Important Questions for CBSE Class 12 Chemistry Chapter 7 - P-Block Elements 2024-25

1 Mark Questions

1. In group 15 elements, there is considerable increase in covalent radius from N to P but a small increase from As to Bi. Why?

(Hint: due to completely filled d- or f-orbitals in As, Sb, Bi)

Ans: because the presence of completely filled d and f orbital in heavier atoms in group 15.

2. The tendency to exhibit -3 oxidation state, decreases down in the group 15 elements. Explain

(Hint: due to increase in size and decrease in electronegativity down the groups)

Ans: There is an increase in the size of the atom and the metallic character and decrease in the electro negativity in the group 15 elements causes a decrease in the tendency to exhibit oxidation state +3.

3. Maximum covalency of Nitrogen is 4, but the heavier elements of group 15 show covalence greater than 4. Why?          

Ans: Because the other heavier elements in group 15 have vacant d orbitals while nitrogen has only 4 orbitals in the valence shell.

4. Nitrogen exists as a diatomic molecule with triple bond between the two atoms, whereas the heavier elements of the group do not exist as ${{E}_{2}}$ at room temperature. Assign a reason.

(Hint: $p\pi -p\pi $ multiple bonds are formed by N due to its small size)

Ans: Because the nitrogen forms $p\pi -p\pi$ linkage in diatomic molecules and other heavier elements forms $p\pi -d\pi $ linkage at room temperature.

5. The ionization enthalpies of group 15 elements are higher than those of corresponding members of  group 14 and 16  elements. Assign the reason.

Ans: Because the extra stable half-filled p orbitals electronic configuration and smaller size in group 15 elements.

6. The boiling point of $P{{H}_{3}}$ is less than $N{{H}_{3}}$ . why?

Ans:

Because there is strong intermolecular hydrogen bonding in $N{{H}_{3}}$causes higher boiling point while in $P{{H}_{3}}$ cannot form hydrogen bonding between themselves because of lesser electro negativity in larger $P{{H}_{3}}$ molecules.

7. ${{N}_{2}}O$ dimerises to form ${{N}_{2}}{{O}_{4}}$ . why?

(Hint: due to the presence of odd electron in N)

Ans: Because of the presence of an unpaired electron on nitrogen makes it very reactive and tries to form dimerize stable compound ${{N}_{2}}{{O}_{4}}$with an even number of electrons.

8. Draw the structure of ${{N}_{2}}{{O}_{5}}$ molecule.

Ans:

9. How does an ammonium solution react with $A{{g}^{+}}$ (aq)? Write the balanced chemical equation.

Ans: 

$A{{g}^{+}}(aq)+2N{{H}_{3}}(aq)\rightleftharpoons {{[Ag{{(N{{H}_{3}})}_{2}}]}^{+}}(aq)$

10. Why does $N{{H}_{3}}$ form intermolecular hydrogen bonds whereas $P{{H}_{3}}$ does not?

(Hint: due to strong electronegativity , small size of N atom  and presence of lone pair of electrons on N atom)

Ans: Because there is strong intermolecular hydrogen bonding in $N{{H}_{3}}$ causes higher boiling point while in $P{{H}_{3}}$ cannot form hydrogen bonding between themselves because of lesser electro negativity in larger $P{{H}_{3}}$ molecules.

11. Write a disproportionation reaction of ${{H}_{3}}P{{O}_{3}}$ .

Ans: $4{{H}_{3}}P{{O}_{3}}\to 3{{H}_{3}}P{{O}_{4}}+P{{H}_{3}}$

12. How does $N{{H}_{3}}$act as a complexing agent?

(Hint: metal hydroxides are dissolved in excess of ammonium hydroxide. Ammonia acts as lewis base)

Ans: Due to the lone pair of nitrogen in ammonia, the molecule acts as a lewis base. Hence ammonia has an ability to donate electrons and form stable complexes.

13. Why HF is the weakest acid and HI is the strongest.

(Hint: ${{K}_{a}}(HF)=7\times {{10}^{-4}}$ , ${{K}_{a}}(HI)=7\times {{10}^{11}}$)

Ans: In HI, iodine is less electronegative and has large size. Thus bonding between H and I is weaker which will easily split in comparison HF.  Hence more liberation of ${{H}^{+}}$ ions, HI acts as a stronger acid than HF.

14. Explain why halogens are strong oxidizing agents.

(Hint: ready acceptance of electron due to more negative electron gain enthalpy)

Ans: Halogens are more electronegative with less dissociation energies and high negative electron gain enthalpies. Hence they act as strong oxidizing agents due to the high tendency to gain electrons.

15. Why Bi (V) a stronger oxidant than Sb (V)?

(Hint: +3 oxidation state is more stable than +5 oxidation state in Bi)

Ans: Because of Bi gains electrons from the reducing agent much faster than the Sb and becomes stable oxidation state +5 from oxidation state +3.

16. Why $S{{F}_{4}}$ is easily hydrolyzed, whereas $S{{F}_{6}}$ is resistant to hydrolysis? 

(Hint: water molecule can attack ‘S’ atom due to steric hindrance and ‘S’ atom is also coordinately saturated in  $S{{F}_{6}}$ molecule)

Ans: Because the Sulphur atom in $S{{F}_{4}}$ is not sterically protected as it is surrounded by only four F atoms, which easily attack by water molecules. In the case of $S{{F}_{6}}$, the S atom is protected by 6F atoms.

17. Bond dissociation enthalpy of ${{F}_{2}}$ is less than that of $C{{l}_{2}}$ . why?

Ans: Because F atom size is small compared to Cl. So, due to this electron-electron repulsions between the lone pairs of F-F are very large.

18. Write the reaction of $PC{{l}_{5}}$ with heavy water.

(Hint: $PC{{l}_{5}}+{{D}_{2}}O\to POC{{l}_{3}}+2DCl$)

Ans: $PC{{l}_{5}}+{{D}_{2}}O\to POC{{l}_{3}}+2 DCl$

19. How many P-O-P bonds are there in cyclotrimetaphosphoric acid?

(Hint: 3 bonds)

Ans: Cyclotrimetaphosphoric acid - ${{(HP{{O}_{3}})}_{3}}$ 

From the above structure, the number of P-O-P bonds = 3

20. In group 16, the stability of +6 oxidation state decreases and that of +4 oxidation state increases down the group. Why?

(Hint: due to inert pair effect)

Ans:  Because of the inert pair effect, the energy required to unpair the s-electrons is high. Hence, +6 oxidation state exhibited by lighter elements.

21. Why we cannot prepare HBr by heating KBr with sulphuric acid.

(Hint: As HBr readily reduces sulphuric acid forming $B{{r}_{2}}$)

Ans: Concentrated ${{H}_{2}}S{{O}_{4}}$ act as strong oxidizing agent and oxidizes KBr to $B{{r}_{2}}$

22. Fluorine exhibits only -1 oxidation state whereas other halogens exhibit +ve oxidation states also. Explain.

Ans: Because F cannot expand its octet due to non availability of d-orbitals.

23. Arrange the following oxoacids of chlorine in increasing order of acidic strength. 

$HOCl,HOClO,HOCl{{O}_{2}},HOCl{{O}_{3}}$ 

Ans: $HOCl < HOClO < HOCl{{O}_{2}} < HOCl{{O}_{3}}$

24. The majority of known noble gas compounds are xenon. Why?

Ans: Among all the noble gases Xe has the largest in size and it has lowest ionization energy.

25. “Hypophosphorous acid is a good reducing agent”. Justify with an example.

(Hint:$4AgN{{O}_{3}}+{{H}_{3}}P{{O}_{2}}+2{{H}_{2}}O\to 4Ag+HN{{O}_{3}}+{{H}_{3}}P{{O}_{4}}$)

Ans: $4AgN{{O}_{3}}+{{H}_{3}}P{{O}_{2}}+2{{H}_{2}}O\to 4Ag+HN{{O}_{3}}+{{H}_{3}}P{{O}_{4}}$

26. Draw the structure of ${{H}_{4}}{{P}_{2}}{{O}_{7}}$ and find out its basicity?

(Hint: Tetrabasic)

Ans:

From the above structure of ${{H}_{4}}{{P}_{2}}{{O}_{7}}$, its basicity is Tetrabasic.

27. Arrange the following triatomic species in the order of increasing bond angle.$N{{O}_{2}},NO_{2}^{+},NO_{2}^{-}$ 

(Hint:

$N{{O}_{2}}$ has a nonbonding electron, $NO_{2}^{-}$ has two non-bonding electrons, $NO_{2}^{+}$ has no nonbonding electron on N atom. Bond angle of $NO_{2}^{+}$is maximum that of $NO_{2}^{-}$is minimum)

Ans: $NO_{2}^{+}>N{{O}_{2}}>NO_{2}^{-}$

28. With what neutral molecule $Cl{{O}^{-}}$ is isoelectronic?

Ans: $Cl{{O}^{-}}$ is isoelectronic with $ClF$ as both compounds contain 26 electrons.

29. Draw the structure of ${{H}_{2}}{{S}_{2}}{{O}_{8}}$ and find the number of S-S bonds if any.

Ans:

From the above structure, there is no S-S bonds

30. What is the cause of the bleaching action of chlorine water? Explain it with a chemical equation?

(Hint: formation of nascent oxygen)

Ans:

\[C{{l}_{2}}+{{H}_{2}}O\to HCl+HClO\] 

\[HClO\to HCl+[O]\] 

[O]=nascent oxygen

31. Electron gain enthalpy of fluorine is more negative than that of chlorine.

(Hint: due to the small size of the F atom, there are strong interelectronic repulsions in the relatively smaller 2p orbitals of fluorine. So the incoming electron does experience less attraction than in Cl)

Ans: Due to the small size of the F atom, there are strong interelectronic repulsions in the relatively smaller 2p orbitals of fluorine. So the incoming electron does experience less attraction than in Cl.

32. Which one of the following is not oxidized by ${{O}_{3}}$ . State the reason. $KI,FeS{{O}_{4}},{{K}_{2}}Mn{{O}_{4}},KMn{{O}_{4}}$ 

(Hint: $KMn{{O}_{4}}$ since Mn is showing maximum oxidation state +7)

Ans: 

$KMn{{O}_{4}}$is not oxidized by ${{O}_{3}}$. Potassium permanganate on separation converts into ${{K}^{+}}$ and $MnO_{4}^{-}$ which exhibits maximum valence and not only be oxidized.

2 Marks Questions

1. Why is red phosphorous denser and less chemically reactive than white phosphorus?   

Ans:

White phosphorus is ${{P}_{4}}$ tetrahedron which is subjected to more angular strain as the angle is ${{60}^{0}}$ while red phosphorous is a polymer of ${{P}_{4}}$ tetrahedral which has less angular strain. Due to high angular strain in white phosphorus leads to unstable and more chemically reactive than red phosphorus.

2. Give chemical reactions in support of the statement that all the bonds in $PC{{l}_{5}}$ molecule are not equivalent.

(Hint: $PC{{l}_{5}}+{{H}_{2}}O\to POC{{l}_{3}}+2HCl$)

Ans:

Due to$PC{{l}_{5}}$has trigonal bipyramidal structure, the two axial bonds are longer and they experience more repulsion than three equatorial bonds.

For example, $PC{{l}_{5}}$dissociates into $PC{{l}_{3}}$ and $C{{l}_{2}}$ on heating.

\[PC{{l}_{5}}\xrightarrow{\Delta }PC{{l}_{3}}+C{{l}_{2}}\]

3. Account the following:

i. $Xe{{F}_{2}}$ has linear structure and not a bent structure.

Ans: According to VSEPR theory, the shape of the molecule is determined by the total number of electron pairs in the valence shell of the central atom. The linear structure of $Xe{{F}_{2}}$has 2 pairs of bond electrons and 3 pairs of lone electrons which causes not a bend angular structure.

ii. Phosphorous show marked tendency for Catenation

Ans: Several Phosphorous atoms exhibit both cyclic and open chain compounds which shows marked tendency for catenation. This is due to greater P-P bond energy.

2. Draw the structures of $Br{{F}_{3}},XeO{{F}_{4}},Xe{{O}_{3}}$ using VSEPR theory.

Ans:

3. Write the conditions that favor the formation of ammonia gas along with reactions involved in Haber’s process.

Ans:  Haber’s process:

\[{{N}_{2}}(g)+3{{H}_{2}}(g)\underset{Mo}{\overset{Fe}{\longleftrightarrow}}2N{{H}_{3}}(g)\] 

High pressure (200 atm), moderate temperature (700K), and presence of catalyst Fe/FeO with small amount of ${{K}_{2}}O$ and $A{{l}_{2}}{{O}_{3}}$ to increase the rate of attainment of equilibrium.

4. Write the chemical equations of the following.

i. Glucose is heated with conc. ${{H}_{2}}S{{O}_{4}}$ 

Ans: $\underset{\text{D-glucose}}{\mathop{{{C}_{6}}{{H}_{12}}{{O}_{6}}}}\,+{{H}_{2}}S{{O}_{4}}\to \underset{\text{Hydroxymethylfurfural}}{\mathop{{{C}_{6}}{{H}_{6}}{{O}_{3}}}}\,+3{{H}_{2}}O$

5. Sodium nitrate is heated with conc. ${{H}_{2}}S{{O}_{4}}$

Ans: $2NaN{{O}_{3}}+{{H}_{2}}S{{O}_{4}}\to N{{a}_{2}}S{{O}_{4}}+2HN{{O}_{3}}$

Complete the following reactions

1.  

i. ${{(N{{H}_{4}})}_{2}}C{{r}_{2}}{{O}_{7}}\xrightarrow{\Delta }$ 

Ans: ${{(N{{H}_{4}})}_{2}}C{{r}_{2}}{{O}_{7}}\xrightarrow{\Delta }C{{r}_{2}}{{O}_{3}}+4{{H}_{2}}O+{{N}_{2}}$

ii. $N{{H}_{4}}Cl+NaN{{O}_{2}}\to $ 

Ans: $N{{H}_{4}}Cl(aq)+NaN{{O}_{2}}(aq)\to {{N}_{2}}+2{{H}_{2}}O+NaCl$
2.

i. $N{{H}_{4}}CON{{H}_{2}}+{{H}_{2}}O\to $ 

Ans: $N{{H}_{4}}CON{{H}_{2}}+{{H}_{2}}O\to C{{O}_{2}}+2N{{H}_{3}}$
 

ii. $FeC{{l}_{3}}+N{{H}_{4}}OH\to $ 

Ans: $FeC{{l}_{3}}(aq)+N{{H}_{4}}OH\to Fe{{(OH)}_{3}}+3N{{H}_{4}}Cl$

3.   

i. $C{{a}_{3}}{{P}_{2}}+{{H}_{2}}O\to $ 

Ans: $C{{a}_{3}}{{P}_{2}}+6{{H}_{2}}O\to 2P{{H}_{3}}+3Ca{{(OH)}_{2}}$ 

ii. ${{I}_{2}}+HN{{O}_{3}}(conc.)\to$

Ans: ${{I}_{2}}+10HN{{O}_{3}}(conc.)\to 2HI{{O}_{3}}+10N{{O}_{2}}+4{{H}_{2}}O$

4.  

i. $Ba{{({{N}_{3}})}_{2}}\xrightarrow{\Delta }$ 

Ans: $Ba{{({{N}_{3}})}_{2}}\xrightarrow{\Delta }Ba+3{{N}_{2}}$ 

ii. ${{H}_{3}}P{{O}_{3}}\xrightarrow{\Delta }$ 

Ans: $4{{H}_{3}}P{{O}_{3}}\xrightarrow{\Delta }3{{H}_{3}}P{{O}_{4}}+P{{H}_{3}}$

5.  

i. $P{{H}_{4}}I+KOH\to $ 

Ans: $P{{H}_{4}}I+KOH\to P{{H}_{3}}+KI+{{H}_{2}}O$ 

ii. $HgC{{l}_{2}}+P{{H}_{3}}\to $ 

Ans: $3HgC{{l}_{2}}+2P{{H}_{3}}\to H{{g}_{3}}{{P}_{2}}+6HCl$

6.  

i. $PC{{l}_{3}}+{{H}_{2}}O\to $ 

Ans: $PC{{l}_{3}}+3{{H}_{2}}O\to {{H}_{3}}P{{O}_{3}}+3HCl$ 

ii. $S+{{H}_{2}}S{{O}_{4}}(conc)\to $ 

Ans: $S+{{H}_{2}}S{{O}_{4}}(\text{conc}\text{.})\to {{H}_{2}}O+S{{O}_{2}}$
7.

i. $A{{l}_{2}}{{O}_{3}}+NaOH+{{H}_{2}}O\to $ 

Ans: $A{{l}_{2}}{{O}_{3}}(s)+2NaOH(aq)+3{{H}_{2}}O(l)\to 2NaAl{{(OH)}_{4}}$

ii. $HCl+{{O}_{2}}\xrightarrow{CuC{{l}_{2}}}$ 

Ans: $4HCl+{{O}_{2}}\xrightarrow{CuC{{l}_{2}}}2C{{l}_{2}}+2{{H}_{2}}O$

8.  

i. $Ca{{(OH)}_{2}}+C{{l}_{2}}\to $ 

Ans: $Ca{{(OH)}_{2}}+C{{l}_{2}}\to CaOC{{l}_{2}}+{{H}_{2}}O$ 

ii. $Xe{{F}_{4}}+{{H}_{2}}O\to $ 

Ans: $6Xe{{F}_{4}}+12{{H}_{2}}O\to 2Xe{{O}_{3}}+4Xe+3{{O}_{2}}+24HF$

9.  

i. $N{{a}_{2}}S{{O}_{3}}+C{{l}_{2}}+{{H}_{2}}O\to $ 

Ans: $N{{a}_{2}}S{{O}_{3}}+C{{l}_{2}}+{{H}_{2}}O\to 2HCl+N{{a}_{2}}S{{O}_{4}}$

ii. $NaHC{{O}_{3}}+HCl\to $ 

Ans: $NaHC{{O}_{3}}+HCl\to NaCl+C{{O}_{2}}+{{H}_{2}}O$

10.
i. $Xe{{F}_{6}}+{{H}_{2}}O\xrightarrow[hydrolysis]{complete}$ 

Ans: $Xe{{F}_{6}}+3{{H}_{2}}O\xrightarrow[hydrolysis]{complete}Xe{{O}_{3}}+6HF$

ii. $Xe{{F}_{6}}+{{H}_{2}}O\xrightarrow[hydrolysis]{partial}$ 

Ans:$Xe{{F}_{6}}+3{{H}_{2}}O\xrightarrow[hydrolysis]{partial}XeO{{F}_{4}}+2HF$

11.  

i. $Zn+HN{{O}_{3}}(conc)\to $ 

Ans: $Zn+4HN{{O}_{3}}(conc)\to Zn{{(N{{O}_{3}})}_{2}}+2N{{O}_{2}}+2{{H}_{2}}O$

ii. ${{P}_{4}}+HN{{O}_{3}}(conc)\to $ 

Ans: ${{P}_{4}}+20HN{{O}_{3}}(conc)\to 4{{H}_{3}}P{{O}_{4}}+20N{{O}_{2}}+4{{H}_{2}}O$

12. 

i. $N{{H}_{3}}+{{O}_{2}}\xrightarrow{Pt/Rh}$ 

Ans: $4N{{H}_{3}}+5{{O}_{2}}\xrightarrow{Pt/Rh}4NO+6{{H}_{2}}O$ 

ii. ${{P}_{4}}+NaOH+{{H}_{2}}O\to $ 

Ans: ${{P}_{4}}+3NaOH+3{{H}_{2}}O\to P{{H}_{3}}+3Na{{H}_{2}}P{{O}_{2}}$

13.  

i. ${{P}_{4}}+SOC{{l}_{2}}\to $ 

Ans: ${{P}_{4}}+8SOC{{l}_{2}}\to 4PC{{l}_{3}}+4S{{O}_{2}}+2{{S}_{2}}C{{l}_{2}}$ 

ii. ${{P}_{4}}+S{{O}_{2}}C{{l}_{2}}\to $

Ans: ${{P}_{4}}+10S{{O}_{2}}C{{l}_{2}}\to 4PC{{l}_{5}}+10S{{O}_{2}}$

14. 

i. $PbS+{{O}_{3}}\to $ 

Ans: $PbS+4{{O}_{3}}\to PbS{{O}_{4}}+4{{O}_{2}}$

ii. $KI+{{H}_{2}}O+{{O}_{3}}\to $ 

Ans: $2KI(aq)+{{O}_{3}}(g)+{{H}_{2}}O(l)\to 2KOH(aq)+{{I}_{2}}(s)+{{O}_{2}}(g)$

15. 

i. $MnO_{4}^{-}+S{{O}_{2}}+{{H}_{2}}O\to $

Ans: $2MnO_{4}^{-}+5S{{O}_{2}}+2{{H}_{2}}O\to 2M{{n}^{+2}}+5SO_{4}^{-2}+4{{H}^{+}}$

ii. $Zn+HN{{O}_{3}}(dil.)\to $ 

Ans: $4Zn+10HN{{O}_{3}}(dil.)\to 4Zn{{(N{{O}_{3}})}_{2}}+N{{H}_{4}}N{{O}_{3}}+3{{H}_{2}}O$ 

$N{{H}_{4}}N{{O}_{3}}\to {{N}_{2}}O+2{{H}_{2}}O$

16.  

i. $C{{l}_{2}}+N{{H}_{3}}(excess)\to $

Ans: $3C{{l}_{2}}+8N{{H}_{3}}(excess)\to 6N{{H}_{4}}Cl+{{N}_{2}}$

ii. $N{{H}_{3}}+C{{l}_{2}}(excess)\to $

Ans: $N{{H}_{3}}+3C{{l}_{2}}(excess)\to NC{{l}_{3}}+3HCl$

17.  

i. $2NaOH(\text{cold and dil})+C{{l}_{2}}\to $ 

Ans: $2NaOH(\text{cold and dil})+C{{l}_{2}}\to NaCl+NaOCl+{{H}_{2}}O$

ii. $3C{{l}_{2}}+6NaOH(\text{hot and conc})\to $ 

Ans: $3C{{l}_{2}}+6NaOH(\text{hot and conc})\to 5NaCl+NaCl{{O}_{3}}+3{{H}_{2}}O$

18.  

i. $Fe(s)+2HCl(l)\to $ 

Ans: $Fe(s)+2HCl(l)\to FeC{{l}_{2}}(s)+{{H}_{2}}(g)$

ii. $C{{l}_{2}}+{{F}_{2}}(excess)\to $ 

Ans: $C{{l}_{2}}+3{{F}_{2}}(excess)\to 2Cl{{F}_{3}}$

19.  

i. $U+3Cl{{F}_{3}}\to $

Ans: $U+3Cl{{F}_{3}}\to U{{F}_{6}}+3ClF$

ii. $FeS{{O}_{4}}+{{H}_{2}}S{{O}_{4}}+C{{l}_{2}}\to $ 

Ans: $2FeS{{O}_{4}}+{{H}_{2}}S{{O}_{4}}+C{{l}_{2}}\to F{{e}_{2}}{{(S{{O}_{4}})}_{3}}+2HCl$

20.  

i. What is the covalency of N in ${{N}_{2}}{{O}_{5}}$ ?

Ans: covalency of N in ${{N}_{2}}{{O}_{5}}$is 4. Covalency is valence characterized by the sharing of electrons in chemical compounds.

ii. Explain why phosphorus forms pentachloride whereas nitrogen and bismuth do not?

Ans: due to the absence of d-orbitals in nitrogen is unable to form pentachloride, while Bi does not form pentachloride due to inert pair effect. Because of vacant d-orbitals to extend its octet, P forms pentachloride.

21. 

i. The acidic character of hydrides of group 15 increases from ${{H}_{2}}O$ to ${{H}_{2}}Te$ . why?

Ans: Down the group, atomic size increases and bond strength decreases. ${{H}_{2}}Te$can release H-ion more easily than ${{H}_{2}}O$, making it more acidic. Hence, the size of the central atom is directly proportional to acidity of its hydride. 

ii. Dioxygen is a gas while sulphur (${{S}_{8}}$) is a solid. Why?

Ans: Due to $p\pi -p\pi $ bonding in ${{O}_{2}}$ molecule, the intermolecular forces are weak Vander Waals forces as it is small in size with high electron density which tends to repel other molecules in it. So, as a result Dioxygen is a gas.

While sulphur does not form an S=S bond hence exists as ${{S}_{8}}$with strong covalent bonds as a polyatomic molecule whose intermolecular forces are strong. So, ${{S}_{8}}$ is solid.

22.

i. Interhalogen compounds are more reactive than halogens except ${{F}_{2}}$ . why?

Ans: because, the interhalogen (A-X) bonds are weaker than the bonds between halogen atoms in dihalogen compounds (X-X).

ii. Give one important use of $Cl{{F}_{3}}$ .

Ans: $Cl{{F}_{3}}$ is used in the semiconductor industry to clean oxides off surfaces.

23.  

i. Write the composition of bleaching powder.

Ans: the composition of bleaching powder is $Ca{{(OCl)}_{2}}.CaC{{l}_{2}}.Ca{{(OH)}_{2}}.2{{H}_{2}}O$

ii. What happens when NaCl is heated with conc.${{H}_{2}}S{{O}_{4}}$ in the presence of $Mn{{O}_{2}}$ . write the chemical equation.

Ans: $4NaCl+Mn{{O}_{2}}+4{{H}_{2}}S{{O}_{4}}\to MnC{{l}_{2}}+4NaHS{{O}_{4}}+2{{H}_{2}}O+C{{l}_{2}}$

24. Arrange the following in the decreasing order of their basicity. Assign the reason: $P{{H}_{3}},N{{H}_{3}},Sb{{H}_{3}},As{{H}_{3}},Bi{{H}_{3}}$ 

Ans: $Bi{{H}_{3}} < Sb{{H}_{3}} < As{{H}_{3}} < P{{H}_{3}} < N{{H}_{3}}$ 

The electron density increases from nitrogen to bismuth down the group, so basicity decreases.

25. A colorless and a pungent smelling gas which easily liquefies to a colorless liquid and freezes to a white crystalline solid, gives dense white fumes with ammonia. Identify the gas and write the chemical equation for its laboratory preparation. [Hint: HCl]

Ans: $NaCl+{{H}_{2}}S{{O}_{4}}(conc)\to NaHS{{O}_{4}}+HCl$

26. Complete the following disproportionation reactions

i. ${{P}_{4}}+NaOH+{{H}_{2}}O\to $ 

Ans: ${{P}_{4}}+3NaOH+3{{H}_{2}}O\to P{{H}_{3}}+3Na{{H}_{2}}P{{O}_{2}}$ 

ii. $HN{{O}_{2}}\xrightarrow{{{H}^{+}}}$ 

Ans: $HN{{O}_{2}}\xrightarrow{{{H}^{+}}}{{H}_{2}}O+NO$

27. Arrange the following trichlorides in decreasing order of bond angel $NC{{l}_{3}},PC{{l}_{3}},AsC{{l}_{3}},SbC{{l}_{3}}$ 

Ans: $NC{{l}_{3}} < PC{{l}_{3}} < AsC{{l}_{3}} < SbC{{l}_{3}}$

28. Suggest reason why only known binary compounds of noble gases are fluorides and oxides of Krypton, Xenon.

(Hint: F and O are the most electronegative elements. Kr and Xe both have low ionization enthalpies)

Ans: The ionization enthalpies of Kr and Xe are lower than other noble gases. F and O are more electronegative elements.

29. Which fluorinating agent are often used instead of ${{F}_{2}}$ ? write two chemical equations showing their use as fluorinating agents.

(Hint: $Br{{F}_{5}}+3{{H}_{2}}O\to HBr{{O}_{3}}+5HF$ 

$2I{{F}_{7}}+Si{{O}_{2}}\to 2IO{{F}_{5}}+Si{{F}_{4}}$ )

Ans: $Br{{F}_{5}}\And I{{F}_{7}}$ are used as fluorinating agents instead of ${{F}_{2}}$

$Br{{F}_{5}}+3{{H}_{2}}O\to HBr{{O}_{3}}+5HF$

$2I{{F}_{7}}+Si{{O}_{2}}\to 2IO{{F}_{5}}+Si{{F}_{4}}$

30.   

i. Hydrolysis of $Xe{{F}_{6}}$ is not regarded as a redox reaction. why?

Ans: Xe do not change its oxidation number during hydrolysis of $Xe{{F}_{6}}$

ii. Write a chemical equation to represent the oxidizing nature of $Xe{{F}_{4}}$ 

(Hint: $Xe{{F}_{4}}+2{{H}_{2}}\to Xe+4HF$ )

Ans: $Xe{{F}_{4}}+2{{H}_{2}}\to Xe+4HF$

31. Write a chemical equation.

i. $Xe{{F}_{2}}$is hydrolysed

Ans: $2Xe{{F}_{2}}+2{{H}_{2}}O\xrightarrow{hydrolysis}2Xe+4HF+{{O}_{2}}$

ii. $Pt{{F}_{6}}$ and Xenon are mixed together

Ans: $Pt{{F}_{6}}+Xe\to X{{e}^{+}}{{[Pt{{F}_{6}}]}^{-}}$

3 Marks Questions

1.  

i. How is $HN{{O}_{3}}$ prepared commercially?

Ans: $HN{{O}_{3}}$ is prepared by Ostwald’s process in a large scale. According to this process nitric acid commercially prepared by catalytic oxidation of ammonia in the presence atmospheric oxygen

ii. Write chemical equations of the reactions involved.

Ans: $4N{{H}_{3}}(g)+5N{{O}_{2}}(g)\xrightarrow[9bar]{Pt,500K}4NO(g)+6{{H}_{2}}O(l)$ 

$2NO(g)+{{O}_{2}}\to 2N{{O}_{2}}(g)$

\[3N{{O}_{2}}(g)+{{H}_{2}}O(l)\to 2HN{{O}_{3}}(aq)+NO(g)\]

iii. What concentration by mass of $HN{{O}_{3}}$ is obtained?

Ans: 68% by mass of $HN{{O}_{3}}$ is obtained.

2.  

i. How does ${{O}_{3}}$ react with lead sulphide? Write a chemical equation.

Ans: $PbS+4{{O}_{3}}\to PbS{{O}_{4}}+4{{O}_{2}}$

ii. What happens when $S{{O}_{2}}$ is passed in acidified $KMn{{O}_{4}}$ solution?

Ans: $5S{{O}_{2}}+2KMn{{O}_{4}}+2{{H}_{2}}O\to {{K}_{2}}S{{O}_{4}}+2MnS{{O}_{4}}+2{{H}_{2}}S{{O}_{4}}$ 

Sulphur dioxide is a strong reducing agent which reduces the purple coloured $KMn{{O}_{4}}$ solution into a colourless solution.

iii. $S{{O}_{2}}$ Behaves with lime water similar to $C{{O}_{2}}$ .

Ans: $S{{O}_{2}}$ turns lime water into milky because of its acidic nature.

3. Assign reason for the following.

i. Sulphur in vapor state exhibits paramagnetism.

Ans: In vapour state, sulphur partially exists as ${{S}_{2}}$ molecule which has two unpaired electrons in the antibonding  ${{\pi }^{*}}$ orbitals.  Hence, Sulphur in vapor state exhibits paramagnetism.

ii. ${{F}_{2}}$ is the strongest oxidizing agent among halogens.

Ans: because ${{F}_{2}}$accepts an electron readily due to its small size, high electronegativity and low dissociation energy.

iii. In spite of having the same electronegativity, oxygen forms hydrogen bonds while chlorine does not. 

Ans: because compared to chlorine, oxygen has a smaller size and as a result higher electron density per unit volume.

4. Give a appropriate reason for the following:

i. Metal fluorides are more ionic than metal chlorides

(Hint: according to Fajan’s rule, bigger ions more are polarized than the smaller ion by a particular ion)

Ans: According to Fajan’s rule, bigger ions are more polarized than the smaller ion by a particular ion. Hence chloride ions are more polarized than fluoride ions which causes more ionic than metal chlorides

ii. Perchloric acid is stronger than sulphuric acid

(Hint: $ClO_{4}^{-}$ is more resonance stabilize than $SO_{4}^{-2}$ since dispersal of negative charge is more effective in $ClO_{4}^{-}$as compared to $SO_{4}^{-2}$)

Ans: $ClO_{4}^{-}$ is more resonance stabilize than $SO_{4}^{-2}$ since dispersal of negative charge is more effective in $ClO_{4}^{-}$as compared to $SO_{4}^{-2}$. For a given non metal central atom, the acid strength increases as the oxidation number of the central atom increases.

iii. Addition of chlorine to KI solution gives it a brown color but excess of $C{{l}_{2}}$ makes it colorless.

(Hint: $2KI+C{{l}_{2}}\to 2KCl+{{I}_{2}}$ 

\[5C{{l}_{2}}(excess)+{{I}_{2}}+6{{H}_{2}}O\to 2HI{{O}_{3}}+10HCl(colourless)\)

Ans: $2KI+C{{l}_{2}}\to 2KCl+{{I}_{2}}$ \[5C{{l}_{2}}(excess)+{{I}_{2}}+6{{H}_{2}}O\to 2HI{{O}_{3}}+10HCl(colourless)\]

5. Explain why:

i. No chemical compound of helium is known.

Ans: He is an inert gas which is extremely stable and does not react with any other element. Thus there are no compound known.

ii. Bond dissociation energy of fluorine is less than that of chlorine.

Ans: due to inter electronic repulsion between F atom, F-F bond length in fluorine molecule is higher than Cl-Cl bond length in chlorine molecule.

iii. Two S-O bonds in $S{{O}_{2}}$ are identical.

Ans: during the formation of $S{{O}_{2}}$molecule, one electron from 3p orbital goes to the 3d orbital and the S undergoes $s{{p}^{2}}$ hybridization.  Two of these orbitals form sigma bonds with two oxygen atoms which indicates two S-O bonds are identical.

6. Out of the following hydrides of group 16 elements, which will have:

i. ${{H}_{2}}O$ 

Ans: highest bond angle

ii. ${{H}_{2}}S$ 

Ans: highest electropositive hydrogen

iii. ${{H}_{2}}Te$ 

Ans: lowest boiling point
7.  

i. How is $Xe{{O}_{3}}$ prepared from $Xe{{F}_{6}}$ ? Write the chemical equation for the reaction.

Ans: $Xe{{F}_{6}}+3{{H}_{2}}O\to Xe{{O}_{3}}+6HF$

ii. Draw the structure of $Xe{{F}_{4}}$ 

Ans:

8. 

i. Thermal stability of hydrides of group 16 elements decreases down the group. Why?

Ans: because of M-H bond dissociation energy decreases down the group with increase in the size of central atom.

ii. Compare the oxidizing powers of ${{F}_{2}}$ and $C{{l}_{2}}$ on the basis of bond dissociation enthalpy, electron gain enthalpy of halogens and hydration enthalpy of halide ions.

Ans: Due to inter electronic repulsion between F atoms, F-F bond length in fluorine molecules is higher than Cl-Cl bond length in chlorine molecules.

iii. Write the chemical equation for the reaction of copper metal with conc.$HN{{O}_{3}}$ 

Ans: $Cu+4HN{{O}_{3}}(conc.)\to Cu{{(N{{O}_{3}})}_{2}}+2N{{O}_{2}}+2{{H}_{2}}O$

9. An unknown salt X reacts with hot conc. ${{H}_{2}}S{{O}_{4}}$  to produce a brown coloured gas which intensifies in addition to copper turnings. On adding dilute ferrous sulphate solution to an aqueous solution of X and then carefully adding conc. ${{H}_{2}}S{{O}_{4}}$ along the sides of the test tube, a brown complex Y is formed at the interface between the solution and ${{H}_{2}}S{{O}_{4}}$. Identify X and Y and write the chemical equation involved in the reaction.

(Hint: X is $NO_{3}^{-}$ salt)

Ans: $NO_{3}^{-}+3F{{e}^{2+}}+4{{H}^{+}}\to NO+3F{{e}^{3+}}+2{{H}_{2}}O$ 

\[{{[Fe{{({{H}_{2}}O)}_{6}}]}^{2+}}+NO\to \underset{brown\text{ }complex}{\mathop{{{[Fe{{({{H}_{2}}O)}_{5}}(NO)]}^{2+}}}}\,+{{H}_{2}}O\] 

X = $NO_{3}^{-}$and Y = ${{[Fe{{({{H}_{2}}O)}_{5}}(NO)]}^{+2}}$

10. Assign reason to the following

i. Noble gases have large positive values of electron gain enthalpy.

Ans: The electron has to enter the next higher principal quantum level leading to a very unstable electronic configuration.

ii. Helium is used by scuba drives

Ans: because to reduce the proportions of nitrogen and oxygen below those of air, allow the gas mixture to be breathed safely on deep drivers.

11. Arrange the following in the order of the property indicated for each set

i. ${{F}_{2}},C{{l}_{2,}}B{{r}_{2}},{{I}_{2}}$ (increasing bond dissociation energy)

(Hint:${{F}_{2}}$ has exceptionally low bond dissociation enthalpy. Lone pairs in ${{F}_{2}}$molecule are much closer to each other in $C{{l}_{2}}$ molecule. Larger electron-electron repulsions among the lone pairs in ${{F}_{2}}$molecule make it bond dissociation enthalpy exceptionally low)

Ans: ${{F}_{2}} < C{{l}_{2}} < B{{r}_{2}} < {{I}_{2}}$ 

ii. HF, HCl, HBr, HI (decreasing acid strength)

(Hint: depends upon H-X bond dissociation enthalpy as the size of atom increases, bond dissociation enthalpy of H-X increases)

Ans: $HF < HCl < HBr < HI$

iii. $N{{H}_{3}},P{{H}_{3}},As{{H}_{3}},Sb{{H}_{3}},Bi{{H}_{3}}$ (decreasing base strength ) 

(Hint: electron availability on the central atom ‘E’ in $E{{H}_{3}}$ decrease down in the group)

Ans: $N{{H}_{3}} > P{{H}_{3}} > As{{H}_{3}} > Sb{{H}_{3}} > Bi{{H}_{3}}$

13. A translucent while waxy solid (A) on heating in an inert atmosphere is

converted to its allotropic form (B), Allotrope (A) on reaction with very dilute aqueous NaOH liberates a highly poisonous gas (C) having a rotten fish smell, with excess of chlorine forms D which hydrolyses to form compound (E). Identify the compounds (A) to (E).

A : White phosphorus, B : Red phosphorus, C : $P{{H}_{3}}$ , D : $PC{{l}_{3}}$ , E : ${{H}_{3}}P{{O}_{4}}$ 

Ans: 

Solid A = white phosphorous, ${{P}_{4}}$ 

On heating white phosphorus converts into red phosphorus.

White phosphorus on heating with alkali NaOH forms phosphorus sodium hypophosphite and phosphine gas which is colorless gas and poisonous gas.

\[{{P}_{4}}+3NaOH+3{{H}_{2}}O\to P{{H}_{3}}+3Na{{H}_{2}}P{{O}_{2}}\] 

White phosphorus reacts with the excess of chlorine to form phosphorus pentachloride

\[{{P}_{4}}+10C{{l}_{2}}\to 4PC{{l}_{5}}\]

Phosphorus pentachloride is hydrolyzed to form $POC{{l}_{3}}$ which on further hydrolysis gives  ${{H}_{3}}P{{O}_{4}}$

\[PC{{l}_{5}}+{{H}_{2}}O\to POC{{l}_{3}}+2HCl\] 

$POC{{l}_{3}}+{{H}_{2}}O\to {{H}_{3}}P{{O}_{4}}+3HCl$

14. Write a balanced equation for the following reactions.

i. Zn is treated with dilute $HN{{O}_{3}}$ 

Ans:  $4Zn+10HN{{O}_{3}}(dil.)\to 4Zn{{(N{{O}_{3}})}_{2}}+N{{H}_{4}}N{{O}_{3}}+3{{H}_{2}}O$         $N{{H}_{4}}N{{O}_{3}}\to {{N}_{2}}O+2{{H}_{2}}O$

ii. NaCl is heated with ${{H}_{2}}S{{O}_{4}}$ in the presence of $Mn{{O}_{2}}$ 

Ans: $4NaCl+Mn{{O}_{2}}+4{{H}_{2}}S{{O}_{4}}\to MnC{{l}_{2}}+4NaHS{{O}_{4}}+2{{H}_{2}}O+C{{l}_{2}}$

iii. Iodine is treated with conc. $HN{{O}_{3}}$

Ans: when iodine is treated with sulphuric acid forming a bright blue paramagnetic solution that contains ${{I}^{2+}}$ cations.

15. ${{X}_{2}}$  is a greenish yellow gas with a pungent offensive smell used in purification of water. It partially dissolves in ${{H}_{2}}O$  to give a solution which turns blue litmus red. When ${{X}_{2}}$  is passed through NaBr Solution, $B{{r}_{2}}$  is obtained.

i. Identify ${{X}_{2}}$, name the group to which it belongs.

Ans: ${{X}_{2}}$= $C{{l}_{2}}$ and belongs to group 17

ii. What are the products obtained when ${{X}_{2}}$ reacts with${{H}_{2}}O$ ? Write a chemical equation.

Ans: $C{{l}_{2}}(g)+2{{H}_{2}}O(l)\to HCl(aq)+HOCl(aq)$ 

iii. What happens when ${{X}_{2}}$ reacts with heat and conc. NaOH? Give an equation.

Ans: $3C{{l}_{2}}+6NaOH\to 5NaCl+NaCl{{O}_{3}}+3{{H}_{2}}O$
16. Assign the appropriate reason for the following:

i. Nitrogen exists as a diatomic molecule and phosphorus as ${{P}_{4}}$ , Why?

(Hint: Due to its small size and high electronegativity N forms $p\pi -p\pi $multiple bond ($N\equiv N$ ). whereas P does not form $p\pi -p\pi $  bonds but forms P – P single bond.)

Ans:

Due to its small size and high electronegativity N forms $p\pi -p\pi $multiple bond             ($N\equiv N$ ). whereas P does not form $p\pi -p\pi $  bonds but forms P – P single bond. Thus nitrogen forms stable diatomic molecule ${{N}_{2}}$ 

ii. Why does ${{R}_{3}}P=0$  exist but ${{R}_{3}}N=0$ does not ? (R = an alkyl group).

(Hint: Due to the absence of d-orbitals, N cannot expand its covalence beyond four. In ${{R}_{3}}N=0$, N should have a covalence of 5 so the compound ${{R}_{3}}N=0$ does not exist since maximum covalence shown by N cannot exceed 4.)

Ans: 

Due to the absence of d-orbitals, N cannot expand its covalence beyond four. In ${{R}_{3}}N=0$, N should have a covalence of 5 so the compound ${{R}_{3}}N=0$ does not exist since maximum covalence shown by N cannot exceed 4.

iii. Explain why fluorine forms only one oxoacid, HOF.

(Hint: F does not form oxoacids in which the oxidation state of F would be +3, +5, +7, it forms one oxoacid, because of unavailability of d orbitals in its valence shell.)

Ans: 

F is highly electronegativity and it shows only +1 oxidation state.

5 Marks Questions

1. How is $P{{H}_{3}}$ prepared in the laboratory? How is it purified? How does the solution of $P{{H}_{3}}$in water react on irradiation with light and on absorption in $CuS{{O}_{4}}$ ? how can you prove that $P{{H}_{3}}$is basic in nature ? write the chemical equations for all the reactions involved.

Ans:

\[{{P}_{4}}+3NaOH+3{{H}_{2}}O\to P{{H}_{3}}+3Na{{H}_{2}}P{{O}_{2}}\] 

It is purified by absorbing in HI to form $P{{H}_{4}}I$ which on treating KOH gives off phosphine.

\[P{{H}_{4}}I+KOH\to KI+{{H}_{2}}O+P{{H}_{3}}\]

$P{{H}_{3}}$in water decomposes into red phosphorus and hydrogen in the presence of light.

\[4P{{H}_{3}}\to {{P}_{4}}+6{{H}_{2}}\] 

Reaction with $CuS{{O}_{4}}$:

\[3CuS{{O}_{4}}+3P{{H}_{3}}\to C{{u}_{3}}{{P}_{2}}+3{{H}_{2}}S{{O}_{4}}\]

2. Assign a possible reason for the following.

i. Stability of +5 oxidation state decreases and that of +3 oxidation state increases down the group 15 elements.

Ans: because of the inert pair effect.

ii. ${{H}_{2}}O$ is less acidic than ${{H}_{2}}S$ 

Ans: Because S-H bond is weaker than O-H bond and also O is more electronegative than S. Due to this the bond dissociation enthalpy of H-S bond is lower than that of O-H bond. Thus, ${{H}_{2}}S$more acidic than ${{H}_{2}}O$

iii. $S{{F}_{6}}$ is inert while $S{{F}_{4}}$ is highly reactive  towards hydrolysis.

Ans: $S{{F}_{4}}$is more reactive because of the available of non-bonding pair of electrons. But in $S{{F}_{6}}$, there are no vacant d-orbital and all the electrons are paired making it more stable.

iv. ${{H}_{3}}P{{O}_{2}}$ and ${{H}_{3}}P{{O}_{3}}$ act as good reducing agents while ${{H}_{3}}P{{O}_{4}}$ does not.

Ans: Incase of ${{H}_{3}}P{{O}_{4}}$, there is no hydrogen atom bonded to P atom. So oxidation state is lower compared to ${{H}_{3}}P{{O}_{2}}$. Thus ${{H}_{3}}P{{O}_{4}}$does not act as a reducing agent.

v. Noble gases have comparatively large size in their respective periods. 

Ans: Because in case of all inert gases are completely filled orbitals and hence inter electronic repulsions are maximum.

3. 

i. How is $Xe{{F}_{6}}$ prepared from $Xe{{F}_{4}}$ ? write the chemical equation for the reaction.

Ans: $Xe{{F}_{4}}+{{O}_{2}}{{F}_{2}}\to Xe{{F}_{6}}+{{O}_{2}}$

ii. Deduce the structure of $Xe{{F}_{6}}$ using VSEPR theory?

Ans: distorted octahedral (6BP+1LP)

iii. How does $Xe{{F}_{2}}$ reacts with $P{{F}_{5}}$ ?

Ans: $Xe{{F}_{2}}+P{{F}_{5}}\to {{[XeF]}^{+}}{{[P{{F}_{6}}]}^{-}}$

iv. Give one use of each helium and neon.

Ans: Helium is used in filling balloons used by scuba divers and Neon is used in discharged tubes, advertisement display purposes.

v. Write the chemical equation for the hydrolysis of $Xe{{F}_{4}}$  

Ans: $6Xe{{F}_{4}}+12{{H}_{2}}O\to 4Xe+2Xe{{O}_{3}}+24HF+3{{O}_{2}}$

4.

i. Why does nitrogen show anomalous behavior? Discuss the trend of chemical reactivity of group 15 elements with oxygen, halogens, and metals.

Ans: Nitrogen shows anomalous behaviour because lone pair to lone pair is minimized.

Reactivity towards oxygen: group 15 elements form oxides ${{E}_{2}}{{O}_{3}}\And {{E}_{2}}{{O}_{5}}$ . in these type of oxides, when the oxidation state of this group element is higher was more acidic than with lower oxidation state. Down the group, acidic character decreases.

Reactivity towards halogens: these group elements form salts $E{{X}_{3}}\And E{{X}_{5}}$ . Except N, all elements in this group form $E{{X}_{5}}$ due to lack of d-orbital in it. All trihalides are stable except nitrogen trihalide.

Reactivity towards metals: these group elements form binary compounds with metals with oxidation state of -3

ii. ${{H}_{3}}P{{O}_{3}}$is a dibasic. Why?

Ans: Due to unionized bond P-H in ${{H}_{3}}P{{O}_{3}}$causes dibasic and it has two P-OH bonds.
5. 

i. Arrange the following in the order of their increasing acid strength.$C{{l}_{2}}{{O}_{7}},S{{O}_{2}},{{P}_{4}}{{O}_{10}}$ 

Ans: $C{{l}_{2}}{{O}_{7}}>S{{O}_{2}}>{{P}_{4}}{{O}_{10}}$

ii. How is ${{N}_{2}}O$ gas prepared and draw its structure.

Ans: reduction of nitric acid by stannic chloride and HCl gives ${{N}_{2}}O$

\[4SnC{{l}_{2}}+8HCl+2HN{{O}_{3}}\to 4SnC{{l}_{4}}+5{{H}_{2}}O+{{N}_{2}}O\] 

iii. Give one chemical reaction to show ${{O}_{3}}$ is an oxidizing agent.

Ans: Ozone reduces hydrogen peroxide to water

\[{{H}_{2}}{{O}_{2}}+{{O}_{3}}\to {{H}_{2}}O+2{{O}_{2}}\]

6. Identify A, B, C, D and E in the following sequence of reactions

(Hint: A=${{P}_{4}}$)

Ans:

7. A white waxy, translucent solid, M, insoluble in water but soluble in $C{{S}_{2}}$ , glows in the dark. M dissolves in NaOH in an inert atmosphere giving a poisonous gas (N). Also M catches fire to give dense white fumes of Q :

i. Identify M, N and Q and write the chemical equations of the reactions involved.

Ans: M is white phosphorus 

\[{{P}_{4}}+3NaOH+3{{H}_{2}}O\to P{{H}_{3}}+3Na{{H}_{2}}P{{O}_{2}}(N)\] 

\[{{P}_{4}}+5{{O}_{2}}\to {{P}_{4}}{{O}_{10}}(Q)\]

ii. M exists in the form of discrete tetrahedral molecules . Draw its structure.

'

Ans:

iii. M on heating at 573 K is changed into other less reactive form, Q, which is non-poisonous, insoluble in water as well as in $C{{S}_{2}}$ and does not glow in dark, Identify Q and draw its structure.

Ans: $white\text{ }phosphorous\xrightarrow{573K}red\text{ }phosphorous$ 

Q= ${{P}_{4}}{{O}_{10}}$

8. Write the structures of A, B, C, D and E in the following sequence of reactions

\[N{{H}_{3}}+{{O}_{2}}\xrightarrow{Pt/Rh}A+{{H}_{2}}O\] 

\[A+{{O}_{2}}\to B(brown\text{ }fumes)\] 

\[B+{{H}_{2}}O\to C+A(C\text{ }is\text{ }an\text{ }oxoacid)\] 

\[C+{{I}^{-}}\to D(violet\text{ }vapours)\] 

Complete reactions of the above mentioned sequence and name the process by which ‘C’ is obtained.

(Hint: A is NO and Ostwald process for manufacturing of $HN{{O}_{3}}$ )

Ans:

\[N{{H}_{3}}+{{O}_{2}}\xrightarrow{Pt/Rh}NO+{{H}_{2}}O\] 

A = NO

\[2NO+{{O}_{2}}\to 2N{{O}_{2}}(brown\text{ }fumes)\] 

B= $N{{O}_{2}}$ 

\[N{{O}_{2}}+{{H}_{2}}O\to HN{{O}_{3}}+NO\] 

C= $HN{{O}_{3}}$ and this process is known as Ostwald’s process

\[HN{{O}_{3}}+{{I}^{-}}\to {{H}_{2}}O+NO+\frac{1}{2}{{I}_{2}}(violet\text{ }vapours)\]

9. Give reason for each of the following:

i. $N{{H}_{3}}$ is more basic than $P{{H}_{3}}$ 

Ans: Due to small atomic size, the density of lone pair electrons on Nitrogen in $N{{H}_{3}}$ is larger than that of P in $P{{H}_{3}}$.

ii. Ammonia is a good complexing agent

Ans: Due to small atomic size, the density of lone pair electrons on Nitrogen in $N{{H}_{3}}$ has a good tendency to donate electrons, and has a good ability to form complex compounds.

iii. Bleaching by $S{{O}_{2}}$ is temporary

Ans: because $S{{O}_{2}}$ involves the process of reduction.

iv. $PC{{l}_{5}}$ is ionic in solid state

Ans: Because $PC{{l}_{5}}$exists as ${{[PC{{l}_{4}}]}^{+}}{{[PC{{l}_{6}}]}^{-}}$ in which the cation has tetrahedral geometry and anion has octahedral geometry.

v. Sulphur in vapour state exhibits paramagnetism

Ans: In vapour state, sulphur partially exists as ${{S}_{2}}$ molecule which has two unpaired electrons in the antibonding  ${{\pi }^{*}}$ orbitals.  Hence, Sulphur in vapor state exhibits paramagnetism.

10. Knowing the electrons gain enthalpy value for $O\to {{O}^{-}}$ and ${{O}^{-}}\to {{O}^{-2}}$ as -141 and 720 kJ/mol respectively, how can you account for the formation of large number of oxides having ${{O}^{2-}}$ species and not ${{O}^{-}}$ ?

(Hint: lattice enthalpy of formation of oxides having ${{O}^{2-}}$more than compensates the second ${{\Delta }_{eg}}H$ of oxygen)

Ans:

Lattice energy is the deciding factor for the stability of ionic compounds. Thus more lattice energy of the compound will be more stable. The charge carried ion is directly proportional to the lattice energy.

When a metal combines with oxygen, the lattice energy of oxide involving ${{O}^{-}}$less than that of oxide involving ${{O}^{2-}}$. Hence the oxide ions having ${{O}^{-}}$less stable than oxides involving${{O}^{2-}}$. Thus, we can say that the formation of ${{O}^{2-}}$is energetically more favourable than the formation of ${{O}^{-}}$

Class 12 Chemistry p-Block Elements Important Questions and Solutions: Summary

As mentioned earlier, Class 12 Chemistry has a brilliant syllabus that prepares the students for their future professional courses. The chapters included in the syllabus help in developing a conceptual foundation among the students. The 7th chapter in this syllabus focuses on describing the specific properties of the elements falling under the p-block division. Let us take a quick look at the chapter and then proceed to the Class 12 Chemistry Chapter 7 Important Questions.

The p-block elements are those elements in the periodic table that fall in groups 13 to 18. The electronic configuration of these elements has a pattern. The basic formula for the electronic configuration is ns2np1–6. If you follow the syllabus, you will find that the elements in Group 13 and Group 14 have already been discussed. In this chapter, we will focus on the concepts related to the elements of Group 15, 16, 17, and 18. The Important Questions of Chapter 7 Chemistry Class 12 will also be based on these elements.

If we proceed further in the chapter, we will find that the concept has been defined and segregated as per the group divisions in the periodic table. Here, you will find out that the first group of Group 15 comprises of nitrogen, phosphorus, antimony, arsenic, and bismuth. Check the increase in the atomic numbers and find out the relevance. This section will describe how the oxidation number of these elements varies from +5 to -3. It means that the valence shell can either take three electrons or can give away 8 electrons to achieve the nearest noble gas electronic configuration.  Follow the p Block Elements Important Questions to find out more about these concepts. You can also refer to the revision notes prepared for this chapter to understand the concepts even better.

In the next segment, you will get the Class 12 Chemistry Chapter 7 Important Questions related to the Group 16 elements. This section describes the physical and chemical properties of Group 16 p-block elements such as oxygen, sulphur, polonium, selenium, and tellurium. Memorize the electronic configuration of the valence shell electron in this group and try to find its relevance with the properties of these elements.

The Group 17 elements comprise halogens. In this group, the p-block elements are fluorine, chlorine, bromine, iodine, and astatine. These elements have a particular valence shell electronic configuration to follow. Study the Important Questions for p Block Elements Class 12 in this segment and learn their physical and chemical properties.

Group 18 elements, as we all know, are noble gases. In this group, the p-block elements are neon, argon, krypton, xenon, and radon. Learn the physical and chemical properties of these noble gases and figure out the relevance of the valence shell electronic configuration of this group. The Important Questions of Chapter p Block Elements Class 12 related to this group must be followed accordingly as these noble gases show different types of activities based on their characteristics.

Why Should You Study p-Block Elements?

The different groups in the modern periodic table have a set of elements that show similar physical and chemical properties. It all depends on the electronic configuration shared by the elements in a group. P-Block elements are particularly different from the elements in the same group. Studying the Important Questions of Chapter 7 Chemistry Class 12 will help you understand these differences even better.

These Important Questions from p Block Elements Class 12 will help you identify the physical and chemical properties of the elements in a group. The tabulated format will assist you to identify the behaviour of the element. You can also easily distinguish between the chemical behaviour of the elements in the same group in a reaction.

Students can easily download the PDF of important questions by clicking on the ‘Download PDF’ button and access the study material anywhere and anytime. Students can also access the other study material from our website.

Students who are preparing for JEE/NEET, can also access the required study materials such as solutions to previous years' question papers, mock tests and other content from our website, at absolutely no cost.

Important Related Links for CBSE Class 12 Chemistry 

Conclusion 

In conclusion, the Important Questions for CBSE Class 12 Chemistry Chapter 7 - P-Block Elements for the academic year 2024-25 provide a targeted and focused approach to help students prepare for their examinations. These questions cover the essential topics and concepts related to p-Block Elements, enabling students to assess their knowledge and practice answering questions in a structured manner.

By practicing these important questions, students can enhance their understanding of p-Block Elements, improve their problem-solving skills, and gain confidence in their ability to tackle questions effectively. These questions serve as a valuable resource for self-assessment, revision, and exam preparation, allowing students to identify areas of strength and areas that require further study.