NCERT Solutions for Class 12 Science Chemistry Chapter 7 - The P Block Elements

Answer:

Colourless fumes of HCl gas are evolved on addition of conc. H₂SO₄ to a chloride salt but in case of iodide salt, HI is formed which is a strong reducing agent. It reduces H₂SO₄ to SO₂ and itself gets oxidised to I₂. Therefore, violet fumes of iodine come out.
H₂SO₄ + 2Cl^- → SO₄^2- + 2HCl ↑ (colourless)
H₂SO₄ + 2I^- → SO₄^2- + 2HI 
2HI + H₂SO₄ → SO₂ + 2H₂O + I₂ ↑ (violet)
Hence, the correct answer is option (iii).

Answer:

 A black precipitate of CuS is formed when H₂S is passed through an aqueous solution of salt acidified with dil. HCl.
$$\begin{gathered} {\mathrm{Cu}}^{2+}+ {\mathrm{H}}_2\mathrm{S} \stackrel{\mathrm{dil}. \mathrm{HCl}}{\to } \mathrm{CuS} \downarrow + 2{\mathrm{H}}^{+} \\ \mathrm{Black} \mathrm{ppt}. \end{gathered}$$
A blue colour solution of Cu(NO₃)₂ is formed on boiling the precipitate with dil. HNO₃.
$$\begin{gathered} \mathrm{CuS} + \mathrm{dil}. {\mathrm{HNO}}_3 \to \mathrm{Cu}{\left({\mathrm{NO}}_3\right)}_2 + \mathrm{NO} + \mathrm{S} + {\mathrm{H}}_2\mathrm{O} \\ \mathrm{Blue} \mathrm{colour} \end{gathered}$$
A deep blue solution of [Cu (NH₃)₄]²⁺ is obtained when an excess of aqueous solution of ammonia is added to this solution.
$$\begin{gathered} \mathrm{Cu}{\left({\mathrm{NO}}_3\right)}_2 + \mathrm{excess} {\mathrm{NH}}_3 \to [\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_4{]}^{2+} \\ \mathrm{Deep} \mathrm{blue} \mathrm{colour} \end{gathered}$$
Hence, the correct answer is option (ii).

Answer:

The structure of cyclotrimetaphosphoric acid molecule is drawn below.

Hence, the correct answer is option (iii).

Answer:

Phosphorus has vacant d-orbitals needed for pπ–dπ bonding due to its large size.
Hence, the correct answer is option (iii).

Answer:

Both CO₃^2- and NO₃^- are isoelectronic because they have 32 electrons each. Also, both of them have trigonal planar shape and therefore, isostructural.
Hence, the correct answer is option (i).
 

Answer:

Bond dissociation enthalpy decreases down the group because bond distance between halogen and hydrogen increases. Since, fluorine is present at top of the halogen group, it has smallest size. Therefore, H-F bond length is smallest and largest energy is required to break H-F bond.
Hence, the correct answer is option (i).

Answer:

As the bond dissociation enthalpy decreases, bond between E—H becomes easier to break and hydrogen is easily released. Since SbH₃ has lowest bond dissociation enthalpy, it will release hydrogen easily and act as strongest reducing agent.
Hence, the correct answer is option D.

Answer:

The gas released is phosphine (PH₃).
P₄ + 3NaOH + 3H₂O → PH₃ ↑ + 3NaH₂PO₂
On moving down the group, electron density decreases on the central atom and therefore, basicity decreases.
Hence, the correct answer is option C.

Answer:

The structure of H₃PO₄ is drawn below. It has three -OH groups and therefore, three ionisable hydrogens due to which it can form three series of salts- H₂PO₄^-, HPO₄^2- and PO₄^3-. 


Hence, the correct answer is option (iii).

Answer:

The structure of H₃PO₂ is drawn below. It has one –OH group and two P–H bonds which are responsible for its strong reducing behaviour.

Hence, the correct answer is option (iii).

Answer:

The following reaction takes place:
$2\mathrm{Pb}{\left({\mathrm{NO}}_3\right)}_2 \stackrel{∆}{\to } 2\mathrm{PbO} + 4{\mathrm{NO}}_2 + {\mathrm{O}}_2$
Hence, the correct answer is option (ii).

Answer:

Allotropy is the property of an element to exist in more than one form due to difference in bonding between atoms of similar type.
Nitrogen does not show allotropy because N-N bond is very weak owing to small size of nitrogen and high interelectronic repulsions of non-bonding electrons.
Hence, the correct answer is option (i).
 

Answer:

Nitrogen does not have vacant d-orbitals to expand its covalency beyond four.
Hence, the correct answer is option (iii).

Answer:

Single N-N bond is weaker than single P-P bond owing to small size of nitrogen and high interelectronic repulsions of non-bonding electrons.
Hence, the correct answer is option (i).

Answer:

Brown ring is formed at the interface between the solution containing nitrate ion with dilute ferrous sulphate solution and sulphuric acid layers due to formation of [Fe(H₂O)₅(NO)]²⁺ complex.
NO₃^- + 3Fe²⁺ + 4H⁺ → NO + 3Fe³⁺ + 2H₂O
[Fe(H₂O)₆]²⁺ + NO → [Fe(H₂O)₅(NO)]²⁺ + H₂O
Hence, the correct answer is option (i).

Answer:

Bismuth usually forms trihalides due to inert pair effect. The only well characterized compound of bismuth in +5 oxidation state is BiF₅ due to small size and high electronegativity of fluorine.
Hence, the correct answer is option (ii).

 

Answer:

Nitrogen gas is obtained by heating ammonium dichromate and barium azide separately.
$$\begin{gathered} {\left({\mathrm{NH}}_4\right)}_2{\mathrm{Cr}}_2{\mathrm{O}}_7 \stackrel{∆}{\to } {\mathrm{N}}_2 + {\mathrm{Cr}}_2{\mathrm{O}}_3 + 4{\mathrm{H}}_2\mathrm{O} \\ \mathrm{Ammonium} \mathrm{dichromate} \\ \mathrm{Ba}{\left({\mathrm{N}}_3\right)}_2 \stackrel{∆}{\to } \mathrm{Ba} + 3{\mathrm{N}}_2 \\ \mathrm{Barium} \mathrm{azide} \end{gathered}$$
Hence, the correct answer is option (i).

Answer:

The following reaction takes place for catalytic oxidation of ammonia:
$2{\mathrm{NH}}_3\left(\mathrm{g}\right) + \frac{5}{2}{\mathrm{O}}_2\left(\mathrm{g}\right) \underset{500 \mathrm{K}, 9 \mathrm{bar}}{\overset{\mathrm{Pt}/\mathrm{Rh} \mathrm{gauge} \mathrm{catalyst}}{\to }} 2\mathrm{NO}\left(\mathrm{g}\right) + 3{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$
Hence, the correct answer is option A.

Answer:

The anion of the compound NaH₂PO₂ will be H₂PO₂^-. The oxidation state of central atom (P) in H₂PO₂^- can be calculated as follows:
2(+1) + P + 2(−2) = −1
2 + P −4 = −1
P = +1
Hence, the correct answer is option (iii).

Answer:

The hybridization of SF₄ is sp³d and it has one lone pair of electron. So, it will have see-saw shape and not tetrahedral.

Hence, the correct answer is option (iii).

Answer:

Peroxoacids contain peroxo (-O-O-) linkage. The structures of peroxoacids of sulphur are drawn below.

Peroxomonosulphuric acid
(H₂SO₅)

 
Hence, the correct answer is option (i).

Answer:

Carbon is oxidized by conc. H₂SO₄ into two gaseous products, carbon dioxide and sulphur dioxide.
C + 2H₂SO₄(conc.) → CO₂ ↑ + 2SO₂ ↑ + 2H₂O
Hence, the correct answer is option (iii).

Answer:

Chlorine gas is produced when manganese dioxide (a black compound of manganese) reacts with a halogen acid (HCl).
MnO₂ + 4HCl → MnCl₂ + Cl₂ ↑ + 2H₂O
When excess of chlorine gas reacts with NH₃ an unstable trihalide (NCl₃) is formed.
NH₃ + excess Cl₂ → HCl + NCl₃
The oxidation state of nitrogen changes from – 3 in NH₃ to +3 in NCl₃.
Hence, the correct answer is option (i).

Answer:

In the preparation of compounds of Xenon, Bartlett had taken ${\mathrm{O}}_2^{+}$ Pt ${\mathrm{F}}_6^{-}$ as a base compound because both O₂ and Xe have almost same ionization enthalpy.
Hence, the correct answer is option (iii).

Answer:

In solid state, PCl₅​ exists as an ionic solid .

Cation,  → Tetrahedral

Anion,  → Octahedral
Hence, the correct answer is option (iv).

Answer:

The greater the reduction potential of an ion, the greater is its tendency to oxidize other substances and itself get reduced.

Hence, the correct answer is option (iii).

Answer:

The species with the same number of electrons are isoelectronic in nature. The number of electrons in BrO₂^- and BrF₂⁺ are the same i.e. 52.

Hence, the correct answer is option (ii).

Answer:

The following reaction takes place when chlorine gas is passed through a hot NaOH solution.

3Cl₂ + 6NaOH → 5NaCl + NaClO₃ + 3H₂O

The oxidation state of chlorine changes from 0 in Cl₂ to −1 in NaCl and +5 in NaClO₃.

Hence, the correct answers are options (i) and (iii).

Answer:

The ionic character of metal halides increases as the size of halogen atoms decreases. So, the correct order will be MI < MBr < MCl < MF.

The bond dissociation enthalpy of halogens decreases down the group with the exception of fluorine because of large electron-electron repulsion among the lone pairs in F₂ molecules. So, the correct order will be Cl₂ > Br₂ > F₂ > I₂.

Hence, the correct answers are options (ii) and (iii).

Answer:

The structure of the P₄ molecule of white phosphorus is drawn below.

It has six single P-P bonds. Since, there are five valence electrons in phosphorus and each atom forms three bonds, there is one lone pair of electrons on each phosphorus atom. So, four lone pairs of electrons are present. 

Hence, the correct answers are options (ii) and (iv).

Answer:

The radius of fluorine is minimum and the radius of iodine is maximum in the halogen group. So, the radius ratio between iodine and fluorine is maximum. Therefore, maximum number of atoms are present in iodine fluoride among interhalogen compounds. Interhalogen compounds are more reactive than halogen compounds because the bond strength of X—X bond is lesser than X—X' bond.

Hence, the correct answers are options (i), (iii) and (iv).

Answer:

SO₂ acts as a bleaching agent due to its reducing nature in moist conditions. Its dilute solution is used as a disinfectant.

Hence, the correct answers are options (i) and (iii).

Answer:

All the three N—O bond lengths in HNO₃ are not equal. All P—Cl bond lengths in PCl₅ molecule in gaseous state are not equal because PCl₅ has trigonal pyramidal shape and it contains two different types of bonds - axial and equatorial. The P₄ molecule in white phosphorus has angular strain, therefore, white phosphorus is very reactive. PCl₅ exists as an ionic solid [PCl₄]⁺[PCl₆]^- in which cation is tetrahedral and anion is octahedral.

Hence, the correct answers are options (iii) and (iv).

Answer:

The acidic strength of oxides decreases down the group and increases across the period left to right. So, the correct order will be As₂O₃ < SiO₂ < P₂O₃ < SO₂.

The enthalpy of vaporization of hydrides increases down the group but NH₃ has high enthalpy of vaporization due to hydrogen bonding. So, the correct order will be PH₃ < AsH₃ < NH₃.

The electron gain enthalpy increases across the period left to right  and decreases down the group but due to small size of oxygen and fluorine, their electron gain enthalpy is less negative than sulphur and chlorine respectively. So, the correct order will be O < S < F < Cl.

The thermal stability of hydrides decreases down the group. So, the correct order will be H₂O > H₂S > H₂Se > H₂Te. 

Hence, the correct answers are options (i) and (iv).

Answer:

The structure of H₂S₂O₆ is drawn below.

Thus,  S–S bond is present in H₂S₂O₆.

In peroxosulphuric acid (H₂SO₅), peroxo linkage is present and therefore, two oxygen atoms are in −1 oxidation state. So, the oxidation state of sulphur is in +6 oxidation state.

Iron oxide is used as a catalyst with small amounts of K₂O and Al₂O₃ to increase the rate of attainment of equilibrium in Haber’s process.

The preparation of SO₃ by catalytic oxidation of SO₂ is an exothermic reaction. So, the change in enthalpy is negative for the reaction.

Hence, the correct answers are options (i) and (ii).

Answer:

Oxidizing reagent oxidizes another substance and itself gets reduced.
In reaction (ii) 2HI + H₂SO₄ → I₂ + SO₂ + 2H₂O,  conc. H₂SO₄ oxidizes iodine present in −1 oxidation state to zero oxidation state and in reaction (iii) Cu + 2H₂SO₄ → CuSO₄ + SO₂ + 2H₂O, it oxidizes copper present in zero oxidation state to +2.

Hence, the correct answers are options (ii) and (iii).

Answer:

The only types of interactions between particles of noble gases are due to weak dispersion forces.
Ionisation enthalpy of molecular oxygen is very close to that of xenon.
Hydrolysis of XeF₆ is not a redox reaction as the oxidation states of all the elements remain the same as it was in the reacting state. Xenon fluorides are very reactive and can be hydrolyzed by even traces of water.

Answer:

During the contact process, SO₃ is not absorbed directly in water to form H₂SO because the reaction is highly exothermic in nature. As a result, acid mist is formed. These are highly corrosive that can corrode the lead pipelines and linings which cover the tower in contact process. Hence, the reaction become difficult to handle.
So, sulphur trioxide is dissolved in concentrated sulphuric acid to produce oleum (H₂S₂O₇) which can be diluted with required amount of water to produce sulphuric acid.

Answer:

The catalytic oxidation of ammonia (NH₃) by atmospheric oxygen in the presence of Rh/Pt gauge at 500 K under the pressure of 9 bar produces nitrous oxide and water.
The reaction involved is as follows:
$4{\mathrm{NH}}_3+5{\mathrm{O}}_2\underset{500 \mathrm{K}, 9 \mathrm{bar}}{\overset{\mathrm{Pt}/\mathrm{Rh} \mathrm{gauge} \mathrm{catalyst}}{\to }}4\mathrm{NO}+6{\mathrm{H}}_2\mathrm{O}$

Answer:

Molecular formula of pyrophosphoric acid = H₄P₂O₇ 

Answer:

The solubility of PH₃ and NH₃ is different in water. It can be explained on the basis of H-bonding.
NH₃ forms H-bond with water molecules. So, it is soluble in water but PH₃ does not form H-bond with water molecules, therefore, remains as gas and forms bubbles in water.

Answer:

PCl₅ has trigonal bipyramidal geometry, in which three Cl atoms occupy equatorial positions and the remaining two Cl atoms occupy axial positions.
All the five P-Cl bonds are not identical.
It involves two type of bond lengths.
(i) Axial bond lengths
(ii) Equatorial bond lengths 
The structure of PCl₅ molecule is as follows:


The difference in bond length is because of the fact that the axial bond pairs experiences greater repulsion as compared to the equatorial bond pairs.

Answer:

NO₂ exists as a monomer in gaseous phase which has one unpaired electron. It dimerises to N₂O₄ in solid state, such that it is left with no unpaired electrons and thus, behaves as diamagnetic species.

Answer:

The existance of ClF₃ and FCl₃ can be explained with respect to size of the central atom. Fluorine is more electronegative than chlorine atom and has a smaller size. Therefore, one larger Cl atom can easily accomodate three F atoms but three chlorine atoms cannot be accomodated easily by one fluorine atom.
Hence, ClF₃ exists but FCl₃ ​does not.

Answer:

The bond angle of H₂O (H–O–H = 104.5º) is larger than that of H₂S (H–S–H = 92º). This is because oxygen is more electronegative than sulphur. As a result of which, the bond pair electron of O-H bond will be closer to oxygen atom and will suffer greater bond pair (b.p) - bond pair (b.p) repulsion between the bond pair of two O-H bonds.


Bond angle of H₂O>Bond angle of H₂S

Answer:

Due to smaller size of the F atoms, six F atoms can be surrounded by a sulphur atom. This case is not possible in case of chlorine atom, due to its much larger size. The 6 Cl atoms cannot be surrounded by one sulphur atom. Therefore, SF₆ is known but SCl₆ is not known due to interionic repulsion between the chlorine atoms.

Answer:

Phosphorus on reaction with Cl₂ forms two types of phosphorus halides 'A' and 'B', where, 'A' is PCl₅ and 'B' is PCl₃.
The reaction is as follows:


When 'A' and 'B' gets hydrolysed

Answer:

The reactions involved in the function of brown ring are as follows:

$$\begin{gathered} [\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6{]}^{+2} + \mathrm{NO} \to [\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_5\mathrm{NO}{]}^{+2} + {\mathrm{H}}_2\mathrm{O} \\ \mathrm{Brown} \mathrm{ring} \end{gathered}$$
The above test is known as the brown ring test of nitrates generally used to identify the presence of nitrate ion in given solution.

Answer:

The electronegativity of oxygen is greater than chlorine, therefore, the dispersal of negative charge present on chlorine increases from ${\mathrm{ClO}}^{–}$ to ${\mathrm{ClO}}_4^{-}$ ion. This is because the number of oxygen atoms attached to chlorine is increasing.
The stability of the ions increases in the order:
${\mathrm{ClO}}^{–}<{\mathrm{ClO}}_2^{-} < {\mathrm{ClO}}_3^{-} < {\mathrm{ClO}}_4^{-}$
Also, the acidic strength of the corresponding acid increases in the same order due to an increase in the stability of the conjugate base.
The stability of oxoacids of chlorine increases in order as given below:
HClO < HClO₂ < HClO₃ < HClO₄

Answer:

The decomposition of ozone into oxygen results in the release of heat $\left(∆\mathrm{H}<0\right)$ and an increase in the entropy $\left(∆\mathrm{S}>0\right).$  As a result of which a large negative value of $∆\mathrm{G}$ is observed. Therefore, ozone is thermodynamically less stable than oxygen.

Answer:

${\mathrm{P}}_4{\mathrm{O}}_6 + 6{\mathrm{H}}_2\mathrm{O} \to 4{\mathrm{H}}_3{\mathrm{PO}}_{3 }........... \left(\mathrm{i}\right)$
For neutralisation,
$[{\mathrm{H}}_3{\mathrm{PO}}_3 + 2\mathrm{NaOH} \to {\mathrm{Na}}_2{\mathrm{HPO}}_3 + 2{\mathrm{H}}_2\mathrm{O}]\times 4 ...... \left(\mathrm{ii}\right)$
On adding equations (i) and (ii), we get

Molar mass of P₄O₆ = (4 × 31) + (6 × 16) = 220
Number of moles of ${\mathrm{P}}_4{\mathrm{O}}_6=\frac{1.1}{220}$
Since, 1 mole of P₄ O₆ is neutralised by 8 moles NaOH
$\therefore$ 5 ×10^–3 moles of P₄O₆ will be neutralised by 8 × 5 × 10^–3, i.e., 40 × 10^–3 moles of NaOH
Given, Molarity of NaOH = 0.1M
$\begin{array}{rcl}\mathrm{Molarity}& =& \frac{\mathrm{Number} \mathrm{of} \mathrm{moles}}{\mathrm{Volume} \mathrm{in} \mathrm{litres}}\\ \mathrm{Volume}& =& \frac{\mathrm{Number} \mathrm{of} \mathrm{moles}}{\mathrm{Molarity}}\\ & =& \frac{40\times 10–3}{0.1}=0.4 \mathrm{L} \mathrm{or} 400 \mathrm{mL}\end{array}$
Therefore, 400 mL of NaOH is required.

Answer:

The reactions involved are as follows:


Adding the two equations, we get,
${\mathrm{P}}_4+ 6{\mathrm{Cl}}_2 + 12{\mathrm{H}}_2\mathrm{O} \to 4{\mathrm{H}}_3{\mathrm{PO}}_3 + 12\mathrm{HCl}$
Since, 1 mol of while phosphorus produces  12 mol of HCl
We know, $\mathrm{Number} \mathrm{of} \mathrm{mole} = \frac{\mathrm{Weight}}{\mathrm{Molecular} \mathrm{mass}}$
$\Rightarrow$ Weight of white phosphorus = 31 × 4 = 124 g
$\Rightarrow$ Weight of HCl = 1 + 35.5 = 36.5 g
$\therefore$ 124 g white phosphorus produces 438 g HCl.
$\Rightarrow$ 62 g of white phosphorus will produce
$\frac{438}{124}\times 62=219 \mathrm{g} \mathrm{of} \mathrm{HCl}.$

 

Answer:

Three oxoacids of nitrogen are as follows:
(i) HNO₂, nitrous acid
(ii) HNO₃, nitric acid
(iii) H₂N₂O₂, hyponitrous acid
Nitrogen has +3 oxidation state in nitric acid, HNO₂
Disproportionation reaction can be written as:
$3{\mathrm{HNO}}_2\stackrel{\mathrm{Disproportionation}}{\to }{\mathrm{HNO}}_3+{\mathrm{H}}_2\mathrm{O}+2\mathrm{NO}$

Answer:

P₄O₁₀ on reaction with HNO₃ form N₂O₅.
The reaction is as follows:

P₄O₁₀, being the dehydrating agent removes a molecule of water on reacting with HNO₃.
The resonating structures of ${\mathrm{N}}_2{\mathrm{O}}_5$ are:


 

Answer:

White phosphorus	Red phosphorus	Black phosphorus
1. It is less stable form of phosphorus.	1. It is more stable than white phosphorus.	1. It is the most stable form of phosphorus.
2. It is highly reactive.	2. It is less reactive than white phosphorus.	2. It is less reactive.
3. It has regular tetrahedron structure.	3. It has polymeric structure.	3. It has a layered structure.

Answer:

The effect of the concentration of nitric acid on the formation of oxidation product can be understood by its reaction with concentrated and dilute nitric acid. Different products are obtained on reaction of dilute and concentrated nitric acid with copper metal.

Answer:

PCl₅ reacts with finely divided silver to produce silver halide.

The product so formed, silver halide, on further reaction with excess aqueous ammonia solution produces a soluble complex of ${\left[\mathrm{Ag}{\left({\mathrm{NH}}_3\right)}_2\right]}^{+}{\mathrm{Cl}}^{-}$.

Answer:

Structure of phosphinic acid is

It has strong reducing property.
Reaction showing its reducing behaviour is as follows:

 

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

N₂ has high value of bond dissociation energy due to presence of triple bond between two N atoms of N₂ molecule. Therefore, N₂ is less reactive than P₄.
Hence, the correct answer is option (iii).

Answer:

HNO₃ makes iron passive by forming a passive layer of oxide on the surface. As a result of which, Fe does not dissolve in concentrated HNO₃ solution.
Hence, the correct answer is option (iii).

Answer:

Reaction of KI with concentrated H₂SO₄ is as follows:
KI + H₂SO₄ → KHSO₄ + HI
HI is formed but it is further oxidised by conc. H₂SO₄ to I₂.
2HI + H₂SO₄ → 2H₂O + I₂ + SO₂
HI has the lowest H-I bond strength and is the strongest halogen acid.

Hence, the correct answer is option (ii).
 

Answer:

Oxygen forms pπ – pπ multiple bond due to its small size and small bond length which is not possible in case of sulphur due to its bigger size as compared to oxygen.
Therefore, both rhombic and monoclinic sulphur exist as S₈ but oxygen exists as O₂.

Hence, the correct answer is option (i).
 

Answer:

NaCl reacts with concentrated H₂SO₄ to give colourless fumes with pungent smell.

The pungent smell is due to formation of HCl. On adding ${\mathrm{MnO}}_2$ to the above reaction, the fumes turn greenish-yellow due to formation of Cl₂ gas.
Hence, the correct answer is option (i).
 

Answer:

SF₄ can be hydrolysed but SF₆ can not. This is due to the fact that six F-atoms in SF₆ prevent the attack of H₂O on the sulphur atom of SF₆.

Hence, the correct answer is option (i).

Answer:

(i) Since gas 'B' is obtained as a by-product during roasting of sulphide, therefore, gas 'B' must be SO₂.
(ii) Since gas 'B' is obtained by burning of amorphous solid 'A', therefore, amorphous solid 'A' must be sulphur, S₈.

(iii) As gas 'B' reduces acidified KMnO₄ solution and reduces Fe³⁺ to Fe²⁺ salts, the reactions can be shown as:


Therefore, solid 'A' is S₈ and gas 'B' is SO₂.
 

Answer:

Lead (II) nitrate, Pb(NO₃)₂ on heating produces a brown colour gas NO₂.

The gas NO₂ on cooling changes to colourless solid N₂O₄.
2NO₂ → N₂O₄
Solid N₂O₄ on heating with NO changes to blue solid N₂O₃.
2NO + N₂O₄ → 2N₂O₃
Therefore, A is NO₂, B is N₂O₄ and C is N₂O₃.
Structure of N₂O3
N₂O3

Structure of N₂O₄


 

Answer:

The main constituents of air are nitrogen (78%) and oxygen (21%). N₂ reacts with three moles of H₂ in the presence of a catalyst to give ammonia which is a gas having basic nature.
Ammonia, on further oxidation gives NO₂ which is a part of acid rain. So, the compounds A to D are.
A = NH₄ NO₂
B = N₂
C = NH₃
D = HNO₃
Reactions that are involved in the process are:
$$\begin{gathered} \left(\mathrm{i}\right) {\mathrm{NH}}_4{\mathrm{NO}}_2 \to {\mathrm{N}}_2 + 2{\mathrm{H}}_2\mathrm{O} \\ \left(\mathrm{ii}\right) {\mathrm{N}}_2 + 3{\mathrm{H}}_2 \to 2{\mathrm{NH}}_3 \\ \left(\mathrm{iii}\right) 4{\mathrm{NH}}_3 + 5{\mathrm{O}}_2 \to 4\mathrm{NO} + 6{\mathrm{H}}_2\mathrm{O} \\ \left(\mathrm{iv}\right) 4\mathrm{NO} + {\mathrm{O}}_2\to 4\mathrm{NO} + 6{\mathrm{H}}_2\mathrm{O} \\ \left(\mathrm{v}\right) 3{\mathrm{NO}}_2 + {\mathrm{H}}_2\mathrm{O} \to 2{\mathrm{HNO}}_3 + \mathrm{NO} \end{gathered}$$