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).

The reaction in each case is,

$ \begin{aligned} 2 \mathrm{NaBH}_4+\mathrm{I}_2 \longrightarrow & \mathrm{~B}_2 \mathrm{H}_6 +2 \mathrm{NaI}+\mathrm{H}_2 \uparrow \end{aligned} $

II. $\mathrm{B}_2 \mathrm{H}_6+3 \mathrm{O}_2 \longrightarrow \mathrm{~B}_2 \mathrm{O}_3+3 \mathrm{H}_2 \mathrm{O}$

III. $\mathrm{BF}_3+\mathrm{NaH} \longrightarrow \mathrm{B}_2 \mathrm{H}_6+6 \mathrm{NaF}$

IV. $\mathrm{B}_2 \mathrm{H}_6+6 \mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{~B}(\mathrm{OH})_3$$ +6 \mathrm{H}_2 \uparrow $

Hence, only in I and IV, $\mathrm{H}_2$ is released.

Statement given in option (d) is incorrect regarding the gas evolved of reaction. The correct form is,

$\mathrm{CO}_2$ is not poisonous gas.

$ \mathrm{CaCO}_3+2 \mathrm{HCl} \rightarrow \mathrm{CaCl}_2+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 \uparrow $

$\mathrm{NaNO}_2$ gives more number of oxides on reacting with HCl .

Among the given statements, option (c) is incorrect. It's correct form is $\mathrm{SeO}_2$ is colourless solid at room temperature.

Step 1: Reaction with Lithium Aluminium Hydride

When boron trifluoride ($\mathrm{BF}_3$) reacts with lithium aluminium hydride ($\mathrm{LiAlH}_4$) in ether, it forms lithium fluoride ($\mathrm{LiF}$), aluminium fluoride ($\mathrm{AlF}_3$), and diborane ($\mathrm{B}_2\mathrm{H}_6$):

$ 4 \mathrm{BF}_3+3 \mathrm{LiAlH}_4 \xrightarrow{\text { Ether }} 3 \mathrm{LiF}+3 \mathrm{AlF}_3 + \mathrm{B}_2 \mathrm{H}_6 $

Step 2: Naming the Product X

So, here $X = \mathrm{B}_2\mathrm{H}_6$ (which is called diborane).

Step 3: Reaction of X with Ammonia

Diborane ($\mathrm{B}_2\mathrm{H}_6$) reacts with ammonia ($\mathrm{NH}_3$) to first form an adduct, which is a combined compound: $\left[\mathrm{B}_2\mathrm{H}_6 \cdot 2\mathrm{NH}_3\right]$.

Step 4: Heating to Get Z

When this adduct is heated, it breaks down to give borazine ($\mathrm{B}_3\mathrm{N}_3\mathrm{H}_6$), which is compound $Z$, and hydrogen gas ($\mathrm{H}_2$):

$ 2~\mathrm{B}_2 \mathrm{H}_6 + 2~\mathrm{NH}_3 \longrightarrow \left[\mathrm{B}_2 \mathrm{H}_6 \cdot 2~\mathrm{NH}_3\right] \longrightarrow 2~\mathrm{B}_3 \mathrm{N}_3 \mathrm{H}_6 (\mathrm{Z}) + \mathrm{H}_2 $

Step 5: Counting Sigma ($\sigma$) and Pi ($\pi$) Bonds in Z

Borazine ($\mathrm{B}_3\mathrm{N}_3\mathrm{H}_6$) is a ring, and its bonds are:

There are 6 sigma ($\sigma$) bonds between B and N in the ring.
There are 3 sigma ($\sigma$) bonds between B and H.
There are 3 sigma ($\sigma$) bonds between N and H.

So, the total number of $\sigma$ bonds ($x$) is: $6 + 3 + 3 = 12$.

There are 3 pi ($\pi$) bonds ($y$) in the ring.

So, $x + y = 12 + 3 = 15$.

The reactions involved are as follows.

$\mathrm{PCl}_5(B)$ in solid state exists as $\left[\mathrm{PCl}_4^{+}\right]\left[\mathrm{PCl}_6^{-}\right]$as ionic lattice. $\mathrm{PCl}_5$ on hydrolysis gives phosphoric acid.

$ \mathrm{PCl}_5+4 \mathrm{H}_2 \mathrm{O} \longrightarrow \underset{\text { (Phosphoric acid) }}{\mathrm{H}_3 \mathrm{PO}_4}+5 \mathrm{HCl} $

In $\mathrm{PCl}_5$, the all bonds are not equivalent.

Trigonal bipyramidal

Paining of electrons takes place.

Thus, option (c) is correct.

The hydrolysis of diborane $\left(\mathrm{B}_2 \mathrm{H}_6\right)$ gives boric acid $\left(\mathrm{H}_3 \mathrm{BO}_3\right)$ as the compound $X$. The reaction is

$ \mathrm{B}_2 \mathrm{H}_6+6 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{H}_3 \mathrm{BO}_3+6 \mathrm{H}_2 $

The properties of boric acid are as follows.

• Acidity : Boric acid is a weak monobasic Lewis acid. It does not donate protons but rather accepts a hydroxide ion from water, making it a monobasic acid. Therefore,

statement II is correct, and statement I (tribasic acid) is incorrect.

• Structure : In the solid state, boric acid has a layered structure, where planar $\mathrm{BO}_3$ units are linked by hydrogen bonds. This structure gives it a slippery feel. Therefore, statement III is correct.

• Solubility : Boric acid has low solubility in cold water, although its solubility increases with temperature. It is not considered highly soluble. Therefore, statement IV is incorrect.

Based on the analysis, the correct statements about compound $X$ are II and III.

Statement I : Graphite's structure is well-known to be layered, with sheets of hexagonal rings of carbon atoms stacked on top of one another. Thus, this statement is correct.

• Statement II : Buckminster fullerene ( $\mathrm{C}_{60}$ ) has a conjugated $\pi$-electron system spread over its spherical surface. This delocalisation of electrons gives it aromatic properties, making it an aromatic compound. Therefore, this statement is incorrect.

• Statement III The distance between the layers in graphite is approximately 335 pm . The value 141.5 pm represents the carbon-carbon bond length within a single layer, not the interlayer distance. Thus, this statement is incorrect.

• Statement IV In both graphite and buckminster fullerene, each carbon atom is bonded to three other carbon atoms, resulting in $s p^2$ hybridisation. The unhybridised $p$-orbitals form delocalised $\pi$-systems. Thus, this statement is correct.

The correct statements are I and IV.

Among the given orders, order given in options (b) i.e.

$ \mathrm{N}_2 \mathrm{O}<\mathrm{NO}<\mathrm{N}_2 \mathrm{O}_3<\mathrm{N}_2 \mathrm{O}_4<\mathrm{N}_2 \mathrm{O}_5 \text { (a } $

(Acidic Nature) is correct.

The correct order for remaining options are

(a) $\mathrm{H}_2 \mathrm{~S}<\mathrm{H}_2 \mathrm{Se}<\mathrm{H}_2 \mathrm{Te}<\mathrm{H}_2 \mathrm{O}$

(Boiling point)

(c) $\mathrm{HF}<\mathrm{HCl}<\mathrm{HBr}<\mathrm{HI}$

(Acidic Nature)

(d) $\mathrm{H}_2 \mathrm{Te}<\mathrm{H}_2 \mathrm{Se}<\mathrm{H}_2 \mathrm{~S}<\mathrm{H}_2 \mathrm{O}$

(Bond angle)

Helium ( $X$ ) is used as a diluent for oxygen in modern dividing apparatus and argon $(Y)$ used to provide an inert atmosphere in high temperature

metallurgical process. Thus $Y$ and $X$ respectively are $\mathrm{Ar}, \mathrm{He}$.

Orthophosphorus acid (or phosphorous acid) on heating disproportionates to orthophosphoric acid (or phosphoric acid) and phosphine. The reaction involved is as follows.

$ 4 \mathrm{H}_3 \mathrm{PO}_3 \rightarrow \underset{(A)}{3 \mathrm{H}_3} \mathrm{PO}_4+\underset{(B)}{\mathrm{PH}_3} $

Thus, $A$ and $B$ are $\mathrm{H}_3 \mathrm{PO}_4$ and $\mathrm{PH}_3$ respectively.

Among the given options, HCl and $\mathrm{H}_2 \mathrm{~S}$ are electron rich hydrides. Electron rich hydrides are those compounds where the central atom has more electron than required to form single covalent bonds with its surrounding atoms.

Among the given statements, statement given in option (c) is incorrect about compounds of group 13 elements. It's correct form is, Diborane $\left(\mathrm{B}_2 \mathrm{H}_6\right)$ is an example of electron deficient hydride.

Among the given statements, statement given in option (c) is incorrect about oxidation state of group 14 elements, it's correct form is lead in +2 state is not a good reducing agent. This is because it readily accepts two more electrons to achieve the stable +4 oxidation state.

Statment given in I, II and IV are correct, while III is incorrect. Its correct form is given in statement II.

The reaction involved are as follows.

Option A: Chlorine oxidises ferrous salts to ferric salts in acidic medium.

• Ferrous (Fe²⁺) has an oxidation state of +2. Ferric (Fe³⁺) has an oxidation state of +3.

• Chlorine (Cl₂) is a strong oxidizing agent.

• In acidic medium, chlorine readily oxidizes Fe²⁺ to Fe³⁺:

  2Fe²⁺ + Cl₂ → 2Fe³⁺ + 2Cl⁻

• This statement is correct.

Option B: Chlorine oxidises iodine to periodic acid in water.

• Iodine (I₂) has an oxidation state of 0.

• Periodic acid (HIO₄) has iodine in a +7 oxidation state.

• When chlorine reacts with iodine in the presence of water, it typically oxidizes iodine to iodic acid (HIO₃), where iodine is in a +5 oxidation state:

  I₂ + 5Cl₂ + 6H₂O → 2HIO₃ + 10HCl

• To obtain periodic acid (HIO₄), a much stronger oxidizing agent than chlorine (like hot concentrated nitric acid, or by electrolysis, or certain peroxo compounds) is generally required, or specific conditions like strongly alkaline medium with hypochlorite. Simple "chlorine in water" does not usually produce periodic acid from iodine.

• Therefore, this statement is incorrect.

Option C: Chlorine acts as a bleaching agent due to oxidation.

• When chlorine dissolves in water, it forms hypochlorous acid (HOCl):

  Cl₂ + H₂O ⇌ HCl + HOCl

• Hypochlorous acid is a powerful oxidizing agent. It decomposes to release nascent oxygen ([O]), which oxidizes colored substances into colorless ones, thereby achieving bleaching.

  HOCl → HCl + [O]

  Colored substance + [O] → Colorless substance

• Thus, the bleaching action of chlorine is indeed due to oxidation.

• This statement is correct.

Option D: Chlorine is manufactured by Deacon's process.

• Deacon's process is an industrial method for the production of chlorine. It involves the catalytic oxidation of hydrogen chloride gas with atmospheric oxygen, usually using copper(II) chloride as a catalyst at high temperatures:

  4HCl(g) + O₂(g) ⇌ 2Cl₂(g) + 2H₂O(g)

• While electrolysis of brine is the predominant modern method, Deacon's process is a known and historically significant method for chlorine manufacturing.

• This statement is correct.

Based on the analysis, Option B is the incorrect statement.

The final answer is $\boxed{B}$

Both the Assertion and the Reason are correct statements, but the Reason does not explain the Assertion properly.

Phosphorus can make both phosphorus(III) chloride and phosphorus(V) chloride because phosphorus has empty $d$-orbitals in its outer shell. This lets phosphorus use these orbitals to form more than four bonds, so it can form compounds like $\mathrm{PCl}_5$ (phosphorus(V) chloride).

Nitrogen cannot form nitrogen(V) chloride ($\mathrm{NCl}_5$) because nitrogen does not have $d$-orbitals in its valence shell. It can only make up to four bonds, so it cannot expand its octet to form five bonds like phosphorus can.

Therefore, the main reason why phosphorus can form $\mathrm{PCl}_5$ but nitrogen cannot form $\mathrm{NCl}_5$ is the availability of $d$-orbitals in phosphorus, not the difference in electronegativity.

Statement given in I and II are correct regarding boron, while statement III and IV are incorrect.

Their correct forms are,

III. in diborane oxidation state of hydrogen is -1 .

IV. boric acid is not a tribasic acid, it is monobasic acid.