Answer

(c) Hydrogen iodide $(HI)$ is more stronger oxidising agent than $H_2 SO_4$. So, it reduces $H_2 SO_4$ to $SO_2$ and itself oxidises to $I_2$. Colour of $I_2$ is violet hence on adding conc. $H_2 SO_4$ to $HI$, it gets oxidised to $I_2$.

$$ H_2 SO_4+2 HI \longrightarrow SO_2+\underset{\substack{\text { (Violet } \\ \text { colour) }}}{I_2} +2 H_2 O $$

Answer

(b) In qualitative analysis when $H_2 S$ is passed through an aqueous solution of salt acidified with dil. $HCl$ a black ppt. of CuS is obtained.

$$ CuSO_4+H_2 S \xrightarrow{\text { dil. } HCl} \underset{\text { black ppt }}{CuS}+H_2 SO_4 $$

On boiling CuS with dil. $HNO_3$ it forms a blue coloured solution and the following reactions occur

$$ \begin{aligned} 3 CuS+8 HNO_3 & \longrightarrow 3 Cu(NO_3)_2+2 NO+3 S+4 H_2 O \\ S+2 HNO_3 & \longrightarrow H_2 SO_4+NO \\ 2 Cu^{2+}+SO_4^{2-}+2 NH_3+2 H_2 O & \longrightarrow Cu(OH)_2 \cdot CuSO_4+2 NH_4 OH \end{aligned} $$

$$ Cu(OH)_2 \cdot CuSO_4+8 NH_3 \longrightarrow \underset{\text { Tetraammine copper (II) (Deep blue solution) }}{2[Cu(NH_3)_4] SO_4+2 OH^{-}+SO_4^{2-}} $$

Answer

(c) Cyclotrimetaphosphoric acid contains three double bonds and 9 single bonds as shown below

$a, b, c$ are three $\pi$ bonds and numerics 1 to 12 are sigma $(\sigma)$ bonds.

Answer

(c) Among given four elements i.e., carbon, nitrogen, phosphorus and boron. Only phosphorus has vacant $d$-orbit so only phosphorus has ability to form $p \pi-d \pi$ bonding.

Answer

(a) Compounds having same value of total number of electrons are known as isoelectronic.

For $CO_3^{2-}$

Total number of electrons

$$ \begin{aligned} & =6+8 \times 3+2 \\ & =6+24+2 \\ & =32 \end{aligned} $$

$$ \text { For } NO_3^{-} $$

Total number of electrons

$$ \begin{aligned} & =7+8 \times 3+1 \\ & =7+25 \\ & =32 \end{aligned} $$

Hence, $CO_3^{2-}$ and $NO_3^{-}$are isoelectronic. These two ions have similar structure so they are isostructural.

Both have triangular planar structure as in both the species carbon and nitrogen are $s p^{2}$ hybridised. Hence, (a) is the correct choice.

Answer

(a)

Answer

(d) On moving top to bottom, size of central atom increases. Bond length of $X-H$ bond increases and bond dissociation energy decreases. Hence, reducing nature increases.

Hence, $SbH_3$ is act as strongest reducing agent among these.

Answer

(c) White phosphorous on reaction with $NaOH$ solution in the presence of inert atmosphere of $CO_2$ it produces phosphine gas which is less basic than $NH_3$.

$$ P_4+3 NaOH+3 H_2 O \longrightarrow PH_3+\underset{\text { (sodium hypophosphite) }}{3 NaH_2 PO_2} $$

Answer

(c) Structure of $H_3 PO_4$ is

$H_3 PO_4$ has $3-OH$ groups i.e., has three ionisable $H$-atoms and hence forms three series of salts. These three possible series of salts for $H_3 PO_4$ are as follows

$NaH_2 PO_4, Na_2 HPO_4$ and $Na_3 PO_4$

Answer

(c) Strong reducing behaviour of $H_3 PO_2$ is due to presence of two $P-H$ bonds and one $P-OH$ bond

Hypophosphorous following (Monobasic)

Answer

(b) On heating lead nitrate it produces brown coloured nitrogen dioxide $(NO_2)$ and lead (II) oxide.

$$ 2 Pb(NO_3)_2 \stackrel{\Delta}{\longrightarrow} 4 NO_2+2 PbO+O_2 $$

Answer

(a) Nitrogen does not show allotropy due to its weak $N-N$ single bond. Therefore, ability of nitrogen to form polymeric structure or more than one structure or form become less. Hence, nitrogen does not show allotropy.

Answer

(c) Maximum covalency of nitrogen is 4 in which one electron is made available by $s$-orbital and 3 electrons are made available by $p$ orbitals. Hence, total four electrons are available for bonding.

Answer

(a) True statement is that single $N-N$ bond is weaker than the single $P-P$ bond. This is why phosphorous show allotropy but nitrogen does not.

(i) $PH_3$ acts as a ligand in the formation of coordination compound due to presence of lone pair of electrons.

(ii) $NO_2$ is paramagnetic in nature due to presence of one unpaired electron. Structure of $NO_2$ is

(iii) Covalency of nitrogen in $N_2 O_5 $ is 4

Answer

(a) When freshly prepared solution of $FeSO_4$ is added in a solution containing $NO_3^{-}$ion, it leads to formation of a brown coloured complex. This is known as brown ring test of nitrate.

$$ \begin{aligned} & NO_3^{-}+3 Fe^{2+}+4 H^{+} \longrightarrow NO+3 Fe^{3+}+2 H_2 O \\ & {[Fe(H_2 O)_6]^{2+}+NO \longrightarrow \underset{\text { Brown ring }}{[Fe(H_2 O)_5(NO)]^{2+}+H_2 O}} \end{aligned} $$

Answer

(b) Stability of +5 oxidation state decreases top to bottom and +3 oxidation state increases top to bottom due to inert pair effect. Meanwhile compound having +5 oxidation state of $Bi$ is $BiF_5$. It is due to smaller size and high electronegativity of fluorine.

Answer

(a) On heating ammonium dichromate and barium azide it produces $N_2$ gas separately.

$$ \begin{aligned} (NH_4)_2 Cr_2 O_7 \stackrel{\Delta}{\longrightarrow} & N_2+4 H_2 O+Cr_2 O_3 \\ Ba(N_3)_2 & \longrightarrow Ba+3 N_2 \end{aligned} $$

Answer

(a) Two moles of $NH_3$ will produce 2 moles of $NO$ on catalytic oxidation of ammonia in preparation of nitric acid.

$$ 4 NH_3+5 O_2 \xrightarrow[\text { Pt } \backslash \text { Rh gauge catalyst }]{\Delta} 4 NO(g)+6 H_2 O(l) $$

Answer

(c) Let oxidation state of $P$ in $NaH_2 PO_2$ is $x$.

$$ \begin{aligned} 1+2 \times 1+x+2 \times-2 & =0 \\ 1+2+x-4 & =0 \\ +x-1 & =0 \\ x & =+1 \end{aligned} $$

Answer

(c) $SF_4$ has sea-saw shaped as shown below

It has trigonal bipyramidal geometry having $s p^{3} d$ hybridisation.

Answer

(a) Peroxoacids of sulphur must contain one- $O-O-$ bond as shown below

Answer

(c) $H_2 SO_4$ is a moderately strong oxidising agent which oxidises both metals and non-metals as shown below

$$ \begin{aligned} & Cu+2 H_2 SO_4 \text { (conc) } \longrightarrow CuSO_4+SO_2+2 H_2 O \\ & S+2 H_2 SO_4 \text { (conc) } \longrightarrow 3 SO_2+2 H_2 O \end{aligned} $$

While carbon on oxidation with $H_2 SO_4$ produces two types of oxides $CO_2$ and $SO_2$.

$C+2 H_2 SO_4$ (conc) $\longrightarrow CO_2+2 SO_2+2 H_2 O$

Answer

(a) Black coloured compound $MnO_2$ reacts with $HCl$ to produce greenish yellow coloured gas of $Cl_2$

$$ \underset{\text { (Black) }}{MnO_2}+4 HCl \longrightarrow MnCl_2+2 H_2 O+\underset{\substack{\text { (greenish } \\ \text { yellow gas) }}}{Cl_2} $$

$Cl_2$ on further treatment with $NH_3$ produces $NCl_3$.

$$ \stackrel{-3}{N} H_3+3 Cl_2 \longrightarrow \stackrel{+3}{N} Cl_3+3 HCl $$

$NH_3(-3)$ changes to $NCl_3(+3)$ in the above reaction. Hence, (a) is the correct choice.

Answer

(c) Bertlett had taken $O_2^{+} PtF_6^{-}$as a base compound because $O_2$ and $Xe$ both have almost same ionisation enthalpy. The ionisation enthalpies of noble gases are the highest in their respective periods due to their stable electronic configurations.

Answer

(d) In solid state $PCl_5$ exists as an ionic solid with $[PCl_4]^{+}$tetrahedral and $[PCl_6]^{-}$ octahedral.

Answer

(c) Greater the SRP value of species higher will be its oxidising power.

Here, $SRP=$ standard reduction potential.

Answer

(b) Isoelectronic pair have same number of electrons

Hence, (b) is the correct choice, while in another cases this value is not equal.

Hence, only (b) is the correct choice.

Answer

$(a, c)$

When chlorine gas is passed through hot $NaOH$ solution it produces $NaCl$ and $NaClO_3$.

Oxidation state varies from 0 to -1 and 0 to +5 .

Hence, (a) and (c) are correct choices.

Answer

$(b, c)$

$F_2>Cl_2>Br_2>I_2$ As ability to gain electron increases oxidising property increases. Here, $F$ is the most electronegative element having highest value of SRP hence it has highest oxidising power.

$$ MI>MBr>MCl>MF $$

This is the incorrect order of ionic character of metal halide.

Correct order can be written as

$$ MI<MBr<MCl<MF $$

As electronegativity difference between metal and halogen increases ionic character increases.

$$ F_2>Cl_2>Br_2>I_2 $$

This is incorrect order of bond dissociation energy. Correct order is $Cl_2>Br_2>F_2>I_2$ due to electronic repulsion among lone pairs in $F_2$ molecule.

Answer

( $b, d)$

Structure of $P_4$ molecule can be represented as

It has total four lone pairs of electrons situated at each P-atom.

It has six $P-P$ single bond.

Answer

$(a, c, d)$

(a) Among halogens, radius ratio between iodine and fluorine is maximum because iodine has maximum radius and fluorine has minimum radius.

(b) It can be correctly stated as in general interhalogen compounds are more reactive than halogen. This is because $X-X^{\prime}$ bond in interhalogen is weaker than $X-X$ bond in halogens except $F-F$ bond.

(c) Among interhalogen compounds maximum number of atoms are present in iodine fluoride because radius ratio of iodine and fluorine has maximum value.

(d) Interhalogen compounds are more reactive than halogen due to weaker $X-X^{\prime}$ bond as compared to $X-X$ of halogen compounds.

Answer

(a, c)

(a) In moist condition $SO_2$ gas acts as a bleaching agent.

e.g., it converts $Fe$ (III) to $Fe$ (II) ion and decolourises acidified $KMnO_4$ (VII).

$$ 2 Fe^{3+}+SO_2+2 H_2 O \longrightarrow 2 Fe^{2+}+SO_4^{2-}+4 H^{+} $$

(b) is incorrect it has bent structure.

(c) Its dilute solution is used as a disinfectant.

(d) It can be prepared by the reaction of $O_2$ with sulphide ore,

$$ 4 FeS_2+11 O_2 \longrightarrow 2 Fe_2 O_3+8 SO_2 $$

while metal on treatment with $H_2 SO_4$ produces $H_2 S$.

Hence, options (a) and (c) are correct choices.

Answer

(c, $d)$

(a) All the three $N-O$ bond lengths in $HNO_3$ are not equal.

(b) All $P-Cl$ bond lengths in $PCl_5$ molecule in gaseous state are not equal. Axial bond is longer than equatorial bond.

(c) $P_4$ molecule in white phosphorous have angular strain therefore white phosphorous is very reactive.

(d) $PCl_5$ is ionic in solid state in which cation is tetrahedral and anion is octahedral.

$$ \begin{aligned} & \text { Cation }-[PCl_4]^{+} \\ & \text {Anion }-[PCl_6]^{-} \end{aligned} $$

Answer

$(a, d)$

(a) $\xrightarrow[\text { acidic strength increases }]{As_2 O_3<SiO_2<P_2 O_3<SO_2}$

(b) Correct order is $\xleftarrow[\text { enthalpy of vaporisation }]{AsH_3 >PH_3 > NH_3}$

(b) Correct order is $\underset{\text { enthalpy of vaporisation }}{\stackrel{AsH_3}{ }>PH_3>NH_3}$

(c) $S<O<Cl<F$ More negative electron gain enthalpy

(d) $H_2 O>H_2 S>H_2 Se>H_2$ Te Thermal stability decreases on moving top to bottom due to increase in its bond length.

Answer

( $a, b$ )

(a) Structure of $H_2 S_2 O_6$ is as shown below

It contains one $S-S$ bond.

(b) In peroxosulphuric acid $(H_2 SO_5)$ sulphur is in +6 oxidation state.

Structure of $H_2 SO_5$ is

Let oxidation state of $S=x$

$$ \begin{aligned} 2 \times(+1)+x+3 \times(-2)+2 \times(-1) & =0 \\ x-6 & =0 \\ x & =6 \end{aligned} $$

(c) During preparation of ammonia, iron oxide with small amount of $K_2 O$ and $Al_2 O_3$ is used as a catalyst to increase the rate of attainment of equilibrium.

(d) Change in enthalpy is negative for preparation of $SO_3$ by catalytic oxidation of $SO_2$.

Answer

$(b, c)$

In the above given four reactions, (b) and (c) represent oxidising behaviour of $H_2 SO_4$. As we know that oxidising agent reduces itself as oxidation state of central atom decreases.

Here,

$$ \begin{aligned} & 2 \stackrel{-1}{HI}+H_2 \stackrel{-6}S_4 \longrightarrow \stackrel{0}{I} I_2+\stackrel{-4}{S} O_2+2 H_2 O \\ & \stackrel{0}{C} u+2 H_2 \stackrel{+6}{S} O_4 \longrightarrow \stackrel{+2}{C} uSO_4+\stackrel{+4}{SO_2}+2 H_2 O \end{aligned} $$

Answer

(a, $b$ )

(a) Only one type of interactions between particles of noble gases are due to weak dispersion forces.

(b) Ionisation enthalpy of molecular oxygen is very close to that of xenon. This is the reason for the formation of xenon oxides.

(c) Hydrolysis of $\mathrm{XeF}_6 ( \mathrm{\stackrel{+6}{Xe}\stackrel{-1}{F_6}} \longrightarrow \mathrm{\stackrel{+6}{Xe}\stackrel{-2}{O_3}} + \mathrm{3 \stackrel{+1}{H} \stackrel{-1}{F} } )$ is not a redox reaction.

(d) Xenon fluorides are highly reactive hydrolysis readily even by traces of water.

Answer

In Contact process $SO_3$ is not absorbed directly in water to from $H_2 SO_4$ because the reaction is highly exothermic, acid mist is formed. Hence, the reaction becomes difficult to handle.

Note

Answer

Ammonia $(NH_3)$ on catalytic oxidation by atmospheric oxygen in presence of $Rh / Pt$ gauge at $500 K$ under pressure of 9 bar produces nitrous oxide.

Balanced chemical reaction can be written as

$$ 4 NH_3+\underset{\text { From air }}{5 O_2} \xrightarrow[500 K ; 9 \text { bar }]{\text { Pt /Rh gauge catalyst }} 4 NO+6 H_2 O $$

Answer

Molecular formula of pyrophosphoric acid is $H_4 P_2 O_7$ and its structure is as follows

Pyrophosphoric acid $(H_4 P_7 O_7)$

Answer

Dissolution of $NH_3$ and $PH_3$ in water can be explained on the basis of $H$-bonding. $NH_3$ forms $H$-bond with water so it is soluble but $PH_3$ does not form $H$-bond with water so it remains as gas and forms bubble in water.

Answer

It has trigonal bipyramidal geometry, in which two $Cl$ atoms occupy axial position while three occupy equatorial positions. All five $P-Cl$ bonds are not identical. There are two types of bond lengths (i) Axial bond lengths (ii) Equatorial bond lengths

Thus, difference in bond length is due to fact that axial bond pairs suffer more repulsion as compared to equatorial bond pairs.

Answer

In gaseous state, $NO_2$ exists as a monomer which has one unpaired electron but in solid state, it dimerises to $N_2 O_4$ so no unpaired electron left. Therefore, $NO_2$ is paramagnetic in gaseous state but diamagnetic in solid state.

f0bd3a00/public)

Answer

Existance of $ClF_3$ and $FCl_3$ can be explained on the basis of size of central atom. Because fluorine is more electronegative as compared to chlorine and has smaller size. Thus, one large $Cl$ atom can accomodate three smaller $F$ atoms but reverse is not true.

Answer

Bond angle of $H_2 O(H-O-H=104.5^{\circ})$ is larger than that of $H_2 S(H-S-H=92^{\circ})$ because oxygen is more electronegative than sulphur therefore, bond pair electron of $O-H$ bond will be closer to oxygen and there will be more bond pair—bond pair repulsion between bond pairs of two $O-H$ bonds.

Answer

Fluorine atom is smaller in size so, six $F^{-}$ions can surround a sulphur atom. The case is not so with chlorine atom due to its large size. $So^{SF_6}$ is known but $SCl_6$ is not known due to interionic repulsion between larger $Cl^{-}$ions.

Answer

Phosphorus on reaction with $Cl_2$ forms two types of halides $A$ and $B$. ’ $A$ ’ is $PCl_5$ and ’ $B$ ’ is $PCl_3$.

When ’ $A$ ’ and ’ $B$ ’ are hydrolysed

$$ \begin{gathered} P_4+10 Cl_2 \longrightarrow 4 PCl_5 \\ P_4+6 Cl_2 \longrightarrow 4 PCl_3 \end{gathered} $$

When ‘A’ and ‘B’ are hydrolysed

$$ (a) \underset{\text{pentachloride}}{\underset{\text{phophorus}}{\underset{[A]}{PCl_5}}} + 4H_2O \longrightarrow \underset{\text{acid}}{\underset{\text{phophoric}}{H_3PO_4}} + 5HCl$$

$$ (a) \underset{\text{trichloride}}{\underset{\text{Phosphorus}}{\underset{[B]}{PCl_3}}} + 3H_2O \longrightarrow \underset{\text{acid}}{\underset{\text{phophorus}}{H_3PO_3}} + 3HCl$$

Answer

$NO_3^{-}+3 Fe^{2+}+4 H^{+} \longrightarrow NO+3 Fe^{3+}+2 H_2 O$

$[Fe(H_2 O)_6]^{2+}+NO \longrightarrow \underset{\text { Brown ring }}{[Fe(H_2 O)_5 NO]^{2+}+H_2 O}$

This test is known as brown ring test of nitrates generally used to identify the presence of nitrate ion in given solution.

Answer

Oxygen is more electronegative than chlorine, therefore dispersal of negative charge present on chlorine increases from $ClO^{-}$to $ClO_4^{-}$ion because number of oxygen atoms attached to chlorine is increasing. Therefore, stability of ions will increase in the order given below

$$ ClO^{-}<ClO_2^{-}<ClO_3^{-}<ClO_4^{-} $$

Due to increase in stability of conjugate base, acidic strength of corresponding acid increases in the same order

$$ HClO<HClO_2<HClO_3<HClO_4 $$

Answer

Ozone is thermodynamically less stable than oxygen because its decomposition into oxygen results in the liberation of heat ( $\Delta H$ is negative) and an increase in entropy ( $\Delta S$ is positive). These two effects reinforce each other, resulting in large negative Gibbs energy change $(\Delta G)$ for its conversion into oxygen.

Answer

$$ P_4 O_6+6 H_2 O \longrightarrow 4 H_3 PO_3 \quad …(i) $$

Neutralisation

$$ .H_3 PO_3+2 NaOH \longrightarrow Na_2 HPO_3+2 H_2 O] \times 4 \quad …(ii) $$

Adding Eqs. (i) and (ii)

$\underset{\text{1 mol}}{\mathrm{P}_4 \mathrm{O}_6}+\underset{8 \mathrm{~mol}}{8 \mathrm{NaOH}} \longrightarrow 4 \mathrm{Na}_2 \mathrm{HPO}_3+2 \mathrm{H}_2 \mathrm{O}$ $…(iii)$

Number of moles of $P_4 O_6$,

$$ n=\frac{m}{M}=\frac{1.1}{220}=\frac{1}{200} mol $$

(Molar mass of $P_4 O_6=(4 \times 31)+(6 \times 16)=220$

$\because$ Product formed by 1 mole of $P_4 O_6$ is neutralised by 8 moles $NaOH$

$\therefore$ Product formed by $\frac{1}{200}$ moles of $P_4 O_6$ will be neutralised by $NaOH$

$$ =8 \times \frac{1}{200}=\frac{8}{200} \text { mole } NaOH $$

Given, $\quad$ Molarity of $NaOH=0.1 M=0.1 mol / L$

$$ \begin{aligned} & \text { Molarity }=\frac{\text { Number of moles }}{\text { Volume in litres }} \\ & \text { Volume }=\frac{\text { Number of moles }}{\text { Molarity }}=\frac{8}{200} \times \frac{1}{0.1}=0.4 L \text { or } 400 mL \end{aligned} $$

$\therefore \quad 400 mL NaOH$ is required.

Answer

Equations for the reactions

$$ \begin{array}{rl} \mathrm{P}_4+6 \mathrm{Cl}_2 & \longrightarrow 4 \mathrm{PCl}_3 \\ \mathrm{PCl}_3+3 \mathrm{H}_2 \mathrm{O}& \longrightarrow \mathrm{H}_3 \mathrm{PO}_3+3 \left. \mathrm{HCl}\right] \times 4 \\ \hline \mathrm{P}_4 +6 \mathrm{Cl}_2+12 \mathrm{H}_2 \mathrm{O} & \longrightarrow 4 \mathrm{H}_3 \mathrm{PO}_3+12 \mathrm{HCl} \\ \end{array} $$

$$1 \mathrm{~mol} \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 12 \mathrm{~mol} $$

$$31 \times 4=124 \mathrm{~g} \quad \quad \quad \quad \quad \quad \quad \quad \quad 12 \times 36.5=438.0 \mathrm{~g} $$

$\because \quad 124 g$ of white phosphorus produces $HCl=438 g$

$\therefore 62 g$ of white phosphorus will produces

$$ HCl=\frac{438}{124} \times 62=219.0 g HCl $$

Answer

Three oxoacids of nitrogen having oxidation state +3 are

(a) $HNO_2$, nitrous acid

(b) $HNO_3$, nitric acid

(c) Hyponitrous acid, $H_2 N_2 O_2$

In $HNO_2, N$ is in +3 oxidation state

Disproportionation reaction

$$ 3 HNO_2 \xrightarrow{\text { Disproportionation }} HNO_3+H_2 O+2 NO $$

Answer

$P_4 O_{10}$ being a dehydrating agent, on reaction with $HNO_3$ removes a molecule of water and forms anhydride of $HNO_3$.

$$ 4 HNO_3+P_4 O_{10} \longrightarrow 4 HPO_3+2 N_2 O_5 $$

Resonating structures of $N_2 O_5$ are

Answer

White phosphorus	Red phosphorus	Black phosphorus
It is less stable form of P	More stable than white P.	It is most stable form of P
It is highly reactive	Less reactive than white P.	It is least reactive
It has regular tetrahedron structure	It has polymeric structure	It has a layered structure.

Answer

Effect of concentration of nitric acid on the formation of oxidation product can be understood by its reaction with conc $HNO_3$. Dilute and concentrated nitric acid give different oxidation products on reaction with copper metal.

$$ \begin{gathered} 3 Cu+8 HNO_3 \text { (Dil.) } \longrightarrow 3 Cu(NO_3)_2+2 NO+4 H_2 O \\ Cu+4 HNO_3 \text { (Conc.) } \longrightarrow Cu(NO_3)_2+2 NO_2+2 H_2 O \end{gathered} $$

Answer

$PCl_5$ on reaction with finely divided silver produced silver halide.

$$ PCl_5+2 Ag \longrightarrow 2 AgCl+PCl_3 $$

$AgCl$ on further reaction with aqueous ammonia solution produces a soluble complex of $[Ag(NH_3)_2]^{+} Cl^{-}$

$$ AgCl+2 NH_3(aq) \longrightarrow \underset{\text { Soluble complex }}{[Ag(NH_3)_2]^{+} Cl^{-}} $$

Answer

Among various forms of oxoacids, phosphinic acid has stronger reducing property.

Structure of phosphinic acid

Reaction showing reducing behaviour of phosphinic acid is as follows $4 AgNO_3+2 H_2 O+H_3 PO_2 \longrightarrow 4 Ag \downarrow+4 HNO_3+H_3 PO_4$

Answer

(a) A. $\rightarrow$ (1) $\quad$

B. $\rightarrow$ (3) $\quad$

C. $\rightarrow$ (4) $\quad$

D. $\rightarrow$ (2)

Answer

(b) A. $\rightarrow$ (4) $\quad$

B. $\rightarrow$ (1) $\quad$

C. $\rightarrow$ (2) $\quad$

D. $\rightarrow$ (3)

$Mn_2 O_7$ on dissolution in water produces acidic solution.

$Bi_2 O_3$ on dissolution in water produces basic solution.

Answer

(a) A. $\rightarrow$ (4)

B. $\rightarrow(3)$

C. $\rightarrow(1)$

D. $\rightarrow(2)$

A. $H_2 SO_4$ is used in storage batteries.

B. $CCl_3 NO_2$ is known as tear gas.

C. $Cl_2$ has highest electron gain enthalpy.

D. Sulphur is a member of chalcogen i.e., ore producing elements.

Answer

(b) A. $\rightarrow$ (3)

B. $\rightarrow$ (4)

C. $\rightarrow$ (2)

D. $\rightarrow$ (1)

Answer

(c) A. $\rightarrow$ (2)

B. $\rightarrow(1)$

C. $\rightarrow(4)$

D. $\rightarrow(3)$

(A) Partial hydrolysis of $XeF_6$ does not change oxidation state of central atom.

$$ \stackrel{+6}{XeF_6}+2 H_2 O \longrightarrow \stackrel{+6}{Xe} O_3+6 HF $$

(B) He is used in modern diving apparatus.

(C) Ar is used to provide inert atmosphere for filling electrical bulbs

(D) Central atom $(Xe)$ of $XeF_4$ is in $s p^{3} d^{2}$ hybridisation.

Answer

(c) Assertion is true, but reason is false.

$N_2$ is less reactive than $P_4$ due to high value of bond dissociation energy which is due to presence of triple bond between two $N$-atoms of $N_2$ molecule.

Answer

(c) Assertion is true, but reason is false.

$HNO_3$ makes iron passive due to formation of passive form of oxide on the surface. Hence, Fe does not dissolve in conc $HNO_3$ solution.

Answer

(b) Both assertion and reason are correct statements, but reason is not the correct explanation of the assertion.

$HI$ cannot be prepared by the reaction of $KI$ with concentrated $H_2 SO_4$ because $HI$ is converted into $I_2$ on reaction with $H_2 SO_4$.

Answer

(a) Both assertion and reason are correct statements, and reason is the correct explanation of the assertion.

Both rhombic and monoclinic sulphur exist as $S_8$ but oxygen exists as $O_2$, because oxygen forms $p \pi-p \pi$ multiple bond due to its small size and small bond length. But $p \pi$ $-p \pi$ bonding is not possible in sulphur due to its bigger size as compared to oxygen.

$p \pi-p \pi bond$

$O=O$

Structure of $O_2$

Answer

(a) Both assertion and reason are correct statements, and reason is the correct explanation of the assertion.

$NaCl$ reacts with concentrated $H_2 SO_4$ to give colourless fumes with pungent smell. Pungent smell is due to formation of $HCl$.

$$ NaCl+H_2 SO_4 \longrightarrow Na_2 SO_4+2 HCl $$

But on adding $MnO_2$ the fumes become greenish yellow due to formation of chlorine gas.

Answer

(a) Assertion and reason both are true and reason is the correct explanation of assertion. $SF_4$ can be hydrolysed but $SF_6$ can not because six $F$-atoms in $SF_6$ prevent the attack of $H_2 O$ on sulphur atoms of $SF_6$.

Answer

Since, the by-product of roasting of sulphide ore is $SO_2$, so $A$ is $S_8$ ’ $A$ ’ $=S_8$; ’ $B$ ’ $=SO_2$ Reactions

(i) $S_8+8 O_2 \xrightarrow{\Delta} 8 SO_2$

(ii) $Ca(OH)_2+SO_2 \longrightarrow CaSO_3+H_2 O$

(iii) $\underset{\text { (Violet) }}{2 MnO_4^{-}}+5 SO_2+2 H_2 O \longrightarrow 5 SO_4^{2-}+4 H^{+}+\underset{\text { (Colourless) }}{2 Mn^{2+}}$

(iv) $2 Fe^{3+}+SO_2+2 H_2 O \longrightarrow 2 Fe^{2-}+SO_4^{2-}+4 H^{+}$

Answer

$Pb(NO_3)_2$ on heating produces a brown coloured gas which may be $NO_2$. Since, on reaction with $N_2 O_4$ and on heating it produces $N_2 O_3$ and $N_2 O_4$ respectively.

Structures

(i) $N_2 O_4$

(ii) $N_2 O_3$

Answer

The main constituents of air are nitrogen (78%) and oxygen (21%). Only $N_2$ reacts with three moles of $H_2$ in the presence of a catalyst to give $NH_3$ (ammonia) which is a gas having basic nature. On oxidation, $NH_3$ gives $NO_2$ which is a part of acid rain. So, the compounds $A$ to $D$ are as

$$ A=NH_4 NO_2 ; B=N_2 ; C=NH_3 ; D=HNO_3 $$

Reactions involved can be given, as

(i) $\underset{(A)}{NH_4 NO_2} \xrightarrow{\Delta}\underset{(B)}{N_2+2 H_2 O} $

(ii) $\underset{[B]}{N_2+3 H_2} \rightleftharpoons \underset{[C]}{2 NH_3}$

(iii) $4 NH_3+5 O_2 \xrightarrow{\text { Oxidation }} 4 NO+6 H_2 O$

(Iv) $2 NO+O_2 \longrightarrow 2 NO_2$

(v) $3 NO_2+H_2 O \longrightarrow \underset{(D)}{2 HNO_3+NO}$