The p Block Elements Class 12 Chemistry Important Questions

Important Questions Class 12

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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.

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 

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?

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
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)

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?

2Fe3+ + SO2 + 2H2O → 2Fe2+ + SO42– + 4H+

Question. What happens when sulphur dioxide reacts with chlorine in the presence of charcoal?
Answer. SO2(g) + Cl2(g) → SO2Cl2(l)

Question. Predict the shape and the asked angle (90° or more or less) in the following case :
SO32– and the angle O — S — O

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

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 :

Question. Draw the structure of the following :

Question. What is the basicity of H3PO3?

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

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

Question. Draw the structure of the following molecules :

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

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) → ……….

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

Question. Pb(NO3)2 on heating gives a brown gas which undergoes dimerisation on cooling?
Identify the 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. Using VSEPR theory predict the probable structure of the following :N2O3

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 NO2– and NO2+.
Answer. In NO2 ‘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.

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 PCl4 and PCl4 is not likely toexist and why?
Answer. PCl4– 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.

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

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.) + Cl2(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

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

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 →
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) + 3ClF3(l) → UF6(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 :
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 :
Answer. XeO3 :
Hybridisation = sp3
Geometry = Tetrahedral
Shape = Pyramidal

Question. Complete the following equation :
XeF4 + O2F2 →

Question. Complete the following equation :
XeF2 + H2O →
Answer. 2XeF2(s) + 2H2O(l) 2Xe(g) + 4HF(aq) + O2(g)

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

Question. Complete the following chemical reactions equations :
XeF6 + H2
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.