The p – Block Elements Class 12 Chemistry Exam Questions

Please refer to the below The p – Block Elements important questions for Class 12 Chemistry. These questions and answers have been prepared as per the latest NCERT Book for Class 12 Chemistry. Students should go through chapter wise Class 12 Chemistry Important Questions designed as per the latest examination pattern issued by CBSE.

Very Short Answer Questions :

Question. Account for the following :
Bond angle in NH⁺₄ is higher than NH₃.
Answer : N in NH₃ in sp³–hybridized. It has three bond pairs and one lone pair around N. Due to stronger lone pair-bond pair repulsions than bond pair-bond pair replusions. the tetrahedral angle decreases from 109° – 28′ to 107.8°. As a result, NH₃ is pyramidal. However, when it reacts with a proton, it forms NH⁺₄ ion which has four bond pairs and no lone pair. Due to the absence of lone pair-bond pair repulsions and presence of four identical bond pair-bond pair interaction, NH⁺₄ assumes tetrahedral geometry with a bond angle of 109° – 28′. This explains why the bond angle in NH⁺₄ is higher than in NH₃.

The p – Block Elements Class 12 Chemistry Exam Questions

Question. Explain the following observations giving appropriate reasons :
The stability of +5 oxidation state decreases down the group in group 15 of the periodic table.
Answer : Due to inert pair effect the stability of +5 oxidation state decreases down the group in group 15. Hence tendency to form pentahalide decreases down the group 15 of the periodic table.

Question. Write a reaction to show the reducing behaviour of H₃PO₂.
Answer : 4AgNO₃ + 2H₂O + H₃PO₂ → 4Ag + 4HNO₃ + H₃PO₄

Question. Arrange the following group of substances in the order of the property indicated against each group :
NH₃, PH₃, AsH₃, SbH₃ – increasing order of boiling points.
Answer : PH₃ < AsH₃ < NH₃ < SbH₃ < BiH₃
The abnormally high boiling point of NH₃ is due to the intramolecular H-bonding. Further as we move from PH₃ to BiH₃ the molecular mass increasing. As a result, the van der walls forces of attraction increase and the boiling points increase regularly from PH₃to BiH₃.

Question. Why are pentahalides of a metal more covalent than its trihalides?
Answer : In + 5 oxidation state charge/radius ratio is higher than that in + 3 oxidation state. Hence, +5 oxidation state has more polarising power than that of + 3 oxidation state and pentahalides (in O.S. + 5) are more covalent than trihalides.

Question. Answer the following.
Of Bi (V) and Sb(V) which may be a stronger oxidising agent and why?
Answer : On moving down the group, the stability of +5 oxidation state decreases while + 3 oxidation state increases due to inert pair effect. Thus +5 oxidation state of Bi is less stable than +5 oxidation state of Sb.
Therefore, Bi(V) is a stronger oxidising agent than Sb(V).

Question. Account for the following :
NH₃ is a stronger base than PH₃.
Answer : Due to presence of a lone pair of electrons on N and P, both NH₃ and PH₃ act as Lewis bases and accept a proton to form an additional N—H and P—H bonds respectively

However, due to smaller size of N over P, N—H bond thus formed is much stronger than the P—H bond. Therefore, NH₃ has higher proton affinity than PH₃. In other words, NH₃ is more basic than PH₃.

Question. Account for the following :
NF₃ is an exothermic compound but NCl₃ is an endothermic compound.
Answer : In case of nitrogen, only NF₃ is known to be stable. N–F bond strength is greater than F–F bond strength, therefore, formation of NF₃ is spontaneous. In case of NCl₃, N—Cl bond strength is lesser than Cl—Cl bond strength. Thus, energy has to be supplied during the formation of NCl₃.

Question. Why is dioxygen a gas but sulphur a solid?
Answer : O₂ molecules are held together by weak van der Waal’s forces because of the small size and high electronegativity of oxygen.
Sulphur shows catenation and the molecule is made up of eight atoms, (S₈) with strong intermolecular attractive forces. Hence, sulphur exists as solid at room temperature.

Question. Explain the following observation :
Ammonia has a higher boiling point than phosphine.
Answer : NH₃ molecules are held together by strong inter molecular hydrogen bonds whereas PH₃ molecules are held together by weak van der Waals bonds. Thus, NH₃ has a higher boiling point than PH₃.

Question. How does ammonia react with a solution of Cu²⁺?
Answer : Cu²⁺_(aq) + 4NH_3(aq) → [Cu(NH₃)₄]²⁺_(aq)
Blue (excess) Deep blue

Question. Give reasons for the following :
Oxygen has less electron gain enthalpy with negative sign than sulphur.
Answer : The electron gain enthalpy of oxygen is less negative than sulphur. This is due to its small size. As a result of which the electron-electron repulsion in the relatively small 2p-subshell are comparatively larger and hence the incoming electrons are not accepted with same ease as in case of other (sulphur) elements of this group.

Question. State reasons for the following :
The N—O bond in NO^–₂ is shorter than theN O bond in NO^–₃.
Answer : In NO^–₂, the average N—O bond order is 1.5 due to two resonating structures whereas in NO^–₃, the average N—O bond order is 1.33 due to three resonating structures. Higher the bond order, shorter is the bond length.

Question. Explain the following observations :
All the bonds in PCl₅ molecule are not equivalent.
Answer :

In PCl₅, there are three equatorial and two axial bonds present. Since, three equatorial bonds are repelled by two bond pairs and two axial bonds are repelled by three bond pairs so, axial bonds are weaker and longer than the equatorial bonds.

Question. How is ammonia manufactured industrially?
Draw flow chart for the manufacture of ammonia. Give any two uses.
Answer : Ammonia is manufactured industrially by Haber’s process.
N_2(g) + 3H_2(g) ⇌ 2NH_3(g); Δ_rH° = – 46.1 kJ mol^–1
This is a reversible exothermic reaction. High pressure about 200 atm, low temperature about 700 K and use of catalyst such as iron oxide with small amounts of Al₂O₃ and K₂O would favour the formation of ammonia according to Le-Chatelier’s principle.
The flow chart for the production of ammonia is shown below :

Uses of ammonia : (i) Ammonia is used to produce various nitrogenous fertilizers – for example : urea, ammonium nitrate, ammonium phosphate etc.
(ii) Liquid ammonia is used as a refrigerant.

Question. Explain the following : NO₂ readily forms a dimer.
Answer : Because NO₂ contains odd number of valence electrons and on dimerisation, it is converted to stable N₂O₄ molecule with even number of electrons.

Question. Draw structures of the following species: NO^–₃Answer :

Question. Mr. Rakesh, a chemistry teacher, observed some suspicious movements in his neighbourhood people and one day he saw packets of ammonium nitrate in their hand. As a chemistry teacher he knew that ammonium nitrate is used in explosives. He immediately informed the police about this. Police immediately took the required action and caught them with 3 kg of ammonium nitrate which they were using in explosives. Comment in brief :
(a) About the value/s displayed by Mr. Rakesh.
(b) Name of gas evolved on heating ammonium nitrate. Write the chemical reaction.
(c) Write two uses of ammonium nitrate.
Answer : (a) Mr. Rakesh displayed values like awareness care, concern alertness.
(b) Nitrous oxide (N₂O)
NH₄NO₃ → N₂O + H₂O
(c) (i) It is used as a fertiliser.
(ii) It is used to modify the detonation rate.

Question. Write balanced equation when ammonia is dissolved in water.
Answer : NH³_(g) + H₂O_(l) ⇌ NH⁺_4(aq) + OH^–_(aq)

Question. PH₃ has lower boiling point than NH₃. Why?
Answer : NH₃ molecules are held together by strong inter molecular hydrogen bonds whereas PH₃ molecules are held together by weak van der Waals bonds. Thus, NH₃ has a higher boiling point than PH₃.

Question. Arrange the following in the order of property indicated against each set :
H₂O, H₂S, H₂Se, H₂Te – increasing acidic character
Answer : H₂O < H₂S < H₂Se < H₂Te
As the atomic size increases down the group, the bond length increases and hence, the bond strength decreases. Consequently, the cleavage of E —H bond (E= O, S, Se, Te, etc.) becomes easier. As a result, the tendency to release hydrogen as proton increases i.e., acidic strength increases down the group.

Question. Using VSEPR theory predict the probable structure of the following : N₂O₃Answer :

Question. What is the covalency of nitrogen in N₂O₅?
Answer : In N₂O₅ covalence of nitrogen is four.

Question. Account for the following :
BiH₃ is the strongest reducing agent amongst all the hydrides of group 15.
Answer : Among hydrides of group-15 elements, the bond length increases from N – H to Bi – H with increasing size of element. Bi – H bond is longest and weakest, it can break more easily and evolve H₂ gas which acts as the reducing agent.

Question. Give reason : Nitric oxide becomes brown when released in air
Answer : Nitric oxide forms brown fumes of nitrogen dioxide (NO₂) instantaneously in the presence of air. 2NO + O₂ → 2NO₂

Question. Assign a reason for each of the following statements :
All the bonds in PCl₅ are not equal in length.
Answer :

Question. Explain the following :
The bond angles (O—N—O) are not of the same value in NO₂^– and NO₂⁺.
Answer : In NO₂^– ‘N’ atom has sp2-hybridisation whereasin NO₂⁺, ‘N’ atom has sp-hybridisation.

Hence, bond angles are not of the same value.

Question. Why does PCl₅ fume in moisture? Give reaction.
Answer : PCl₅ hydrolyses in the presence of moisture giving fumes of HCl.
PCl₅ + H₂O → POCl₃ + 2HCl

Question. What is the basicity of H₃PO₃?
Answer :

It is dibasic due to the presence of two replaceable hydrogen atoms.

Question. Why does R₃P=O exist but R₃N=O does not?
(R = alkyl group)
Answer : R₃N=O molecule has five covalent bonds with N atom. The octet in N cannot be extended as it does not have d orbitals for the formation of pπ-dπ bond.
In the case of R₃P=O, P can extend its octet since it has empty d-orbitals in its valence shell and form pπ-dπ bond.

Question. Arrange the following in the increasing order of property mentioned :
H₃PO₃, H₃PO₄, H₃PO₂ (Reducing character)
Answer : Reducing character of oxyacids of phosphorus depends on the number of P–H bonds. More the number of P–H bonds in oxyacid, more is the reducing character. H₃PO₂ has two P–H bonds, H₃PO₃ has one P–H bond and H₃PO₄ has no P–H bond. Thus, order of reducing character is H₃PO₂ > H₃PO₃ > H₃PO₄

Question. Give reason :
Nitrogen does not form pentahalide.
Answer : Nitrogen can not expand its octet due to absence of d-orbitals.

Question. Pb(NO₃)₂ on heating gives a brown gas which undergoes dimerisation on cooling?
Identify the gas.
Answer : 2Pb(NO³)₂ →^{673 K} 4 NO₂ + 2PbO + O₂
NO₂ is the brown gas

Question. Draw the structures of the following compounds: N₂O₅
Answer : The structure of N₂O₅ is

Question. Why does NO₂ dimerise?
Answer : Because NO₂ contains odd number of valence electrons and on dimerisation, it is converted to stable N₂O₄ molecule with even number of electrons.

Question. Draw the structure of the following :
H₄P₂O₇ (Pyrophosphoric acid)
Answer :

Question. H₃PO₂ is a stronger reducing agent than H₃PO₃.
Answer : The acids which contain P—H bond, have strong reducing properties. Hypophosphorus acid (H₃PO₂) contains two P—H bonds, whereas orthophosphorus acid (H₃PO₃) has one P—H bond. Hence, H₃PO₂ is stronger reducing agent than H₃PO₃.

Question. Why is NH₃ more basic than PH₃?
Answer : Lewis basic nature of NH₃ and PH₃ molecules is due to the presence of lone pairs on N and Bi atoms, respectively. P atom is much larger than N atom and also has empty d orbitals. Electron density due to lone pair on P gets diffused because of the presence of d-orbitals and so the lone pair is not easily available for donation. Hence PH₃ is less basic than NH₃.

Question. Give reasons for the following :
(CH₃)₃P O exists but (CH₃)₃N O does not.
Answer : R₃N=O molecule has five covalent bonds with N atom. The octet in N cannot be extended as it does not have d orbitals for the formation of pπ-dπ bond.
In the case of R₃P=O, P can extend its octet since it has empty d-orbitals in its valence shell and form pπ-dπ bond.

Question. Why is the single N N bond weaker than the single P P bond?
Answer : The single N—N bond is weaker than the single P—P bond because of high interelectronic repulsion of the non-bonding electrons, occurring due to the small bond length.

Question. Account for the following :
Bi is a strong oxidizing agent in the +5 state.
Answer : On moving down the group, the stability of +5 oxidation state decreases while +3 oxidation state increases due to inert pair effect.
Thus +5 oxidation state of Bi is less stable and Bi(V) is a stronger oxidising agent.

Question. Write the structural difference between white phosphorus and red phosphorus.
Answer : White phosphorus consists of discrete tetrahedral P₄ molecule.
Red phosphorus is polymeric, consisting of chains of P₄ tetrahedra linked together.

Question. Complete the following equation :
HgCl₂ + PH₃→
Answer : 3HgCl₂ + 2PH₃ → Hg₃P₂ + 6HCl

Question. Give reasons for the following : PH₃ has lower boiling point than NH₃.
Answer : Unlike NH₃, PH₃ molecules are not associated through hydrogen bonding in liquid state. Therefore, the boiling point of PH₃ is lower than NH₃.

Question. Complete the following chemical equations :
I₂ + HNO₃ →
Answer : I₂ + 10HNO_3(conc.) → 2HIO₃ + 10NO₂ + 4H₂O

Question. Complete the following equations :
P₄ + H₂O →
Answer : P₄ + H₂O → No reaction

Question. Elements of group 16 generally show lower value of first ionization enthalpy compared to the corresponding elements of group 15 Why?
Answer : The first ionization enthalpy of group 16 elements is lower than those of group 15 elements despite their smaller atomic radii and higher nuclear charge. This is due to the relatively symmetrical and more stable configuration of the elements of group15 as compared to those of the elements of group 16.

Question. Assign reasons for the following :
SF₆ is kinetically inert.
Answer : In SF₆, S atom is sterically protected by six F atoms and does not allow any reagent to attack on the S atom. Due to these reasons, SF₆ is kinetically an inert substance.

Question. Why is red phosphorus, less reactive than white phosphorus?
Answer : White phosphorus is more reactive than red phosphorus under normal conditions because of angular strain in the P₄ molecule where the angles are only 60°.

Question. Draw the structure of the following :
Red P₄Answer :

White phosphorus is more reactive due to its discrete tetrahedral structure and angular strain.

Question. Draw the structure of each of the following :
Solid PCl₅
Answer :

Question. Why is nitrogen gas very unreactive?
Answer : The bond dissociation enthalpy of triple bond in N≡N is very high due to pπ – pπ overlap.
Hence, N₂ is less reactive at room temperature.

Question. Assign reasons for the following :
Sulphur has a greater tendency for catenation than oxygen.
Answer : The property of catenation depends upon E – E bond strength of the element. As S – S bond is much stronger (213 kJ mol^–1) than O – O bond (138 kJ mol^–1), sulphur has greater tendency for catenation than oxygen.

Question. Why does PCl₃ fume in moisture?
Answer : PCl₅ hydrolyses in the presence of moisture giving fumes of HCl.
PCl₅ + H₂O → POCl₃ + 2HCl

Question. Draw the structure of PCl_5(s) molecule.
Answer :

Question. Complete the following chemical reaction equations ;
Cu + HNO_3(dilute)→
Answer : 3 Cu + 8HNO_{3 (dilute)} → 3Cu(NO₃)₂ + 2NO + 4H₂O

Question. White phosphorus is more reactive than red phosphorus.
Answer : White phosphorus consists of discrete P₄ molecules in which each phosphorus atom is tetrahedrally bonded to other three phosphorus atoms. So, white phosphorus is highly reactive. In red phosphorus, P₄ molecules are linked in an extended chain structure. So, red phosphorus is much less reactive.

Question. Complete the following chemical equations :
P₄ + SOCl₂ →
Answer : P₄ + 10SO₂Cl₂ → 4 PCl₅ + 10SO₂

Question. Give reasons :
SO₂ is reducing while TeO₂ is an oxidising agent.
Answer : The +6 oxidation state of S is more stable than +4 therefore, SO₂ acts as a reducing agent. Further, since the stability of +6 oxidation decreases from S to Te therefore, the reducing character of the dioxides decreases while their oxidising character increases. Thus, TeO₂ acts as an oxidising agent.

Question. What is the basicity of H₃PO₄?
Answer : Basicity of oxoacids of P is equal to the number of P—OH bonds in the molecule.

It is tribasic due to the presence of three replaceable hydrogen atoms.

Question. Explain the following giving appropriate reasons :
Red phosphorus is less reactive than white phosphorus.
Answer : White phosphorus consists of discrete P₄ molecules in which each phosphorus atom is tetrahedrally bonded to other three phosphorus atoms. So, white phosphorus is highly reactive. In red phosphorus, P₄ molecules are linked in an extended chain structure. So, red phosphorus is much less reactive.

Question. Why is the bond angle in PH₃ molecule lesser than that in NH₃ molecule?
Answer : The bond angle in PH₃ is much lower [93.6°] than that in NH₃ [107.8°] due to less repulsion between bond pairs.

Question. Assign reasons for the following :
NF₃ is an exothermic compound whereas NCl₃ is not.
Answer : In case of nitrogen, only NF₃ is known to be stable. N–F bond strength is greater than F–F bond strength, therefore, formation of NF₃ is spontaneous. In case of NCl₃, N—Cl bond strength is lesser than Cl—Cl bond strength. Thus, energy has to be supplied during the formation of NCl₃.

Question. Which one of PCl₄⁺ and PCl₄^– is not likely to exist and why?
Answer : PCl₄^– because PCl₃ cannot form bond with Cl^– ions.

Question. Write chemical equations for the following processes : orthophosphorous acid is heated.
Answer : 4H₃PO₃ →^Δ 3H₃PO₄ + PH₃

Short Answer Questions :

Question. Account for the following :
PCl₅ is known but NCl₅ is not known.
Answer : Nitrogen cannot expand its valency beyond 4 due to absence of d-orbitals whereas phosphorus show pentavalency due to presence of d-orbitals.

Question. Explain the following observation :
Phosphorus is much more reactive than nitrogen.
Answer : Since nitrogen forms triple bond between the two N-atoms and the phosphorus forms single bond between two P-atoms, bond dissociation energy of nitrogen (941.4 kJ mol^–1) is larger than the bond dissociation energy of phosphorus (213 kJ mol^–1). Hence, phosphorus is much more reactive than nitrogen.

Question. Account for the following :
Nitrogen is found in gaseous state.
Answer : Nitrogen exists as a diatomic moleucle with a triple bond between two atoms. These N₂ molecules are held together by weak van der Waals force of attraction which can be easily broken by the collision of the molecules at room temperature. Therefore N₂ is a gas at room temperature.

Question. Account for the following :
Tendency to form pentahalides decreases down the group in group 15 of the periodic table.
Answer : Due to inert pair effect the stability of +5 oxidation state decreases down the group in group 15. Hence tendency to form pentahalide decreases down the group 15 of the periodic table.

Question. What happens when H₃PO₃ is heated? Write the reactions involved.
Answer : 4H₃PO₃ →^Δ 3H₃PO₄ + PH₃

Question. Draw the structure of the following molecules :
(HPO₃)₃
Answer :

Question. Draw the structure of the following molecule :
H₃PO₃Answer :

It is dibasic due to the presence of two replaceable hydrogen atoms.

Question. Account for the following :
NH₃ is clearly basic while PH₃ is only feebly basic.
Answer : Lewis basic nature of NH₃ and PH₃ molecules is due to the presence of lone pairs on N and Bi atoms, respectively. P atom is much larger than N atom and also has empty d orbitals. Electron density due to lone pair on P gets diffused because of the presence of d-orbitals and so the lone pair is not easily available for donation. Hence PH₃ is less basic than NH₃.

Question. Bismuth is a strong oxidising agent in the pentavalent state. Explain.
Answer : On moving down the group, the stability of +5 oxidation state decreases while +3 oxidation state increases due to inert pair effect.
Thus +5 oxidation state of Bi is less stable and Bi(V) is a stronger oxidising agent.

Question. Draw the structure of the following molecule :
NF₃
Answer : Total no. of electrons around the central
N atom = 5
No. of bond pairs = 3
No. of lone pairs = 1
Hybridisation = sp³
Therefore, according to VSEPR theory; NF₃ should be pyramidal.

Question. Though nitrogen exhibits +5 oxidation state, it does not form pentahalide. Why?
Answer : Nitrogen can not expand its octet due to absence of d-orbitals.

Question. Arrange the following in the increasing order of their basic character :
NH₃, PH₃, AsH₃, SbH₃, BiH₃Answer : Increasing (Lewis) base strength order is :
BiH₃ < SbH₃ < AsH₃ < PH₃ < NH₃
The reason for this order is that as we move from NH₃ to BiH₃, the size of the central atom increases. Hence lone pair is not easily available for donation. The electron density on the central atom decreases on moving from NH₃ to BiH₃ and so the basic strength also decreases.

Question. Explain the following observations :
The molecules NH₃ and NF₃ have dipole moments which are of opposite direction.
Answer :

Question. Explain the following :
+3 oxidation state becomes more and more stable from As to Bi in the group.
Answer : On moving down the group, the stability of + 5 oxidation state decreases while that of + 3 oxidation state increases due to inert pair effect.

Question. Write the reaction of thermal decomposition of sodium azide.
Answer : Thermal decomposition of sodium azide gives nitrogen gas.
2NaN₃ → 2Na + 3N₂

Question. Assign reasons for the following :
Ammonia (NH₃) has greater affinity for protons than phosphine (PH₃).
Answer : PH₃ and NH₃ both are Lewis bases, since they have a lone pair of electrons on ‘N’ and ‘P’ atom respectively.

Because size of P is larger than N atom, therefore N atom carries more negative charge density than carried by P. Hence NH₃ has more proton affinity than PH₃.

Question. Why is Bi(V) stronger oxidant than Sb(V)?
Answer : On moving down the group, the stability of +5 oxidation state decreases while + 3 oxidation state increases due to inert pair effect. Thus +5 oxidation state of Bi is less stable than +5 oxidation state of Sb.
Therefore, Bi(V) is a stronger oxidising agent than Sb(V).

Question. Explain the following observation :
Phosphorus has greater tendency for catenation than nitrogen.
Answer : The property of catenation depends upon the strength of the element – element bond. Since, P – P (213 kJ mol^–1) bond strength is much more than N – N (159 kJ mol^–1) bond strength so, phosphorus shows marked catenation properties than nitrogen.

Question. Account for the following : H₃PO₂ has reducing nature.
Answer : The acids which contain P—H bond, have strong reducing properties. Hypophosphorus acid (H₃PO₂) contains two P—H bonds, whereas orthophosphorus acid (H₃PO₃) has one P—H bond. Hence, H₃PO₂ is stronger reducing agent than H₃PO₃.

Question. Account for the following :
PCl₅ acts as an oxidising agent.
Answer : Te oxidation state of phosphorus in PCl₅ is +5. As P has five electrons in its valence shell, it cannot increase its oxidation state beyond +5 by donating electrons. It can decrease its oxidation number from +5 to +3 or some lower value. So, PCl₅ acts as an oxidising agent.

Question. Account for the following :
BiCl₃ is less covalent that PCl₃.
Answer : BiCl₃ is less covalent than PCl₃ because the size of Bi³⁺ is much larger than P³⁺ (According to Fajan’s rule)

Question. On adding NaOH to ammonium sulphate, a colourless gas with pungent odour is evolved which forms a blue coloured complex with Cu²⁺ ion. Identify the gas.
Answer :

Question. Draw the structure of H₃PO₂ molecule.
Answer : The acids which contain P—H bond, have strong reducing properties. Hypophosphorus acid (H₃PO₂) contains two P—H bonds, whereas orthophosphorus acid (H₃PO₃) has one P—H bond. Hence, H₃PO₂ is stronger reducing agent than H₃PO₃.

Question. What is the basicity of H₃PO₂ acid and why?
Answer :

It is monobasic acid due to the presence of one replaceable hydrogen.

Question. Why is BiH₃ the strongest reducing agent amongst all the hydrides of group 15 elements?
Answer : Among hydrides of group-15 elements, the bond length increases from N – H to Bi – H with increasing size of element. Bi – H bond is longest and weakest, it can break more easily and evolve H₂ gas which acts as the reducing agent.