Answer:(c) $HI$ gets oxidised to $I_2$

Explanation:

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 colour) }}}{I_2} +2 H_2 O $

Now, consider the incorrect options:

(a) $H_2 SO_4$ reduces $HI$ to $I_2$: This statement is incorrect because it is actually the $HI$ that reduces $H_2 SO_4$ to $SO_2$, not the other way around. $HI$ is oxidized to $I_2$ in the process.

(b) $HI$ is of violet colour: This statement is incorrect because $HI$ itself is not violet in color. It is a colorless gas. The violet color is due to the formation of $I_2$.

(d) $HI$ changes to $HIO_3$: This statement is incorrect because $HI$ does not change to $HIO_3$ in this reaction. Instead, $HI$ is oxidized to $I_2$ while reducing $H_2 SO_4$ to $SO_2$.

Answer:(b) Deep blue solution of $[Cu(NH_3)_4]^{2+}$

Explanation:

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

$ \mathrm{CuS}+2 \mathrm{HNO}_3 \rightarrow \underset{\text { blue solution }}{\mathrm{Cu}(\mathrm{NO}_3)_2}{+\mathrm{H}_2 \mathrm{~S}} $

$ \mathrm{Cu}\left(\mathrm{NO}_3\right)_2+4 \mathrm{NH}_3 \rightarrow \underset{\substack{\text { deep blue soution }}}{\mathrm{Cu}\left(\mathrm{NH}_3\right)4^{2+}}+2 \mathrm{NO}{3}^{-}$

Now, consider the incorrect options:

(a) Deep blue precipitate of $Cu(OH)_2$: This option is incorrect because the addition of excess aqueous ammonia to the blue solution of $Cu(NO_3)_2$ does not result in the formation of a precipitate of $Cu(OH)_2$. Instead, ammonia acts as a ligand and forms a complex ion with copper, leading to the formation of a deep blue solution of $[Cu(NH_3)_4]^{2+}$.

(c) Deep blue solution of $Cu(NO_3)_2$: This option is incorrect because $Cu(NO_3)_2$ itself does not form a deep blue solution. The deep blue color is specifically due to the formation of the complex ion $[Cu(NH_3)_4]^{2+}$ when excess ammonia is added to the solution containing $Cu^{2+}$ ions.

(d) Deep blue solution of $Cu(OH)_2 \cdot Cu(NO_3)_2$: This option is incorrect because $Cu(OH)_2 \cdot Cu(NO_3)_2$ is not a known compound that forms a deep blue solution. The deep blue color is due to the formation of the tetraammine copper(II) complex, $[Cu(NH_3)_4]^{2+}$, and not due to any compound like $Cu(OH)_2 \cdot Cu(NO_3)_2$.

Answer:(c) 3 double bonds; 12 single bonds

Explanation:

According to the structure of cyclotrimetaphosphoric there are 12 single bonds and 3 double bonds.

Now, consider the incorrect options:

(a) 3 double bonds; 9 single bonds: This option is incorrect because, while it correctly identifies the presence of 3 double bonds, it underestimates the number of single bonds. Cyclotrimetaphosphoric acid actually contains 12 single bonds, not 9.

(b) 6 double bonds; 6 single bonds: This option is incorrect because it overestimates the number of double bonds and underestimates the number of single bonds. Cyclotrimetaphosphoric acid has only 3 double bonds and 12 single bonds, not 6 double bonds and 6 single bonds.

(d) Zero double bond; 12 single bonds: This option is incorrect because it fails to recognize the presence of any double bonds. Cyclotrimetaphosphoric acid contains 3 double bonds in addition to the 12 single bonds.

Answer:(c) Phosphorus

Explanation:

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.

Now, consider the incorrect options:

Carbon: Carbon does not have vacant $d$-orbitals, so it cannot participate in $p \pi-d \pi$ bonding.

Nitrogen: Nitrogen also lacks vacant $d$-orbitals, making it unable to form $p \pi-d \pi$ bonds.

Boron: Boron does not possess vacant $d$-orbitals, thus it cannot engage in $p \pi-d \pi$ bonding.

Answer:(a) $CO_3^{2-}, NO_3^{-}$

Explanation:

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

For $CO_3^{2-}$

Total number of electrons

$ =6+8 \times 3+2 $

$ =6+24+2 $

$ =32 $

For $ NO_3^{-}$

Total number of electrons

$ =7+8 \times 3+1 $

$ =7+25 $

$ =32 $

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.

Now, consider the incorrect options:

(b) $ClO_3^{-}, CO_3^{2-}$:

These ions are not isoelectronic.

For $ClO_3^{-}$:

Total number of electrons = 17 (Cl) + 3 $ \times $ 8 (O) + 1 (charge) = 17 + 24 + 1 = 42 electrons.

For $CO_3^{2-}$:

Total number of electrons = 6 (C) + 3 $ \times $ 8 (O) + 2 (charge) = 6 + 24 + 2 = 32 electrons.

Since they do not have the same number of electrons, they are not isoelectronic and thus cannot be isostructural.

(c) $SO_3^{2-}, NO_3^{-}$:

These ions are not isoelectronic.

For $SO_3^{2-}$:

Total number of electrons = 16 (S) + 3 $ \times $ 8 (O) + 2 (charge) = 16 + 24 + 2 = 42 electrons.

For $NO_3^{-}$:

Total number of electrons = 7 (N) + 3 $ \times $ 8 (O) + 1 (charge) = 7 + 24 + 1 = 32 electrons.

Since they do not have the same number of electrons, they are not isoelectronic and thus cannot be isostructural.

(d) $ClO_3^{-}, SO_3^{2-}$:

These ions are not isoelectronic.

For $ClO_3^{-}$:

Total number of electrons = 17 (Cl) + 3 $ \times $ 8 (O) + 1 (charge) = 17 + 24 + 1 = 42 electrons.

For $SO_3^{2-}$:

Total number of electrons = 16 (S) + 3 $ \times $ 8 (O) + 2 (charge) = 16 + 24 + 2 = 42 electrons.

Although they have the same number of electrons and are isoelectronic, they are not isostructural.

$ClO_3^{-}$ has a trigonal pyramidal structure due to the presence of a lone pair on chlorine, while $SO_3^{2-}$ has a trigonal planar structure.

Answer:(a) $HF$

Explanation:

Now, consider the incorrect options:

(b) HCl: The bond dissociation enthalpy of HCl is lower than that of HF because the bond between hydrogen and chlorine is weaker than the bond between hydrogen and fluorine. This is due to the larger atomic radius of chlorine compared to fluorine, which results in a longer and weaker H-Cl bond.

(c) HBr: The bond dissociation enthalpy of HBr is lower than that of HF because the bond between hydrogen and bromine is even weaker than the bond between hydrogen and chlorine. Bromine has a larger atomic radius than chlorine, leading to an even longer and weaker H-Br bond.

(d) HI: The bond dissociation enthalpy of HI is the lowest among the given options because iodine has the largest atomic radius of all the halogens listed. This results in the longest and weakest H-I bond, making it easier to dissociate compared to HF, HCl, and HBr.

Answer:(d) $SbH_3$

Explanation:

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.

Now, consider the incorrect options:

(a) $NH_3$: The bond dissociation enthalpy of $NH_3$ is the highest among the given compounds (389 kJ/mol). This means that the $N-H$ bond is the strongest and requires the most energy to break. As a result, $NH_3$ is the least likely to donate electrons and act as a reducing agent.

(b) $PH_3$: The bond dissociation enthalpy of $PH_3$ is 322 kJ/mol, which is lower than that of $NH_3$ but still relatively high compared to $AsH_3$ and $SbH_3$. This indicates that the $P-H$ bond is stronger than the $As-H$ and $Sb-H$ bonds, making $PH_3$ a weaker reducing agent than $AsH_3$ and $SbH_3$.

(c) $AsH_3$: The bond dissociation enthalpy of $AsH_3$ is 297 kJ/mol, which is lower than that of $NH_3$ and $PH_3$ but higher than that of $SbH_3$. This means that the $As-H$ bond is weaker than the $N-H$ and $P-H$ bonds but stronger than the $Sb-H$ bond. Therefore, $AsH_3$ is a stronger reducing agent than $NH_3$ and $PH_3$, but not as strong as $SbH_3$.

Answer:(c) It is more basic than $NH_3$

Explanation:

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} $

Now, consider the correct options:

(a) This statement is correct. Phosphine $(PH_3)$ is indeed highly poisonous and has a smell similar to rotten fish.

(b) This statement is correct. The solution of phosphine in water decomposes in the presence of light.

(d) This statement is correct. Phosphine $(PH_3)$ is less basic than ammonia $(NH_3)$, which aligns with the correct answer provided.

Answer:(c) $H_3 PO_4$

Explanation:

Structure of $H_3 PO_4$ is

image

$H_3 PO_4$ 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 $Na_3 PO_4$

Now, consider the incorrect options:

(a) $H_3PO_2$: This acid, also known as hypophosphorous acid, has the structure $H_2PO(OH)$. It contains only one ionizable hydrogen atom, which means it can only form one series of salts.

(b) $H_3BO_3$: This acid, also known as boric acid, has the structure $B(OH)_3$. It behaves as a monobasic acid in water, meaning it can only donate one proton and thus forms only one series of salts.

(d) $H_3PO_3$: This acid, also known as phosphorous acid, has the structure $HPO(OH)_2$. It contains two ionizable hydrogen atoms, which means it can form only two series of salts.

Answer:(c) presence of one $-OH$ group and two $P-H$ bonds

Explanation:

The presence of $ \mathrm{P}-\mathrm{H}$ bonds in $ \mathrm{H}_3 \mathrm{PO}_2$ is crucial. The $ \mathrm{P}-\mathrm{H}$ bonds can donate electrons, which is a characteristic of reducing agents.

Now, consider the incorrect options:

(a) low oxidation state of phosphorus: The P-H bonds can easily be oxidized, leading to the reduction of other species. While the low oxidation state of phosphorus in $H_3PO_2$ does contribute to its reducing nature, it is not the primary reason for its strong reducing behavior. The presence of $P-H$ bonds is more directly responsible for this property.

The strong reducing behavior $ H_3PO_2 $ is due to the presence of two P-H bonds, which can easily donate electrons, making it a strong reducing agent.

(b) Presence of two $-OH$ groups and one $P-H$ bond: This option is incorrect because $H_3PO_2$ actually has two $P-H$ bonds and only one $-OH$ group. The presence of two $P-H$ bonds is crucial for its strong reducing behavior.

(d) High electron gain enthalpy of phosphorus: This option is incorrect because the electron gain enthalpy of phosphorus is not the primary factor influencing the reducing behavior of $H_3PO_2$. The reducing behavior is mainly due to the presence of $P-H$ bonds, which can donate electrons easily.

Answer:(b) $NO_2, PbO$

Explanation:

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 $

Now, consider the incorrect options:

(a) $N_2O, PbO$: This option is incorrect because heating lead nitrate does not produce nitrous oxide ($N_2O$). The thermal decomposition of lead nitrate specifically produces nitrogen dioxide ($NO_2$), not $N_2O$.

(c) $NO, PbO$: This option is incorrect because heating lead nitrate does not produce nitric oxide ($NO$). The correct nitrogen oxide produced is nitrogen dioxide ($NO_2$), not $NO$.

(d) $NO, PbO_2$: This option is incorrect because heating lead nitrate does not produce nitric oxide ($NO$) or lead dioxide ($PbO_2$). The correct products are nitrogen dioxide ($NO_2$) and lead (II) oxide ($PbO$).

Answer:(a) Nitrogen

Explanation:

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.

Now, consider the incorrect options:

(b)Bismuth: Bismuth does not exhibit allotropy because it exists in only one stable form under normal conditions. It has a rhombohedral crystal structure and does not form different structural modifications.

(c)Antimony: Antimony exhibits allotropy, with several allotropes including metallic antimony and amorphous antimony. Therefore, it is incorrect to say that antimony does not show allotropy.

(d)Arsenic: Arsenic also exhibits allotropy, with several allotropes including gray arsenic (metallic), yellow arsenic, and black arsenic. Therefore, it is incorrect to say that arsenic does not show allotropy.

Answer:(c) 4

Explanation:

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.

Now, consider the incorrect options:

(a) 3: This option is incorrect because while nitrogen commonly forms three bonds (as in $NH_3$), it can also form a fourth bond by utilizing its lone pair, leading to a covalency of 4.

(b) 5: This option is incorrect because nitrogen does not have available d-orbitals in its valence shell to expand its covalency beyond 4. Therefore, it cannot form five bonds.

(d) 6: This option is incorrect because nitrogen lacks the necessary d-orbitals in its valence shell to accommodate six bonds. Its maximum covalency is limited to 4.

Answer:(a) Single $N-N$ bond is stronger than the single $P-P$ bond.

Explanation:

$ \mathrm{N}-\mathrm{N}$ sigma bond (single bond) is weaker than P-P sigma bond (single bond) due to the small bond length between the nitrogen atoms. The lone pair of electrons of both the atoms nitrogen repel each other making the single bond between $ \mathrm{N}-\mathrm{N}$ weaker than $ \mathrm{P}-\mathrm{P}$ sigma bond.

In $ \mathrm{P}-\mathrm{P}$ bond the repulsion between sigma bond and lone pair electrons is less as the size of phosphorous is large.

Now, consider the correct options:

(b): $PH_3$ can indeed act as a ligand in the formation of coordination compounds with transition elements due to the presence of a lone pair of electrons on the phosphorus atom.

(c): $NO_2$ is paramagnetic in nature because it has an odd number of electrons, resulting in one unpaired electron.

(d): The covalency of nitrogen in $N_2O_5$ is indeed four, as each nitrogen forms four covalent bonds in the structure of $N_2O_5$.

Answer:(a) $[Fe(H_2 O)_5(NO)]^{2+}$

Explanation:

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.

$ 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}} $

Now, consider the incorrect options:

(b) $FeSO_4 \cdot NO_2$: This compound is not formed during the brown ring test. The brown ring test specifically involves the formation of a complex between $Fe^{2+}$ and $NO$, not $NO_2$.

(c) $[Fe(H_2 O)_4(NO)_2]^{2+}$: This complex does not form in the brown ring test. The correct complex formed is $[Fe(H_2 O)_5(NO)]^{2+}$, which involves only one $NO$ molecule coordinated to the iron center.

(d) $FeSO_4 \cdot HNO_3$: This compound is not relevant to the brown ring test. The test involves the formation of a complex ion, not a simple mixture or compound of $FeSO_4$ and $HNO_3$.

Answer:(b) $BiF_5$

Explanation:

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.

Now, consider the incorrect options:

$Bi_2 O_5$: Bismuth does not form a stable oxide in the +5 oxidation state due to the inert pair effect and the relatively larger size and lower electronegativity of oxygen compared to fluorine.

$BiCl_5$: Bismuth does not form a stable chloride in the +5 oxidation state because chlorine is less electronegative than fluorine and cannot stabilize the +5 oxidation state of bismuth effectively.

$Bi_2 S_5$: Bismuth does not form a stable sulfide in the +5 oxidation state due to the much larger size and lower electronegativity of sulfur, which makes it unable to stabilize the +5 oxidation state of bismuth.

Answer:(a) $N_2$ in both cases

Explanation:

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

$ (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 $

Now, consider the incorrect options:

(b) is incorrect because ammonium dichromate does not produce $N_2$; it produces $N_2O$ instead. Additionally, barium azide does not produce $NO$; it produces $N_2$.

(c) is incorrect because ammonium dichromate does not produce $N_2O$; it produces $N_2$. However, barium azide correctly produces $N_2$.

(d) is incorrect because ammonium dichromate does not produce $N_2O$; it produces $N_2$. Additionally, barium azide does not produce $NO_2$; it produces $N_2$.

Answer:(a) 2

Explanation:

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) $

Now, consider the incorrect options:

(b) is incorrect because the balanced chemical equation shows that 4 moles of $NH_3$ produce 4 moles of $NO$. Therefore, 2 moles of $NH_3$ would produce 2 moles of $NO$, not 3 moles.

(c) is incorrect because the balanced chemical equation shows that 4 moles of $NH_3$ produce 4 moles of $NO$. Therefore, 2 moles of $NH_3$ would produce 2 moles of $NO$, not 4 moles.

(d) is incorrect because the balanced chemical equation shows that 4 moles of $NH_3$ produce 4 moles of $NO$. Therefore, 2 moles of $NH_3$ would produce 2 moles of $NO$, not 6 moles.

Answer:(c) +1

Explanation:

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

$ 1+2 \times 1+x+2 \times-2=0 $

$ 1+2+x-4=0 $

$ +x-1=0 $

$ x=+1 $

Now, consider the incorrect options:

(a) +3: This is incorrect because the calculation of the oxidation state of phosphorus (P) in $NaH_2PO_2$ does not result in +3. The correct calculation shows that the oxidation state of P is +1.

(b)+5: This is incorrect because the calculation of the oxidation state of phosphorus (P) in $NaH_2PO_2$ does not result in +5. The correct calculation shows that the oxidation state of P is +1.

(d) -3: This is incorrect because the calculation of the oxidation state of phosphorus (P) in $NaH_2PO_2$ does not result in -3. The correct calculation shows that the oxidation state of P is +1.

Answer:(c) $SF_4$

Explanation:

$SF_4$ has sea-saw shaped as shown below :

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

Now, consider the incorrect options:

(a) $NH_4^{+}$: The ammonium ion ($NH_4^{+}$) has a tetrahedral shape because it has four hydrogen atoms symmetrically arranged around the central nitrogen atom. The nitrogen atom undergoes $sp^3$ hybridization, resulting in a tetrahedral geometry.

(b) $SiCl_4$: Silicon tetrachloride ($SiCl_4$) has a tetrahedral shape because the silicon atom is surrounded by four chlorine atoms. The silicon atom undergoes $sp^3$ hybridization, leading to a tetrahedral geometry.

(d) $SO_4^{2-}$: The sulfate ion ($SO_4^{2-}$) has a tetrahedral shape because the sulfur atom is surrounded by four oxygen atoms. The sulfur atom undergoes $sp^3$ hybridization, resulting in a tetrahedral geometry.

Answer:(a) $H_2 SO_5$ and $H_2 S_2 O_8$

Explanation:

Peroxoacids of sulphur must contain one- $O-O-$ bond as shown below:

Now, consider the incorrect options:

(b) $H_2 SO_5$ and $H_2 S_2 O_7$: $H_2 S_2 O_7$ (disulfuric acid) does not contain a peroxo bond ($O-O$ bond). It is not a peroxoacid of sulfur.

(c) $H_2 S_2 O_7$ and $H_2 S_2 O_8$: $H_2 S_2 O_7$ (disulfuric acid) does not contain a peroxo bond ($O-O$ bond). It is not a peroxoacid of sulfur.

(d) $H_2 S_2 O_6$ and $H_2 S_2 O_7$: Neither $H_2 S_2 O_6$ (dithionic acid) nor $H_2 S_2 O_7$ (disulfuric acid) contain a peroxo bond ($O-O$ bond). They are not peroxoacids of sulfur.

Answer:(c) $C$

Explanation:

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

$ 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 $

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 $

Now, consider the incorrect options:

(a) $Cu$: Copper ($Cu$) reacts with hot concentrated sulfuric acid ($H_2SO_4$) to produce copper sulfate ($CuSO_4$), sulfur dioxide ($SO_2$), and water ($H_2O$). Only one gaseous product, $SO_2$, is formed in this reaction.

(b) $S$: Sulfur ($S$) reacts with hot concentrated sulfuric acid ($H_2SO_4$) to produce sulfur dioxide ($SO_2$) and water ($H_2O$). Only one gaseous product, $SO_2$, is formed in this reaction.

(d) $Zn$: Zinc ($Zn$) reacts with hot concentrated sulfuric acid ($H_2SO_4$) to produce zinc sulfate ($ZnSO_4$), sulfur dioxide ($SO_2$), and water ($H_2O$). Only one gaseous product, $SO_2$, is formed in this reaction.

Answer:(a) -3 to +3

Explanation:

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 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.

Now, consider the incorrect options:

(b) -3 to 0:

This option is incorrect because in the reaction between $NH_3$ and $Cl_2$, the nitrogen in $NH_3$ changes its oxidation state from -3 to +3, not to 0. There is no intermediate step where nitrogen reaches an oxidation state of 0 in this reaction.

(c) -3 to +5:

This option is incorrect because the reaction between $NH_3$ and $Cl_2$ results in the formation of $NCl_3$, where nitrogen has an oxidation state of +3. There is no formation of a compound where nitrogen has an oxidation state of +5 in this reaction.

(d) 0 to -3:

This option is incorrect because the initial oxidation state of nitrogen in $NH_3$ is -3, not 0. The reaction involves the change of nitrogen’s oxidation state from -3 to +3, not from 0 to -3.

Answer:(c) both $O_2$ and $Xe$ have almost same ionisation enthalpy.

Explanation:

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.

Now, consider the incorrect options:

(a) Both $O_2$ and $Xe$ do not have the same size. $O_2$ is a diatomic molecule with a smaller size compared to the monatomic noble gas $Xe$.

(b) Both $O_2$ and $Xe$ do not have the same electron gain enthalpy. $O_2$ has a higher tendency to gain electrons compared to $Xe$, which is a noble gas with a stable electronic configuration and very low electron affinity.

(d) While it is true that both $Xe$ and $O_2$ are gases, this is not the reason Bartlett chose $O_2^{+} PtF_6^{-}$ as a base compound. The key factor was the similarity in ionisation enthalpy, not their physical state.

Answer:(d) ionic solid with $[PCl_4]^{+}$ tetrahedral and $[PCl_6]^{-}$ octahedral

Explanation:

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

Now, consider the incorrect options:

(a) covalent solid: This option is incorrect because in the solid state, $PCl_5$ does not exist as a covalent solid. Instead, it dissociates into ionic species, forming $[PCl_4]^{+}$ and $[PCl_6]^{-}$ ions.

(b) octahedral structure: This option is incorrect because $PCl_5$ in the solid state does not form a simple octahedral structure. Instead, it forms an ionic lattice with $[PCl_4]^{+}$ ions having a tetrahedral geometry and $[PCl_6]^{-}$ ions having an octahedral geometry.

(c) ionic solid with $[PCl_6]^{+}$ octahedral and $[PCl_4]^{-}$ tetrahedral: This option is incorrect because the charges and geometries of the ions are reversed. In the solid state, $PCl_5$ forms $[PCl_4]^{+}$ ions with a tetrahedral geometry and $[PCl_6]^{-}$ ions with an octahedral geometry, not the other way around.

Answer:(c) $BrO_4^{-}>I O_4^{-}>ClO_4^{-}$

Explanation:

Higher the reduction potential value greater is the tendency to undergo redution and stronger is the oxidising power.

$\quad \quad \quad BrO_4^{-}>I O_4^{-}>ClO_4^{-}$

$E^{\circ} /V$ $\quad 1.74 \quad \quad 1.65 \quad 1.19 $

Now, consider the incorrect options:

(a) $ClO_4^{-}>IO_4^{-}>BrO_4^{-}$: This option is incorrect because it suggests that $ClO_4^{-}$ has the highest oxidizing power, followed by $IO_4^{-}$ and then $BrO_4^{-}$. However, the standard reduction potentials indicate that $BrO_4^{-}$ has the highest value (1.74 V), followed by $IO_4^{-}$ (1.65 V), and then $ClO_4^{-}$ (1.19 V). Therefore, $ClO_4^{-}$ should have the lowest oxidizing power, not the highest.

(b) $IO_4^{-}>BrO_4^{-}>ClO_4^{-}$: This option is incorrect because it suggests that $IO_4^{-}$ has a higher oxidizing power than $BrO_4^{-}$. However, the standard reduction potentials indicate that $BrO_4^{-}$ (1.74 V) has a higher value than $IO_4^{-}$ (1.65 V), meaning $BrO_4^{-}$ should have a higher oxidizing power than $IO_4^{-}$.

(d) $BrO_4^{-}>ClO_4^{-}>IO_4^{-}$: This option is incorrect because it suggests that $ClO_4^{-}$ has a higher oxidizing power than $IO_4^{-}$. However, the standard reduction potentials indicate that $IO_4^{-}$ (1.65 V) has a higher value than $ClO_4^{-}$ (1.19 V), meaning $IO_4^{-}$ should have a higher oxidizing power than $ClO_4^{-}$.

Answer:(b) $BrO_2^{-}$, $BrF_2^{+}$

Explanation:

Isoelectronic pair have same number of electrons

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

Now, consider the incorrect options:

(a) $ICl_2$ and $ClO_2$:

$ICl_2$: $53 + 2 \times 17 = 87$ electrons

$ClO_2$: $17 + 16 \times 2 = 49$ electrons

The total number of electrons is not equal, hence they are not isoelectronic.

(c) $ClO_2$ and $BrF$:

$ClO_2$: $17 + 16 \times 2 = 49$ electrons

$BrF$: $35 + 9 = 44$ electrons

The total number of electrons is not equal, hence they are not isoelectronic.

(d) $CN^{-}$ and $O_3$:

$CN^{-}$: $6 + 7 + 1 = 14$ electrons

$O_3$: $8 \times 3 = 24$ electrons

The total number of electrons is not equal, hence they are not isoelectronic.

Multiple Choice Questions (More Than One Options)

Answer:(a, c)

Explanation:

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.

Now, consider the incorrect options:

(b) 0 to +3 is incorrect because chlorine does not exhibit an oxidation state of +3 in the reaction between chlorine gas and hot NaOH solution. The oxidation states involved are 0, -1, and +5.

(d) 0 to +1 is incorrect because chlorine does not exhibit an oxidation state of +1 in the reaction between chlorine gas and hot NaOH solution. The oxidation states involved are 0, -1, and +5.

Answer:(b, c)

Explanation:

(a) Halogens readily accepts electrons, So they have strong oxidizing power. $F_2$ is the strongest oxidizing Halogen and this oxidizing power decreases down the group

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

(b) Halogens react with metals to form metal halides. The Ionic character of the halides decreases in the order of

$ \mathrm{MF}>\mathrm{MCl}>\mathrm{MBr}>\mathrm{MI} $

(c) Bond dissociation Enthalpy of Halogens decreases in the order of $ \mathrm{Cl}_2>\mathrm{F}_2>\mathrm{Br}_2>I_2$. Bond dissociation enthalpy of $ \mathrm{Cl}_2$ is greater than $ \mathrm{F}_2$ because of large Electron-electron repulsion among lone pairs of $f_2$ molecule.

(d) Hydrogen-halogen bond strength decreases down the group due to decrease in done dissociation Enthalpy.

$ \mathrm{HF}>\mathrm{HCl}>\mathrm{HBr}>\mathrm{HI} $

Hence, (b) and (c) are not in accordence with the property mentioned against them.

Answer:(b, d)

Explanation:

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.

Now, consider the incorrect options:

(a) is incorrect because the $P_4$ molecule has a total of 4 lone pairs of electrons, not 6. Each phosphorus atom in the $P_4$ molecule has one lone pair, resulting in a total of 4 lone pairs.

(c) is incorrect because the $P_4$ molecule has six $P-P$ single bonds, not three. Each phosphorus atom forms three single bonds with three other phosphorus atoms, resulting in a total of six $P-P$ single bonds.

Answer:(a, c, d)

Explanation:

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

(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.

Now, consider the incorrect options:

(b) The statement is incorrect because it implies that all halogens have weaker $X-X$ bonds than $X-X^{\prime}$ bonds in interhalogens, except for the $F-F$ bond. However, this is not universally true for all halogens. The correct statement should be that interhalogen compounds are generally more reactive than halogen compounds because the $X-X^{\prime}$ bond in interhalogens is typically weaker than the $X-X$ bond in halogens, with the exception of the $F-F$ bond.

Answer:(a, c)

Explanation:

(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^{+} $

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

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

Now, consider the incorrect options:

(b) is incorrect because the $SO_2$ molecule has a bent structure, not a linear geometry.

(d) is incorrect because $SO_2$ is not prepared by the reaction of dilute $H_2SO_4$ with metal sulphide. Instead, this reaction produces $H_2S$. $SO_2$ 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 $

Answer:(c, d)

Explanation:

(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.

$ \text { Cation }-[PCl_4]^{+} $

$ \text {Anion }-[PCl_6]^{-} $

Now, consider the incorrect options:

(a) All the three $N-O$ bond lengths in $HNO_3$ are not equal because $HNO_3$ has one $N=O$ double bond and two $N-O$ single bonds, leading to different bond lengths.

(b) All $P-Cl$ bond lengths in $PCl_5$ molecule in gaseous state are not equal because the molecule has a trigonal bipyramidal structure, where the axial $P-Cl$ bonds are longer than the equatorial $P-Cl$ bonds.

Answer:(a, d)

Explanation:

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

(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.

Now, consider the incorrect options:

(b) The given order is incorrect because the enthalpy of vaporization generally decreases as we move down the group due to the increase in molecular size and weaker intermolecular forces. The correct order should be $( \text{AsH}_3 > \text{PH}_3 > \text{NH}_3 )$.

(c) The given order is incorrect because the electron gain enthalpy becomes more negative as we move from left to right across a period and less negative as we move down a group. The correct order should be $( \text{F} > \text{Cl} > \text{O} > \text{S} )$.

Answer:( a, b )

Explanation:

(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} $

Now, consider the incorrect options:

(c) During the preparation of ammonia by the Haber process, iron oxide $(Fe_2 O_3)$ is used as the primary catalyst, with small amounts of $K_2O$ and $Al_2 O_3$ added as promoters to increase the efficiency of the catalyst. Iron powder is not used as the catalyst.

(d) The preparation of $SO_3$ by the catalytic oxidation of $SO_2$ is an exothermic reaction, meaning that the change in enthalpy $(\Delta H)$ is negative, not positive.

Answer:(b, c)

Explanation:

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,

$ 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 $

Now, consider the incorrect options:

(a) $CaF_2+H_2 SO_4 \longrightarrow CaSO_4+2 HF$, conc. $H_2 SO_4$ is acting as an acid, not as an oxidizing agent. There is no change in the oxidation state of sulfur.

(d) $NaCl+H_2 SO_4 \longrightarrow NaHSO_4+HCl$, conc. $H_2 SO_4$ is also acting as an acid, not as an oxidizing agent. There is no change in the oxidation state of sulfur.

Answer:(a, b )

Explanation:

(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.

Now, consider the incorrect options:

(c) Hydrolysis of $XeF_6$ is not a redox reaction because there is no change in the oxidation states of xenon and fluorine during the reaction. The oxidation state of xenon remains +6 and that of fluorine remains -1.

(d) Xenon fluorides are highly reactive and hydrolyze readily even in the presence of traces of water.

Short Answer Type Questions

Answer:

In contact process $ \mathrm{SO}_3$ is not absorbed directly in water to form $ \mathrm{H}_2 \mathrm{SO}_4$ because the direct reaction is highly exothermic, leading to the formation of acid mist or fog, which is hazardous. Instead, the process is designed to convert $ \mathrm{SO}_3$ into oleum first, which can then be safely diluted with water.

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 $(H_4 P_2 O_7)$

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.

Answer:

Existence 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.

Or

Chlorine has vacant d-orbitals. Which become excited upon bonding when electrons from the 3p-orbitals are promoted to the 3d-orbital giving it a covalency of three.

Fluorine cannot expand it octet due to the absence of empty d-orbitals in 2nd energy shell. Thus it cannot exhibit convalency more than 1.

Hence $ClF_3$ exists but $FCl_3$ does not.

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:

This is due to small size of fluorine and high electronegativity of fluorine. Fluorine being highly electronegative can oxidize sulphur to ( +6 ) oxidation state and can be accomodated easily due to its small size.

Chlorine on the contrary, cannot be accomodated due to its large size. Due to this reason $ SF_6$ is known and $SCl_6$ do not.

Answer:

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

A is $ \mathrm{PCl}_5$ (It is yellowish white powder)

$ \mathrm{P}_4+10 \mathrm{Cl}_2 \longrightarrow 4 \mathrm{PCl}_5 $

B is $ \mathrm{PCl}_3$ (It is a colourless oily liquid)

$ \mathrm{P}_4+6 \mathrm{Cl}_2 \longrightarrow 4 \mathrm{PCl}_3 $

Hydrolysis products are formed as follows:

$ \mathrm{PCl}_3+3 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{H}_3 \mathrm{PO}_3+3 \mathrm{HCl} $

$ \mathrm{PCl}_5+4 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{H}_3 \mathrm{PO}_4+5 \mathrm{HCl} $

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 Reaction

$ 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 $NaOH=0.1 M=0.1 mol / L$

$ \text { Molarity }=\frac{\text { Number of moles }}{\text { Volume in litres }}$

$\text { Volume }=\frac{\text { Number of moles }}{\text { Molarity }}$

$=\frac{8}{200} \times \frac{1}{0.1}$

$=0.4 L \text { or } 400 mL$

$\therefore \quad 400 mL NaOH$ 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 \\ \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 mol of white phosphorus produces 12 mol of HCl

62 g of white phosphorus has been taken which is equivalent to $\frac{62}{124}=\frac{1}{2} \mathrm{~mol}$.

Therefore 6 mol HCl will be formed.

Mass of $6 \mathrm{~mol} \mathrm{HCl}=6 \times 36.5\=219.0 \mathrm{~g} \mathrm{HCl}$

Answer:

Three oxoacids of nitrogen 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 of oxoacid of nitrogen in which nitrogen is in +3 oxidation state.

$ 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:

Structure of phosphinic acid (Hypophosphorous acid) is as follows :

Structure of phosphinic acid

Reducing behaviour of phosphinic acid is observable in the reaction with silver nitrate given below :

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

Matching The Columns

Answer:(a)

A. $\rightarrow (1) $

B. $\rightarrow (3) $

C. $\rightarrow (4) $

D. $\rightarrow (2) $

Answer:(b)

A. $\rightarrow (4) $

B. $\rightarrow (1) $

C. $\rightarrow (2) $

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.

Assertion and Reason

In the following questions a statement of Assertion (A) followed by a statement of Reason ( $R$ ) is given. Choose the correct answer out of the following choices.

(a) Both Assertion and Reason are correct statements, and Reason is the correct explanation of the Assertion.

(b) Both Assertion and Reason are correct statements, and Reason is not the correct explanation of the Assertion.

(c) Assertion is correct, but Reason is wrong statement.

(d) Assertion is wrong but Reason is correct statement.

(e) Both Assertion and Reason are wrong statements.

Answer:(c) Assertion is correct, but Reason is wrong statement.

$N_2$ 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 correct, but Reason is wrong statement.

$HNO_3$ 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, and 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.

$O=O$

$p \pi-p \pi $ bond

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) Both Assertion and Reason are correct statements, and Reason is the correct explanation of the 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:

Amorphous solid A gives B is a gas which turns lime water milky and also produced as a by product during roasting of sulphide ore. This gas decolourises acidified aqueous $KMnO_4$ solution and reduces $Fe^{3+}$ to $Fe^{2+}$. Hence, compound $B(g)$ must be $SO_2$.

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:

(i) Since lead (II) nitrate on heating gives a brown gas ‘A’ therefore, gas ‘A’ must be nitrogen dioxide $\left(\mathrm{NO}_2\right)$.

$ 2 \mathrm{~Pb}\left(\mathrm{NO}_3\right)_2 \xrightarrow{{\Delta},673 \mathrm{~K}} \quad 2 \mathrm{PbO}+\underset{\text { Brown color }(\mathrm{A})}{4 \mathrm{NO}_2}+\mathrm{O}_2 $

(ii) The brown gas ’ A ’ on cooling dimerises to give a colourless solid ’ B ’ therefore ‘B’ must be $ \mathrm{N}_2 \mathrm{O}_4$ (dinitrogen tetroxide).

$ 2 \mathrm{NO}_2 \xleftrightharpoons [heating]{\text{on cooling }} \underset{\text { Colourless solid}(\mathrm{B})}{ \mathrm{~N}_2 \mathrm{O}_4 } $

(iii) Since colourles solid ’ B ’ on heating with NO , gives a blue solid ’ C ’ therefore ‘C’ must be dinitrogen trioxide.

$ 2 \mathrm{NO}+\underset{\text { Colourless solid }(B)}{\mathrm{N}_2 \mathrm{O}_4}\xrightarrow{{\Delta},250 \mathrm{~K}}\underset{\text { Blue solid }(C)}{2 \mathrm{~N}_2 \mathrm{O}_3} $

Thus, $A=\mathrm{NO}_2, B=\mathrm{N}_2 \mathrm{O}_4$ and $ \mathrm{C}=\mathrm{N}_2 \mathrm{O}_3$.

The structure of B and C given below:

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) $3 NO_2+H_2 O \longrightarrow \underset{(D)}{2 HN}{O_3+NO}$