p-Block Elements

$\mathrm{H}_3 \mathrm{PO}_3$ and $\mathrm{H}_2 \mathrm{SO}_4$ both have basicity of $2 . \mathrm{H}_3 \mathrm{PO}_3$ has two $\mathrm{P}-\mathrm{OH}$ bond meaning it can donate two $\left[\mathrm{OH}^{-}\right]$ions when in solution.

$\mathrm{H}_2 \mathrm{SO}_4$ also have two ionisation H -atom attached to oxygen, giving it a basicity of 2 .

The conjugate base of phosphorus acid is $\mathrm{H}_2 \mathrm{PO}_3^{-}$conjugate base of oleum is, $\mathrm{HS}_2 \mathrm{O}_7^{-}$

$ \begin{aligned} & \mathrm{H}_3 \mathrm{PO}_3(\mathrm{aq}) \rightleftharpoons \mathrm{H}_2 \mathrm{PO}_3^{-}+\mathrm{H}^{+} \\ & \mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_7 \rightleftharpoons \mathrm{HS}_2 \mathrm{O}_7^{-}+\mathrm{H}^{+} \end{aligned} $

Statement I:

CCl₄ does not undergo hydrolysis but SiCl₄ undergoes hydrolysis.

Explanation:

• In SiCl₄, the central atom Si has vacant 3d orbitals.

  → Hence, water (a nucleophile) can attack the Si atom forming hydrolyzed products like silicic acid (via intermediate Si(OH)₄).

• In CCl₄, the central atom C does not have vacant d orbitals (carbon is limited to 2s and 2p orbitals only).

  → Therefore, nucleophilic attack by water is not possible, so CCl₄ does not undergo hydrolysis.

✅ Statement I is correct.

Statement II:

Thermal and chemical stability of GeX₄ is more than GeX₂.

Explanation:

• Germanium belongs to Group 14.

  → As we go down the group, +2 oxidation state becomes more stable than +4 (due to the inert pair effect).

• However, for Ge, the +4 oxidation state is still more stable than the +2 state (the inert pair effect is not very pronounced yet).

So:

• GeX₄ (Ge⁴⁺) is more stable thermally and chemically than GeX₂ (Ge²⁺).

✅ Statement II is also correct.

✅ Final Answer:

Option A: Both Statement-I and Statement-II are correct.

$ \text { The disproportion reaction is, } $

$ 4 \mathrm{H}_3 \mathrm{PO}_3 \longrightarrow \underset{\text { (Phosphoric acid) }}{3 \mathrm{H}_3 \mathrm{PO}_4}+\mathrm{PH}_3 $

Statements given in $C$ and $D$ are correct, while $A$ and $B$ are incorrect.

Their correct forms are

$ \begin{aligned} (A)& 2 \mathrm{Al}(s)+6 \mathrm{HCl}(a q) \longrightarrow 2 \mathrm{Al}^{3+}(a q)+ 6 \mathrm{Cl}^{-}(a q)+3 \mathrm{H}_2(g) \\ & 2 \mathrm{Al}(s)+2 \mathrm{NaOH}(a q)+6 \mathrm{H}_2 \mathrm{O}(l) \longrightarrow 2 \mathrm{Na}^{+}\left[\mathrm{Al}\left(\mathrm{OH}_4\right)\right]^{-}(a q)+3 \mathrm{H}_2(g) \end{aligned} $

(It Liberate $\mathrm{H}_2$ with NaOH )

(B) Sodium metaborate formula is $\mathrm{NaBO}_2$

The number of amphoteric oxides is 4 i.e.

$\mathrm{SnO}_2, \mathrm{PbO}_2, \mathrm{PbO}, \mathrm{SnO}$

Acidic oxide : $\mathrm{CO}_2, \mathrm{GeO}_2, \mathrm{GeO}$

Neutral oxide : CO

The complete hydrolysis reaction is,

$ \mathrm{XeF}_6+3 \mathrm{H}_2 \mathrm{O} \longrightarrow \underset{(\text { Product } X)}{\mathrm{XeO}_3+6 \mathrm{Hg}} $

$ \mathrm{XeO}_3 \text { : Xenon trioxide }=s p^3 $

The correct order of electronegativity of group 13 elements is,

$ \mathrm{B}>\mathrm{Tl}>\mathrm{In}>\mathrm{Ga}>\mathrm{Al} $

Electronegativity of 13th group generally decreases down the group with some exception due to inert pair effect.

Statement I: CO reduces the oxygen-carrying ability of blood.

• True.

  Carbon monoxide (CO) binds strongly with hemoglobin to form carboxyhemoglobin, which prevents hemoglobin from carrying oxygen effectively.

  ⇒ ✅ True

Statement II: Producer gas contains CO and N₂.

• True.

  Producer gas is produced by passing air over red-hot coke:

  $ \text{C} + \text{O}_2 \rightarrow \text{CO}_2 \quad \text{and} \quad \text{C} + \text{CO}_2 \rightarrow 2\text{CO} $

  Since air is used (not pure oxygen), nitrogen is also present.

  Components ≈ CO + N₂.

  ⇒ ✅ True

Statement III: C–O bond length in CO₂ is 115 pm.

• True.

  In CO₂, there are two equivalent C=O bonds due to resonance.

  Experimental bond length ≈ 116 pm — very close to 115 pm.

  ⇒ ✅ True

✅ All three statements (I, II, III) are correct.

✅ Correct option: Option B — I, II and III

$ \begin{aligned} &\text { } \mathrm{P}_4+\mathrm{SOCl}_2=\mathrm{PCl}_3 \text { and }\\ &\begin{aligned} & \mathrm{P}_4+\mathrm{SO}_2 \mathrm{Cl}_2 \longrightarrow \mathrm{PCl}_3 \\ & \mathrm{PCl}_3=\text { central atom }=\mathrm{P} \end{aligned} \end{aligned} $

P has 5 valence electron 3

Statement (d) is incorrect regarding silica. The correct form is, While graphite has a two dimension layered structure, $\mathrm{SiO}_2$ has three dimensional network structure.

$ \text{The oxoacid made when red } P_4 \text{ reacts with alkali is :} $

Name of the acid :

It is called Hypophosphoric acid, with the formula $\mathrm{H}_4\mathrm{P}_2\mathrm{O}_6$.

Counting bonds :

There are two P=O (double) bonds in this acid.

There is one P–P (single) bond in this acid.

Among the given options, statement given in option (b) is incorrect regarding group 13 elements. Its the correct form is, orthoboric acid is a weak monobasic acid.

Statement I and II are true, while III is incorrect.

The correct form of statement III is, $\mathrm{GeCl}_2$ is less stable the $\mathrm{GeCl}_4$.

When sodium nitrate react with HCl , two nitrogen oxide (nitrogen dioxide $\left(\mathrm{NO}_2\right)$ and nitric oxide $\left.(\mathrm{NO})\right]$ are formed. $\mathrm{NO}_2$ is acid and NO is neutral.

The complete reaction is,

$ \begin{array}{r} 2 \mathrm{NaNO}_2+2 \mathrm{HCl} \longrightarrow 2 \mathrm{NaCl}+\underset{X}{\mathrm{NO}_2(g)} +\underset{Y}{\mathrm{NO}(g)}+\mathrm{H}_2 \mathrm{O} \end{array} $

The correct reaction to produce diborane on industrial scale is, Reaction of $\mathrm{BF}_3$ with NaH at 450 K The reaction involved is,

$ 2 \mathrm{BF}_3+6 \mathrm{NaH} \xrightarrow{450 \mathrm{~K}} \mathrm{~B}_2 \mathrm{H}_6+6 \mathrm{NaF} $

The reducing property is incorrectly ordered.

The correct order is, $\mathrm{H}_2 \mathrm{~S}<\mathrm{H}_2 \mathrm{Se}<\mathrm{H}_2 \mathrm{Te}<\mathrm{H}_2 \mathrm{Po}$ Reducing strength increases down the group in chalcogen due to decreasing bond dissociation enthalpy.

Statement I:

$\mathrm{Al_2O_3}$ is amphoteric in nature.

• Explanation:

  Aluminum oxide reacts both with acids and with bases, forming salts in each case.

  $ \mathrm{Al_2O_3 + 6HCl → 2AlCl_3 + 3H_2O} $

  $ \mathrm{Al_2O_3 + 2NaOH + 3H_2O → 2Na[Al(OH)_4]} $

  Hence, aluminum oxide is amphoteric. ✅

So, Statement I is correct.

Statement II:

$\mathrm{Tl_2O_3}$ is more basic than $\mathrm{Ga_2O_3}$.

We are comparing Group 13 element oxides:

$ B_2O_3 < Al_2O_3 < Ga_2O_3 < In_2O_3 < Tl_2O_3 $

→ Trend: From B to Tl, metallic character increases, so basic character of oxides increases (acidic → amphoteric → basic).

Thus, Tl₂O₃ (Thallium(III) oxide) is more basic than Ga₂O₃ (Gallium oxide). ✅

So, Statement II is also correct.

✅ Final Answer:

Option A: Both Statement I and Statement II are correct.

When white phosphorus ($ P_4 $) is heated with concentrated NaOH solution in an inert atmosphere of CO₂, we are told it gives a salt 'X' and a gas 'Y'.

Step 1: Recall the known reaction

White phosphorus reacts with hot concentrated NaOH (and no oxygen) to produce phosphine gas (PH₃) and sodium hypophosphite (NaH₂PO₂) as the salt:

$ P_4 + 3NaOH + 3H_2O \longrightarrow PH_3 + 3NaH_2PO_2 $

Step 2: Identify 'X' and 'Y'

• Salt $ X = \text{NaH}_2\text{PO}_2 $ (sodium hypophosphite)

• Gas $ Y = \text{PH}_3 $ (phosphine)

Step 3: Find oxidation states of phosphorus in each

For $ X = \text{NaH}_2\text{PO}_2 $

Let oxidation state of P = $ x $.

$ (+1) + 2(+1) + x + 2(-2) = 0 $

$ 1 + 2 + x - 4 = 0 $

$ x = +1 $

So in NaH₂PO₂, P has oxidation state +1.

For $ Y = \text{PH}_3 $

Let oxidation state of P = $ x $.

$ x + 3(+1) = 0 $

$ x = -3 $

Thus, in PH₃, P has oxidation state −3.

✅ Answer:

Oxidation state of central atom in $ X $ and $ Y $ are respectively:

$ +1 ~\text{and}~ -3 $

✅ Correct option: (B) $ +1, -3 $

Statement given in option (a) is incorrect and it's correct form is, when aluminium react with conc. $\mathrm{HNO}_3$, it forms a protective oxide layer on surface.

In buckminster fullerene $\left(C_{60}\right)$, there are 20 six member rings ( $X$ ) and 12 five member rings $(Y)$ are present. $X+Y=20+12=32$

Among given, statement (a) is incorrect. It's correct form is, On heating potassium permanganate, it produces.

Potassium manganate, maganese oxide and oxygen gas

$ \begin{aligned} 2 \mathrm{KMnO}_4 \xrightarrow{\Delta} & \mathrm{~K}_2 \mathrm{MnO}_4+\mathrm{MnO}_2 + \mathrm{O}_2 \uparrow \end{aligned} $

$ \text { Diborane : } \mathrm{B}_2 \mathrm{H}_6 $

There are four terminal $\mathrm{B}-\mathrm{H}$ bonds. These are 2-center-2-electron bonds Thus $X=4$.

There are two bridging $\mathrm{B}-\mathrm{H}-\mathrm{B}$ bonds. These are 3-centre-2-electron bonds. $Y=2 \Rightarrow X+Y=4+2=6$

The complete reaction is as follows

(i) $\mathrm{P}_4 \mathrm{O}_{10}+4 \mathrm{HNO}_3 \longrightarrow 4 \mathrm{HPO}_3 +2 \mathrm{~N}_2 \mathrm{O}_5$

$ \begin{aligned}(ii) \mathrm{P}_4+2 \mathrm{OHNO}_3 \rightarrow & 4 \mathrm{H}_3 \mathrm{PO}_4 +2 \mathrm{ONO}_2+4 \mathrm{H}_2 \mathrm{O} \end{aligned} $

Hence, oxides are $\mathrm{N}_2 \mathrm{O}_5$ and $\mathrm{NO}_2$.

Both statement I and statement II are incorrect. Their correct form are The order of electronegativity is

$ \mathrm{B}>\mathrm{Tl}>\mathrm{In}>\mathrm{Al} $

Boric acid is not a protonic acid, because it does not donate $\left(\mathrm{H}^{+}\right)$proton, it accept a lone pair of electrons from water.

$\left[\mathrm{SiCl}_6\right]^{2-}$ does not exist.

This is due to the larger size of Cl atom, which make it difficult for six chloride ions to be accomodated around the relatively small silicon ions leading to steric hindrance and instability.

The complete reaction is as follows,

Deacon's process

$ \begin{aligned} 4 \mathrm{HCl}+\mathrm{O}_2 & \longrightarrow 2 \mathrm{H}_2 \mathrm{O}+2 \mathrm{Cl}_2(\mathrm{~g}) \uparrow 5 \mathrm{Cl}_2+6 \mathrm{H}_2 \mathrm{O}+\mathrm{I}_2 \longrightarrow 10 \mathrm{HCl}+2 \mathrm{HIO}_3 \end{aligned} $

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."

$\mathrm{Cr{O_2} + 4NaOH \to \mathop {N{a_2}Cr{O_4}}\limits_{(A)} + 2NaCl + 2{H_2}O}$

$\mathrm{N{a_2}Cr{O_4} + 2{H_2}{O_2} + 2HCl \to \mathop {Cr{O_5}}\limits_{(B)} + \mathop {2NaCl}\limits_{(Mis\sin g\,from\,balanced\,equation)} + 3{H_2}O}$

On moving down the group, bond strength of $\mathrm{M}-\mathrm{H}$ bond decreases, which reduces the thermal stability but increases reducing nature of hydrides, hence A and B are correct statements.

If nitrogen or sulphur is also present in the compound, the sodium fusion extract is first boiled with concentrated nitric acid to decompose cyanide or sulphide of sodium during Lassaigne's test

Nessler's Reagent Reaction :

$\underset{\text { (Nessler's Reagent) }}{2 \mathrm{~K}_2 \mathrm{HgI}_4}+\mathrm{NH}_3+3 \mathrm{KOH} \rightarrow \underset{\substack{\text { (lodine of Millon's base } \\ \text { (Brown precipitate }}}{\mathrm{HgO} . \mathrm{Hg}\left(\mathrm{NH}_2\right) \mathrm{I}}+7 \mathrm{KI}+2 \mathrm{H}_2 \mathrm{O}$