p-Block Elements

(A) Compound X is used for sterilizing drinking water.

Chlorine gas ($\mathrm{Cl_2}$) is widely used as a disinfectant in water treatment. It kills harmful bacteria and microorganisms, making the water safe for drinking.

(B) Compound Y is $\mathrm{NCl_3}$ which has a pyramidal shape.

This is because the nitrogen atom has one lone pair of electrons. Due to the repulsion between the bond pairs and lone pair, the molecular shape becomes pyramidal (similar to $\mathrm{NH_3}$).

(C) Compound Z is used in the enrichment of ${}^{235}\mathrm{U}$.

Chlorine trifluoride ($\mathrm{ClF_3}$) is a very strong fluorinating agent. In the nuclear industry, it converts uranium into uranium hexafluoride ($\mathrm{UF_6}$). This compound is important in the gaseous diffusion or centrifugation process used to separate ${}^{235}\mathrm{U}$ from ${}^{238}\mathrm{U}$.

$ \mathrm{U(s)} + 3\,\mathrm{ClF_3} \rightarrow \mathrm{UF_6(g)} + 3\,\mathrm{ClF(g)} $

(D) $\mathrm{NH_3}$ is a stronger Lewis base than $\mathrm{NCl_3}$.

In $\mathrm{NH_3}$, nitrogen’s lone pair is more available for donation because hydrogen is less electronegative. In $\mathrm{NCl_3}$, the chlorine atoms pull electron density away from nitrogen, making its lone pair less available. Hence, $\mathrm{NH_3}$ acts as the stronger Lewis base.

$ \begin{aligned} & \mathrm{NH}_4 \mathrm{NO}_2 \xrightarrow{\text { heat, } 60-70^{\circ} \mathrm{C}} X \\ & \mathrm{NH}_4 \mathrm{NO}_3 \xrightarrow{\text { heat, } 200-250^{\circ} \mathrm{C}} Y \end{aligned} $ $\mathrm{NH}_4 \mathrm{NO}_2$ on heating at $60-70^{\circ} \mathrm{C}$ gives decomposition reaction. The reactant ammonium nitrite is unstable and readily decomposes into nitrogen gas $\left(\mathrm{N}_2\right)$ and water $\mathrm{H}_2 \mathrm{O}$ ). The decomposition is faster at higher temperatures, and in concentrated aqueous solutions. The chemical equation for the heating of $\mathrm{NH}_4 \mathrm{NO}_2$ can be written as $ \mathrm{NH}_4 \mathrm{NO}_2 \frac{\Delta}{6 \mathrm{O}-70^{\circ} \mathrm{C}} \mathrm{~N}_2+2 \mathrm{H}_2 \mathrm{O} $ So, the product $x$ is $N_2$ Heating ammonium nitrate $\left(\mathrm{NH}_4 \mathrm{NO}_3\right)$ at $200^{\circ} \mathrm{C}$ - $250^{\circ} \mathrm{C}$ leads to decomposition reaction. The product of heating $\mathrm{NH}_4 \mathrm{NO}_3$ are nitrous oxide ( $\mathrm{N}_2 \mathrm{O}$ ) and water ( $\mathrm{H}_2 \mathrm{O}$ ). The reaction is exothermic and the salt leaves no residue on heating. The chemical reaction can be written as $ \mathrm{NH}_4 \mathrm{NO}_3 \xrightarrow[200-250 \mathrm{C}]{\Delta} \mathrm{N}_2 \mathrm{O}+2 \mathrm{H}_2 \mathrm{O} $ So, the product $y$ is $\mathrm{N}_2 \mathrm{O}$ Answer: Option (A) $\mathrm{N}_2$ and $\mathrm{N}_2 \mathrm{O}$

When $ \text{PCl}_5 $ is fluorinated in a polar organic solvent, it forms ionic isomers. The formation of these ionic species can be understood as follows:

1. $ \text{PCl}_5 $ in a polar organic solvent can dissociate into ionic forms, such as $ \left[\text{PCl}_4\right]^+ $ and $ \left[\text{PCl}_6\right]^- $.


2. Upon fluorination, $ \text{PCl}_5 $ can be converted into $ \left[\text{PF}_6\right]^- $ and $ \left[\text{PCl}_4\right]^+ $.


3. Further fluorination can lead to the formation of mixed ionic compounds where $ \text{PCl}_5 $ can form isomers like $ \left[\text{PCl}_4\right]^+ \left[\text{PCl}_4 \text{F}_2\right]^- $.

The two ionic isomers formed are:

1. $ \left[\text{PCl}_4\right]^+ \left[\text{PCl}_4 \text{F}_2\right]^- $, which are colorless crystals.


2. $ \left[\text{PCl}_4\right]^+ \left[\text{PF}_6\right]^- $, which are white crystals.

Hence, the ionic isomers formed when $ \text{PCl}_5 $ is fluorinated in a polar organic solvent are :

Option B : $ \left[\text{PCl}_4\right]^+ \left[\text{PCl}_4 \text{F}_2\right]^- $ and $ \left[\text{PCl}_4\right]^+ \left[\text{PF}_6\right]^- $

To determine the products of the disproportionation of nitrous acid ($ \mathrm{HNO}_2 $) at room temperature, let's analyze the chemical behavior of $ \mathrm{HNO}_2 $ in aqueous solution.

Disproportionation is a redox reaction in which a single substance is simultaneously oxidized and reduced, yielding two different products. In the case of nitrous acid ($ \mathrm{HNO}_2 $), the reaction can be represented as follows:

$ 3 \mathrm{HNO}_2 \rightarrow \mathrm{HNO}_3 + \mathrm{H_2O} + 2 \mathrm{NO} $

Breaking this down, we see that nitrous acid disproportionates into nitric acid ($ \mathrm{HNO}_3 $), water ($ \mathrm{H_2O} $), and nitrogen monoxide ($ \mathrm{NO} $). Additionally, in an aqueous solution, nitric acid dissociates to form $ \mathrm{H_3O}^+ $ (hydronium ion) and $ \mathrm{NO_3}^- $ (nitrate ion).

Therefore, the species formed from this disproportionation reaction are $ \mathrm{H_3O}^+ $, $ \mathrm{NO_3}^- $, and $ \mathrm{NO} $.

From the given options, the correct answer is:

Option A

$ \mathrm{H_3O}^+ $, $ \mathrm{NO_3}^- $, and $ \mathrm{NO} $

To match the reactions given in List-I with the products in List-II, let's analyze each reaction individually based on its reactants and expected products.

• Reaction (P): $\mathrm{P}_2 \mathrm{O}_3 + 3 \mathrm{H}_2 \mathrm{O} \rightarrow$ involves the hydration of phosphorus trioxide. The product of this reaction is phosphorous acid:

$ \mathrm{P}_2 \mathrm{O}_3 + 3 \mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{H}_3 \mathrm{PO}_3 $

Hence, P matches with (2) $\mathrm{H}_3 \mathrm{PO}_3$.

• Reaction (Q): $\mathrm{P}_4 + 3 \mathrm{NaOH} + 3 \mathrm{H}_2 \mathrm{O} \rightarrow$ involves the reaction of phosphorus with a base (NaOH) and water, leading to the production of phosphine and sodium hypophosphite, not exactly reflected in the products listed, but given the context and possible products, we align with phosphorus chemistry principles. Typically, reactions involving phosphorus and strong bases can generate phosphine:

$ \mathrm{P}_4 + \mathrm{NaOH} + 3 \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{PH}_3 +\mathrm{NaH}_2 \mathrm{PO}_2 $

Thus, Q matches best with (3) $\mathrm{PH}_3$ given the options, acknowledging the simplicity of the given answer choices.

• Reaction (R): $\mathrm{PCl}_5 + \mathrm{CH}_3 \mathrm{COOH} \rightarrow$ involves phosphorus pentachloride and acetic acid. The key reaction here is that $\mathrm{PCl}_5$ reacts with $\mathrm{CH}_3\mathrm{COOH}$ to replace $\mathrm{OH}$ groups in acetic acid with $\mathrm{Cl}$, creating acetyl chloride $\mathrm{CH}_3\mathrm{COCl}$ and $\mathrm{POCl}_3$:

$ \mathrm{PCl}_5 + \mathrm{CH}_3 \mathrm{COOH} \rightarrow \mathrm{CH}_3\mathrm{COCl} + \mathrm{POCl}_3 $

Therefore, R matches with (4) $\mathrm{POCl}_3$.

• Reaction (S): $\mathrm{H}_3 \mathrm{PO}_2 + 2 \mathrm{H}_2 \mathrm{O} + 4 \mathrm{AgNO}_3 \rightarrow$ involves hypophosphorous acid reacting in an oxidative environment given by silver nitrate, leading to the production of orthophosphoric acid among other products:

$ \mathrm{H}_3 \mathrm{PO}_2 + 2 \mathrm{H}_2 \mathrm{O} + 4 \mathrm{AgNO}_3 \rightarrow \mathrm{H}_3 \mathrm{PO}_4 +\mathrm{Ag}+\mathrm{HNO}_3 $

S, therefore, matches with (5) $\mathrm{H}_3 \mathrm{PO}_4$.

Given these reactions and the best fits for the products listed,

• P matches with 2,


• Q matches with 3,


• R matches with 4,


• S matches with 5.

Thus, the correct option is Option D: $\mathrm{P} \rightarrow 2 ; \mathrm{Q} \rightarrow 3 ; \mathrm{R} \rightarrow 4 ; \mathrm{S} \rightarrow 5$.

$\mathrm{HClO}_3$ reacts with $\mathrm{HCl}$ according to the following equation,

$ 2 \mathrm{HClO}_3+2 \mathrm{HCl} \longrightarrow 2 \mathrm{ClO}_2+\mathrm{Cl}_2+2 \mathrm{H}_2 \mathrm{O} $

$\mathrm{ClO}_2$ molecule is paramagnetic, as it contains odd number of electrons.

$ 2 \mathrm{ClO}_2+2 \mathrm{O}_3 \longrightarrow \mathrm{Cl}_2 \mathrm{O}_6+2 \mathrm{O}_2 $

$\mathrm{Pb}^{2+}$ on reaction with $\mathrm{Cl}^{-}$, produces white precipitate of $\mathrm{PbCl}_2$

$ \mathrm{Pb}^{2+}+2 \mathrm{Cl}^{-} \longrightarrow \mathrm{PbCl}_2 \downarrow $ (White)

This precipitate is soluble in concentrated hydrochloric acid due to formation of tetrachloroplumbate (II) ion

$ \mathrm{PbCl}_2 \downarrow+2 \mathrm{Cl}^{-} \longrightarrow\left[\mathrm{PbCl}_4\right]^{2-} $

The compound(s) which react with $\mathrm{NH}_3$ to form Boron nitride $(\mathrm{BN})$ is/are:

(A) Amorphous $\mathrm{B}+\mathrm{NH}_3 \underset{\text { Tempreature }}{\stackrel{\text { Very high }}{\longrightarrow}} \mathrm{BN}+\mathrm{H}_2$

(B) $\mathrm{B}_2 \mathrm{H}_6+\mathrm{NH}_3 \stackrel{\text { High Temp. }}{\longrightarrow}(\mathrm{BN})_{\mathrm{x}}$

(C) $\mathrm{B}_2 \mathrm{O}_3+2 \mathrm{NH}_3 \stackrel{900^{\circ} \mathrm{C}}{\longrightarrow} 2 \mathrm{BN}+3 \mathrm{H}_2 \mathrm{O}$

(D) $\mathrm{HBF}_4+\mathrm{NH}_3 \longrightarrow\left[\mathrm{NH}_4^{\oplus}\right]\left[\mathrm{BF}_4^{\ominus}\right]$

Hence, $(A, B, C)$ are correct

(a) H₃PO₃ gives H₃PO₄ and PH₃ on heating. This reaction is known as disproportionation reaction.

$4{H_3}P{O_3}\buildrel \Delta \over \longrightarrow 3{H_3}P{O_4} + P{H_3}$

(b) P in H₃PO₄ is in its highest oxidation state i.e. + 5 hence, it cannot act as reducing agent. But P in H₃PO₃ is in oxidation state, + 3 hence, it can act as reducing agent.

(c) H₃PO₃ contains two $-$OH groups, hence it is a dibasic acid.


(d) Hydrogen attached to P, does not ionise in water. Therefore, options (a), (b) and (d) are correct.

X : Ag Y : Pb P : AgCl Q : PbCl₂



Hence, the correct option is (a).

(a) Order of acid strength different oxyacids of chlorine are :

$\mathop {HClO}\limits_{(Hypochlorous\,acid)}^{( + 1)} < \mathop {HCl{O_3}}\limits_{(Chloric\,acid)}^{( + 5)} < \mathop {HCl{O_4}}\limits_{(Perchloric\,acid)}^{( + 7)} $

Weak acid have strong conjugate base thus hypochlorite ion has strongest conjugate base. Therefore, statement (a) is correct.

(b) Hypochlorite ion is linear and perchlorate ion is tetrahedral and there is no effect of lone pair on hypochlorite ion. Thus, statement (b) is correct.



(c) In the disproportionation reaction, chlorate ion Cl(+5) is oxidised to perchlorate, Cl(+7) and reduce to chloride, Cl($-$1).

$4ClO_3^ - \mathrel{\mathop{\kern0pt\longrightarrow} \limits_{}} 3ClO_4^ - + C{l^ - }$

While in hypochlorite ion, chlorite ion Cl(+1) is oxidised to chlorate, Cl(+5) and reduced to chloride, Cl($-$1) ion.

$3ClO_{}^ - \mathrel{\mathop{\kern0pt\longrightarrow} \limits_{}} ClO_3^ - + 2C{l^ - }$

Thus, statement (c) is incorrect.

(d) The hypochlorite ion oxidises the sulphite ion to sulphate ion, because HOCl is the strongest oxidising Cloxyacids,

$Cl{O^ - } + SO_3^{2 - }\buildrel {} \over \longrightarrow SO_4^{2 - } + C{l^ - }$

Thus, statement (d) is correct.

When Zn react with hot conc. H₂SO₄ then SO₂ is released and ZnSO₄ is obtained.



R(SO₂) molecule is V-shaped



Thus, option (b) is correct.

When Zn is react with conc. NaOH then H₂ gas is evolved and Na₂ZnO₂ is obtained.

$Zn + 2NaOH(conc.)\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{}} \mathop {N{a_2}Zn{O_2}}\limits_{(T)} + \mathop {{H_2}}\limits_{(Q)} \uparrow $

In ground state, H$-$H (Q)

(bond order = 1)

Thus, option (c) is correct.

The oxidation state of Zn in T(Na₂ZnO₂) is +2

Thus, option (a) is incorrect.

$\mathop {ZnS{O_4}}\limits_{(G)} + {H_2}S + N{H_4}OH\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{}} \mathop {ZnS}\limits_{(Z)} \downarrow + \mathop {2{H_2}O}\limits_{(X)} + \mathop {{{(N{H_4})}_2}S{O_4}}\limits_{(Y)} $

ZnS (Z) compound is dirty white coloured.

Thus, option (d) is correct.

Sn can exist in +2 or +4 oxidation state. So, Q in the given reactions can be SnCl₂ or SnCl₄. Since X is monoanion having trigonal pyramidal geometry (sp³ with one lone pair as per VSEPR), so Q is SnCl₂, and the first reaction is

The second reaction is Stephen’s reaction, where nitrile is converted to aldehyde via formation of iminium salt.


There is a coordinate bond in the Cl₂Sn.N(CH₃)₃ in between nitrogen and Sn metal.

Z is oxidised product and oxidation state of Sn is +4 in Z compound. Structure of SnCl₄ (Z) is

Option (A) : Correct. The basicity of oxides usually increases on descending a group. Therefore, Bi₂O5 is more basic than N₂O₅.

Option (B) : Correct. Covalent nature of a molecule depends on the electronegativity difference between bonded atoms.

Option (C) : Correct. Boiling point of NH₃ is more than that of PH₃ due to hydrogen bonding.

Option (D) : Incorrect. P$-$P single bond is stronger than N$-$N single bond. This is due to the fact that N is small in size, due to smaller size of atoms lone pair of repulsion will be more.

(A) $N{H_4}N{O_3}\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{below\,300^\circ C}} {N_2}O + 2{H_2}O$

${N_2}O\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{above\,600^\circ C}} {N_2} + {O_2}$

(B) ${(N{H_4})_2}C{r_2}{O_7}\buildrel \Delta \over \longrightarrow {N_2} + C{r_2}{O_3} + 4{H_2}O$

(C) $Ba{({N_3})_2}\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{180^\circ C}^\Delta } Ba + 3{N_2}$

(D) $M{g_3}{N_2}\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{above\,700^\circ C}} 3Mg + {N_2}$

Hence, only ${(N{H_4})_2}C{r_2}{O_7}$ and $Ba{({N_3})_2}$ can provide N₂ gas on heating below $300^\circ C$

(i) Potassium chlorate $\left(\mathrm{KClO}_3\right)$ is decomposed in presence of $\mathrm{MnO}_2$ as catalyst to form potassium chloride and oxygen gas.

$2 \mathrm{KClO}_3(s) \underset{\mathrm{MnO}_2}{\stackrel{\Delta}{\longrightarrow}} 2 \mathrm{KCl}+\underset{(w)}{3 \mathrm{O}_2(g)}$

The gas $(w)$ is oxygen.

(ii) Reaction of white phosphorous with excess of gas $w$ (i.e., $\mathrm{O}_2$ ) gives $\mathrm{P}_4 \mathrm{O}_{10}$ (a dimer of phosphorous pentaoxide).

$ \underset{\text{White}}{\mathrm{P}_4(s)}+\underset{\text{(excess)}}{5 \mathrm{O}_2(g)} \rightarrow \underset{(X)}{\mathrm{P}_4 \mathrm{O}_{10}(s)}$

phosphorous oxygen

The compound $(X)$ is $\mathrm{P}_4 \mathrm{O}_{10}$.

Reaction of $\mathrm{P}_4 \mathrm{O}_{10}$ with pure nitric acid $\left(\mathrm{HNO}_3\right)$ gives dinitrogen pentaoxide $\left(\mathrm{N}_2 \mathrm{O}_5\right)$ and metaphosphoric acid $\left(\mathrm{HPO}_3\right)$

$ \mathrm{P}_4 \mathrm{O}_{10}+4 \mathrm{HNO}_3 \rightarrow \underset{Z}{4 \mathrm{HPO}_3}+\underset{Y}{2 \mathrm{~N}_2 \mathrm{O}_5} $

The compound $\mathrm{Y}$ is dinitrogen pentaoxide $\left(\mathrm{N}_2 \mathrm{O}_5\right)$ and compound $\mathrm{Z}$ is metaphosphoric $\operatorname{acid}\left(\mathrm{HPO}_3\right)$.

Option (A): Correct.

The aluminium compounds are unusual because they have dimeric structures, and appear to have three-centre bonds involving $s p^3$ hybrid orbitals on $\mathrm{Al}$ and $\mathrm{C}$ in $\mathrm{Al}-\mathrm{C}-\mathrm{Al}$ bridges.


Option (B): Correct.

In diborane $\left(\mathrm{BH}_3\right)$ there are 12 valency electrons, three from each $B$ atom and six from the $\mathrm{H}$ atoms. An $s p^3$ hybrid orbital from each boron atom overlaps with the $1 s$ orbital of the hydrogen. This gives a delocalised molecular orbital covering all three nuclei, containing one pair of electrons and making up one of the bridges. This is a three-centre two-electron bond $(3 c-2 e)$.


Option (D): Group 13 elements have only three valency electrons. When these are used to form three covalent bonds, the atom has a share in only six electrons. The compounds are therefore electron deficient. In $\mathrm{AlCl}_3$, effective $\pi$ overlap takes place between $p$ orbitals of $\mathrm{Al}$ and $\mathrm{Cl}$ due to their comparable size while in $\mathrm{BCl}_3$, the $\pi$ overlap is not effective as $p$ orbital of boron is smaller than that of $p$ orbital of chlorine, hence, the acidity of $\mathrm{BCl}_3$ is greater than that of $\mathrm{AlCl}_3$.

Halogens exist as diatomic molecule of different colours:

$ \begin{array}{|c|c|c|} \hline \text { S.No. } & \text { Halogen } & \text { Colour } \\ \hline 1 . & \text { Fluorine } & \text { Pale yellow } \\ \hline 2 . & \text { Chlorine } & \text { Yellow green } \\ \hline 3 . & \text { Bromine } & \text { Red brown } \\ \hline 4 . & \text { Iodine } & \text { Purple } \\ \hline \end{array} $

(i) The molecular orbital energy level diagram explains the appearance of colour by halogens. The MOT for halogens is represented.


(ii) It represents molecular orbital energy level diagram for fluorine. Similar molecular energy level diagram exist. Other halogens.

(iii) Antibonding $\pi$ orbitals, i.e., $\pi^*{ }_{2 p x}$ and $\pi^*{ }_{2 p y}$ forms the highest occupied molecular orbital (HOMO) and antibonding sigma* orbitals forms the lowest unoccupied molecular orbital (LUMO) for halogens.

(iv) Absorption of energy of suitable wavelength (or colour) results in transition of electron from HOMO to LUMO. As electron returns back to ground state, i.e, HOMO, it releases energy corresponding to a different wavelength (colour complementary to the colour absorbed). This gives halogens their characteristic colour.

(v) As we move down the group 17, size of halogen atom increases and nuclear force of attraction for the outermost shell electrons decrease. This affects the energy gap between $\mathrm{HOMO}$ and LUMO.

$ \mathrm{F}_2>\mathrm{Cl}_2>\mathrm{Br}_2>\mathrm{I}_2 $

HOMO-LUMO energy gap decreases $\rightarrow$

Wavelength of light emitted decreases $\rightarrow$

(vi) The energy gap between HOMO and LUMO keeps on decreasing as we move down the group. As a result, the energy required for transition of electron from HOMO to LUMO decreases and less energy (or light of lower wavelength is absorbed).

(vii)When electron moves back to HOMO, energy is emitted. It corresponds to the light of complementary colour to the colour of the light absorbed.

Option (A): Correct. The structures of the ions formed are shown in below figure. All these structures are based on a tetrahedron. The $s p^3$ hybrid orbitals used for bonding form only weak $\sigma$ bonds, because the $s$ and $p$ levels differ appreciably in energy. The ions are stabilised by strong $p \pi-d \pi$ bonding between full $2 p$ orbitals on oxygen with empty $d$ orbitals on the halogen atoms.


Option (B): $\mathrm{HClO}_4$ is an extremely strong acid, while $\mathrm{HOCl}$ is a very weak acid. Oxygen is more electronegative than chlorine. The more oxygen atoms that are bonded, the more the electrons will be pulled away from the $\mathrm{O}-\mathrm{H}$ bond, and the more this bond will be weakened. Thus $\mathrm{HClO}_4$ requires the least energy to break the $\mathrm{O}-\mathrm{H}$ bond and form $\mathrm{H}^{+}$.

Option (C): The reaction is $\mathrm{Cl}_2+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{HOCl}+\mathrm{HCl}$

Option (D): $\mathrm{As} ~\mathrm{HClO}_4$ is a stronger acid than $\mathrm{H}_2 \mathrm{O}$, therefore, its conjugate base $\left(\mathrm{ClO}_4^{-}\right)$will be weaker than $\left(\mathrm{OH}^{-}\right)$that of $\mathrm{H}_2 \mathrm{O}$.

$ \mathrm{HClO}_4+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{ClO}_4^{-} $

The increasing order of atomic radii is as follows:

Ga < Al < In < Tl

The atomic radius generally increases on moving down a group in the periodic table. As an anomaly, the atomic radius of Ga is less than that of aluminium because of poor shielding of nuclear charge by 10 number of 3d electrons. As a result, the outershell electrons are held more firmly by the nucleus and contraction of radius is observed. This contraction is also called d-block contraction. The size of Tl is similarly affected by 14 number of 4f electrons (lanthanoid contraction) and the atomic radius of Tl is almost similar in size to In.

The reaction of HNO₃ and P₄O₁₀ produces N₂O₅.

4HNO₃ + P₄O₁₀ $\to$ 2N₂O₅ + 4HPO₃

The reaction of HNO₃ with P₄ does not yield N₂O₅.

P₄ + 20 HNO₃ $\to$ 4H₃PO₄ + 20 NO₂ + 4 H₂O

The structure of N₂O₅ has one N$-$O$-$N bond, but no N$-$N bond. It is diamagnetic in nature.

N₂O₅ reacts with sodium metal to produce NO₂ (brown gas)

N₂O₅ + Na $\to$ NaNO₃ + $\mathop {N{O_2} \uparrow }\limits_{Brown\,gas} $

Structure of crystalline form of borax


(a) Borax contains four boron atom; hence, it is tetranuclear.

(b) Only 2 boron atoms lie in the same plane.

(c) Two boron atoms are $s p^3$ hybridised other two borons are $s p^2$ hybridised.

(d) Each of the boron has hydroxyl group attached to it with two hydroxyl groups present as hydroxyl salts.

In all the oxyacids of chlorine : Cl undergoes sp³ hybridization HClO₄ is the strongest acid among the given compounds.

The reaction involved is

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

The reaction involved is

(a) It does not undergo self ionisation in water but accepts an electron pair from water, so it behaves as weak monobasic acid

H₃BO₃ + H₂O $\rightleftharpoons$ B(OH)$_4^ - $ + H⁺

Hence, (a) is incorrect.

(b) When treated with 1, 2-dihydroxy or polyhydroxy compounds, they form chelate (ring complex) which effectively remove [B(OH)₄]^$-$ species from solution and thereby produce maximum number of H₃O⁺ or H⁺ ions, i.e., results in increased acidity.

(c) Baric acid crystallises in a layer structure in which planar triangular BO$_3^{3 - }$ ions are bonded together through hydrogen bonds.

(d) In water the pK_a value of H₃BO₃ is 9.25

H₃BO₃ + H₂O $\rightleftharpoons$ B(OH)$_4^ - $ + H⁺ ; pK_a = 9.25

So, it is weak electrolyte in water.

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.