Multiple Choice Questions (MCQs)

1. on addition of conc. $H_2 SO_4$ to a chloride salt, colourless fumes are evolved but in case of iodide salt, violet fumes come out. This is because

(a) $H_2 SO_4$ reduces $HI$ to $I_2$

(b) $HI$ is of violet colour

(c) HI gets oxidised to $I_2$

(d) $HI$ changes to $HIO_3$

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 $I_2$.

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

• (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 $SO_2$.

2. In qualitative analysis when $H_2 S$ is passed through an aqueous solution of salt acidified with dil. $HCl$, a black precipitate is obtained. On boiling the precipitate with dil. $HNO_3$, it forms a solution of blue colour. Addition of excess of aqueous solution of ammonia to this solution gives…… .

(a) deep blue precipitate of $Cu(OH)_2$

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

(c) deep blue solution of $Cu(NO_3)_2$

(d) deep blue solution of $Cu(OH)_2 \cdot Cu(NO_3)_2$

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

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

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

• (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 $Cu(OH)_2 \cdot Cu(NO_3)_2$.

3. In a cyclotrimetaphosphoric acid molecule, how many single and double bonds are present?

(a) 3 double bonds; 9 single bonds

(b) 6 double bonds; 6 single bonds

(c) 3 double bonds; 12 single bonds

(d) Zero double bond; 12 single bonds

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.

• Option (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.

• Option (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.

• Option (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.

4. Which of the following elements can be involved in $p \pi-d \pi$ bonding?

(a) Carbon

(b) Nitrogen

(c) Phosphorus

(d) Boron

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.

• 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.

5. Which of the following pairs of ions are isoelectronic and isostructural?

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

(b) $CIO_3^{-}, CO_3^{2-}$

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

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

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.

• (b) $ClO_3^{-}, CO_3^{2-}$:

  • These ions are not isoelectronic.
  • For $ClO_3^{-}$:
    • Total number of electrons = 17 (Cl) + 3 × 8 (O) + 1 (charge) = 17 + 24 + 1 = 42 electrons.
  • For $CO_3^{2-}$:
    • Total number of electrons = 6 (C) + 3 × 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 × 8 (O) + 2 (charge) = 16 + 24 + 2 = 42 electrons.
  • For $NO_3^{-}$:
    • Total number of electrons = 7 (N) + 3 × 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 × 8 (O) + 1 (charge) = 17 + 24 + 1 = 42 electrons.
  • For $SO_3^{2-}$:
    • Total number of electrons = 16 (S) + 3 × 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.

6. Affinity for hydrogen decreases in the group from fluorine to iodine. Which of the halogen acids should have highest bond dissociation enthalpy?

(a) $HF$

(b) $HCl$

(c) $HBr$

(d) $HI$

Answer

(a)

• (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.

7. Bond dissociation enthalpy of $E-H(E=$ element) bonds is given below. Which of the compounds will act as strongest reducing agent?

(a) $NH_3$

(b) $PH_3$

(c) $AsH_3$

(d) $SbH_3$

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.

• (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 $SbH_3$.

8. On heating with concentrated $NaOH$ solution in an inert atmosphere of $CO_2$, white phosphorus gives a gas. Which of the following statement is incorrect about the gas?

(a) It is highly poisonous and has smell like rotten fish

(b) It’s solution in water decomposes in the presence of light

(c) It is more basic than $NH_3$

(d) It is less basic than $NH_3$

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 $NH_3$.

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

• (a) This statement is correct. Phosphine (PH₃) 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 incorrect. Phosphine (PH₃) is less basic than ammonia (NH₃), which aligns with the correct answer provided.

9. Which of the following acids forms three series of salts?

(a) $H_3 PO_2$

(b) $H_3 BO_3$

(c) $H_3 PO_4$

(d) $H_3 PO_3$

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 $Na_3 PO_4$

• (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.

10. Strong reducing behaviour of $H_3 PO_2$ is due to

(a) low oxidation state of phosphorus

(b) presence of two $-OH$ groups and one $P-H$ bond

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

(d) high electron gain enthalpy of phosphorus

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)

• (a) Low oxidation state of phosphorus: 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.

• (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.

11. On heating lead nitrate forms oxides of nitrogen and lead. The oxides formed are…… .

(a) $N_2 O, PbO$

(b) $NO_2, PbO$

(c) $NO, PbO$

(d) $NO, PbO_2$

Answer

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

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

• Option (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 $N_2O$.

• Option (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 $NO$.

• Option (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 ($PbO$).

12. Which of the following elements does not show allotropy?

(a) Nitrogen

(b) Bismuth

(c) Antimony

(d) Arsenic

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.

• 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.

• 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.

• 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.

13. Maximum covalency of nitrogen is…… .

(a) 3

(b) 5

(c) 4

(d) 6

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.

• (a) 3: This option is incorrect because while nitrogen commonly forms three bonds (as in NH₃), 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.

14. Which of the following statements is wrong?

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

(b) $PH_3$ can act as a ligand in the formation of coordination compound with transition elements.

(c) $NO_2$ is paramagnetic in nature.

(d) Covalency of nitrogen in $N_2 O_5$ is four.

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

• (b) Incorrect Reason: $PH_3$ 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) Incorrect Reason: $NO_2$ is paramagnetic in nature because it has an odd number of electrons, resulting in one unpaired electron.

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

15. $ A$ brown ring is formed in the ring test for $NO_3^{-}$ion. It is due to the formation of

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

(b) $FeSO_4 \cdot NO_2$

(c) $[Fe(H_2 O)_4(NO)_2]^{2+}$

(d) $FeSO_4 \cdot HNO_3$

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.

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

• (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 $HNO_3$.

16. Elements of group- 15 form compounds in +5 oxidation state. However, bismuth forms only one well characterised compound in +5 oxidation state. The compound is

(a) $Bi_2 O_5$

(b) $BiF_5$

(c) $BiCl_5$

(d) $Bi_2 S_5$

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.

• $Bi_2 O_5$: 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.

17. On heating ammonium dichromate and barium azide separately we get

(a) $N_2$ in both cases

(b) $N_2$ with ammonium dichromate and $NO$ with barium azide

(c) $N_2 O$ with ammonium dichromate and $N_2$ with barium azide

(d) $N_2 O$ with ammonium dichromate and $NO_2$ with barium azide

Answer

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

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

• Option (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 $N_2$.

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

• Option (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 $N_2$.

18. In the preparation of $HNO_3$, we get $NO$ gas by catalytic oxidation of ammonia. The moles of NO produced by the oxidation of two moles of $NH_3$ will be…… .

(a) 2

(b) 3

(c) 4

(d) 6

Answer

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

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

• Option (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.

• Option (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.

• Option (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.

19. The oxidation state of central atom in the anion of compound $NaH_2 PO_2$ will be…… .

(a) +3

(b) +5

(c) +1 $($ d) -3

Answer

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

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

• Option (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.

• Option (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.

• Option (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.

20. Which of the following is not tetrahedral in shape?

(a) $NH_4^{+}$

(b) $SiCl_4$

(c) $SF_4$

(d) $SO_4{ }^{2-}$

Answer

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

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

• (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.

21. Which of the following are peroxoacids of sulphur?

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

(b) $H_2 SO_5$ and $H_2 S_2 O_7$

(c) $H_2 S_2 O_7$ and $H_2 S_2 O_8$

(d) $H_2 S_2 O_6$ and $H_2 S_2 O_7$

Answer

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

• Option (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.

• Option (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.

• Option (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.

22. Hot conc. $H_2 SO_4$ acts as moderately strong oxidising agent. It oxidises both metals and non-metals. Which of the following element is oxidised by conc. $H_2 SO_4$ into two gaseous products?

(a) $Cu$

(b) $S$

(c) $C$

(d) $Zn$

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) $\longrightarrow CO_2+2 SO_2+2 H_2 O$

• (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.

23. A black compound of manganese reacts with a halogen acid to give greenish yellow gas. When excess of this gas reacts with $NH_3$ an unstable trihalide is formed. In this process the oxidation state of nitrogen changes from…… .

(a) -3 to +3

(b) -3 to 0

(c) -3 to +5

(d) 0 to -3

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.

• Option (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.

• Option (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.

• Option (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.

24. In the preparation of compounds of $Xe$, Bartlett had taken $O_2^{+} Pt F_6^{-}$as a base compound. This is because

(a) both $O_2$ and $Xe$ have same size.

(b) both $O_2$ and $Xe$ have same electron gain enthalpy.

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

(d) both $Xe$ and $O_2$ are gases.

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.

• (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.

25. In solid state $PCl_5$ is a…… .

(a) covalent solid

(b) octahedral structure

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

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

Answer

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

• (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.

26. Reduction potentials of some ions are given below. Arrange them in decreasing order of oxidising power.

(a) $ClO_4^{-}>IO_4^{-}>BrO_4^{-}$

(b) $IO_4^{-}>BrO_4^{-}>ClO_4^{-}$

(c) $BrO_4^{-}>1 O_4^{-}>ClO_4^{-}$

(d) $BrO_4^{-}>ClO_4^{-}>IO_4^{-}$

Thinking Process

This problem is based on concept of standard reduction potential of species and oxidising property.

Answer

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

Here, $SRP=$ standard reduction potential.

• Option (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.

• Option (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 $IO_4^{-}$.

• Option (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 $ClO_4^{-}$.

27. Which of the following is isoelectronic pair?

(a) $ICl_2, ClO_2$

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

(c) $ClO_2, BrF$

(d) $CN^{-}, O_3$

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.

• For option (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.

• For option (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.

• For option (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)

28. If chlorine gas is passed through hot $NaOH$ solution, two changes are observed in the oxidation number of chlorine during the reaction. These …… are and……

(a) 0 to +5

(b) 0 to +3

(c) 0 to -1

(d) 0 to +1

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.

• Option (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.

• Option (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.

29. Which of the following options are not in accordance with the property mentioned against them?

(a) $F_2>Cl_2>Br_2>I_2$ Oxidising power

(b) $MI>MBr>MCl>MF$ Ionic character of metal halide

(c) $F_2>Cl_2>Br_2>I_2$ Bond dissociation enthalpy

(d) $HI<HBr<HCl<HF$ Hydrogen-halogen bond strength

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.

• For option (c) $F_2>Cl_2>Br_2>I_2$ Bond dissociation enthalpy: The given order is incorrect because the bond dissociation enthalpy does not follow this trend. The correct order is $Cl_2 > Br_2 > F_2 > I_2$. This is due to the electronic repulsion among lone pairs in the $F_2$ molecule, which makes the $F-F$ bond weaker than the $Cl-Cl$ bond. Additionally, the bond length increases from $Cl_2$ to $I_2$, making the bonds easier to break.

• For option (d) $HI<HBr<HCl<HF$ Hydrogen-halogen bond strength: The given order is incorrect because the bond strength of hydrogen-halogen bonds does not follow this trend. The correct order is $HF > HCl > HBr > HI$. This is because the bond strength decreases as the size of the halogen atom increases, making the bond longer and weaker. Hence, the $HF$ bond is the strongest due to the small size and high electronegativity of fluorine, while the $HI$ bond is the weakest due to the large size of iodine.

30. Which of the following is correct for $P_4$ molecule of white phosphorus?

(a) It has 6 lone pairs of electrons

(b) It has six $P-P$ single bonds

(c) It has three $P-P$ single bonds

(d) It has four lone pairs of electrons

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.

• Option (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.

• Option (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.

31. Which of the following statements are correct?

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

(b) Leaving $F-F$ bond, all halogens have weaker $X-X$ bond than $X-X^{\prime}$ bond in interhalogens.

(c) Among interhalogen compounds maximum number of atoms are present in iodine fluoride.

(d) Interhalogen compounds are more reactive than halogen compounds.

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.

• (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.

32. Which of the following statements are correct for $SO_2$ gas?

(a) It acts as bleaching agent in moist conditions.

(b) Its molecule has linear geometry.

(c) Its dilute solution is used as disinfectant.

(d) It can be prepared by the reaction of dilute $H_2 SO_4$ with metal sulphide.

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.

• (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.

33. Which of the following statements are correct?

(a) All the three $N-O$ bond lengths in $HNO_3$ are equal.

(b) All $P-Cl$ bond lengths in $PCl_5$ molecule in gaseous state are equal

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

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

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.

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

• (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.

34. Which of the following orders are correct as per the properties mentioned against each?

$ \begin{array}{lll} \text{(a) } & As_2 O_3 < SiO_2 < P_2 O_3 < SO_2 & \text{ Acid strength.} \\ \text{(b)} & AsH_3 < PH_3 < NH_3 & \text{ Enthalpy of vaporisation.} \\ \text{(c) } & S < O < Cl < F & \text{ More negative electron gain enthalpy.} \\ \text{(d)} & H_2 O > H_2 S > H_2 Se > H_2 Te & \text{ Thermal stability.} \end{array} $

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.

• (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} ).

35. Which of the following statements are correct?

(a) $S-S$ bond is present in $H_2 S_2 O_6$

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

(c) Iron powder along with $Al_2 O_3$ and $K_2 O$ is used as a catalyst in the preparation of $NH_3$ by Haber’s process

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

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 $SO_2$.

• (c) During the preparation of ammonia by the Haber process, iron oxide (Fe₂O₃) is used as the primary catalyst, with small amounts of K₂O and Al₂O₃ added as promoters to increase the efficiency of the catalyst. Iron powder is not used as the catalyst.

• (d) The preparation of SO₃ by the catalytic oxidation of SO₂ is an exothermic reaction, meaning that the change in enthalpy (ΔH) is negative, not positive.

36. In which of the following reactions conc. $H_2 SO_4$ is used as an oxidising reagent?

(a) $CaF_2+H_2 SO_4 \longrightarrow CaSO_4+2 HF$

(b) $2 HI+H_2 SO_4 \longrightarrow I_2+SO_2+2 H_2 O$

(c) $Cu+2 H_2 SO_4 \longrightarrow CuSO_4+SO_2+2 H_2 O$

(d) $NaCl+H_2 SO_4 \longrightarrow NaHSO_4+HCl$

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,

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

• In reaction (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.

• In reaction (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.

37. Which of the following statements are true?

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

(b) lonisation enthalpy of molecular oxygen is very close to that of xenon.

(c) Hydrolysis of $XeF_6$ is a redox reaction.

(d) Xenon fluorides are not reactive.

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.

• (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

38. In the preparation of $H_2 SO_4$ by Contact process, why is $SO_3$ not absorbed directly in water to form $H_2 SO_4$ ?

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

~~ 39. Write a balanced chemical equation for the reaction showing catalytic oxidation of $NH_3$ by atmospheric oxygen.

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

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

~~ 40. Write the structure of pyrophosphoric acid.

Answer

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

Pyrophosphoric acid $(H_4 P_7 O_7)$

~~ 41. $ \mathbf{P H}_3$ forms bubbles when passed slowly in water but $NH_3$ dissolves. Explain why?

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.

~~ 42. In $PCl_5$, phosphorus is in $sp^{3} d$ hybridised state but all its five bonds are not equivalent. Justify your answer with reason.

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.

~~ 43. Why is nitric oxide paramagnetic in gaseous state but the solid obtained on cooling it is diamagnetic?

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.

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~~ 44. Give one reason to explain why $ClF_3$ exists but $FCl_3$ does not exist?

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.

~~ 45. Out of $H_2 O$ and $H_2 S$, which one has higher bond angle and why?

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.

~~ 46. $ SF_6$ is known but $SCl_6$ is not. Why?

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.

~~ 47. on reaction with $Cl_2$, phosphorus forms two types of halides ’ $A$ ’ and ’ $B$ ‘. Halide ’ $A$ ’ is yellowish-white powder but halide ’ $B$ ’ is colourless oily liquid. Identify $A$ and $B$ and write the formulae of their hydrolysis products.

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

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

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

~~ 48. In the ring test of $NO_3^{-}$ion, $Fe^{2+}$ ion reduces nitrate ion to nitric oxide, which combines with $Fe^{2+}$ (aq) ion to form brown complex. Write the reactions involved in the formation of brown ring.

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.

~~ 49. Explain why the stability of oxoacids of chlorine increases in the order given below.

$$ HClO<HClO_2<HClO_3<HClO_4 $$

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

$$ HClO<HClO_2<HClO_3<HClO_4 $$

~~ 50. Explain why ozone is thermodynamically less stable than oxygen?

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.

~~ 51. $ P_4 O_6$ reacts with water according to equation $P_4 O_6+6 H_2 O \longrightarrow 4 H_3 PO_3$. Calculate the volume of $0.1 M NaOH$ solution required to neutralise the acid formed by dissolving $1.1 g$ of $P_4 O_6$ in $H_2 O$.

Thinking Process

This problem includes conceptual mixing of chemical properties of oxides of phosphorus, mole concept and stoichiometry.

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.

~~ 52. White phosphorus reacts with chlorine and the product hydrolyses in the presence of water. Calculate the mass of $HCl$ obtained by the hydrolysis of the product formed by the reaction of $62 g$ of white phosphorus with chlorine in the presence of water.

Thinking Process

This problem is based on concept of chemical reaction of phosphorus and stoichiometry. Write balanced chemical reaction and then calculate the amount of $HCl$ produced.

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

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

~~ 53. Name three oxoacids of nitrogen. Write the disproportionation reaction of that oxoacid of nitrogen in which nitrogen is in +3 oxidation state.

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

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

~~ 54. Nitric acid forms an oxide of nitrogen on reaction with $P_4 O_{10}$. Write the reaction involved. Also write the resonating structures of the oxide of nitrogen formed.

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

~~ 55. (i) white phosphorus (ii) red phosphorus and (iii) black phosphorus. Write the difference between white red and black phosphorus on the basis of their structure and reactivity.

Phosphorus has three allotropic forms -

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.

~~ 56. Give an example to show the effect of concentration of nitric acid on the formation of oxidation product.

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.

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

~~ 57. $ PCl_5$ reacts with finely divided silver on heating and a white silver salt is obtained, which dissolves on adding excess aqueous $NH_3$ solution.

Write the reactions involved to explain what happens.

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 $[Ag(NH_3)_2]^{+} Cl^{-}$

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

~~ 58. Phosphorus forms a number of oxoacids. Out of these oxoacids, phosphinic acid has strong reducing property. Write its structure and also write a reaction showing its reducing behaviour.

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 $4 AgNO_3+2 H_2 O+H_3 PO_2 \longrightarrow 4 Ag \downarrow+4 HNO_3+H_3 PO_4$

~~

Matching The Columns

59. Match the compounds given in Column I with the hybridisation and shape given in Column II and mark the correct option.

Codes

Answer

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

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

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

D. $\rightarrow$ (2)

~~ 60. Match the formulas of oxides given in Column I with the type of oxide given in Column II and mark the correct option.

Codes

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.

~~ 61. Match the items of Columns I and II and mark the correct option.

Codes

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.

~~ 62. Match the species given in Column I with the shape given in Column II and mark the correct option.

Codes

Answer

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

B. $\rightarrow$ (4)

C. $\rightarrow$ (2)

D. $\rightarrow$ (1)

~~ 63. Match the items of Columns I and II and mark the correct option.

Codes

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.

64. Assertion (A) $N_2$ is less reactive than $P_4$.

Reason (R) Nitrogen has more electron gain enthalpy than phosphorus.

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.

~~ 65. Assertion (A) $HNO_3$ makes iron passive.

Reason (R) $HNO_3$ forms a protective layer of ferric nitrate on the surface of iron.

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.

~~ 66. Assertion (A) HI cannot be prepared by the reaction of KI with concentrated $H_2 SO_4$.

Reason (R) HI has lowest $H-X$ bond strength among halogen acids.

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 $H_2 SO_4$.

~~ 67. Assertion (A) Both rhombic and monoclinic sulphur exist as $S_8$ but oxygen exists as $O_2$.

Reason (R) Oxygen forms $p \pi-p \pi$ multiple bond due to small size and small bond length but $p \pi-p \pi$ bonding is not possible in sulphur.

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 $O_2$

~~ 68. Assertion (A) $NaCl$ reacts with concentrated $H_2 SO_4$ to give colourless fumes with pungent smell. But on adding $MnO_2$ the fumes become greenish yellow.

Reason (R) $MnO_2$ oxidises $HCl$ to chlorine gas which is greenish yellow.

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.

~~ 69. Assertion (A) $SF_6$ cannot be hydrolysed but $SF_4$ can be.

Reason (R) Six F-atoms in $SF_6$ prevent the attack of $H_2 O$ on sulphur atom of $SF_6$.

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 $SF_6$.

~~

Long Answer Type Questions

70. An amorphous solid " $A$ " burns in air to form a gas " $B$ " which turns lime water milky. The gas is also produced as a by-product during roasting of sulphide ore. This gas decolourises acidified aqueous $KMnO_4$ solution and reduces $Fe^{3+}$ to $Fe^{2+}$. Identify the solid " $A$ " and the gas " $B$ " and write the reactions involved.

Thinking Process

This problem is based on concept of properties of sulphur and its oxide. $A \xrightarrow[\substack{\text { air } \\ \text { (amorphous solid) } }]{\text { Burn in }}$ (gas)

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 $SO_2$.

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

~~ 71. On heating lead (II) nitrate gives a brown gas " $A$ “. The gas " $A$ " on cooling changes to colourless solid " $B$ “. Solid " $B$ " on heating with NO changes to a blue solid ’ $C$ ‘. Identify ’ $A$ ‘, ’ $B$ ’ and ’ $C$ ’ and also write reactions involved and draw the structures of ’ $B$ ’ and ’ $C$ ‘.

Thinking Process

This problem is based on preparation and properties of $NO_2$.

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) $N_2 O_3$

~~ 72. On heating compound (A) gives a gas (B) which is a constituent of air. This gas when treated with 3 moles of hydrogen $(H_2)$ in the presence of a catalyst gives another gas (C) which is basic in nature. Gas $C$ on further oxidation in moist condition gives a compound (D) which is a part of acid rain. Identify compounds (A) to (D) and also give necessary equations of all the steps involved.

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