mathrm{kJ\ mol^{-1}}$).

Reaction 1: Finding A

$\mathrm{PbCl_2 + K_2CrO_4 \rightarrow A + 2KCl}$

This is a double displacement reaction (precipitation reaction).

• $\mathrm{Pb^{2+}}$ from $\mathrm{PbCl_2}$ combines with $\mathrm{CrO_4^{2-}}$ from $\mathrm{K_2CrO_4}$

• The product formed is lead chromate

$\boxed{\mathrm{A = PbCrO_4}}$

(This is a yellow precipitate)

Reaction 2: Finding B

$\mathrm{A + NaOH \rightleftharpoons B + Na_2CrO_4}$

Substituting A:

$\mathrm{PbCrO_4 + NaOH \rightleftharpoons B + Na_2CrO_4}$

Lead chromate reacts with excess NaOH. Since lead is amphoteric, it forms a plumbate complex.

Balancing the equation:

$\mathrm{PbCrO_4 + 4NaOH \rightarrow Na_2[Pb(OH)_4] + Na_2CrO_4}$

Here, $\mathrm{Pb^{2+}}$ is oxidised to $\mathrm{Pb^{4+}}$ in the plumbate ion, or more accurately, lead forms the tetrahydroxoplumbate(II) complex which gets oxidised to tetrahydroxoplumbate(IV).

$\boxed{\mathrm{B = Na_2[Pb(OH)_4]}}$

Reaction 3: Finding X

$\mathrm{PbSO_4 + 4CH_3COONH_4 \rightarrow (NH_4)_2SO_4 + X}$

Here, ammonium acetate acts as a complexing agent.

• 4 acetate ions ($\mathrm{CH_3COO^-}$) coordinate with $\mathrm{Pb^{2+}}$

• This forms a tetraacetatoplumbate(II) complex

$\mathrm{PbSO_4 + 4CH_3COONH_4 \rightarrow (NH_4)_2SO_4 + (NH_4)_2[Pb(CH_3COO)_4]}$

$\boxed{\mathrm{X = (NH_4)_2[Pb(CH_3COO)_4]}}$

$\Rightarrow \mathrm{SiO}_2, \mathrm{CO}_2, \mathrm{GeO}, \mathrm{GeO}_2$ are acidic oxides.

$\Rightarrow \mathrm{SnO}, \mathrm{SnO}_2, \mathrm{PbO}, \mathrm{PbO}_2$ are amphoteric oxides; they can react with both acids and bases.

Counting the required pairs

1. $[\mathrm{SiO}_2,\mathrm{CO}_2]$ – both acidic

2. $[\mathrm{SnO},\mathrm{SnO}_2]$ – both amphoteric

3. $[\mathrm{PbO},\mathrm{PbO}_2]$ – both amphoteric

4. $[\mathrm{GeO},\mathrm{GeO}_2]$ – both acidic

Only two pairs (2 and 3) contain oxides that are both amphoteric, so Statement I is correct.

$\Rightarrow \mathrm{BF}_3$ is electron-deficient (only six electrons around B) and therefore acts as a Lewis acid.

The boron atom in $\mathrm{BF}_3$ is $\mathrm{sp}^2$ hybridised, giving the molecule a trigonal planar geometry.

$\mathrm{BF}_3$ accepts the lone pair from $\mathrm{NH}_3$ to form an adduct, confirming its Lewis-acid behaviour. Hence Statement II is also correct.

For group 15 hydrides, the hydrides are:

$ \mathrm{NH_3,\ PH_3,\ AsH_3,\ SbH_3,\ BiH_3} $

Now let us check each statement one by one.

A. Reducing nature increases from $\mathrm{NH_3}$ to $\mathrm{BiH_3}$

Down the group, the $E-H$ bond strength decreases because the size of the central atom increases.

So, the hydrides can release hydrogen more easily and hence their reducing character increases down the group.

Thus,

$ \mathrm{NH_3 < PH_3 < AsH_3 < SbH_3 < BiH_3} $

in reducing nature.

So, A is correct.

B. Tendency to donate lone pair of electrons decreases from $\mathrm{NH_3}$ to $\mathrm{BiH_3}$

All group 15 hydrides have one lone pair on the central atom.

But down the group, the atomic size increases and the lone pair becomes more diffuse and less available for donation.

Hence, basic character decreases down the group:

$ \mathrm{NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3} $

So, the tendency to donate lone pair decreases from $\mathrm{NH_3}$ to $\mathrm{BiH_3}$.

Therefore, B is correct.

C. The stability of hydrides decreases from $\mathrm{NH_3}$ to $\mathrm{BiH_3}$

As we move down the group, the size of the central atom increases and the overlap between $E$ and $H$ orbitals becomes poorer.

Therefore, the $E-H$ bond becomes weaker and thermal stability decreases.

So,

$ \mathrm{NH_3 > PH_3 > AsH_3 > SbH_3 > BiH_3} $

in stability.

Hence, C is correct.

D. $H-E-H$ bond angle decreases from $\mathrm{NH_3}$ to $\mathrm{SbH_3}$

The bond angles are approximately:

$ \mathrm{NH_3} \approx 107.5^\circ $

$ \mathrm{PH_3} \approx 93.5^\circ $

$ \mathrm{AsH_3} \approx 92^\circ $

$ \mathrm{SbH_3} \approx 91^\circ $

So, the bond angle indeed decreases from $\mathrm{NH_3}$ to $\mathrm{SbH_3}$.

Therefore, D is correct.

Since all four statements are correct, the correct option is:

$ \boxed{\text{Option C: A, B, C and D}} $

The reaction described is the well-known brown ring test, which is a confirmatory test for nitrate ions ($NO_3^-$). Let's understand the chemistry behind it.

1. Understanding the Reactants:

   • The test uses a freshly prepared solution of ferrous sulphate ($FeSO_4$). It needs to be fresh because the ferrous ion ($Fe^{2+}$) can be easily oxidized to ferric ion ($Fe^{3+}$) by the oxygen in the air.

   • The question mentions a gas 'X' is treated with this solution. In a standard brown ring test, this gas is produced in situ (within the reaction mixture). Let's see how.

2. Formation of Gas 'X':

   The test is for nitrate ions ($NO_3^-$). In an acidic medium (usually provided by adding concentrated $H_2SO_4$), the nitrate ion oxidizes the ferrous ion ($Fe^{2+}$) to ferric ion ($Fe^{3+}$), and in the process, the nitrate ion itself gets reduced to nitric oxide ($NO$).

   The balanced ionic equation is:

   $NO_3^- + 3Fe^{2+} + 4H^+ \longrightarrow NO + 3Fe^{3+} + 2H_2O$

   So, the gas 'X' that is formed is nitric oxide, $NO$.

3. Formation of the Brown Ring (Compound 'Y'):

   The nitric oxide ($NO$) gas produced then reacts with the excess ferrous ions ($Fe^{2+}$) present in the solution to form a coordination complex. This complex has a characteristic brown colour.

   The reaction is:

   $[Fe(H_2O)_6]^{2+} + NO \longrightarrow [Fe(H_2O)_5(NO)]^{2+} + H_2O$

   This brown-coloured complex is pentaaquanitrosyliron(II) ion. The complete compound 'Y' is pentaaquanitrosyliron(II) sulphate, with the formula $[Fe(H_2O)_5(NO)]SO_4$.

   For simplicity in the options, the water ligands are often omitted, and the compound is written as $[Fe(NO)]SO_4$.

Conclusion:

• The gas 'X' is nitric oxide, $NO$.

• The brown compound 'Y' is the complex $[Fe(NO)]SO_4$.

Looking at the options, we can see that Option A matches our findings.

Therefore, the correct option is A.

Statement I says:

“Aluminium is more electropositive than thallium as the standard electrode potential value of $ E^\circ_{\text{Al}^{3+}/\text{Al}} $ is negative and $ E^\circ_{\text{Tl}^{3+}/\text{Tl}} $ is positive.”

According to standard reduction potentials:

$ E^\circ_{\text{Al}^{3+}/\text{Al}} = -1.66\, \text{V} $

$ E^\circ_{\text{Tl}^{3+}/\text{Tl}} = +1.25\, \text{V} $

A more negative standard reduction potential indicates a greater tendency to lose electrons, meaning the element is more electropositive.

Therefore, aluminium (with a more negative value) is indeed more electropositive than thallium (with a positive value).

So, Statement I is true.

Statement II says:

“The sum of the first three ionization enthalpies of boron is very high when compared to that of aluminium. Due to this reason boron forms covalent compounds only and aluminium forms $ \text{Al}^{3+} $ ion.”

This is correct according to periodic trends:

• Boron, being a small atom with high effective nuclear charge, requires very high energy to remove three electrons. Hence, it cannot easily form $ \text{B}^{3+} $ ions, and thus forms covalent compounds.

• Aluminium, being larger in size, has lower ionization enthalpies, and can readily form $ \text{Al}^{3+} $ ions in its compounds.

Therefore, Statement II is also true.

Both the statements are correct.

$ \boxed{\text{Option A: Both Statement I and Statement II are true}} $

Statement I:

The complex formed is a compound with coordination number 4 not 6 .

$ \text { Statement II: } $

$ \text { } \begin{aligned}Steric ,\,No. \,\, & =4 \mathrm{bp}+2 \mathrm{lp} \\ & =6 \end{aligned} $

$ \begin{aligned} \text { Stervic No. } & =3 bp+1 lp \\ & =4 \end{aligned} $

$ \begin{aligned} \text { Steric No. } & =2 \mathrm{bp}+3 \mathrm{lp} \\ & =5 \end{aligned} $

Statement A: Nitrogen in oxidation states from $+1$ to $+4$ disproportionates in acid medium.

This statement is correct. All oxidation states of nitrogen from $+1$ to $+4$ tend to undergo disproportionation reactions in an acidic solution. For example, nitrous acid, in which nitrogen is in the $+3$ oxidation state, disproportionates into nitric acid ($+5$) and nitric oxide ($+2$):

$ 3\mathrm{HNO}_2 \rightarrow \mathrm{HNO}_3 + \mathrm{H}_2\mathrm{O} + 2\mathrm{NO} $

Statement B: Nitrogen has the ability to form $\mathrm{d} \pi$ - $\mathrm{p} \pi$ multiple bonds with itself and other elements with small size and high electronegativity.

This statement is incorrect. The valence shell electronic configuration of nitrogen is $2\mathrm{s}^2 2\mathrm{p}^3$. Since nitrogen belongs to the second period, it does not possess any empty d-orbitals in its valence shell. Due to the absence of d-orbitals, nitrogen can readily form $\mathrm{p} \pi$ - $\mathrm{p} \pi$ multiple bonds, but it is physically impossible for it to form $\mathrm{d} \pi$ - $\mathrm{p} \pi$ bonds.

Statement C: N-N single bond is stronger than P-P single bond.

This statement is incorrect. The single N-N bond is actually weaker than the single P-P bond. Nitrogen atoms are very small, which results in a short N-N bond length. This brings the non-bonding electron pairs (lone pairs) on the two bonded nitrogen atoms very close to each other, leading to high interelectronic repulsion that weakens the chemical bond.

Statement D: Nitrogen has highest density in its group due to small size.

This statement is incorrect. As we move down Group 15 from nitrogen to bismuth, the atomic mass increases at a greater rate than the atomic volume. As a result, density smoothly increases down the group. Nitrogen is a diatomic gas at room temperature and has the lowest density in the group, whereas bismuth is a solid and possesses the highest density.

Statement E: The maximum covalency of nitrogen is four since it has only four valence orbitals for bonding.

This statement is correct. The valence shell of nitrogen consists of only four orbitals available for bonding: one $\mathrm{s}$ orbital ($2\mathrm{s}$) and three $\mathrm{p}$ orbitals ($2\mathrm{p}_x, 2\mathrm{p}_y, 2\mathrm{p}_z$). Without any d-orbitals available to expand its octet, nitrogen is strictly restricted to a maximum covalency of four.

Through our step-by-step analysis, we have found that only statements A and E are correct.

Correct Answer:

Option D (A and E Only)

For the group 13 element $E$, we are told that its electronegativity is same as that of Ge.

In group 13, the element whose electronegativity is nearly equal to Ge is thallium, Tl.

So, here

$ E = \mathrm{Tl} $

and therefore

$ E^{3+} = \mathrm{Tl}^{3+} $

Now let us examine the statements one by one.

Important concept: Inert pair effect

In group 13, due to inert pair effect, the outer $ns^2$ electrons are less available for bonding in heavier elements.

So for Tl:

• $+1$ oxidation state is more stable

• $+3$ oxidation state is less stable

Thus, $\mathrm{Tl}^{3+}$ tends to gain electrons and get reduced to $\mathrm{Tl}^{+}$.

Statement A: It can act as a reducing agent.

A reducing agent causes reduction of another species and itself gets oxidized.

But $\mathrm{Tl}^{3+}$ is already in a high oxidation state and tends to get reduced, not oxidized further.

So, $\mathrm{Tl}^{3+}$ does not act as a reducing agent.

Therefore, A is false.

Statement B: It can act as an oxidizing agent.

Since $\mathrm{Tl}^{3+}$ is less stable and readily gets reduced to $\mathrm{Tl}^{+}$, it can oxidize other substances.

Hence, $\mathrm{Tl}^{3+}$ acts as an oxidizing agent.

Therefore, B is true.

Statement C: $E^{3+}$ is more stable than $E^{+}$.

For Tl, because of inert pair effect:

$ \mathrm{Tl}^{+} \text{ is more stable than } \mathrm{Tl}^{3+} $

So this statement is false.

Therefore, C is false.

Statement D: The standard electrode potential value for $E^{3+}/E$ is positive.

For thallium, the standard reduction potential for

$ \mathrm{Tl}^{3+} + 3e^- \rightarrow \mathrm{Tl} $

is positive.

This is because $\mathrm{Tl}^{3+}$ is readily reduced.

So, D is true.

Thus, the correct statements are:

$ \boxed{B \text{ and } D \text{ only}} $

So the correct option is:

$ \boxed{\text{Option C}} $

For a monoatomic anion $A^-$:

• Number of electrons in $A^-$ $= 36$

• Since it is an anion with charge $-1$, the neutral atom $A$ must have one electron less.

So, number of electrons in neutral atom $A$:

$ 36 - 1 = 35 $

Hence, atomic number of $A$ is $35$.

This element is bromine.

Now, number of neutrons $= 45$

So, mass number:

$ \text{Mass number} = \text{protons} + \text{neutrons} = 35 + 45 = 80 $

Thus, atomic mass number is:

$ 80 $

Now let us find the group.

Bromine $(Z = 35)$ belongs to group 17 in the periodic table.

Also, bromine is a liquid at room temperature.

So, the correct answer is:

$ \boxed{\text{Option A: } 80,\ 17,\ \text{liquid}} $

Analysis of Statement I:

1. Covalency of Oxygen: The electronic configuration of oxygen is $1s^2 2s^2 2p^4$. In its outermost shell ($n = 2$), it possesses exactly four orbitals available for bonding: one $2s$ orbital and three $2p$ orbitals. Because oxygen is a Period 2 element, it completely lacks empty $d$-orbitals. Consequently, its maximum covalency is restricted to a maximum of 4. While oxygen generally shows a normal covalency of 2 (as seen in $\mathrm{H_2O}$), it can exceed this completely up to 4 in certain scenarios (co-ordinating to maximum four atoms as seen theoretically, and a covalency of 3 is commonly observed in the hydronium ion, $\mathrm{H_3O^+}$). So, the first part of the statement is correct.

2. Oxidation States:

   • In $\mathrm{SO}_2$, oxygen is more electronegative than sulphur. Here, sulphur has an oxidation state of +4, and each oxygen atom securely takes an oxidation state of -2.

   • In $\mathrm{OF}_2$, fluorine is the most electronegative element in the periodic table. Because fluorine is more electronegative than oxygen, each fluorine atom takes an oxidation state of -1. To maintain neutrality in the $\mathrm{OF}_2$ molecule, the oxidation state of oxygen must be +2.

   Therefore, the second part of the statement is perfectly correct as well.

Conclusion: Statement I is true.

Analysis of Statement II:

In the $p$-block elements, the first member of every group always differs significantly from the heavier members of the same group. According to the NCERT, the anomalous behavior of oxygen compared to the rest of the group 16 elements (like Sulphur, Selenium, etc.) is prominently due to:

1. Its exceptionally small atomic size.

2. Its high electronegativity.

3. The non-availability of $d$-orbitals in its valence shell.

Because of its small size and high electronegativity, it readily forms strong $p \pi - p \pi$ double bonds and extensive hydrogen bonds, uniquely distinguishing it from other group elements. The statement accurately lists the main reasons, so it is correct.

Conclusion: Statement II is true.

Final Conclusion:

Since both the provided statements are factually correct based on standard chemical principles, we determine the correct option.

Correct Answer:

Option A

Both Statement I and Statement II are true

For boron family elements, the order of second ionization enthalpy is:

$ \underset{(2427 \mathrm{KJ} / \mathrm{mol})}{\mathrm{B}}>\underset{(1979 \mathrm{KJ} / \mathrm{mol})}{\mathrm{Ga}}>\underset{(1816 \mathrm{KJ} / \mathrm{mol})}{\mathrm{Al}} $

So, the given order $ \mathrm{B} > \mathrm{Al} > \mathrm{Ga} $ is incorrect.

For carbon family elements, the order of first ionization enthalpy is:

$ \underset{(786 \mathrm{KJ} / \mathrm{mol})}{\mathrm{Si}}>\underset{(761 \mathrm{KJ} / \mathrm{mol})}{\mathrm{Ge}}>\underset{(715 \mathrm{KJ} / \mathrm{mol})}{\mathrm{Pb}}>\underset{(708 \mathrm{KJ} / \mathrm{mol})}{\mathrm{Sn}} $

So, the given order $ \mathrm{Si} < \mathrm{Ge} < \mathrm{Pb} < \mathrm{Sn} $ is also incorrect.

Hence, both Statement I and Statement II are false.

(i) True: Tl¹⁺ is more stable than Tl³⁺ due to the inert pair effect. Therefore, Tl³⁺ acts as a powerful oxidizing agent.

(ii) True: The standard reduction potential for Al^3+/Al is -1.66 V, indicating that Al³⁺ is difficult to reduce and thus highly stable.

(iii) False: Tl³⁺ is not stable in solution.

(iv) True: Tl¹⁺ is more stable than Tl³⁺.

(v) True: Both Al³⁺ and Tl¹⁺ are highly stable in their respective oxidation states.

Statement 1 is correct since left to right 1 E increases in general in periodic table.

Statement 2 is incorrect since M.P. of group 14 elements is more than group 13 elements.

In oxide of nitrogen it can achieve +5 oxidation state because it can form $\mathrm{p} \pi-\mathrm{p} \pi$ bond with oxygen e.g. $\mathrm{N}_2 \mathrm{O}_5$

Nitrogen cannot form halide in +5 oxidation state because it does not contain d-orbital.

e.g. $\quad \mathrm{NX}_5$ does not exist

X = halide

Size order

$\begin{aligned} & \mathrm{Al}>\mathrm{Ga} \\ & \mathrm{Al}^{3+}<\mathrm{Ga}^{3+} \end{aligned}$

Atomic number of Ga is 31

single bond weaker than

due to more $\ell \mathrm{p}-\ell \mathrm{p}$ repulsion.

Bond length $\Rightarrow d_{p-p}>d_{\mathrm{N}-\mathrm{N}}(\operatorname{size} \uparrow$, B.L. $\uparrow)$

In group 15 elements only N \& P show disproportionation in +3 oxidation state, As, $\mathbf{S b} \&$ Bi have almost inert for disproportionation in +3 oxidation state.

So both statements are false.

Tellurium dioxide ($\mathrm{TeO}_2$) acts as an oxidizing agent. This is because it can accept electrons and be reduced from its +4 oxidation state to a lower oxidation state.

Tellurium hydride ($\mathrm{TeH}_2$) is considered acidic. This is due to its relatively low bond dissociation energy, which makes it prone to breaking, releasing protons (H$^+$).

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.

$\mathrm{PH_3} \text{ shows lower proton affinity than } \mathrm{NH_3}.$

Ammonia (NH₃) has a more available lone pair on nitrogen than PH₃ on phosphorus due to smaller size and higher electronegativity. Thus, NH₃ is a stronger base and has a higher proton affinity than PH₃.

This statement is correct.

$\mathrm{SO_2} \text{ can act as an oxidizing agent, but not as a reducing agent.}$

Sulfur dioxide (SO₂) is quite versatile in redox chemistry. In many reactions, it actually behaves as a reducing agent by being oxidized (e.g., to sulfate, where sulfur goes from +4 to +6 oxidation state). While under some conditions it may act as an oxidizing agent, it is well known and widely used for its reducing properties.

Hence, the claim that it “cannot act as a reducing agent” is incorrect.

$\mathrm{NO_2} \text{ can dimerise easily.}$

Nitrogen dioxide (NO₂) is known to dimerise to form dinitrogen tetroxide (N₂O₄), especially at lower temperatures.

This statement is correct.

$\mathrm{PF_3} \text{ exists but } \mathrm{NF_5} \text{ does not.}$

Phosphorus trifluoride (PF₃) is a known stable compound. In contrast, a compound like nitrogen pentafluoride (NF₅) is not observed, largely due to the limitations of nitrogen’s size and bonding capabilities in forming such a structure.

This statement is correct.

Based on the analysis, the incorrect statement is:

$\textbf{Option B.}$

Among the group 15 elements (N, P, As, Sb, Bi), the lightest two (N and P) are commonly considered nonmetals. Of these, nitrogen (N) is known to have the weakest single bond to itself (N–N).

The N–N single‐bond enthalpy ($\approx 160$\,kJ/mol) is lower (weaker) than the P–P single‐bond enthalpy ($\approx 200$\,kJ/mol).

Therefore, the “nonmetallic group 15 element with the weakest E–E bond” is nitrogen.

Maximum Covalency of Nitrogen

Although nitrogen typically forms three covalent bonds in neutral compounds (e.g., $\mathrm{NH_3}$), it can expand to four bonds in certain cationic species such as ammonium $\mathrm{NH_4}^+$ or $\mathrm{NF_4}^+$. In such species, nitrogen has a formal positive charge but is still forming four covalent bonds.

Hence, the maximum covalency of nitrogen is 4.

Answer: 4 (Option C).

Oxygen exists as $\mathrm{O}_2$

Sulphur exists as $\mathrm{S}_8$

Because of small size of oxygen, oxygen can form $2 \mathrm{p}_\pi-2 \mathrm{p}_\pi$ bond.

$\Rightarrow$ Assertion (A) is correct.

$\Rightarrow$ Reason (R) is correct.

Reason (R) is correct explanation of Assertion (A).

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 ($N_2$) 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 $-3,+3$ and +5.

This statement is also correct. The group 15 elements commonly exhibit oxidation states of $-3$ (in nitrides), $+3$ (as in $\mathrm{NH}_3$), and +5 (as in $\mathrm{HNO_3}$).

(C) Nitrogen has unique ability to form $\mathrm{p}\pi-\mathrm{p}\pi$ multiple bonds.

This statement is correct. Nitrogen has a unique ability to form $p\pi-p\pi$ multiple bonds, which is why it can form a triple bond ($N_2$).

(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 $\mathrm{NH_4^+}$). 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:

$\mathrm{{H_3}P{O_4} + 12{(N{H_4})_2}Mo{O_4} + 21HN{O_3} \to \mathop {{{(N{H_4})}_3}\,.\,12Mo{O_3}}\limits_{(Yellow\,ppt)} + 21N{H_4}N{O_3} + 12{H_2}O}$

To evaluate the statements provided, let's first understand the concepts involved:

Assertion A: The stability order of the +1 oxidation state of $\mathrm{Ga}$, In, and $\mathrm{Tl}$ is Ga < In < Tl.

Reason R: The inert pair effect stabilizes the lower oxidation state down the group.

The elements $\mathrm{Ga}$ (Gallium), $\mathrm{In}$ (Indium), and $\mathrm{Tl}$ (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 $\mathrm{Tl}$ compared to $\mathrm{Ga}$ and $\mathrm{In}$. 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 $\mathrm{F}_2, \mathrm{Cl}_2, \mathrm{Br}_2$, and $\mathrm{I}_2$ 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:

$ \mathrm{X}_2 + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{HX} + \mathrm{HXO} $

Here, X represents a halogen.

Let's analyze each halogen to see if disproportionation is possible:

Fluorine ($\mathrm{F}_2$): 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 $-1$ oxidation state and does not form higher oxidation states easily. Therefore, $\mathrm{F}_2$ cannot undergo a disproportionation reaction.

Chlorine ($\mathrm{Cl}_2$): Chlorine can undergo disproportionation. In water, chlorine disproportionates as follows:

$ \mathrm{Cl}_2 + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{HCl} + \mathrm{HOCl} $

Here, chlorine is both reduced to $\mathrm{HCl}$ (chloride, -1 oxidation state) and oxidized to $\mathrm{HOCl}$ (hypochlorite, +1 oxidation state).

Bromine ($\mathrm{Br}_2$): Bromine can also undergo disproportionation. In water, bromine disproportionates as follows:

$ \mathrm{Br}_2 + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{HBr} + \mathrm{HOBr} $

Here, bromine is both reduced to $\mathrm{HBr}$ (bromide, -1 oxidation state) and oxidized to $\mathrm{HOBr}$ (hypobromite, +1 oxidation state).

Iodine ($\mathrm{I}_2$): Iodine can also undergo disproportionation. In alkaline solutions, iodine disproportionates as follows:

$ \mathrm{I}_2 + \mathrm{OH}^- \rightarrow \mathrm{I}^- + \mathrm{IO}^- $

Here, iodine is both reduced to $\mathrm{I}^-$ (iodide, -1 oxidation state) and oxidized to $\mathrm{IO}^-$ (hypoiodite, +1 oxidation state).

Based on the above analysis, all halogens except $\mathrm{F}_2$ can undergo disproportionation. Therefore, the correct answer is:

Option B

$\mathrm{Cl}_2, \mathrm{Br}_2$ and $\mathrm{I}_2$

(A) Covalent radius of group 14 elements increases from $\mathrm{C}$ to $\mathrm{Pb}$.

(B) Electronegativity decreases from $\mathrm{C}$ to $\mathrm{Si}$ and thereafter it remains more or less constant.

(C) Maximum covalency of $\mathrm{C}$ is 4 as the outermost shell of $\mathrm{C}$ is $2^{\text {nd }}$ 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 $\mathrm{p} \pi-\mathrm{p} \pi$ bonds due to their larger atomic size.

(E) Carbon can show negative oxidation states when it is covalently bonded to less electronegative elements like $\mathrm{CH}_4$.

S-1 is correct as gallium is used in manufacturing of thermometers.

S-2 is incorrect

To determine which of the following materials is not a semiconductor, we must understand what a semiconductor is and then look at the properties of each material listed.

Semiconductors are materials that have a conductivity level between conductors (such as metals) and insulators (such as most ceramic materials). They can conduct electricity under certain conditions or when doped with impurities. The most common semiconductors are silicon and germanium, but some compound materials like gallium arsenide and silicone carbide also behave as semiconductors.

Now, analyzing each option given:

Option A: Copper oxide - Copper oxide can behave as a semiconductor, particularly in some specific structures or when used in thin-film solar cells. Thus, it can exhibit semiconductor properties.

Option B: Graphite - Graphite is a form of carbon where the carbon atoms are bonded together in a sheet-like form. It is known for its excellent electrical conductivity due to the delocalization of pi electrons across the carbon layers. Its behavior is more characteristic of a conductor rather than a semiconductor.

Option C: Silicon - Silicon is the most well-known and widely used semiconductor material in electronic devices, including transistors, diodes, and integrated circuits. Its semiconducting properties are due to its band structure, which can be altered by doping with other elements.

Option D: Germanium - Germanium is another classic semiconductor material used in the early development of electronic devices. Though not as commonly used as Silicon in modern electronics, due to its similar crystal structure and semiconducting properties, Germanium remains an important semiconductor material.

Based on the provided information, Option B: Graphite is not a semiconductor. Its properties align more with those of a conductor rather than a semiconductor, distinguishing it from the other options listed which all have semiconducting capabilities under certain conditions.

(A) Group-13 size : $\mathrm{B}<\mathrm{Ga}<\mathrm{Al}<\mathrm{In} \simeq$ $\mathrm{TI}$

(B) In Group-13, Al has exceptional lower EN

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

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

(D) +1 oxidation state is observed mainly for heavier Group-13 element like $\mathrm{TI}$.

$\mathrm{B}$ and $\mathrm{Al}$ do not show +1 oxidation state

(E) $\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+} \rightarrow$ Hybridisation : $s p^3 d^2$

The most abundant isotope of Boron is ${ }_5 \mathrm{~B}^{11}$.

No. of neutrons in it $=x=6$

$2 \mathrm{B}(\mathrm{s})+\mathrm{N}_2(\mathrm{g}) \xrightarrow{\Delta} 2 \mathrm{BN}(\mathrm{s})$

Oxidation state of boron in $\mathrm{B N=y=+3}$

So, $x+y=6+3$

$=9$

Both statements need to be evaluated to determine which (if any) are true:

Statement (I) : $\mathrm{SiO}_2$ and $\mathrm{GeO}_2$ are acidic while $\mathrm{SnO}$ and $\mathrm{PbO}$ are amphoteric in nature.

This statement is generally true. The oxides of silicon ($\mathrm{SiO}_2$) and germanium ($\mathrm{GeO}_2$) are indeed acidic. These elements belong to group 14 of the periodic table and as you move down the group, the acidic character of the oxides decreases while the basic character increases. Tin oxide ($\mathrm{SnO}$) and lead oxide ($\mathrm{PbO}$) are lower in the group and they are amphoteric, which means they show both acidic and basic properties depending on the reacting substance.

Statement (II) : Allotropic forms of carbon are due to the property of catenation and $\mathrm{p} \pi-\mathrm{d} \pi$ bond formation.

This statement is false. Catenation is indeed the property that carbon has to form long chains and rings with other carbon atoms, which is responsible for the vast number of organic compounds. However, in the context of pure carbon allotropes—such as diamond, graphite, and fullerene—the different physical structures and properties are mainly due to the $\mathrm{sp}^3$, $\mathrm{sp}^2$, and $\mathrm{sp}$ hybridizations of carbon, not to $\mathrm{p} \pi-\mathrm{d} \pi$ bonding. Carbon does not use d-orbitals in forming the allotropes. The $\mathrm{p} \pi-\mathrm{d} \pi$ bonding is typically discussed in the context of transition metal complexes, not allotropic forms of carbon.

Thus, the correct answer is Option D : Statement I is true but Statement II is false.

The reducing strength of a substance is generally associated with its ability to donate electrons to other substances. In the context of hydrides of Group 15 elements, such as $\mathrm{SbH}_3$, $\mathrm{NH}_3$, $\mathrm{BiH}_3$, and $\mathrm{PH}_3$, the reducing strength can be considered based on their chemical reactivity, stability, and tendency to donate electrons.

In general, the stability of hydrides decreases as we move down the group in the periodic table, which means that heavier hydrides are less stable and hence act as stronger reducing agents. This trend is primarily due to the increase in the size of the central atom as we move down the group, leading to weaker bonds between the central atom and hydrogen in the hydride. As the bond strength decreases, the hydrides can more readily release hydrogen in the form of protons (H⁺) or hydride ions (H⁻), making them stronger reducing agents.

The order of reducing strength for Group 15 hydrides is typically as follows:

$\mathrm{BiH}_3 > \mathrm{SbH}_3 > \mathrm{PH}_3 > \mathrm{NH}_3$

Among the options given, $\mathrm{BiH}_3$ (bismuthine) is the strongest reducing agent. It is the least stable of the listed hydrides, due to the large size and low electronegativity of bismuth compared to antimony, phosphorus, and nitrogen. This makes $\mathrm{BiH}_3$ more prone to releasing electrons and acting as a reducing agent.

In contrast, $\mathrm{NH}_3$ (ammonia) is the most stable among the given hydrides and, therefore, the weakest reducing agent. Its bond with hydrogen is comparatively stronger, and nitrogen's higher electronegativity makes the molecule more resistant to donating electrons.

Thus, the correct answer is Option C, $\mathrm{BiH}_3$.

In trivalent state most of the compounds being covalent are hydrolysed in water. Trichlorides on hydrolysis in water form tetrahedral $\left[\mathrm{M}(\mathrm{OH})_4\right]^{-}$ species, the hybridisation state of element $\mathrm{M}$ is $\mathrm{sp}^3$.

In case of aluminium, acidified aqueous solution forms octahedral $\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ ion.

$\mathrm{S}_1: \mathrm{S}_8+12 \mathrm{OH}^{\ominus} \rightarrow 4 \mathrm{~S}^{2-}+2 \mathrm{~S}_2 \mathrm{O}_3^{2-}+6 \mathrm{H}_2 \mathrm{O}$

$\mathrm{S}_2: \mathrm{ClO}_4^{\ominus}$ cannot undergo disproportionation reaction as chlorine is present in it's highest oxidation state.

(A) All group 16 elements form oxides of the $\mathrm{EO}_2$ and $\mathrm{EO}_3$ type where $\mathrm{E}=\mathrm{S}, \mathrm{Se}, \mathrm{Te}$ or $\mathrm{Po}$.

(B) $\mathrm{SO}_2$ is reducing while $\mathrm{TeO}_2$ is an oxidising agent.

(C) The reducing property increases from $\mathrm{H}_2 \mathrm{S}$ to $\mathrm{H}_2$ Te down the group.

(D)

SnO$_2$ and PbO$_2$ are amphoteric.

Option A is the correct answer. Both Statement I and Statement II are false.

Noble gases actually have low boiling points, not high. This is because noble gases are indeed monoatomic gases that exist as single atoms. They have a completely filled valence shell and do not easily form chemical bonds with other atoms. As a result, the intermolecular forces present in noble gases are only temporary dipole-induced dipole attractions, also known as dispersion forces (or London forces). However, these dispersion forces are very weak compared to other types of intermolecular forces like hydrogen bonding or dipole-dipole interactions that occur in other substances.

The dispersion forces in noble gases are weak because these gases are non-polar and since they are monoatomic, the electron cloud around the atom is symmetrical, meaning that there are no permanent dipoles within the molecules. As the atomic size of noble gases increases down the group, the dispersion forces do increase in strength due to the larger electron cloud that can be more easily distorted, resulting in higher boiling points for the heavier noble gases. Still, compared to other substances, the overall strength of the dispersion forces in noble gases is low, which means they can be liquefied at low temperatures and hence their boiling points are low, not high.

Therefore, the correct statement would be: "Noble gases are monoatomic gases. They are held together by weak dispersion forces. Because of this they are liquefied at very low temperatures. Hence, they have very low boiling points."