The p-Block Elements

Multiple Choice Questions (MCQs)

1. on addition of conc. H2SO4 to a chloride salt, colourless fumes are evolved but in case of iodide salt, violet fumes come out. This is because

(a) H2SO4 reduces HI to I2

(b) HI is of violet colour

(c) HI gets oxidised to I2

(d) HI changes to HIO3

Answer

(c) Hydrogen iodide (HI) is more stronger oxidising agent than H2SO4. So, it reduces H2SO4 to SO2 and itself oxidises to I2. Colour of I2 is violet hence on adding conc. H2SO4 to HI, it gets oxidised to I2.

H2SO4+2HISO2+I2 (Violet  colour) +2H2O

  • (a) H2SO4 reduces HI to I2: This statement is incorrect because it is actually the HI that reduces H2SO4 to SO2, not the other way around. HI is oxidized to I2 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 I2.

  • (d) HI changes to HIO3: This statement is incorrect because HI does not change to HIO3 in this reaction. Instead, HI is oxidized to I2 while reducing H2SO4 to SO2.

2. In qualitative analysis when H2S is passed through an aqueous solution of salt acidified with dil. HCl, a black precipitate is obtained. On boiling the precipitate with dil. HNO3, 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(NH3)4]2+

(c) deep blue solution of Cu(NO3)2

(d) deep blue solution of Cu(OH)2Cu(NO3)2

Answer

(b) In qualitative analysis when H2S is passed through an aqueous solution of salt acidified with dil. HCl a black ppt. of CuS is obtained.

CuSO4+H2S dil. HClCuS black ppt +H2SO4

On boiling CuS with dil. HNO3 it forms a blue coloured solution and the following reactions occur

3CuS+8HNO33Cu(NO3)2+2NO+3S+4H2OS+2HNO3H2SO4+NO2Cu2++SO42+2NH3+2H2OCu(OH)2CuSO4+2NH4OH

Cu(OH)2CuSO4+8NH32[Cu(NH3)4]SO4+2OH+SO42 Tetraammine copper (II) (Deep blue solution) 

  • (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(NO3)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(NH3)4]2+.

  • (c) deep blue solution of Cu(NO3)2: This option is incorrect because Cu(NO3)2 itself does not form a deep blue solution. The deep blue color is specifically due to the formation of the complex ion [Cu(NH3)4]2+ when excess ammonia is added to the solution containing Cu2+ ions.

  • (d) deep blue solution of Cu(OH)2Cu(NO3)2: This option is incorrect because Cu(OH)2Cu(NO3)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(NH3)4]2+, and not due to any compound like Cu(OH)2Cu(NO3)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 π bonds and numerics 1 to 12 are sigma (σ) 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 pπdπ 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 pπdπ bonding.

  • Carbon: Carbon does not have vacant d-orbitals, so it cannot participate in pπdπ bonding.
  • Nitrogen: Nitrogen also lacks vacant d-orbitals, making it unable to form pπdπ bonds.
  • Boron: Boron does not possess vacant d-orbitals, thus it cannot engage in pπdπ bonding.

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

(a) CO32,NO3

(b) CIO3,CO32

(c) SO32,NO3

(d) ClO3,SO32

Answer

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

For CO32

Total number of electrons

=6+8×3+2=6+24+2=32

 For NO3

Total number of electrons

=7+8×3+1=7+25=32

Hence, CO32 and NO3are isoelectronic. These two ions have similar structure so they are isostructural.

alt text

Both have triangular planar structure as in both the species carbon and nitrogen are sp2 hybridised. Hence, (a) is the correct choice.

  • (b) ClO3,CO32:

    • These ions are not isoelectronic.
    • For ClO3:
      • Total number of electrons = 17 (Cl) + 3 × 8 (O) + 1 (charge) = 17 + 24 + 1 = 42 electrons.
    • For CO32:
      • 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) SO32,NO3:

    • These ions are not isoelectronic.
    • For SO32:
      • Total number of electrons = 16 (S) + 3 × 8 (O) + 2 (charge) = 16 + 24 + 2 = 42 electrons.
    • For NO3:
      • 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) ClO3,SO32:

    • These ions are not isoelectronic.
    • For ClO3:
      • Total number of electrons = 17 (Cl) + 3 × 8 (O) + 1 (charge) = 17 + 24 + 1 = 42 electrons.
    • For SO32:
      • 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.
    • ClO3 has a trigonal pyramidal structure due to the presence of a lone pair on chlorine, while SO32 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)

alt text

  • (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 EH(E= element) bonds is given below. Which of the compounds will act as strongest reducing agent?

Compound NH3 PH3 AsH3 SbH3
Δdiss (EH)/kJmol1 389 322 297 255

(a) NH3

(b) PH3

(c) AsH3

(d) SbH3

Answer

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

Hence, SbH3 is act as strongest reducing agent among these.

  • (a) NH3: The bond dissociation enthalpy of NH3 is the highest among the given compounds (389 kJ/mol). This means that the NH bond is the strongest and requires the most energy to break. As a result, NH3 is the least likely to donate electrons and act as a reducing agent.

  • (b) PH3: The bond dissociation enthalpy of PH3 is 322 kJ/mol, which is lower than that of NH3 but still relatively high compared to AsH3 and SbH3. This indicates that the PH bond is stronger than the AsH and SbH bonds, making PH3 a weaker reducing agent than AsH3 and SbH3.

  • (c) AsH3: The bond dissociation enthalpy of AsH3 is 297 kJ/mol, which is lower than that of NH3 and PH3 but higher than that of SbH3. This means that the AsH bond is weaker than the NH and PH bonds but stronger than the SbH bond. Therefore, AsH3 is a stronger reducing agent than NH3 and PH3, but not as strong as SbH3.

8. On heating with concentrated NaOH solution in an inert atmosphere of CO2, 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 NH3

(d) It is less basic than NH3

Answer

(c) White phosphorous on reaction with NaOH solution in the presence of inert atmosphere of CO2 it produces phosphine gas which is less basic than NH3.

P4+3NaOH+3H2OPH3+3NaH2PO2 (sodium hypophosphite) 

  • (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) H3PO2

(b) H3BO3

(c) H3PO4

(d) H3PO3

Answer

(c) Structure of H3PO4 is

POHOHOOH

H3PO4 has 3OH groups i.e., has three ionisable H-atoms and hence forms three series of salts. These three possible series of salts for H3PO4 are as follows

NaH2PO4,Na2HPO4 and Na3PO4

  • (a) H3PO2: This acid, also known as hypophosphorous acid, has the structure H2PO(OH). It contains only one ionizable hydrogen atom, which means it can only form one series of salts.

  • (b) H3BO3: 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) H3PO3: 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 H3PO2 is due to

(a) low oxidation state of phosphorus

(b) presence of two OH groups and one PH bond

(c) presence of one OH group and two PH bonds

(d) high electron gain enthalpy of phosphorus

Answer

(c) Strong reducing behaviour of H3PO2 is due to presence of two PH bonds and one POH bond

Hypophosphorous following (Monobasic)

  • (a) Low oxidation state of phosphorus: While the low oxidation state of phosphorus in H3PO2 does contribute to its reducing nature, it is not the primary reason for its strong reducing behavior. The presence of PH bonds is more directly responsible for this property.

  • (b) Presence of two OH groups and one PH bond: This option is incorrect because H3PO2 actually has two PH bonds and only one OH group. The presence of two PH 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 H3PO2. The reducing behavior is mainly due to the presence of PH bonds, which can donate electrons easily.

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

(a) N2O,PbO

(b) NO2,PbO

(c) NO,PbO

(d) NO,PbO2

Answer

(b) On heating lead nitrate it produces brown coloured nitrogen dioxide (NO2) and lead (II) oxide.

2Pb(NO3)2Δ4NO2+2PbO+O2

  • Option (a) N2O,PbO: This option is incorrect because heating lead nitrate does not produce nitrous oxide (N2O). The thermal decomposition of lead nitrate specifically produces nitrogen dioxide (NO2), not N2O.

  • 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 (NO2), not NO.

  • Option (d) NO,PbO2: This option is incorrect because heating lead nitrate does not produce nitric oxide (NO) or lead dioxide (PbO2). The correct products are nitrogen dioxide (NO2) 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 NN 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 NN bond is stronger than the single PP bond.

(b) PH3 can act as a ligand in the formation of coordination compound with transition elements.

(c) NO2 is paramagnetic in nature.

(d) Covalency of nitrogen in N2O5 is four.

Answer

(a) True statement is that single NN bond is weaker than the single PP bond. This is why phosphorous show allotropy but nitrogen does not.

(i) PH3 acts as a ligand in the formation of coordination compound due to presence of lone pair of electrons.

(ii) NO2 is paramagnetic in nature due to presence of one unpaired electron. Structure of NO2 is

alt text

(iii) Covalency of nitrogen in N2O5 is 4

alt text

  • (b) Incorrect Reason: PH3 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: NO2 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 N2O5 is indeed four, as each nitrogen forms four covalent bonds in the structure of N2O5.

15. A brown ring is formed in the ring test for NO3ion. It is due to the formation of

(a) [Fe(H2O)5(NO)]2+

(b) FeSO4NO2

(c) [Fe(H2O)4(NO)2]2+

(d) FeSO4HNO3

Answer

(a) When freshly prepared solution of FeSO4 is added in a solution containing NO3ion, it leads to formation of a brown coloured complex. This is known as brown ring test of nitrate.

NO3+3Fe2++4H+NO+3Fe3++2H2O[Fe(H2O)6]2++NO[Fe(H2O)5(NO)]2++H2O Brown ring 

  • (b) FeSO4NO2: This compound is not formed during the brown ring test. The brown ring test specifically involves the formation of a complex between Fe2+ and NO, not NO2.

  • (c) [Fe(H2O)4(NO)2]2+: This complex does not form in the brown ring test. The correct complex formed is [Fe(H2O)5(NO)]2+, which involves only one NO molecule coordinated to the iron center.

  • (d) FeSO4HNO3: 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 FeSO4 and HNO3.

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) Bi2O5

(b) BiF5

(c) BiCl5

(d) Bi2S5

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 BiF5. It is due to smaller size and high electronegativity of fluorine.

  • Bi2O5: 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.
  • BiCl5: 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.
  • Bi2S5: 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) N2 in both cases

(b) N2 with ammonium dichromate and NO with barium azide

(c) N2O with ammonium dichromate and N2 with barium azide

(d) N2O with ammonium dichromate and NO2 with barium azide

Answer

(a) On heating ammonium dichromate and barium azide it produces N2 gas separately.

(NH4)2Cr2O7ΔN2+4H2O+Cr2O3Ba(N3)2Ba+3N2

  • Option (b) is incorrect because ammonium dichromate does not produce N2; it produces N2O instead. Additionally, barium azide does not produce NO; it produces N2.

  • Option (c) is incorrect because ammonium dichromate does not produce N2O; it produces N2. However, barium azide correctly produces N2.

  • Option (d) is incorrect because ammonium dichromate does not produce N2O; it produces N2. Additionally, barium azide does not produce NO2; it produces N2.

18. In the preparation of HNO3, we get NO gas by catalytic oxidation of ammonia. The moles of NO produced by the oxidation of two moles of NH3 will be…… .

(a) 2

(b) 3

(c) 4

(d) 6

Answer

(a) Two moles of NH3 will produce 2 moles of NO on catalytic oxidation of ammonia in preparation of nitric acid.

4NH3+5O2 Pt  Rh gauge catalyst Δ4NO(g)+6H2O(l)

  • Option (b) is incorrect because the balanced chemical equation shows that 4 moles of NH3 produce 4 moles of NO. Therefore, 2 moles of NH3 would produce 2 moles of NO, not 3 moles.

  • Option (c) is incorrect because the balanced chemical equation shows that 4 moles of NH3 produce 4 moles of NO. Therefore, 2 moles of NH3 would produce 2 moles of NO, not 4 moles.

  • Option (d) is incorrect because the balanced chemical equation shows that 4 moles of NH3 produce 4 moles of NO. Therefore, 2 moles of NH3 would produce 2 moles of NO, not 6 moles.

19. The oxidation state of central atom in the anion of compound NaH2PO2 will be…… .

(a) +3

(b) +5

(c) +1 ( d) -3

Answer

(c) Let oxidation state of P in NaH2PO2 is x.

1+2×1+x+2×2=01+2+x4=0+x1=0x=+1

  • Option (a) +3: This is incorrect because the calculation of the oxidation state of phosphorus (P) in NaH2PO2 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 NaH2PO2 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 NaH2PO2 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) NH4+

(b) SiCl4

(c) SF4

(d) SO42

Answer

(c) SF4 has sea-saw shaped as shown below

It has trigonal bipyramidal geometry having sp3d hybridisation.

  • (a) NH4+: The ammonium ion (NH4+) has a tetrahedral shape because it has four hydrogen atoms symmetrically arranged around the central nitrogen atom. The nitrogen atom undergoes sp3 hybridization, resulting in a tetrahedral geometry.

  • (b) SiCl4: Silicon tetrachloride (SiCl4) has a tetrahedral shape because the silicon atom is surrounded by four chlorine atoms. The silicon atom undergoes sp3 hybridization, leading to a tetrahedral geometry.

  • (d) SO42: The sulfate ion (SO42) has a tetrahedral shape because the sulfur atom is surrounded by four oxygen atoms. The sulfur atom undergoes sp3 hybridization, resulting in a tetrahedral geometry.

21. Which of the following are peroxoacids of sulphur?

(a) H2SO5 and H2S2O8

(b) H2SO5 and H2S2O7

(c) H2S2O7 and H2S2O8

(d) H2S2O6 and H2S2O7

Answer

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

  • Option (b) H2SO5 and H2S2O7: H2S2O7 (disulfuric acid) does not contain a peroxo bond (OO bond). It is not a peroxoacid of sulfur.

  • Option (c) H2S2O7 and H2S2O8: H2S2O7 (disulfuric acid) does not contain a peroxo bond (OO bond). It is not a peroxoacid of sulfur.

  • Option (d) H2S2O6 and H2S2O7: Neither H2S2O6 (dithionic acid) nor H2S2O7 (disulfuric acid) contain a peroxo bond (OO bond). They are not peroxoacids of sulfur.

22. Hot conc. H2SO4 acts as moderately strong oxidising agent. It oxidises both metals and non-metals. Which of the following element is oxidised by conc. H2SO4 into two gaseous products?

(a) Cu

(b) S

(c) C

(d) Zn

Answer

(c) H2SO4 is a moderately strong oxidising agent which oxidises both metals and non-metals as shown below

Cu+2H2SO4 (conc) CuSO4+SO2+2H2OS+2H2SO4 (conc) 3SO2+2H2O

While carbon on oxidation with H2SO4 produces two types of oxides CO2 and SO2.

C+2H2SO4 (conc) CO2+2SO2+2H2O

  • (a) Cu: Copper (Cu) reacts with hot concentrated sulfuric acid (H2SO4) to produce copper sulfate (CuSO4), sulfur dioxide (SO2), and water (H2O). Only one gaseous product, SO2, is formed in this reaction.

  • (b) S: Sulfur (S) reacts with hot concentrated sulfuric acid (H2SO4) to produce sulfur dioxide (SO2) and water (H2O). Only one gaseous product, SO2, is formed in this reaction.

  • (d) Zn: Zinc (Zn) reacts with hot concentrated sulfuric acid (H2SO4) to produce zinc sulfate (ZnSO4), sulfur dioxide (SO2), and water (H2O). Only one gaseous product, SO2, 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 NH3 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 MnO2 reacts with HCl to produce greenish yellow coloured gas of Cl2

MnO2 (Black) +4HClMnCl2+2H2O+Cl2 (greenish  yellow gas) 

Cl2 on further treatment with NH3 produces NCl3.

N3H3+3Cl2N+3Cl3+3HCl

NH3(3) changes to NCl3(+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 NH3 and Cl2, the nitrogen in NH3 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 NH3 and Cl2 results in the formation of NCl3, 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 NH3 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 O2+PtF6as a base compound. This is because

(a) both O2 and Xe have same size.

(b) both O2 and Xe have same electron gain enthalpy.

(c) both O2 and Xe have almost same ionisation enthalpy.

(d) both Xe and O2 are gases.

Answer

(c) Bertlett had taken O2+PtF6as a base compound because O2 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 O2 and Xe do not have the same size. O2 is a diatomic molecule with a smaller size compared to the monatomic noble gas Xe.
  • (b) Both O2 and Xe do not have the same electron gain enthalpy. O2 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 O2 are gases, this is not the reason Bartlett chose O2+PtF6 as a base compound. The key factor was the similarity in ionisation enthalpy, not their physical state.

25. In solid state PCl5 is a…… .

(a) covalent solid

(b) octahedral structure

(c) ionic solid with [PCl6]+octahedral and [PCl4]tetrahedral

(d) ionic solid with [PCl4]+tetrahedral and [PCl6]octahedral

Answer

(d) In solid state PCl5 exists as an ionic solid with [PCl4]+tetrahedral and [PCl6] octahedral.

alt text

  • (a) covalent solid: This option is incorrect because in the solid state, PCl5 does not exist as a covalent solid. Instead, it dissociates into ionic species, forming [PCl4]+ and [PCl6] ions.

  • (b) octahedral structure: This option is incorrect because PCl5 in the solid state does not form a simple octahedral structure. Instead, it forms an ionic lattice with [PCl4]+ ions having a tetrahedral geometry and [PCl6] ions having an octahedral geometry.

  • (c) ionic solid with [PCl6]+ octahedral and [PCl4] tetrahedral: This option is incorrect because the charges and geometries of the ions are reversed. In the solid state, PCl5 forms [PCl4]+ ions with a tetrahedral geometry and [PCl6] 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.

Ion ClO4 IO4 BrO4
Reduction potential E/V E=1.19V Es=1.65V Es=1.74V

(a) ClO4>IO4>BrO4

(b) IO4>BrO4>ClO4

(c) BrO4>1O4>ClO4

(d) BrO4>ClO4>IO4

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.

alt text

Here, SRP= standard reduction potential.

  • Option (a) ClO4>IO4>BrO4: This option is incorrect because it suggests that ClO4 has the highest oxidizing power, followed by IO4 and then BrO4. However, the standard reduction potentials indicate that BrO4 has the highest value (1.74 V), followed by IO4 (1.65 V), and then ClO4 (1.19 V). Therefore, ClO4 should have the lowest oxidizing power, not the highest.

  • Option (b) IO4>BrO4>ClO4: This option is incorrect because it suggests that IO4 has a higher oxidizing power than BrO4. However, the standard reduction potentials indicate that BrO4 (1.74 V) has a higher value than IO4 (1.65 V), meaning BrO4 should have a higher oxidizing power than IO4.

  • Option (d) BrO4>ClO4>IO4: This option is incorrect because it suggests that ClO4 has a higher oxidizing power than IO4. However, the standard reduction potentials indicate that IO4 (1.65 V) has a higher value than ClO4 (1.19 V), meaning IO4 should have a higher oxidizing power than ClO4.

27. Which of the following is isoelectronic pair?

(a) ICl2,ClO2

(b) BrO2, BrF2+

(c) ClO2,BrF

(d) CN,O3

Answer

(b) Isoelectronic pair have same number of electrons

BrO2 BrF2+
Total number of electrons =35+2×8+1=52 =35+9×21=52

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

ICl2 ClO2
53+2×17=87 17+16=33
ClO2 BrF
17+16=33 35+9=44
CN O3
=6+7+1=14 =8×3=24

Hence, only (b) is the correct choice.

  • For option (a) ICl2 and ClO2:

    • ICl2: 53+2×17=87 electrons
    • ClO2: 17+16×2=49 electrons
    • The total number of electrons is not equal, hence they are not isoelectronic.
  • For option (c) ClO2 and BrF:

    • ClO2: 17+16×2=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 O3:

    • CN: 6+7+1=14 electrons
    • O3: 8×3=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 NaClO3.

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) F2>Cl2>Br2>I2 Oxidising power

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

(c) F2>Cl2>Br2>I2 Bond dissociation enthalpy

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

Answer

(b,c)

F2>Cl2>Br2>I2 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.

F2>Cl2>Br2>I2

This is incorrect order of bond dissociation energy. Correct order is Cl2>Br2>F2>I2 due to electronic repulsion among lone pairs in F2 molecule.

  • For option (c) F2>Cl2>Br2>I2 Bond dissociation enthalpy: The given order is incorrect because the bond dissociation enthalpy does not follow this trend. The correct order is Cl2>Br2>F2>I2. This is due to the electronic repulsion among lone pairs in the F2 molecule, which makes the FF bond weaker than the ClCl bond. Additionally, the bond length increases from Cl2 to I2, 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 P4 molecule of white phosphorus?

(a) It has 6 lone pairs of electrons

(b) It has six PP single bonds

(c) It has three PP single bonds

(d) It has four lone pairs of electrons

Answer

( b,d)

Structure of P4 molecule can be represented as

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

It has six PP single bond.

  • Option (a) is incorrect because the P4 molecule has a total of 4 lone pairs of electrons, not 6. Each phosphorus atom in the P4 molecule has one lone pair, resulting in a total of 4 lone pairs.

  • Option (c) is incorrect because the P4 molecule has six PP single bonds, not three. Each phosphorus atom forms three single bonds with three other phosphorus atoms, resulting in a total of six PP single bonds.

31. Which of the following statements are correct?

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

(b) Leaving FF bond, all halogens have weaker XX bond than XX 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 XX bond in interhalogen is weaker than XX bond in halogens except FF 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 XX bond as compared to XX of halogen compounds.

  • (b) The statement is incorrect because it implies that all halogens have weaker XX bonds than XX bonds in interhalogens, except for the FF 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 XX bond in interhalogens is typically weaker than the XX bond in halogens, with the exception of the FF bond.

32. Which of the following statements are correct for SO2 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 H2SO4 with metal sulphide.

Answer

(a, c)

(a) In moist condition SO2 gas acts as a bleaching agent.

e.g., it converts Fe (III) to Fe (II) ion and decolourises acidified KMnO4 (VII).

2Fe3++SO2+2H2O2Fe2++SO42+4H+

(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 O2 with sulphide ore,

4FeS2+11O22Fe2O3+8SO2

while metal on treatment with H2SO4 produces H2S.

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

  • (b) is incorrect because the SO2 molecule has a bent structure, not a linear geometry.

  • (d) is incorrect because SO2 is not prepared by the reaction of dilute H2SO4 with metal sulphide. Instead, this reaction produces H2S. SO2 can be prepared by the reaction of O2 with sulphide ore.

33. Which of the following statements are correct?

(a) All the three NO bond lengths in HNO3 are equal.

(b) All PCl bond lengths in PCl5 molecule in gaseous state are equal

(c) P4 molecule in white phosphorus have angular strain therefore white phosphorus is very reactive

(d) PCl5 is ionic in solid state in which cation is tetrahedral and anion is octahedral.

Answer

(c, d)

(a) All the three NO bond lengths in HNO3 are not equal.

(b) All PCl bond lengths in PCl5 molecule in gaseous state are not equal. Axial bond is longer than equatorial bond.

(c) P4 molecule in white phosphorous have angular strain therefore white phosphorous is very reactive.

(d) PCl5 is ionic in solid state in which cation is tetrahedral and anion is octahedral.

 Cation [PCl4]+Anion [PCl6]

  • (a) All the three NO bond lengths in HNO3 are not equal because HNO3 has one N=O double bond and two NO single bonds, leading to different bond lengths.

  • (b) All PCl bond lengths in PCl5 molecule in gaseous state are not equal because the molecule has a trigonal bipyramidal structure, where the axial PCl bonds are longer than the equatorial PCl bonds.

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

(a) As2O3<SiO2<P2O3<SO2 Acid strength.(b)AsH3<PH3<NH3 Enthalpy of vaporisation.(c) S<O<Cl<F More negative electron gain enthalpy.(d)H2O>H2S>H2Se>H2Te Thermal stability.

Answer

(a,d)

(a)  acidic strength increases As2O3<SiO2<P2O3<SO2

(b) Correct order is  enthalpy of vaporisation AsH3>PH3>NH3

(b) Correct order is AsH3>PH3>NH3 enthalpy of vaporisation 

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

(d) H2O>H2S>H2Se>H2 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) SS bond is present in H2S2O6

(b) In peroxosulphuric acid (H2SO5) sulphur is in +6 oxidation state

(c) Iron powder along with Al2O3 and K2O is used as a catalyst in the preparation of NH3 by Haber’s process

(d) Change in enthalpy is positive for the preparation of SO3 by catalytic oxidation of SO2

Answer

( a,b )

(a) Structure of H2S2O6 is as shown below

It contains one SS bond.

(b) In peroxosulphuric acid (H2SO5) sulphur is in +6 oxidation state.

Structure of H2SO5 is

OSOOHOOH

Let oxidation state of S=x

2×(+1)+x+3×(2)+2×(1)=0x6=0x=6

(c) During preparation of ammonia, iron oxide with small amount of K2O and Al2O3 is used as a catalyst to increase the rate of attainment of equilibrium.

(d) Change in enthalpy is negative for preparation of SO3 by catalytic oxidation of SO2.

  • (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. H2SO4 is used as an oxidising reagent?

(a) CaF2+H2SO4CaSO4+2HF

(b) 2HI+H2SO4I2+SO2+2H2O

(c) Cu+2H2SO4CuSO4+SO2+2H2O

(d) NaCl+H2SO4NaHSO4+HCl

Answer

(b,c)

In the above given four reactions, (b) and (c) represent oxidising behaviour of H2SO4. As we know that oxidising agent reduces itself as oxidation state of central atom decreases.

Here,

2HI1+H2S64I0I2+S4O2+2H2OC0u+2H2S+6O4C+2uSO4+SO2+4+2H2O

  • In reaction (a) CaF2+H2SO4CaSO4+2HF, conc. H2SO4 is acting as an acid, not as an oxidizing agent. There is no change in the oxidation state of sulfur.

  • In reaction (d) NaCl+H2SO4NaHSO4+HCl, conc. H2SO4 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 XeF6 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 XeF6(Xe+6F61Xe+6O32+3H+1F1) is not a redox reaction.

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

  • (c) Hydrolysis of XeF6 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 H2SO4 by Contact process, why is SO3 not absorbed directly in water to form H2SO4 ?

Answer

In Contact process SO3 is not absorbed directly in water to from H2SO4 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 NH3 by atmospheric oxygen.

Answer

Ammonia (NH3) on catalytic oxidation by atmospheric oxygen in presence of Rh/Pt gauge at 500K under pressure of 9 bar produces nitrous oxide.

Balanced chemical reaction can be written as

4NH3+5O2 From air 500K;9 bar  Pt /Rh gauge catalyst 4NO+6H2O

~~ 40. Write the structure of pyrophosphoric acid.

Answer

Molecular formula of pyrophosphoric acid is H4P2O7 and its structure is as follows

OHPOOHOPOOHOH

Pyrophosphoric acid (H4P7O7)

~~ 41. PH3 forms bubbles when passed slowly in water but NH3 dissolves. Explain why?

Answer

Dissolution of NH3 and PH3 in water can be explained on the basis of H-bonding. NH3 forms H-bond with water so it is soluble but PH3 does not form H-bond with water so it remains as gas and forms bubble in water.

~~ 42. In PCl5, phosphorus is in sp3d 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 PCl 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, NO2 exists as a monomer which has one unpaired electron but in solid state, it dimerises to N2O4 so no unpaired electron left. Therefore, NO2 is paramagnetic in gaseous state but diamagnetic in solid state.

f0bd3a00/public)

~~ 44. Give one reason to explain why ClF3 exists but FCl3 does not exist?

Answer

Existance of ClF3 and FCl3 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 H2O and H2S, which one has higher bond angle and why?

Answer

Bond angle of H2O(HOH=104.5) is larger than that of H2S(HSH=92) because oxygen is more electronegative than sulphur therefore, bond pair electron of OH bond will be closer to oxygen and there will be more bond pair—bond pair repulsion between bond pairs of two OH bonds.

~~ 46. SF6 is known but SCl6 is not. Why?

Answer

Fluorine atom is smaller in size so, six Fions can surround a sulphur atom. The case is not so with chlorine atom due to its large size. SoSF6 is known but SCl6 is not known due to interionic repulsion between larger Clions.

~~ 47. on reaction with Cl2, 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 Cl2 forms two types of halides A and B. ’ A ’ is PCl5 and ’ B ’ is PCl3.

When ’ A ’ and ’ B ’ are hydrolysed

P4+10Cl24PCl5P4+6Cl24PCl3

When ‘A’ and ‘B’ are hydrolysed

(a)PCl5[A]phophoruspentachloride+4H2OH3PO4phophoricacid+5HCl

(a)PCl3[B]Phosphorustrichloride+3H2OH3PO3phophorusacid+3HCl

~~ 48. In the ring test of NO3ion, Fe2+ ion reduces nitrate ion to nitric oxide, which combines with Fe2+ (aq) ion to form brown complex. Write the reactions involved in the formation of brown ring.

Answer

NO3+3Fe2++4H+NO+3Fe3++2H2O

[Fe(H2O)6]2++NO[Fe(H2O)5NO]2++H2O Brown ring 

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<HClO2<HClO3<HClO4

Answer

Oxygen is more electronegative than chlorine, therefore dispersal of negative charge present on chlorine increases from ClOto ClO4ion because number of oxygen atoms attached to chlorine is increasing. Therefore, stability of ions will increase in the order given below

ClO<ClO2<ClO3<ClO4

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

HClO<HClO2<HClO3<HClO4

~~ 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 ( ΔH is negative) and an increase in entropy ( ΔS is positive). These two effects reinforce each other, resulting in large negative Gibbs energy change (ΔG) for its conversion into oxygen.

~~ 51. P4O6 reacts with water according to equation P4O6+6H2O4H3PO3. Calculate the volume of 0.1MNaOH solution required to neutralise the acid formed by dissolving 1.1g of P4O6 in H2O.

Thinking Process

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

Answer

P4O6+6H2O4H3PO3(i)

Neutralisation

.H3PO3+2NaOHNa2HPO3+2H2O]×4(ii)

Adding Eqs. (i) and (ii)

P4O61 mol+8NaOH8 mol4Na2HPO3+2H2O (iii)

Number of moles of P4O6,

n=mM=1.1220=1200mol

(Molar mass of P4O6=(4×31)+(6×16)=220

Product formed by 1 mole of P4O6 is neutralised by 8 moles NaOH

Product formed by 1200 moles of P4O6 will be neutralised by NaOH

=8×1200=8200 mole NaOH

Given, Molarity of NaOH=0.1M=0.1mol/L

 Molarity = Number of moles  Volume in litres  Volume = Number of moles  Molarity =8200×10.1=0.4L or 400mL

400mLNaOH 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 62g 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

P4+6Cl24PCl3PCl3+3H2OH3PO3+3HCl]×4P4+6Cl2+12H2O4H3PO3+12HCl

1 mol12 mol

31×4=124 g12×36.5=438.0 g

124g of white phosphorus produces HCl=438g

62g of white phosphorus will produces

HCl=438124×62=219.0gHCl

~~ 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) HNO2, nitrous acid

(b) HNO3, nitric acid

(c) Hyponitrous acid, H2N2O2

In HNO2,N is in +3 oxidation state

Disproportionation reaction

3HNO2 Disproportionation HNO3+H2O+2NO

~~ 54. Nitric acid forms an oxide of nitrogen on reaction with P4O10. Write the reaction involved. Also write the resonating structures of the oxide of nitrogen formed.

Answer

P4O10 being a dehydrating agent, on reaction with HNO3 removes a molecule of water and forms anhydride of HNO3.

4HNO3+P4O104HPO3+2N2O5

Resonating structures of N2O5 are

alt text

~~ 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 HNO3. Dilute and concentrated nitric acid give different oxidation products on reaction with copper metal.

3Cu+8HNO3 (Dil.) 3Cu(NO3)2+2NO+4H2OCu+4HNO3 (Conc.) Cu(NO3)2+2NO2+2H2O

~~ 57. PCl5 reacts with finely divided silver on heating and a white silver salt is obtained, which dissolves on adding excess aqueous NH3 solution.

Write the reactions involved to explain what happens.

Answer

PCl5 on reaction with finely divided silver produced silver halide.

PCl5+2Ag2AgCl+PCl3

AgCl on further reaction with aqueous ammonia solution produces a soluble complex of [Ag(NH3)2]+Cl

AgCl+2NH3(aq)[Ag(NH3)2]+Cl Soluble complex 

~~ 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.

alt text

Structure of phosphinic acid

Reaction showing reducing behaviour of phosphinic acid is as follows 4AgNO3+2H2O+H3PO24Ag+4HNO3+H3PO4

~~

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.

Column I Column II
A. XeF6 1. sp3d3-distorted octahedral
B. XeO3 2. sp3d2-square planar
C. XeOF4 3. sp3-pyramidal
D. XeF4 4. sp3d2-square pyramidal

Codes

A B C D
(a) 1 3 4 2
(b) 1 2 4 3
(c) 4 3 1 2
(d) 4 1 2 3

Answer

(a) A. (1)

B. (3)

C. (4)

D. (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.

Column I Column II
A. Pb3O4 1. Neutral oxide
B. N2O 2. Acidic oxide
C. Mn2O7 3. Basic oxide
D. Bi2O3 4. Mixed oxide

Codes

A B C D
(a) 1 2 3 4
(b) 4 1 2 3
(c) 3 2 4 1
(d) 4 3 1 2

Answer

(b) A. (4)

B. (1)

C. (2)

D. (3)

Formulas of the compound Type of oxide
A. Pb3O4(PbOPb2O3) Mixed oxide
B. N2O Neutral oxide
C. Mn2O7 Acidic oxide
D. Bi2O3 Basic oxide

Mn2O7 on dissolution in water produces acidic solution.

Bi2O3 on dissolution in water produces basic solution.

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

Column I Column II
A. H2SO4 1. Highest electron gain enthalpy
B. CCl3NO2 2. Chalcogen
C. Cl2 3. Tear gas
D. Sulphur 4. Storage batteries

Codes

A B C D
(a) 4 3 1 2
(b) 3 4 1 2
(c) 4 1 2 3
(d) 2 1 3 4

Answer

(a) A. (4)

B. (3)

C. (1)

D. (2)

A. H2SO4 is used in storage batteries.

B. CCl3NO2 is known as tear gas.

C. Cl2 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.

Column I Column II
A. SF4 1. Tetrahedral
B. BrF3 2. Pyramidal
C. BrO3 3. Sea-saw shaped
D. NH4+ 4. Bent T-shaped

Codes

A B C D
(a) 3 2 1 4
(b) 3 4 2 1
(c) 1 2 3 4
(d) 1 4 3 2

Answer

(b) A. (3)

B. (4)

C. (2)

D. (1)

alt text

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

Column I Column II
A. Its partial hydrolysis does not change oxidation state of central atom. 1. He
B. It is used in modern diving apparatus. 2. XeF6
C. It is used to provide inert atmosphere for filling electrical bulbs. 3. XeF4
D. Its central atom is in sp3d2 hybridisation. 4. Ar

Codes

A B C D
(a) 1 4 2 3
(b) 1 2 3 4
(c) 2 1 4 3
(d) 1 3 2 4

Answer

(c) A. (2)

B. (1)

C. (4)

D. (3)

(A) Partial hydrolysis of XeF6 does not change oxidation state of central atom.

XeF6+6+2H2OXe+6O3+6HF

(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 XeF4 is in sp3d2 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) N2 is less reactive than P4.

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

Answer

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

N2 is less reactive than P4 due to high value of bond dissociation energy which is due to presence of triple bond between two N-atoms of N2 molecule.

~~ 65. Assertion (A) HNO3 makes iron passive.

Reason (R) HNO3 forms a protective layer of ferric nitrate on the surface of iron.

Answer

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

HNO3 makes iron passive due to formation of passive form of oxide on the surface. Hence, Fe does not dissolve in conc HNO3 solution.

~~ 66. Assertion (A) HI cannot be prepared by the reaction of KI with concentrated H2SO4.

Reason (R) HI has lowest HX 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 H2SO4 because HI is converted into I2 on reaction with H2SO4.

~~ 67. Assertion (A) Both rhombic and monoclinic sulphur exist as S8 but oxygen exists as O2.

Reason (R) Oxygen forms pπpπ multiple bond due to small size and small bond length but pπpπ 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 S8 but oxygen exists as O2, because oxygen forms pπpπ multiple bond due to its small size and small bond length. But pπ pπ bonding is not possible in sulphur due to its bigger size as compared to oxygen.

pπpπbond

O=O

Structure of O2

~~ 68. Assertion (A) NaCl reacts with concentrated H2SO4 to give colourless fumes with pungent smell. But on adding MnO2 the fumes become greenish yellow.

Reason (R) MnO2 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 H2SO4 to give colourless fumes with pungent smell. Pungent smell is due to formation of HCl.

NaCl+H2SO4Na2SO4+2HCl

But on adding MnO2 the fumes become greenish yellow due to formation of chlorine gas.

~~ 69. Assertion (A) SF6 cannot be hydrolysed but SF4 can be.

Reason (R) Six F-atoms in SF6 prevent the attack of H2O on sulphur atom of SF6.

Answer

(a) Assertion and reason both are true and reason is the correct explanation of assertion. SF4 can be hydrolysed but SF6 can not because six F-atoms in SF6 prevent the attack of H2O on sulphur atoms of SF6.

~~

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 KMnO4 solution and reduces Fe3+ to Fe2+. 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 air  (amorphous solid)  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 KMnO4 solution and reduces Fe3+ to Fe2+. Hence, compound B(g) must be SO2.

Answer

Since, the by-product of roasting of sulphide ore is SO2, so A is S8A=S8; ’ B=SO2 Reactions

(i) S8+8O2Δ8SO2

(ii) Ca(OH)2+SO2CaSO3+H2O

(iii) 2MnO4 (Violet) +5SO2+2H2O5SO42+4H++2Mn2+ (Colourless) 

(iv) 2Fe3++SO2+2H2O2Fe2+SO42+4H+

~~ 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 NO2.

Answer

Pb(NO3)2 on heating produces a brown coloured gas which may be NO2. Since, on reaction with N2O4 and on heating it produces N2O3 and N2O4 respectively.

Structures

(i) N2O4

(ii) N2O3

~~ 72. On heating compound (A) gives a gas (B) which is a constituent of air. This gas when treated with 3 moles of hydrogen (H2) 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 N2 reacts with three moles of H2 in the presence of a catalyst to give NH3 (ammonia) which is a gas having basic nature. On oxidation, NH3 gives NO2 which is a part of acid rain. So, the compounds A to D are as

A=NH4NO2;B=N2;C=NH3;D=HNO3

Reactions involved can be given, as

(i) NH4NO2(A)ΔN2+2H2O(B)

(ii) N2+3H2[B]2NH3[C]

(iii) 4NH3+5O2 Oxidation 4NO+6H2O

(Iv) 2NO+O22NO2

(v) 3NO2+H2O2HNO3+NO(D)