p-Block Elements

The reaction involved is

P₄ + 8SOCl₂ $\to$ 4PCl₃ + 4SO₂ + 2S₂Cl₂

The reaction involved is

Option A: O₃ molecule is bent

✅ True

• The structure of ozone (O₃) is angular (bent) with a bond angle of about 117°.

• It has resonance structures and sp²-hybridized central oxygen.

Option B: Ozone is violet-black in solid state

✅ True

• In solid state, ozone is violet-black.

• In liquid state, ozone is dark blue.

• In gas state, it is pale blue.

Option C: Ozone is diamagnetic gas

✅ True

• Although O₂ is paramagnetic, ozone (O₃) has all paired electrons, hence diamagnetic.

Option D: ONCl and ONO⁻ are isoelectronic

❌ False

Let’s check the total number of electrons:

• ONCl:

  O (8) + N (7) + Cl (17) = 32 electrons

• ONO⁻:

  O (8×2 = 16) + N (7) + 1 (extra for negative charge) = 24 electrons

→ They are not isoelectronic.

✅ Correct answer: Option D is the wrong statement.

The reactions involved are

Cl₂ + 2NaOH (cold, dil.) $\to$ $\mathop {NaOCl}\limits_{(P)} $ + NaCl + H₂O

3Cl₂ + 6NaOH (hot, conc.) $\to$ $\mathop {NaCl{O_3}}\limits_{(Q)} $ + 5NaCl + 3H₂O

where NaOCl and NaClO₃ are salts of hypochlorous acid (HOCl) and chloric acid (HClO₃), respectively.

The reactions involved are

$S{O_2} + C{l_2}\buildrel {Charcoal} \over \longrightarrow \mathop {S{O_2}C{l_2}}\limits_{(R)} $

$\mathop {10S{O_2}C{l_2}}\limits_{(R)} + {P_4} \to \mathop {4PC{l_5}}\limits_{(S)} + 10S{O_2}$

$\mathop {PC{l_5}}\limits_{(S)} + 4{H_2}O \to \mathop {{H_3}P{O_4}}\limits_{(T)} + 5HCl$

The reactions involved are as follows:

(P) $Pb{O_2} + {H_2}S{O_4}\buildrel {Warm} \over \longrightarrow PbS{O_4} + {H_2}O + {1 \over 2}{O_2}$

(Q) $2N{a_2}{S_2}{O_3} + 2{H_2}O + C{l_2} \to 2NaCl + 2NaHS{O_4} + 2S$

(R) ${N_2}{H_4} + 2{I_2} \to {N_2} + 4HI$

(S) $Xe{F_2} + 2NO \to Xe + 2NOF$

The reaction involved is 4HNO₃ $\to$ 4NO₂ + 2H₂O + O₂. The yellow-brown color appears due to the formation of NO₂.

As all electrons are paired, ozone is diamagnetic in nature. The structure is bent or V-shaped. The structure of ozone is resonance hybrid of the two structures with a delocalised p-orbital which covers all three atoms. Because of this, the two O-O bond lengths are equal.

According to the reaction

P₄ + NaOH + 3H₂O $\to$ PH₃ + 3NaH₂PO₂

oxidation state of phosphorus in P₄ is zero while in PH₃ it is $-$3 and in NaH₂PO₂, it is +1. This shows that this is a type of disproportionation reaction because there is an increase as well as decrease in the oxidation state of phosphorus.

Bleaching powder is CaOCl₂ which contains HOCl as an oxoacid and the anhydride of which is Cl₂O.

Consider the titration reaction

CaOCl₂ + 2KI $\to$ I₂ + Ca(OH)₂ + KCl

According to this reaction, 25 mL of CaOCl₂ reacts with 30 mL of 0.50 M KI

I₂ + 2Na₂S₂O₃ $\to$ Na₂S₄O₆ + 2NaI

Given that 48 mL of 0.25 N Na₂S₂O₃ was used to reach the end point. So, the number of moles of I₂ produced = 48 $\times$ 0.25/2 = 6.

According to the reaction,

Number of millimoles of bleaching powder = Number of moles of I₂ = (1/2) $\times$ Number of moles of Na₂S₂O₃ = 6

So, the molarity of CaOCl₂ is

${{Number\,of\,moles\,of\,bleaching\,powder} \over {Volume\,of\,solution}} = {{6\,mmol} \over {25\,mL}} = 0.24M$

In HNO₃, the oxidation state of N is +5 while in NO, it is +2. In N₂, it is zero and in NH₄Cl, the oxidation state of nitrogen is $-$3. So, the decreasing order of oxidation state of nitrogen is

HNO₃ > NO > N₂ > NH₄Cl

Diamond is hard and graphite is soft. Graphite is a good conductor of electricity as it has one free electron which is responsible for the conduction.

Diamond has higher thermal conductivity than graphite because the structure of diamond is precise and thus the transfer of heat is faster in it.

In case of graphite, the C-C bond has a double bond character so its bond order becomes higher than that of diamond which has only single C-C bonds.

Only HF does not react with AgNO₃. All others react with AgNO₃ to give precipitate of AgX which is soluble in Na₂S₂O₃

$HX + AgN{O_3} \to AgX \downarrow + HN{O_3}(X = Cl,Br,I)$

$AgX + N{a_2}{S_2}{O_3} \to \mathop {N{a_3}[Ag{{({S_2}{O_3})}_2}]}\limits_{Soluble} \, + NaX$

Extra pure N₂ can be obtained by heating Ba(N₃)₂.

Explanation

Thermal Decomposition of Barium Azide

Thermal decomposition of barium azide produces very pure nitrogen gas (N₂).

$\text{Ba(N}_3)_2 \xrightarrow{\Delta} \text{Ba}(s) + 3\text{N}_2(g)$

This reaction yields pure nitrogen gas.

Decomposition of Ammonium Dichromate

Decomposition of ammonium dichromate ((NH₄)₂Cr₂O₇) produces chromium (III) oxide as impurities.

$(\text{NH}_4)_2\text{Cr}_2\text{O}_7 \xrightarrow{\Delta} \text{N}_2(g) + 4\text{H}_2\text{O} + \text{Cr}_2\text{O}_3(s)$

While nitrogen gas is produced, chromium oxide impurities also form.

Heating Ammonium Nitrate

Heating ammonium nitrate (NH₄NO₃) results in the formation of nitrous oxide and water, not pure nitrogen gas.

$\text{NH}_4\text{NO}_3 \xrightarrow{\Delta} \text{N}_2\text{O}(g) + \text{H}_2\text{O}$

No nitrogen gas is produced in this reaction.

Reaction of Ammonia with Copper Oxide

The reaction between ammonia (NH₃) and copper oxide (CuO) gives nitrogen gas along with copper as an impurity.

$2\text{NH}_3(g) + 3\text{CuO}(s) \rightarrow \text{N}_2(g) + 3\text{Cu}(s) + 3\text{H}_2\text{O}(l)$

While nitrogen gas is produced, copper impurities remain.

(A) Dimethyl silyl chloride undergoes hydration followed by polymerisation owing to the loss of water molecule.

(A) → P, S

$ \begin{aligned} & \text { (B) } 6 \mathrm{XeF}_4+12 \mathrm{H}_2 \mathrm{O} \rightarrow 4 \mathrm{Xe}+2 \mathrm{XeO}_3 +24 \mathrm{HF} \text { (Hydrogen fluoride) }+3 \mathrm{O}_2 \text { (Oxygen) } \\ & \end{aligned} $

Here, xenon undergoes disproportionation reaction (which is a kind of redox reaction) with xenon (in +4 oxidation state) gets oxidized to xenon trioxide $\left(\mathrm{XeO}_3\right)$ and reduced to xenon (Xe) at the same time.

HF reacts with glass as shown :

$ \mathrm{Na}_2 \mathrm{SiO}_3+6 \mathrm{HF} \rightarrow \mathrm{Na}_2 \mathrm{SiF}_6+3 \mathrm{H}_2 \mathrm{O} $

(B) → P, Q, R, T

(C)

$2 \mathrm{Cl}_2+2 \mathrm{H}_2 \mathrm{O} \longrightarrow {4 \mathrm{HCl} \text { (Hydrogen } \text { halide )}+\mathrm{O}_2}$

(C) $\rightarrow$ P, Q

(D) $\mathrm{VCl}_5+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{VOCl}_3+2 \mathrm{HCl}$ (Hydrogen chloride)

(D) $\rightarrow P$

To determine which of the given reactions of Xenon compounds is not feasible, we can analyze each option based on the chemical properties and reactivity of the substances involved :

1. Option A :
   $ \text{XeO}_3 + 6\text{HF} \to \text{XeF}_6 + 3\text{H}_2\text{O} $ This reaction involves the conversion of xenon trioxide to xenon hexafluoride. While both compounds are known, the direct conversion of XeO₃ to XeF₆ is not typically feasible due to the different oxidation states of xenon in these compounds and the reaction conditions required.

2. Option B :
   $ \text{XeF}_6 + \text{RbF} \to \text{Rb}[\text{XeF}_7] $ This reaction represents the formation of a rubidium heptafluoroxenate(VII) compound. XeF₆ can react with RbF to form Rb[XeF₇], which is a known compound. This reaction is feasible.

3. Option C :
   $ 2\text{XeF}_2 + 2\text{H}_2\text{O} \to 2\text{Xe} + 4\text{HF} + \text{O}_2 $ In this reaction, xenon difluoride is reacting with water to produce xenon, hydrofluoric acid, and oxygen. XeF₂ is known to hydrolyze in the presence of water to produce these products, making this reaction feasible.

4. Option D :
   $ 3\text{XeF}_4 + 6\text{H}_2\text{O} \to 2\text{Xe} + \text{XeO}_3 + 12\text{HF} + 1.5\text{O}_2 $ This reaction suggests that xenon tetrafluoride reacts with water to form xenon, xenon trioxide, hydrofluoric acid, and oxygen. The formation of XeO₃ and O₂ in this reaction is feasible as XeF₄ can undergo hydrolysis under certain conditions.

Given these considerations, Option A is the least feasible reaction among the ones listed. The direct conversion of XeO₃ to XeF₆ is not commonly reported or expected under normal chemical reaction conditions.

The reactions are as follows:

(A) 3Cu + 8HNO$_3$ (dil.) $\to$ 3Cu(NO$_3$)$_2$ + 4H$_2$O + 2NO

(B) Cu + 4HNO$_3$ (conc.) $\to$ Cu(NO$_3$)$_2$ + 4H$_2$O + 2NO$_2$

(C) 4Zn + 10HNO$_3$ (dil.) $\to$ 4Zn(NO$_3$)$_2$ + 5H$_2$O + N$_2$O

(D) Zn + HNO$_3$ (conc.) $\to$ Zn(NO$_3$)$_2$ + 2H$_2$O + 2NO$_2$

Phosphorus reacts with limited supply of oxygen to form the given product.

P$_4$ + 3O$_2$ $\to$ P$_4$O$_6$

N$_2$ is used to retard the further oxidation.

Tertiary amine will not react due to the bulkiness of trimethyl groups. Tertiary amines instead react with diborane to give simple addition compounds through symmetrical cleavage of diborane.

The bond formation in the given oxides of nitrogen is as follows.

Thus N-N bond exists in N$_2$O, N$_2$O$_3$ and N$_2$O$_4$.

In p-block elements, inert pair effect is observed, due to which lower oxidation state becomes more stable on going down the group. In 14^th group (C, Si, Ge, Sn, Pb) Pb is placed at lower position than Sn, that's why lower oxidation state is more stable for lead (Pb$^{2+}$) and Sn$^{+4}$ is more stable than Pb$^{+4}$. So, Pb$^{+4}$ compounds are stronger oxidising agents than Sn$^{+4}$ compounds. Lower oxidation for the group 14^th elements are more stable for the heavier members.

Due to greater solubility and nature to be prone to microbial action, nitrates are less abundant in earth's crust. Nitrates are soluble. In nitrates, nitrogen has maximum oxidation state and cannot be further oxidised.

Both NH$_3$ and PH$_3$ have one lone pair of electron on it. The lone pair in case of ammonia occupy sp$^3$ orbital is directional.

Hence, the lone pair in case of ammonia is easily available and can be easily donated. But for PH$_3$ the lone pair of electrons occupy spherical s-orbital which is less directional and thus its lone pair is not easily available neither it can be donated easily.

Also the lone pair of electrons in sp$^3$ orbital has lesser s-character and less attracted by nucleus and available for easy donation than in PH$_3$ with pure s orbital, the lone pair in which lone pair are more strongly attracted by nucleus. As we go down a group, the p-character in bond pair increases and the s-character in lone pair increases.

Thus ammonia is a better base than PH$_3$.

White phosphorus on reaction with NaOH gives PH$_3$ as one of the products.

P$_4$ + 3NaOH + 3H$_2$O $\to$ 3NaH$_2$PO$_2$ + PH$_3$

The oxidation state of P in P$_4$, NaH$_2$PO$_2$ and PH$_3$ is 0, +1 and $-$3 respectively. It is an example of disproportionation reaction.

The colourless salt H is either NH$_4$NO$_3$ or NH$_4$NO$_2$. The colourless salt produces NH$_3$ gas (non-inflammable) on boiling with excess of NaOH. On addition of zinc dust (a reducing agent) to this solution, sodium nitrite or sodium nitrate will liberate NH$_3$ gas again.

NH$_4$NO$_3$ + NaOH $\to$ NH$_3$ + NaNO$_3$ + H$_2$O

7NaOH + NaNO$_3$ + 4Zn $\to$ 4Na$_2$ZnO$_2$ + NH$_3$ + 2H$_2$O

NH$_4$NO$_2$ + NaOH $\to$ NaNO$_2$ + NH$_3$ + H$_2$O

3Zn + 5NaOH + NaNO$_2$ $\to$ 3Na$_2$ZnO$_2$ + NH$_3$

To identify the incorrect statement among the given options, let's analyze each one :

1. Option A : Ozone reacts with SO₂ to give SO₃.

   
   
   • Ozone (O₃) can react with sulfur dioxide (SO₂) to form sulfur trioxide (SO₃). This reaction is a part of the atmospheric chemistry where ozone acts as an oxidizing agent.



2. Option B : Silicon reacts with NaOH(aq) in the presence of air to give Na₂SiO₃ and H₂O.

   
   
   • Silicon reacts with sodium hydroxide (NaOH) in the presence of air (oxygen) to form sodium silicate (Na₂SiO₃) and water (H₂O). This reaction is feasible as silicon can be oxidized and dissolved by NaOH, especially in the presence of oxygen.



3. Option C : Cl₂ reacts with excess of NH₃ to give N₂ and HCl.

   
   
   • When chlorine (Cl₂) reacts with an excess of ammonia (NH₃), the main products are nitrogen (N₂) and hydrogen chloride (HCl). This reaction is a well-known example of a redox reaction where chlorine is reduced and ammonia is oxidized.



4. Option D : Br₂ reacts with hot and strong NaOH solution to give NaBr, NaBrO₄, and H₂O.

   
   
   • Bromine (Br₂) reacts with hot and concentrated sodium hydroxide (NaOH) to form sodium bromide (NaBr), sodium bromate (NaBrO₃), and water (H₂O). The formation of sodium perbromate (NaBrO₄) is not typical in this reaction under normal conditions. Usually, NaBr and NaBrO₃ are the primary products.

Based on this analysis, the incorrect statement is Option D. The expected products of the reaction between Br₂ and hot, concentrated NaOH are NaBr and NaBrO₃, not NaBr, NaBrO₄, and H₂O.

Statement‑1:

“Band gap in germanium is small.”

✅ True.

Germanium is a semiconductor with an energy band gap of about 0.66 eV at 300 K, which is much smaller than that of silicon (~1.1 eV).

Statement‑2:

“The energy gap of each germanium atomic energy level is infinitesimally small.”

❌ False.

In a single germanium atom, the energy levels are discrete with finite energy differences between them — not infinitesimally small.

In a solid, when many atoms come together, each discrete atomic level splits into closely spaced energy states due to interactions between atoms, forming continuous bands; the gap between valence and conduction bands (the band gap) is small but finite.

So, the “infinitesimally small” part is incorrect.

✅ Correct Option: C

Statement‑1 is True, Statement‑2 is False.

A molecule is said to be chiral if it cannot be super imposed on its mirror image by any combination of rotations or translation. A chiral molecule exists as two stereoisomers that are mirror images of each other called enantiomers. Chirality is based on molecular symmetry. A chiral molecule cannot have an improper axis of rotation which includes planes of symmetry and inversion center.

Statement- 1 is true as molecules that are not superimposable on their mirror images are a symmetric and hence chiral. Such molecules are known as enantiomers.

Statement- 2: All chiral molecules don’t have to have a stereogenic center. A molecule can be chiral even if it does not have a stereocenter. A molecule’s chirality depends entirely on whether it is a symmetrical. Example-Biphenyls do not have a chiral carbon but they cannot rotate freely due to steric crowding. Hence, they are a symmetrical. Therefore, we can conclude that Statement-2 is false.

P = 5 Value electrons

Steric number = 3 + 1 = 4

Geometry - Tetrahedral hybridisation $-$ sp$^3$

P character $=\frac{3}{4}\times100=75\%$

H$_3$BO$_3$ (orthoboric acid) is a weak Lewis acid.

H$_3$BO$_3$ + H$_2$O $\rightleftharpoons$ B(OH)$^{-}_{4}$ + H$^ \oplus $

It does not donate proton rather it accepts OH$^-$ form water.

So, Statement 1 is True, Statement 2 is False.

Argon is an inert gas; hence, less reactive. It shield metal during welding. Argon is frequently blended with carbon dioxide (CO$_2$), hydrogen (H$_2$), helium (He) or oxygen (O$_2$) to enhance the arc characteristics or facilitate metal transfer in gas metal arc welding.

Hybridisation, H = $\frac{1}{2}$(V + X $-$ C + A) = $\frac{1}{2}$ (8 + 0 $-$ 0 + 0 ) = 4. So, sp$^3$ hybridised and has tetrahedral structure.

No. of lone pair = H $-$ X $-$ D = 4 $-$ 0 $-$ 3 = 1. As lone pair is present the electron pair geometry will be distorted tetrahedral and structure is pyramidal.

H = hybridisation no. (if 2 then sp, if 3 then sp$^2$ etc.)

V = valence electrons of central atom.

X = no. of monovalent atom.

C = cationic charge, A = anionic charge.

D = no. of divalent atom

6XeF$_4$ + 12H$_2$O $\to$ 4Xe + 2XeO$_3$ + 24HF + 3O$_2$

XeF$_6$ + 3H$_2$O $\to$ XeO$_3$ + 6HF

XeF$_4$ and XeF$_6$ are expected to be oxidizing because in both the reactions XeF$_4$ and XeF$_6$ are themselves reduced to xenon (Xe) and oxidises water to oxygen.