The p Block Elements Class 12 Chemistry Important Questions

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Important Questions of The p Block Elements Class 12

Very Short Answer Questions

Question. Complete the following reactions :
(i) C₂H₄ + O₂ →
(ii) 4Al + 3O₂ →
Answer. (i) C₂H₄ + 3O₂ → 2CO₂ + 2H₂O
(ii) 4Al + 3O₂ → 2Al₂O₃

Question. Ozone is thermodynamically unstable?
Answer. Ozone is thermodynamically unstable and decomposes into oxygen.

The above conversion is exothermic i.e., ΔH is negative. Also, entropy increases i.e., ΔS = + ve.
Thus, ΔG for the decomposition of ozone is negative.
Hence, it is thermodynamically unstable

Question. How the supersonic jet aeroplanes are responsible for the depletion of ozone layers?
Answer. Nitrogen oxide emitted from the exhausts of supersonic jet aeroplanes readily combines with ozone to form nitrogen dioxide and diatomic oxygen. Since supersonic jets fly in the stratosphere near the ozone layer, they are responsible for the depletion of ozone layer

Question. Account for the following :
The two O — O bond lengths in the ozone molecule are equal.
Answer. The two O — O bond lengths in the ozone molecule are equal as it is a resonance hybrid of two main forms :

Question. Account for the following :
O₃ acts as a powerful oxidising agent.
Answer. 

Ozone is a powerful oxidising agent because ozone has higher energy content than dioxygen hence, decomposes to give dioxygen and atomic oxygen.

The atomic oxygen thus liberated brings about the oxidation while molecular oxygen is set free.

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

Question. Account for the following :
Solid PCl₅ is ionic in nature. 
Answer. Phosphorus pentachloride is a salt containing the tetrahedral cation [PCl₄]⁺ and the octahedral anion [PCl₆]^– therefore, it is ionic in solid state.

Question. Complete the following equation :
Ag + PCl₅ →
Answer. 2Ag + PCl₅ → 2AgCl + PCl₃

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

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

Question. What happens when PCl₅ is heated?
Answer. 

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

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

Question. Account for the following :
BiH₃ is the strongest reducing agent amongst
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. Why does R₃P O exist but R₃N O does not?
(R = alkyl group)
Answer. R₃N = O molecule has ve 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 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. 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. What happens when ammonium chloride is treated with Ca(OH)₂?
Answer. 2NH₄Cl + Ca(OH)₂ → 2NH₃ + 2H₂O + CaCl₂

Question. Why does NH₃ act as a Lewis base?
Answer. NH₃ has a lone pair of electrons on the N-atom which it can donate to an electron acceptor. Hence, NH₃ acts as a Lewis base.

Question. Mention the optimum conditions for the industrial manufacture of ammonia by Haber’s process.
Answer. Optimum conditions for the production of ammonia are :
Temperature = ~ 700 K
Pressure = about 200 atm (200 × 10⁵ Pa)
Catalyst = iron oxide with small amounts of K2O and Al₂O₃ (as promoters).

Question. Complete the following reactions :
NH₃ + NaOC l →
Answer. NH₃ + NaOCl → NaNH₂ + HClO

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 reaction equation :
P₄ + NaOH + H₂O →
Answer. P₄ + 3NaOH + 3H₂O → PH₃ + 3NaH₂PO₂

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

Question. Complete the following chemical equations :
Ca₃P₂ + H₂O →
Answer. Ca₃P₂ + 6H₂O → 3Ca(OH)₂ + 2PH₃

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. What happens when :
SO₂ gas is passed through an aqueous solution Fe³⁺ salt?
Answer.

2Fe³⁺ + SO₂ + 2H₂O → 2Fe²⁺ + SO₄^2– + 4H⁺

Question. What happens when sulphur dioxide reacts with chlorine in the presence of charcoal?
Answer. SO_2(g) + Cl_2(g) → SO₂Cl_2(l)

Question. Predict the shape and the asked angle (90° or more or less) in the following case :
SO₃^2– and the angle O — S — O
Answer. 

Question. Why are the two S — O bonds in SO₂ molecule of equal strength?
Answer. 

Due to resonance, the two p-bonds are equal (143 pm) and are of equal strength

Question. Excess of SO₂ reacts with sodium hydroxide solution.
Answer. 2NaOH + SO₂ → Na₂SO₃ + H₂O

Question. Write the structure of the following molecule :
H₂SO₃
Answer. 

Question. Draw the structure of the following :
H₂SO₄
Answer. 

Question. What is the basicity of H₃PO₃?
Answer. 

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

Question. What happens when H₃PO₃ is heated? Write the reactions involved.
Answer. 

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

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

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

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

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. Complete the following chemical equation :
Cu²⁺_(aq) + NH³_(aq) → ……….
                     (excess)
Answer.

Question. Write balanced equation when ammonia is dissolved in water.
Answer.

Question. Pb(NO₃)₂ on heating gives a brown gas which undergoes dimerisation on cooling?
Identify the gas.
Answer.

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. Using VSEPR theory predict the probable structure of the following :N₂O₃
Answer.

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

Question. Explain the following :
The bond angles (O –N –O) are not of the same value in NO^2– and NO²⁺.
Answer. In NO^–₂ ‘N’ atom has sp2-hybridisation whereas in NO₊² ‘N’ atom has sp-hybridisation.

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. 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. 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 PH₃ to BiH₃.

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. 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. 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. Explain the following :
BiCl₃ is more stable than BiCl₅.
Answer. BiCl₃ is more stable than BiCl₅. On moving down the group, the stability of + 5 oxidation state decreases while + 3 oxidation state increases due to inert pair effect.

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

Question. Explain the following observation :
Phosphorus has greater tendency for catenation than nitrogen. 
Answer. Th 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 :
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. Which one of PCl^–₄ and PCl^–₄ is not likely toexist and why?
Answer. PCl^4– because PCl₃ cannot form bond with Cl– ions.

Question. Account for the following :
PCl₅ acts as an oxidising agent.
Answer. The 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. 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. 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. Draw the structure of O₃ molecule.
Answer. 

Question. Write balanced equation for the following reactions :
Chlorine reacts with dry slaked lime.
Answer. 

Question. Name two poisonous gases which can be prepared from chlorine gas.
Answer. (i) Phosgene
(ii) Mustard gas

Question. Account for the following :
Bleaching of flowers by Cl₂ is permanent while that of SO₂ is temporary? 
Answer. The bleaching action of Cl₂ is due to oxidation of coloured substances to colourless substances by nascent oxygen. Since, the bleaching action of Cl₂ is due to oxidation and that of SO₂ is due to reduction, therefore, bleaching effect of Cl₂ is permanent while that of SO₂ is temporary.

Question. Complete the following chemical equation :
NaOH + Cl₂ →
(hot and conc.)
Answer. 6NaOH + 3Cl₂ → 5NaCl + NaClO₃ + H₂O
(hot and conc.)

Question. Account for the following :
Chlorine water loses its yellow colour on standing.
Answer. Chlorine water on standing loses its yellow colour due to the formation of HCl and HClO.

Question. Complete the following reaction equation :
NaOH (cold & dilute) + Cl₂ →
Answer. 2NaOH _(dil.) + Cl_2(aq) → NaCl_(aq) + NaClO_(aq.) + H₂O_(l)

Question. Complete the following reaction equation :
l₂ + H₂O + Cl₂ →
Answer. I₂ + 6H₂O + 5Cl₂ → 2HIO₃ + 10HCl
                                          Iodic acid

Question. Account for the following
HClO₄ is stronger acid than HClO.
Answer. As the number of oxygen bonded to the central atom increases, the oxidation number of the oxidation atom increases causing a weakening of the O—H bond strength and an increase in the acidity. Hence, HClO₄ is stronger acid than HClO.

Question. Draw the structures of the following : HClO₄
Answer. 

Question. How would you account for the following :
The oxidising power of oxoacids of chlorine follows the order :
HClO₄ < HClO₃ < HClO₂ < HClO
Answer. As the stability of the oxoanion increases, its tendency to decompose to give O₂ decreases and hence its oxidising power decreases. Since the stability of the oxoanion decreases in the order :
ClO₄^– > ClO₃^– > ClO₂^– > ClO^– Therefore oxidising power of their oxoacids increases in the reverse order :
HClO₄ < HClO₃ < HClO₂ < HClO.

Question. Draw the structure of the following : HClO₃
Answer.

Question. Answer the following :
Which neutral molecule would be isoelectronic with ClO^–?
Answer. ClO^– has 17 + 8 + 1 = 26 electrons.
A neutral molecule with 26 electrons is OF₂ (8 + 2 × 9) = 18 + 8 = 26 electrons.

Question. Arrange HClO₃, HClO₂, HOCl and HClO₄ in order of increasing acid strength. Give reason for your answer. Acid strength of oxoacids of the same halogen increases with increase in oxidation number of the halogen.
Thus the increasing order of acid strength is
HOCl < HClO₂ < HClO₃ < HClO₄
 +1        +3           +5         +7

Question. Complete the following chemical equations:
Br₂   +     F₂ →
           (excess)
Answer. Br₂ + 5F₂ (excess) → 2BrF₅

Question. Give one use of ClF₃.
Answer. ClF₃ is used for the production of UF₆ in enrichment of U²³⁵.
U_(s) + 3ClF_3(l) → UF_6(g) + 3ClF_(g)

Question. Give reasons for the following :
Helium is used in diving apparatus as a diluent for oxygen.
Answer. Helium is used in diving apparatus as diluent for oxygen because of its low solubility (as compared to N₂) in blood, a mixture of oxygen and helium is used in diving apparatus used by deep sea divers.
244. XeF₂ :

Question. Draw the structures of the following :
XeF₂
Answer. Total valence electron pair

Bond pairs = 2
Lone pairs = 5 – 2 = 3
Hybridisation = sp3d
Geometry = Trigonal bipyramidal
Shape = Linear

Question. Draw the structure of the following :
XeO₃
Answer. XeO₃ :
Hybridisation = sp³
Geometry = Tetrahedral
Shape = Pyramidal

Question. Complete the following equation :
XeF₄ + O2F₂ →
Answer. 

Question. Complete the following equation :
XeF₂ + H₂O →
Answer. 2XeF_2(s) + 2H2O_(l) → 2Xe_(g) + 4HF_(aq) + O_2(g)

Question. Draw the structures of the following molecules:
XeF₆ 
Answer. XeF₆

Question. Complete the following chemical reactions equations :
XeF₆ + H₂O →
Answer. XeF₆ + H₂O → XeOF₄ + 2HF

Question. Explain the following observations:
Helium forms no real chemical compound.
Answer. Helium has completely filled ns² electronic configurations in its valence shell. Due to its small size and high IE, helium is chemically unreactive. That’s why it forms no real chemical compound.

Question. Noble gases have low boiling points. Why?
Answer. Noble gases being monoatomic gases are held together by weak London dispersion forces, therefore they have low boiling points.

Question. Complete the following reaction equation :
XeF₂ + PF₅
Answer. XeF₂ + PF₅ → [XeF]+[PF₆]^–

Question. Explain the following :
XeF₂ is linear molecule without a bend.
Answer. Since there are two Xe—F covalent bonds and three one pairs in XeF₂. According to VSEPR theory, the shape of XeF₂ is linear.

Question. Explain the following observations :
The majority of known noble gas compounds are those of Xenon.
Answer. Since, xenon (Xe) has least ionization energy among noble gases hence it readily forms chemical compounds particularly with O₂ and F₂.

Question. Complete the following reaction :
XeF₆ + KF →
Answer. XeF₆ + KF → K+[XeF₇]^–

Question. Write the chemical equation for the following process : PtF₆ and xenon are mixed together
Answer. PtF₆ + Xe → Xe+[PtF₆]^–

Short Answer Questions

Question. How is ammonia prepared on the large scale?
Name the process and mention the optimum conditions for the production of ammonia by this process.
Answer. Ammonia is manufactured industrially by Haber’s process.

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.

Question. Name the two most important allotropes of sulphur. Which one of the two is stable at room temperature? What happens when the stable form is heated about 370 K?
Answer. Sulphur exists in numerous allotropes of which yellow rhombic (a-sulphur) and monoclinic (b-sulphur) is most important. The stable form is rhombic, which transform to monoclinic sulphur, when heated above 369 K.

Question. Write the balanced chemical equations for obtaining XeO₃ and XeOF₄ from XeF₆.
Answer. XeF₆ + 3H₂O → XeO₃ + 6HF
XeF₆ + H₂O → XeOF₄ + 2HF

Question. With the help of chemical equation explain the principle of Contact process in brief for the manufacture of sulphuric acid by Contact process.
Answer. (a) Contact process : It involves three stages :
(i) Burning of sulphur or sulphide ore in air to generate SO₂.
S + O₂ → SO₂
4FeS₂ + 11O₂ → 8SO₂ + 2Fe₂O₃
(ii) Conversion of SO₂ to SO₃ by reaction with
oxygen in the presence of V₂O₅ catalyst.

(iii) The SO₃ gas from the catalytic converter is absorbed in conc. H₂SO₄ to form oleum
(H₂S₂O₇). Dilution of oleum with water gives
H₂SO₄ of desired concentration.
SO3 + H₂SO₄ → H₂S₂O₇ (Oleum)
H₂S₂O₇ + H₂O → 2H₂SO₄

Long Answer Questions

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)

(c) (i) It is used as a fertiliser.
(ii) It is used to modify the detonation rate.