Important Questions for Class 12 Chemistry Chapter 7 The p-Block Elements Class 12 Important Questions

Important Questions for Class 12 Chemistry Chapter 7 The p-Block Elements Class 12 Important Questions

The p-Block Elements Class 12 Important Questions Very Short Answer Type

Question 1.
Why is Bi(v) a stronger oxidant than Sb(v)? (Delhi 2009)
Answer:
The stability of +5 oxidation state decreases and that of +3 state increases due to inert pair effect down the group therefore Bi(v) accepts two electrons and gets reduced to Bi (v).
Bi⁵⁺ + 2e^– → Bi³⁺

Question 2.
Which is a stronger oxidizing agent Bi(v) or Sb(v)? (Delhi 2009)
Answer:
Bi(v) is stronger oxidizing agent due to inert pair effect.

Question 3.
Why is red phosphorus less reactive than white phosphorus? (All India 2009)
Answer:
Because white phosphorus has angular strain in its P4 molecules where the angle is only 60°.

Question 4.
Why does NO₂ dimerise? (Delhi 2010)
Answer:
NO₂ contains 7 + 2 × 8 i.e. 23 odd electrons. In the valence shell N has seven electrons and hence less stable. To acquire stability it dimerizes to form N₂O₄

Question 5.
What is the oxidation number of phosphorus in H₃PO₂ molecule? (Delhi 2010)
Answer:
H₃PO₂
3 + x- 4 = 0 or x – 1 = 0 ∴ x = + 1
Thus oxidation number of P in H₃PO₂ = +1.

Question 6.
Draw the structure of 03 molecule. (Delhi 2010)
Answer:
Structure of Ozone (Os) : Angular structure

Question 7.
Fluorine does not exhibit any positive oxidation state. Why? (All India 2010)
Answer:
Since fluorine is the most electronegative element and does not have d-orbitals in its valence shell, therefore, it cannot expand its octet and hence does not show positive oxidation state (O.S.) while other halogens have d-orbitals and therefore exhibit many oxidation states.

Question 8.
Give the IUPAC name of the following compound : (All India 2010)

Answer:
IUPAC name : 2-Bromo-3-methylpent-3-ene

Question 9.
Nitrogen is relatively inert as compared to phosphorus. Why? (All India 2010)
Answer:
Because P -P single bond is much weaker than N = N triple bond and the bond length of nitrogen is small and bond dissociation energy is very large which makes it inert and urtreactive and thus phosphorus becomes more reactive.

Question 10.
Draw the structure of XeF₂ molecule. (Delhi 2011)
Answer:
XeF₂:

Question 11.
Draw the structure of XeF₄ molecule. (Delhi 2011)
Answer:
XeF₄ : sp³d² hybridization
Shape → Square planar

Question 12.
Draw the structure of BrF3 molecule. (Delhi 2011)
Answer:
BrF₃

Question 13.
Which one of \(\mathbf{P C l}_{4}^{+}\) and \(\mathbf{P C l}_{4}^{-}\) is not likely to exist and why? (Delhi 2012)
Answer:
\(\mathbf{P C l}_{4}^{-}\), because P has 10 electrons which cannot be accommodated in sp³ hybrid orbitals.

Question 14.
Of PH₃ and H₂S which is more acidic and why? (Delhi 2012)
Answer:
H₂S, because of higher electronegativity of sulphur.

Question 15.
Which is a stronger reducing agent, SbH₃ or BiH₃, and why? (All India 2012)
Answer:
BiH₃ : Because it is stronger reducing agent as its tendency to liberate H is maximum.

Question 16.
What is the basicity of H₃PO₂ acid and why? (All India 2012)
Answer:
H₃PO₂ has one replaceable H atom so it is monobasic.

Question 17.
Though nitrogen exhibits +5 oxidation state, it does not form pentahalide. Why? (Comptt. Delhi 2012)
Answer:
Due to non-availability of d-orbitals in its valence electronic configuration nitrogen does not form pentahalide.

Question 18.
Noble gases have low boiling points. Why? (Comptt. Delhi 2012)
Answer:
Noble gases being monoatomic have no interatomic forces except weak dispersion force and therefore, they are liquified at very low temperature. Hence, they have low boiling point.

Question 19.
Write a reaction to show the reducing behaviour of H₃PO₂. (Comptt. Delhi 2012)
Answer:
H₃PO₂ reduces AgNO₃ to metallic Ag :
4AgNO₃ + 2H₂O + H₃PO₂ → 4Ag + 4HNO₃ + H₃PO₄

Question 20.
Why does PCl₃ fume in moisture? (Comptt. Delhi 2012)
Answer:
PCl₃ hydrolyses in moisture giving fumes of HCl.
PCl₃ + 3H₂O → H₃PO₃ + 3HCl

Question 21.
Why does NO₂ dimerise ? (Comptt. All India 2012)
Answer:
NO₃ contains odd number of valence electrons. It behaves as a typical odd molecule. On dimerisation, it is converted to stable N₂O₄ molecule with even number of electrons.

Question 22.
Why is ICl more reactive than I₂? (Comptt. All India 2012)
Answer:
ICl is more reactive than I₂ because I-Cl bond is weaker than I-I bond of I₂.

Question 23.
Why is BiH₃ the strongest reducing agent amongst all the hydrides of group 15 elements? (Comptt. All India 2012)
Answer:
Reducing nature depends upon the stability of M- H bond. As the stability of the bond decreases from N to Bi hydrides, BiH₃ is the strongest reducing agent.

Question 24.
What is the covalency of nitrogen in N₂O₅? (Delhi 2013)
Answer:
The covalency of nitrogen in N₂O₅ is 4 because each nitrogen atom has four shared pairs of electrons.

Question 25.
What inspired N.Bartlett for carrying out reaction between Xe and PtF₆? (Delhi 2013)
Answer:
Since PtF6 oxidises Osub>2 to O₂⁺, Bartlett thought that PtF₆ should also oxidise Xe to Xe⁺ because the ionization enthalpies of O₂ (1175 kj mol^-1) and Xe (1170 kj mol^-1) are quite close.

Question 26.
What happens when ethyl chloride is treated with aqueous KOH? (Delhi 2013)
Answer:

Question 27.
Name two poisonous gases which can be prepared from chlorine gas. (All India 2013)
Answer:
(i) Phosgene gas (COCl₂) and (ii) Chloropicrin or tear gas (CCl₃NO₂).

Question 28.
Which aerosol depletes ozone layer? (All India 2013)
Answer:
Aerosols like foams, sprays etc. contain freons which are responsible for depletion of ozone layer.

Question 29.
What is the basicity of H₃PO₅ and why? (All India 2013)
Answer:
The basicity of H₃PO₅ is 2 because it contains only two ionizable H-atoms which are present as OH groups.

Questions 30.
Why does PCl₃ fume in moisture? (Comptt. Delhi 2013)
Answer:
Phosphorus trichloride reacts readily with water giving phosphorus acid and hydrochloric acid. Reaction is very quick and exothermic

Question 31.
Draw the structure of H₃PO₂ molecule. (Comptt. Delhi 2013)
Answer:
Cl₂ + H₂O → [HCl + HOCl] → 2HCl + [O]

Question 32.
Fluorine exhibits only -1 oxidation state whereas other halogens exhibit +1, +3, +5 and +7 oxidation states also. Why is it so? (Comptt. Delhi 2013)
Answer:
It is because fluorine is the most electronegative element and it does not have d-orbitals.

Question 33.
Though nitrogen exhibits +5 oxidation state, it does not form pentahalide. Why? (Comptt. All India 2013)
Answer:
Due to absence of d-subshell in N atom.

Question 34.
Bond enthalpy of fluorine is lower than that of chlorine. Why? (Comptt. All India 2013)
Answer:
Because F₂ is very small and its interelectronic repulsions between the lone pairs of electrons are very large.

Question 35.
Write the structural formula of PCl₅(s). (Comptt. All India 2013)
Answer:
PCl₅(s)

Question 36.
HF is a weaker acid than HCI. Why? (Comptt. All India 2013)
Answer:
Because of higher bond dissociation energy and strong H-bonding in HF.

Question 37.
Draw the structure of XeF2 molecule. (Comptt. All India 2013)
Answer:
XeF₂ :

Question 38.
What is the basicity of H₃PO₃? (All India 2014)
Answer:
Basicity of H₃PO₃ = 2
Because basicity is the number of replaceable H⁺ ions in an acid and H₃PO₃ is a Dibasic acid.

Question 39.
Why does NOz dimerise? (All India 2014)
Answer:
NO₂ contains 7 + 2 × 8 i.e. 23 odd electrons. In the valence shell N has seven electrons and hence less stable. To acquire stability it dimerizes to form
N₂O₄.

Question 40.
Why does NH₃ act as a Lewis base? (All India 2014)
Answer:
Due to presence of lone pair on nitrogen NH₃ acts as .a Lewis base.

Question 41.
Why is F₂ a stronger oxidising agent than Cl₂? (Comptt. All India 2014)
Answer:
Due to low bond dissociation enthalpy and high electronegativity of Fluorine, it has strong tendency to accept electrons and thus get reduced.
F + e^– → F^–
Therefore F₂ acts as strong oxidising agent, while Cl₂ is weak oxidising agent due to low electronegativity.

Question 42.
What is the basicity of H₃PO₄? (Delhi 2015)
Answer:
Since there are 3 OH groups present in II H₃PO₄, its basicity is 3.

Question 43.
Write the formulae of any two oxoacids of sulphur. (All India 2015)
Answer:
H₂SO₃ and H₂SO₄

Question 44.
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. (Delhi 2016)
Answer:
The gas with a pungent odour is Ammonia (NH₃) and the blue coloured complex is Tetra-ammine copper (II) sulphate monohydrate.

Question 45.
Pb(NO₃)₂ on heating gives a brown gas which undergoes dimerization on cooling. Identify the gas. (All India 2016)
Answer:
The brown gas is nitrogen dioxide (NO₂) which can dimerize to N₂O₄
2Pb(NO₃)₂ \(\stackrel{\Delta}{\longrightarrow}\) 2PbO + 4NO₂ + O₂

Question 46.
Write the formula of the compound of phosphorus which is obtained when cone. HNO₃ oxidises P₄. (All India 2017)
Answer:

Question 47.
Write the formula of the compound of sulphur which is obtained when cone. HNO₃ oxidises S₈ (All India 2017)
Answer:
S₈ + 48HNO₃ → 8H₃SO₄ + 48NO₃ + 16H₃O

Question 48.
Write the formula of the compound of iodine which is obtained when cone. HNO₃ oxidises I₂ . (All India 2017)
Answer:
I₂ – 10HNO₃ → 2HIO₃ + 10NO₂ + 4H₃ O

The p-Block Elements Class 12 Important Questions Short Answer Type -I [SA – I]

Question 49.
Draw the structures of the following molecules :
(i) XeF₄(it) BrF₃ (Delhi 2009)
Answer:
(i) XeF₄

(ii) BrF₃

Question 50.
Complete the following chemical reaction equations : (Delhi 2009)
(i) P₄ (s) + NaOH (aq) + H₂O (1) →
(ii) I^– (aq) + H₂O (1) + O₃ (g) →
Answer:

Question 51.
Complete the following chemical reaction equations : (All India 2009)
(i) XeF₂ (s) + H₂O (l) →
(ii) PH₃ + HgCl₂ →
Answer:

Question 52.
Draw the structures of white phosphorus and red phosphorus, phosphorus is more reactive and why? Which one of these two types of (Delhi 2010)
Answer:

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

Question 53.
Draw the structural formulae of molecules of following compounds :
(i) BrF₃ and (ii) XeF₄ (Delhi 2010)
Answer:
(i) BrF₃

(ii) XeF₄

Question 54.
Complete the following chemical reaction equations : (All India 2010)
(i) I₂ + HNO₃(Conc.) → (ii) HgCl₂ + PH₃ →
Answer:

Question 55.
Draw the structural formulae of the following compounds :
(i) H₄P₂O₅
(ii) XeF₄ (All India 2010)
Answer:

Question 56.
Complete the following chemical reaction equations : (All India 2010)

Answer:

Question 57.
State reasons for each of the following :
(i) The N-O bond in \(\mathrm{NO}_{2}^{-}\) is shorter than the N-O bond in \(\mathrm{NO}_{3}^{-}\).
(ii) SF₆ is kinetically an inert substance. (Delhi 2011)
Answer:
(i) The resonating structure of \(\mathrm{NO}_{2}^{-}\) and \(\mathrm{NO}_{3}^{-}\) show that in \(\mathrm{NO}_{2}^{-}\) two bonds are sharing a double bond while in \(\mathrm{NO}_{3}^{-}\), 3 bonds are sharing a double bond. That’s why \(\mathrm{NO}_{2}^{-}\) has shorter bond than that of \(\mathrm{NO}_{3}^{-}\).
Answer:

(ii) Because SF₆ is showing steric hindrance due to 6 (six) fluorine atoms which make it unable to react further with any other atom.

Question 58.
State reasons for each of the following :
(i) All the P-Cl bonds in PCl₅ molecule are not equivalent.
(ii) Sulphur has greater tendency for catenation than oxygen. (Delhi 2011)
Answer:
(i) The PCl₅ molecule has sp³d hybridization and trigonal bipyramidal geometry. Therefore it has 3 equatorial P – Cl bonds and two axial P-Cl bonds. Since two axial P-Cl bonds are repelled by 3 bond pairs while 3 equatorial bonds are repelled by two bond pairs, so axial bonds are longer than equatorial bonds.

(ii) The greater catenation tendency of sulphur is due to two reasons :
(a) The lone pair of electrons feels more repulsion in 0-0 bond than S-S bond due to its small size and thus S-S forms strong bond.
(b) As the size of atom increases down the group from O – PO, the strength of bond increases and therefore catenation tendency also increases.

Question 59.
Explain the following giving an appropriate reason in each case. (Delhi 2012)
(i) O₂ and F₂ both stabilize higher oxidation states of metals but O₂ exceeds F₂ in doing so.
(ii) Structures of Xenon fluorides cannot be explained by Valence Bond approach.
Answer:
(i) This is due to the ability of oxygen to form multiple bonds to metals.
(ii) This is because the energy required for the promotion of electrons in Xenon is very high. Energy factor does not favour VB approach.

Question 60.
Explain the following facts giving appropriate reason in each case :
(i) NF₃ is an exothermic compound whereas NCl₃ is not.
(ii) All the bonds in SF₄ are not equivalent. (All India 2012)
Answer:
(i) The bond energy of F – F bond is lower than that of N-F bond so NF₃ is an exothermic . compound. On the other hand bond energy
of Cl – Cl bond is higher than that of N – Cl bond so NCl₃ is an endothermic as well as unstable compound.

(ii) SF₄ has trigonal bipyramidal structure in which one position of equitorial is occupied by a lone pair of electrons. Sulphur undergoes sp³d hybridisation and has sea-saw geometry.

Question 61.
Explain the following :
(a) Xenon does not form such fluorides as XeF₃ and XeF₅.
(b) Out of noble gases, only Xenon is known to form real chemical compounds. (Comptt. Delhi 2012)
Answer:
(a) By impairing of one paired orbital, two singly occupied orbitals come into existence.Thus, either two or four or six singly occupied orbitals can be formed instead of one, three or five singly occupied orbitals. Hence XeF, XeF₃ or XeF₅ are not formed.
(b) Xe atom has a large size and lower ionisation potential and hence the force of nucleus over the electrons is weak and hence very small energy can excite the electrons and hence it is easier for Xenon to form compounds than other noble gases.

Question 62.
(a) Which form of sulphur shows paramagnetic behaviour and why ?
(b) Fluorine exhibits only -1 oxidation state whereas other halogens exhibit +1, +3, +5 or +7 oxidation states also. Explain as to why. (Comptt All India 2012)
Answer:
(a) In vapour state sulphur partly exists as S₂ molecule which has two unpaired electrons in the antibonding π* orbitals like O₂ and hence exhibits paramagnetism.
(b) It is because fluorine is the most electronegative element and it does not have d-orbitals.

Question 63.
How is XeO₃ obtained? Write the related chemical equations. Draw the structure of XeO₃ (Comptt. All India 2012)
Answer:
Hydrolysis of XeF₄ and XeF₆ with water gives XeO₃
6XeF₄ + 12H₂O → 4Xe + 2XeO₃ + 24HF + 3O₂
XeF₆ + 3H₂O → XeO₃ + 6HF

Question 64.
What happens when
(i) PCl₅ is heated? (ii) H₃PO₃ is heated? Write the reactions involved. (Delhi 2013)
Answer:
(i) On heating PCl₅ decomposes into PCl₃ + Cl₂

(ii) Orthophosphorous acid on heating disproportionates to give orthophosphoric acid and phosphine.

Question 65.
Draw the structures of the following molecules: (All India 2013)
(i) XeOF₄ (ii) H₃PO₃
Answer:
(i) XeOF₄

Question 66.
How are interhalogen compounds formed? What general compositions can be assigned to them? (All India 2013)
Answer:
Interhalogen compounds: Halogens react with each other to form a number of compounds called interhalogen compounds, whose general formula is XX’n.
Where X = less electronegative atom (have larger size)
X’ = more electronegative atom (have smaller size)
n = no. of more electronegative atoms/high
They are of four types :
XX’ = ClF, BrF, IF, BrCl, ICl, IBr
XX’₃ = ClF₃ BrF₃, IF₃, ICl₃
XX’₅ = ClF₅, BrF₅, IF₅
XX’₇ = IF₇
Naming: The halogen with positive oxidation state named first and with negative oxidation state named after first with suffix ‘ide’.
Example : BrCl₃ → Bromine trichloride
IF₇ → Iodine heptafluoride
Preparation of Interhalogen Compounds :
By direct combination ;
Example:

Question 67.
Draw the structures of the following molecules :
(i) XeF₆ (ii) H₂S₂O₇ (All India 2013)
Answer:
(i) XeF₆ :

(ii) H₂S₂O₇ :

Question 68.
Draw the structures of the following molecules :
(i) N₂O₅ (ii) XeF₂ (All India 2013)
Answer:
(i) N₂O₅:

(ii) XeF₂ :

Shape : Linear
Angle : F-Xe – F > 90°

Question 69.
Explain the following :
(a) NO₂ readily forms a dimer.
(b) BiClj is more stable than BiCl₅. (Comptt. Delhi 2013)
Answer:
(a) NO₂ contains 7 + 2 × 8 i.e. 23 odd electrons. In the valence shell N has seven electrons and hence less stable. To acquire stability it dimerizes to form N₂O₄

(b) BiCl₃ is more stable than BiCl₅ due to inert pair effect because as we move down the group, the stability of +3 oxidation state increases and of +5 decreases.

Question 70.
Complete the following chemical equations :
(i) Ca₃P₂ + H₂O →
(ii) Cu + H₂SO₄ (cone.) → (Delhi 2014)
Answer:
(i) Ca₃P₂ + 6H₂O → 2PH₃ + 3Ca(OH)₂
(ii) Cu + 2H₂SO₄ (cone.) → CuS0₄ + 2H₂O + SO₂

Question 71.
Arrange the following in the order of property indicated against each set :
(i) HF, HCl, HBr, HI – increasing bond dissociation enthalpy.
(ii) H₂O, H₂S, H₂Se, H₂Te – increasing acidic character. (Delhi 2014)
Answer:
(i) HI < HBr < HCl < HF
(ii) H₂O < H₂S < H₂Se < H₂Te

Question 72.
Complete the following equations :
(i) P₄ + H₂O →
(ii) XeF₄ + O₂F₂ → (All India 2014)
Answer:
(i) P₄ + 6H₂O → 2PH₃+ 2H₃PO₃
(ii) XeF₄ + O₂F₂ \(\underrightarrow { 143K } \) XeF₆ + O₂

Question 73.
Draw the structures of the following :
(i) XeF₂ (ii) BrF₃ (All India 2014)
Answer:
(i) XeF₂ : Refer to Q. 84 (ii), Page 125
(ii) BrF₃ (SP³d hybridization)

Question 74.
Complete the following equations :
(i) Ag + PCl₅ →
(ii) CaF₂ + H₂SO₄ → (All India 2014)
Answer:
(i) Ag + PCl₅ → 2AgCl + PCl₃
(ii) CaF₂ + H₂SO₄ → CaSO₄+ 2HF

Question 75.
Draw the structures of the following :
(i) XeF₄
(ii) HClO₄ (All India 2014)
Answer:
(i) XeF₄:

(ii) HClO₄:

Question 76.
Complete the following equations :
(i) C + conc. H₂SO₄ →
(ii) XeF₂ + H₂O → (All India 2014)
Answer:

Question 77.
Draw the structures of the following :
(i) XeO₃ (ii) H₂SO₄ (All India 2014)
Answer:
(i) XeO₃ :

(ii) H₂SO₄:

Question 78.
Draw the structure of each of the following:
(i) H₂SO₄
(ii) Solid PCl₅ (Comptt. Delhi 2014)
Answer:
(i) H₂SO₄:

(ii) PCl₅ (s):

Question 79.
Draw the structures of the following compounds :
(i) H₂SO₃
(ii) N₂O₅ (Comptt. Delhi 2014)
Answer:

Question 80.
Complete the following chemical equations :
(i) PCl₅ \(\underrightarrow { Heat } \)
(ii) NaHCO₃ + HCl → (Comptt. Delhi 2014)
Answer:

Question 81.
Complete the following chemical equations : (Comptt. All India 2014)
(i) SO₂ + MnO₄^– + H₂O →
(ii) F₂ (g) + H₂O (l) →
Answer:

Question 82.
Complete the following chemical reaction equations : (Comptt. All India 2014)

Answer:

Question 83.
Complete the following chemical equations : (Comptt. All India 2014)

Answer:

Question 84.
Draw the structures of the following :
(i) H₂SO₄
(ii) XeF₂ (Comptt. Delhi 2015)
Answer:

Question 85.
Explain the following :
(i) Nitrogen is much less reactive than phosphorus.
(ii) NF₃ is an exothermic compound but NCl₃ is an endothermic compound. (Comptt All India 2015)
Answer:
(i) Due to presence of weak single bond in P – P than N = N, phosphorous is more reactive than nitrogen and also because of high bond dissociation enthalpy of N = N.

(ii) Due to smaller size of F as compared to Cl, the N – F bond is much stronger than N – Cl bond while bond dissociation energy of F2 is much lower than that of Cl₂. Therefore, energy released during the formation of NF₃ molecule is more than the energy needed to break N₂ and F₂ molecules into individual atoms. In other words, formation of NF₃ is an exothermic reaction.
The energy released during the formation of NCl₃ molecule is less than the energy needed to break N₂ and Cl₂ molecules into individual atoms. Thus formation of NCl₃ is an endothermic reaction.

Question 86.
What happens when:
(i) SO₂ gas is passed through an aqueous solution of Fe³⁺ salt?
(ii) XeF₄ reacts with SbF₅? (All India 2016)
Answer:
(i) In this sulphur dioxide acts as a reducing agent and reduces Fe³⁺ to Fe²⁺.
2Fe³⁺ + SO₂ + 2H₂O → 2Fe²⁺ + SO₄^2- + 4H⁺
(ii) XeF₄ + SbF₅ → [XeF₃]⁺ [SbF₆]^–

Question 87.
Draw the structures of the following compounds
(i) BrF₃ (ii) XeF₄ (All India, Comptt. Delhi 2016)
Answer:

Question 88.
What happens when :
(i) Concentrated H₂SO₄ is added to calcium fluoride?
(ii) SO₃ is passed through water? (Comptt. All India 2016)
Answer:
(i) Cone. H₂SO₄ reacts with CaF₂ giving Hydrogen Fluoride
CaF₂ + H₂SO₄ → CaSO₄ + 2HF
(ii) SO₃ passed in water giving Sulphuric Acid
SO₃ + H₂O → H₂SO₄

Question 89.
Complete the following reactions:
(i) NH₃ + 3CI₃ (excess) →
(ii) XeF₆ + 2H₂O → (Delhi 2017)
Answer:

Question 90.
What happens when H₃PO₃ is heated? (Delhi 2017)
Answer:

Question 91.
Draw the structures of the following:
(i) H₂S₂O₇ (ii) XeF₆
Answer:
(i) H₂S₂O₇ (Pyrosulphuric acid or Oleum)

(ii)

Question 92.
Draw the structures of the following:
(i) H₃PO₂ (ii) XeF₄ (Delhi 2017)
Answer:

Question 93.
Complete the following reactions: (Delhi 2017)
(i) Cl₂ + H₂O →
(ii) XeF₆ + 3H₂O →
Answer:
(i) Cl₂ + H₂O → [HCl + HOCl] → 2HCl + [O]
(ii) XeF₆ + 3H₂O → XeO₃ + 6HF

Question 94.
What happens when
(i) cone. H₂SO₄ is added to Cu?
(ii) SO₃ is passed through water? Write the equations. (Delhi 2017)
Answer:
(i) Cu + 2H₂SO₄ (cone) → CuSO₄ + 2H₂O+SO₂
(ii) SO₃ + H₂O → H₂SO₄

Question 95.
Draw the structures of the following:
(i) H₄P₂O₇
(ii) XeOF₄
Answer:
(i) H₄P₂O₇

(ii)

Question 96.
Complete the following chemical equations:
(i) F₂ + 2Cl^– →
(ii) 2XeF₂ + 2H₄O → (Delhi 2017)
Answer:
(i) F₂+ 2Cl^– → Cl₂ + 2F^–
(ii) 2XeF₂ + 2H₂O → 2Xe + 4HF + O₂

Question 97.
What happens when
(i) HCl is added to MnO₂?
(it) PCl₅is heated? Write the equations involved. (Delhi 2017)
Answer:
(i) When HCl is added to MnO₂, chlorine gas is formed, along with other products.
MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂
(ii) PCl₅ \(\stackrel{\Delta}{\longrightarrow}\) PCl₃ + Cl₂

Question 98.
Draw the structures of the following:
(i) H₂SO₃ (ii) HClO₃ (All India 2017)
Answer:
(i) Structure of H₂SO₃ (Sulphurous acid):

Question 99.
Draw the structures of the following:
(a) H₂S₂O₈ (b) ClF₃ (All India 2017)
Answer:
(a) Structure of H₂S₂O₈:

Question 100.
Draw the structures of the following:
(a) XeF₄ (b) BrF₅ (All India 2017)
Answer:
(i) Structure of BrF₅

Question 101.
“Orthophosphoric acid (H₃PO₄) is non-reducing whereas hypophosphorus acid (H₃PO₂) is a strong reducing agent.” Explain and justify the above statement with suitable example. (Comptt. Delhi 2017)
Answer:
Orthophosphoric add (H₃PO₄) is not a reducing agent because it doesn’t contain any P-H bond whereas hypophosphorus acid (H₃PO₂) is a strong reducing agent as it contains two P-H bonds. H₃PO₂ can reduce silver nitrate (AgNO₃) into metallic silver which H₃PO₄ can not.
4AgNO₃ + H₃PO₂ + 2H₂O → 4Ag ↓ + H₃PO₄ + 4HNO₃

Question 102.
(a) What is the covalence of nitrogen in N₂O₅?
(b) BiH₃ is a stronger reducing agent than SbH₃, why? (Comptt. Delhi 2017)
Answer:
(a) The covalency of nitrogen in N₂O₅ is 4 because each nitrogen atom has four shared pairs of electrons.
(b) BiH₃: Because it is a stronger reducing agent as its tendency to liberate H is maximum.

Question 103.
Account for the following:
(i) The two oxygen-oxygen bond lengths in ozone molecule are identical.
(ii) Most of the reactions of fluorine are exothermic. (Comptt. Delhi 2017)
Answer:
(i) Due to resonance the two oxygen atoms have partial double bond character and thus have same bond length i.e. 128 pm

(ii) Due to much higher electrode potential, high electro-negativity and low bond dissociation enthalpy of F₂.

Question 104.
Account for the following :
(i) Two S-O bond lengths in SO₂ are equal.
(ii) Fluorine shows only -1 oxidation state in its compounds. (Comptt. Delhi 2017)
Answer:
(i) Due to resonance in SO₂ the double bond (π) electrons are distributed equally in both resonating structures as a result of which the bond length of two S-O becomes equal.

Because it is most electronegative element and does not have d-orbitals for octet expansion, therefore it shows only a negative oxidation state of -1.

Question 105.
Account for the following:
(i) Bond angle is \(\mathrm{NH}_{4}^{+}\) is higher than that in NH₃.
(ii) ICl is more reactive than I₂. (Comptt. Delhi 2017)
Answer:
(i) Because in \(\mathrm{NH}_{4}^{+}\) ion there is no lone pair of electrons which is present in NH₃ due to which lone pair-bond pair repulsion occurs and bond angle decreases from 109°28′ to 107.3°.
(ii) Because I-Cl bond is weaker than I-1 bond as a result of which ICl breaks easily to form halogen atoms which readily bring about the reaction, hence more reactive.

Question 106.
“Orthophosphoric acid (H₃PO₄) is not a reducing agent whereas hypophosphorus acid (H₃PO₂) is a strong reducing agent.” Explain and justify the above statement with the help of a suitable example. (Comptt. All India 2017)
Answer:
Orthophosphoric add (H₃PO₄) is not a reducing agent because it doesn’t contain any P-H bond whereas hypophosphorus acid (H₃PO₂) is a strong reducing agent as it contains two P-H bonds. H₃PO₂ can reduce silver nitrate (AgNO₃) into metallic silver which H₃PO₄ can not.
4AgNO₃ + H₃PO₂ + 2H₂O → 4Ag ↓ + H₃PO₄ + 4HNO₃

The p-Block Elements Class 12 Important Questions Short Answer Type -II [SA-II]

Question 107.
Account for the following :
(i) NH₃ is a stronger base than PH₃.
(ii) Sulphur has a greater tendency for catenation than oxygen.
(iii) Bond dissociation energy of F₂ is less than that of Cl₃. (Delhi 2009)
Answer:
(i) Since both P and N contain lone pairs of electrons but due to small size and high electronegativity of Nitrogen in NH₃, the electron density is much higher than PH₃, therefore it can easily donate electrons and acts as strong Lewis base than PH₃.

(ii) The greater catenation tendency of sulphur is due to two reasons :
(a) The lone pair of electrons feels more repulsion in 0-0 bond than S-S bond due to its small size and thus S-S forms strong bond.
(b) As the size of atom increases down the group from O – PO, the strength of bond increases and therefore catenation tendency also increases.

(iii) Due to smaller size of F than Cl as a result of which electron-electron repulsions between the lone pairs of electrons are very large than that of Cl, hence bond dissociation energy of F₂ is less than that of Cl₂.

Question 108.
Explain the following situations :
(i) In the structure of HNO₃ molecule, the N-O bond (121 pm) is shorter than N – OH bond (140 pm).
(ii) SF₄ is easily hydrolysed whereas SF₆ is not easily hydrolysed.
(iii) XeF₂ has a straight linear structure and not a bent angular structure. (Delhi 2009)
Answer:
(i)

In the structure the bond length of N-O is shorter due to formation of coordinate bond and double bond while in N-OH the bond is single covalent due to which its bond length is greater than other N-O bond.
(ii) In SF₄, due to less steric hindrance by four F atoms, H₂O molecules can attack easily while in SF₆ tire S atom is completely protected by six F atoms and does not allow H₂O molecules to attack the S atom.
(iii) In XeF₂ there are 2 bond pairs and 3 lone pairs and thus show sp³ d hybridization. It has linear geometry.

Question 109.
Explain the following observations :
(i) Fluorine does not exhibit any positive oxidation state.
(ii) The majority of known noble gas compounds are those of Xenon.
(iii) Phosphorus is much more reactive than nitrogen. (Delhi 2009)
Answer:
(i) Because it is most electronegative element and does not have d-orbitals for octet expansion, therefore it shows only a negative oxidation state of -1.
(ii) Because xenon has least ionization energy among noble gases and hence it readily forms chemical compounds particularly with oxygen and fluorine.
(iii) Because P-P single bond is much weaker than N = N triple bond and the bond length of nitrogen is small and bond dissociation energy is very large which makes it inert and unreactive and thus phosphorus becomes more reactive.

Question 110.
How would you account for the following :
(i) NCl₃ is an endothermic compound while NF₃ is an exothermic one.
(ii) XeF₂ is a linear molecule without a bend.
(iii) The electron gain enthalpy with negative sign for fluorine is less than that for chlorine, still fluorine is a stronger oxidising agent than chlorine. (All India 2010)
Answer:
(i) F is more electronegative than Cl. The difference in the electronegativity between N and F is much more than the difference between electronegativity of N and Cl. So there is need of much more energy to break the N-F bond.
(ii) In XeF₂ there are 2 bond pairs and 3 lone pairs and thus show sp³ d hybridization. It has linear geometry.

(iii) Because of small size of flourine atom and strong electron-electron repulsions in its compact 2p orbitals.

Question 111.
How would you account for the following :
(i) The electron gain enthalpy with negative sign is less for oxygen than that for sulphur.
(ii) Phosphorus shows greater tendency for catenation than nitrogen.
(iii) Fluorine never acts as the central atom in polyatomic interhalogen compounds. (All India 2010)
Answer:
(i) The least negative electron gain enthalpy of oxygen is due to small size and more interelectronic repulsion with coming electron.
(ii) The bond strength of P-P is more than N-N, therefore phosphorus shows more tendency for catenation than nitrogen.
(iii) Because F being smaller, it cannot accomodate larger sized other halogen atoms around it. Due to the absence of d-orbitals, F does not show positive oxidation state of +3, +5, +7 needed for the formation of polyatomic interhalogen compounds.

Question 112.
How would you account for the following :
(i) H₂S is more acidic than H₂O.
(ii) The N-O bond in \(\mathbf{N O}_{2}^{-}\) is shorter than the N-O bond in \(\mathbf{N O}_{3}^{-}\).
(iii) Both O₂ and F₂ stabilize high oxidation states but the ability of oxygen to stabilize the higher oxidation state exceeds that of fluorine. (All India 2011)
Answer:
(i) Since the size of sulphur is more than oxygen, S-H bond length increases and hence bond dissociation energy of S-H is less than O-H. Therefore S-FI easily loses H+ and thus is more acidic than H₂O.
(ii) The resonating structure of \(\mathrm{NO}_{2}^{-}\) and \(\mathrm{NO}_{3}^{-}\) show that in \(\mathrm{NO}_{2}^{-}\) two bonds are sharing a double bond while in \(\mathrm{NO}_{3}^{-}\), 3 bonds are sharing a double bond. That’s why \(\mathrm{NO}_{2}^{-}\) has shorter bond than that of \(\mathrm{NO}_{3}^{-}\).
Answer:

(iii) Oxygen stabilizes the highest oxidation state even more than fluorine.
Example : Highest fluoride of Mn is MnF₄ whereas highest oxide is Mn₂O₇. It is due to ability of oxygen to form multiple bonds with the metal atoms.

Question 113.
How would you account for the following :
(i) NF₃ is an exothermic compound but NCl₃ is not.
(ii) The acidic strength of compounds increases in the order :
PH₃ < H₂S < HCl.
(iii) SF₆ is kinetically inert. (All India 2011)
Answer:
(i) F is more electronegative than Cl. The difference in the electronegativity between N and F is much more than the difference between electronegativity of N and Cl. So there is need of much more energy to break the N-F bond.
(ii) As the electronegativity increases in the same period from left to right so their electronegativity are in the increasing order, P < S < Cl.
In the same way the acid strength is also in the increasing order i.e. PH₃ < H₂S < HCl.
(iii) Because SF₆ is showing steric hindrance due to 6 (six) fluorine atoms which make it unable to react further with any other atom.

Question 114.
Give reasons for the following:
(i) Where R is an alkyl group, R₃P = O exists but R₃N = O does not.
(ii) PbCl₄ is more covalent than PbCl₂.
(iii) At room temperature, N₂ is much less reactive. (All India 2013)
Answer:
(i) Due to presence of d-orbitals in P, it can form pπ -dπ bonds and can extend its covalency beyond 4 while N cannot do so due to absence of d-orbitals.
(ii) According to Fajan’s rule, highly charged Pb⁴⁺ can polarise the anion i.e., Cl ^– more effectively than Pb²⁺ and hence PbCl₄ becomes more covalent than PbCl₂.
(iii) Due to presence of triple bonds between 2 N atoms, their bond length decreases and hence bond dissociation energy increases which makes N₂ lesser reactive. While in phosphorus due to presence of single bond, more bond length, bond dissociation energy is low, hence very reactive.

Question 115.
Give reasons for the following :
(i) Though nitrogen exhibits +5 oxidation state, it does not form pentahalide.
(ii) Electron gain enthalpy with negative sign of fluorine is less than that of chlorine.
(iii) The two oxygen-oxygen bond lengths in ozone molecule are identical. (All India 2013)
Answer:
(i) Due to absence of empty d-orbitals, N₂ does not form pentahalide.
(ii) Because of small size of flourine atom and strong electron-electron repulsions in its compact 2p orbitals.
(iii) Due to resonance the two oxygen atoms have partial double bond character and thus have same bond length i.e. 128 pm

Question 116.
(a) Name the gas evolved on heating ammonium nitrate. Write the chemical reaction.
(b) Write two uses of ammonium nitrate. (Comptt. All India 2013)
Answer:
The gas evolved on heating is Nitrous oxide

(b) Uses of NH₄NO₃

• It is used in fertilizers.
• It is used in explosives.

Question 117.
Account for the following :
(i) NF₃ is an exothermic compound but NCl₃ is an endothermic compound.
(ii) HF is not stored in glass bottles but is kept in wax-coated bottles.
(iii) Bleaching of flowers by Cl₂ is permanent while that of SO₂ is temporary. (Comptt. All India 2013)
Answer:
(i) F is more electronegative than Cl. The difference in the electronegativity between N and F is much more than the difference between electronegativity of N and Cl. So there is need of much more energy to break the N-F bond.
(ii) HF is highly corrosive and etches glass hence it is kept in wax-coated bottles.
(iii) Chlorine bleaches the material by oxidation hence it is permanent while SO₂ bleaches the material by reduction and as the material is exposed to air, it gets oxidised and the colour is restored, hence it is temporary.

Question 118.
(a). With the help of chemical equations explain the principle of contact process in brief for the manufacture of sulphuric acid by contact process.
(b) Bismuth is a strong oxidizing agent in the pentavalent state. Explain. (Comptt. All India 2013)
Answer:
(a) Contact Process : Burning sulphur in an excess of air
S + O₂ → SO₂ (g)
or, By heating sulphide ores like pyrites in an excess of air :
4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂
In either case, an excess of air is used so that the SO₂ produced is already mixed with oxygen for the next stage. This is reversible reaction and the formation of SO₃ is exothermic in the presence of catalyst V₂O₅ at 720 K
2SO₂ + O₂ ⇌ 2SO₃ ΔH = -196 KJ/mol
This cannot be done by simply adding water to the S03. The reaction is so uncontrollable that it creates a fog of H2S04. Instead, the S03 is first dissolved in cone. H2S04.
H₂SO₄ + SO₃ → H₂S₂O₇
The product is known as fuming sulphuric acid or oleum to which water is added to get H₂SO₄
H₂S₂O₇ + H₂O → 2H₂SO₄

(b) The stability of +5 oxidation state decreases and that of +3 state increases due to inert pair effect down the group therefore Bi (V) accepts two electrons and gets reduced to Bi (III).
Bi⁵⁺ + 2e^– → Bi³⁺ :
So, Bi(V) is more stronger oxidising agent.

Question 119.
(a) Draw the structures of the following molecules :
(i) XeOF₄ (ii) H₂SO₄
(b) Write the structural difference between white phosphorus and red phosphorus. (Delhi 2014)
Answer:
(a)

(b) White phosphorus exists as discrete P4 units with SP3 hybridized phosphorus atom, arranged tetrahedrally but in red phosphorus all P₄ tetrahedral units are linked with each other to form polymeric structure.

Question 120.
Account for the following :
(i) PCl₅ is more covalent than PCl₃.
(ii) Iron on reaction with HCl forms FeCl₂ and not FeCl₃.
(iii) The two 0-0 bond lengths in the ozone molecule are equal. (Delhi 2014)
Answer:
(i) In PCl₅, phosphorus has +5 oxidation state and has less tendency to loose electrons than in +3 of PCl₃. Therefore, PCl₅ has more tendency to share e^-1s than PCl₃.
(ii) Because HCl on reaction with iron liberates H₂ gas which prevents the formation of ferric chloride.
(iii) Due to resonance the two oxygen atoms have partial double bond character and thus have same bond length i.e. 128 pm

Question 121.
Account for the following :
(i) Bi(V) is a stronger oxidizing agent than Sb(V).
(ii) N – N single bond is weaker than P – P single bond.
(iii) Noble gases have very low boiling points. (Delhi 2014)
Answer:
(i) Bi(V) is a stronger oxidizing agent than Sb(V) due to inert pair effect as the stability of lower oxidation state (+3) increases down the group.
(ii) Due to smaller size of Nitrogen, their lone pairs repel the bond pair of N – N bond while P – P due to bigger size does not show more repulsion.
(iii) Due to presence of weak Van der waal forces of attraction, noble gases have very low boiling point.

Question 122.
Account for the following :
(i) Sulphur in vapour form exhibits paramagnetic behaviour.
(ii) SnCl₄ is more covalent than SnCl₂.
(iii) H₃PO₂ is a stronger reducing agent than H₃PO<sub3. (Delhi 2014)
Answer:
(i) In vapour state sulphur partly exists as S2 molecule which has two unpaired electrons in the antibonding II orbitals and hence exhibits paramagnetism.
(ii) Sn⁺⁴ in SnCl₄ has more polarising power than SnCl₂
(iii) H₃PO₂ contains two P-H bonds while H₃PO₃ contains only one P-H bond therefore H₃PO₂ is stronger reducing agent.

Question 123.
Give reasons for the following :
(i) (CH₃)₃ P = O exists but (CH₃)₃ N = O does not.
(ii) Oxygen has less electron gain enthalpy with negative sign than sulphur.
(iii) H₃PO₂ is a stronger reducing agent than H₃PO₃. (All India 2014)
Answer:
(i) (CH₃)₃P = 0 exists due to presence of empty d-orbitals and thus can expand its covalency upto 6 but (CH₃)₃ N = O cannot expand its covalency due to absence of d-orbitals.
(ii) The least negative electron gains enthalpy of oxygen is due to small size and more interelectronic repulsion with coming electron.
(iii) H₃PO₂ contains two P-H bonds while H₃PO₃ contains only one P-H bond therefore H₃PO₂ is a stronger reducing agent.

Question 124.
Give reasons:
(i) SO₂ is reducing while TeO₂ is an oxidizing agent.
(ii) Nitrogen does not form pentahalide.
(iii) ICl is more reactive than I₂. (All India 2016)
Answer:
(i) SO₂ is reducing while TeO₂ is an oxidising agent because sulphur can expand its covalency upto +6 from +4 due to presence of empty d-orbital but as we move down the group the stability of +6 oxidation state decreases and of +4 oxidation state increases due to inert pair effect. Hence SO₂ acts as reducing agent while TeO₂ acts as an oxidising agent.
(ii) Due to absence of empty d-orbitals, N₂ does not form pentahalides.
(iii) Because ICl bond is weaker than I -I bond as a result of which ICl breaks easily to form halogen atoms which readily bring about the reaction, hence more reactive.

Question 125.
Give reasons:
(i) Thermal stability decreases from H₂O to H₂Te.
(ii) Fluoride ion has higher hydration enthalpy than chloride ion.
(iii) Nitrogen does not form pentahalide. (Delhi 2017)
Answer:
(i) Thermal stability decreases from H₂O to H₂Te due to weakening of bond between hydrogen and the atom from O to T_e as size is increasing down the group.
(ii) Fluoride ion has higher hydration enthalpy than chloride ion due to stronger attractions of smaller in size fluoride ion.
(iii) Nitrogen does not contain’d’ orbitals.

Question 126.
Give reasons for the following:
(a) Red phosphorus is less reactive than white phosphorus.
(b) Electron gain enthalpies of halogens are largely negative.
(c) N₂O₅ is more acidic than N₂O₃.(All India 2017)
Answer:
(a) Red phosphorus is less reactive than white phosphorus because white phosphorus possess angle strain where long angles are only 60° making it more reactive. Also, red phosphorus being polymeric is less reactive than white phosphorus which has discrete tetrahedral structure.
(b) Electron gain enthalpies of halogens are largely negative due to high effective nuclear charge and smaller size among period. They readily accept an electron to attain noble gas configuration.
(c) N₂O₅ is more acidic than N₂O₃ because higher the oxidation state, higher will be acidic character. N₂O₅ has +5 oxidation state and N₂O₃ has +3 oxidation state.

Question 127.
(a) Arrange the hydrides of group 16 in increasing order of their acidic character. Justify your answer.
(b) Draw structure of XeOF4. (Comptt. Delhi 2017)
Answer:
(a) H₂O < H₂S < H₂Se < H₂Te
As we move down the group the bond dissociation enthalpy decreases due to increase in bond length and size of the central atom.
(b)

Question 128.
(a) Account for the the following :
(i) PCl₅ is more covalent than PCl₃
(ii) Iron on reaction with HC1 forms FeCl₂ and not FeCl₃
(b) Draw structure of XeO₃ (Comptt. Delhi 2017)
Answer:
(a) (i) In PCl₅, phosphorus has +5 oxidation state and has less tendency to loose electrons than in +3 of PCl₃. Therefore, PCl₅ has more tendency to share e^-1s than PCl₃.
(ii) Because HCl on reaction with iron liberates H₂ gas which prevents the formation of ferric chloride.
(iii) Due to resonance the two oxygen atoms have partial double bond character and thus have same bond length i.e. 128 pm

(b)

The p-Block Elements Class 12 Important Questions long Answer Type [LA]

Question 129.
(a) Draw the structures of the following :
(i) H₂S₂O₈ (ii) HClO₄
(b) How would you account for the following :
(i) NH₃ is a stronger base than PH₃
(ii) Sulphur has a greater tendency for catenation than oxygen.
(iii) F₂ is a stronger oxidising agent than Cl₂. (All India 2009)
Answer:
(a) (i) H₂S₂O₈ (Peroxodisulphuric acid) or Marshall’s acid :

(b)
(i) Since both P and N contain lone pairs of electrons but due to small size and high electronegativity of Nitrogen in NH₃, the electron density is much higher than PH₃, therefore it can easily donate electrons and acts as strong Lewis base than PH₃.

(ii) The greater catenation tendency of sulphur is due to two reasons :
(a) The lone pair of electrons feels more repulsion in 0-0 bond than S-S bond due to its small size and thus S-S forms strong bond.
(b) As the size of atom increases down the group from O – PO, the strength of bond increases and therefore catenation tendency also increases.

(iii) Due to low bond dissociation enthalpy and high electronegativity of Fluorine, it has strong tendency to accept electrons and thus get reduced.
F + e^– → F^–
Therefore F₂ acts as strong oxidising agent, while Cl₂ is weak oxidising agent due to low electronegativity.

Question 130.
(a) Draw the structures of the following :
(i) H₂S₂O₇ (ii) HClO₃
(b) Explain the following observations :
(i) In the structure of HNO₃ the N-O bond (121 pm) is shorter than the N- OH bond (140 pm).
(ii) All the P-Cl bonds in PCl₅ are not equivalent.
(iii) ICl is more reactive than I₂. (All India 2009)
Answer:
(a) (i) H₂S₂O₇ (Pyrosulphuric acid) or oleum :

(b) (i) The N-O bond has partial double bond character while the N-OH bond is a single bond in both resonance of HNO₃
(ii) All the P-Cl bonds in PCl₅ are not equivalent due to the fact that the axial bond pairs suffer more repulsion as compared to equatorial bond pairs.
(iii) Because ICl bond is weaker than I-I bond as a result of which ICl breaks easily to form halogen atoms which readily bring about the reaction, hence more reactive.

Question 131.
(a) Draw the structures of the following :
(i) H₃PO₂ (ii) BrF₃
(b) How would you account for the following observations :
(i) Phosphorus has a greater tendency for catenation than nitrogen.
(ii) Bond dissociation energy of fluorine is less than that of chlorine.
(iii) No chemical compound of helium is known. (All India 2009)
Answer:
(a) (i) H₃PO₂ (Hypophosphorous acid)

(b) (i) The bond strength of P-P is more than N-N, therefore phosphorous shows more tendency for catenation than nitrogen.
(ii) Due to smaller size of F than Cl as a result of which electron-electron repulsions between the lone pairs of electrons are very large than that of Cl, hence bond dissociation energy of F₂ is less than that of Cl₂.
(iii) Because the ionization energy of Helium is very high and the empty d-orbitals are also absent in it.

Question 132.
(a) Draw the structures of the following :
(i) N₂O₅ (ii) XeOF₄
(b) Explain the following observations :
(i) The electron gain enthalpy of sulphur atom has a greater negative value than that of oxygen atom.
(ii) Nitrogen does not form pentahalides.
(iii) In aqueous solutions HI is a stronger acid than HCl. (All India 2009)
Answer:
(a) (i) N₂O₅

(ii) XeOF₄

(b) (i) Because enthalpy of dissociation of S-S bond is higher than 0-0 bond and the hydration energy of S^2- is less than that of O^2- ion.
(ii) Due to absence of empty d-orbitals, N₂ does not form pentahalides.
(iii) Due to lower bond dissociation energy and higher degree of ionization, HI acts as stronger acid than HCl in aqueous solution.

Question 133.
(a) Draw the structures of the following :
(i) XeF₄ (ii) H₂S₂O₇
(b) Explain the following observations :
(i) Phosphorus has a greater tendency for catenation than nitrogen.
(ii) The negative value of electron gain enthalpy is less for fluorine than that for chlorine.
(iii) Hydrogen fluoride has a much higher boiling point than hydrogen chloride. (All India 2009)
Answer:
(a) (i) XeF₄ :

(b) (i) The bond strength of P-P is more than N-N, therefore phosphorous shows more tendency for catenation than nitrogen.
(ii) Because of small size of flourine atom and strong electron-electron repulsions in its compact 2p orbitals.
(iii) Hydrogen fluoride (HF) has higher boiling point than HC1 due to extensive intermolecular hydrogen bonding while HCl doesn’t show this H-bonding.

Question 134.
(a) Draw the structures of the following :
(i) PCl₅(s)
(ii) \(\mathrm{SO}_{3}^{2-}\)
(b) Explain the following observations :
(i) Ammonia has a higher boiling point than phosphine.
(ii) Helium does not form any chemical compound.
(iii) Bi (V) is a stronger oxidising agent than Sb (V). (All India 2009)
Answer:
(a) (i) PCl₅ (s)

Angle: The angle O – S – O is greater than 90°
(b) (i) Due to intermolecular H-bonding in NH₃ it has higher boiling point than PH₃ which does not have any H-bonding.
(ii) Because the ionization energy of Helium is very high and very high positive electrons gain enthalpy.
(iii) The stability of +5 oxidation state decreases and that of +3 state increases due to inert pair effect down the group therefore Bi(v) accepts two electrons and gets reduced to Bi (v).
Bi⁵⁺ + 2e^– → Bi³⁺

Question 135.
(a) Complete the following chemical equations :
(i) NaOH (aq) + Cl₂ (g) →
(Hot and cone.)
(ii) XeF₆ (s) + H₂O(l) →
(b) How would you account for the following?
(i) The value of electron gain enthalpy with negative sign for sulphur is higher than that for oxygen.
(ii) NF₃ is an exothermic compound but NClj is endothermic compound.
(iii) ClF₃ molecule has a T-shaped structure and not a trigonal planar one. (Delhi 2010)
Answer:
(a) (i) 6NaOH + 3Cl₂ → 5NaCl + 1NaClO₃ + 3H₂O
(ii) XeF₆ (s) + 3H₂O (1) → XeO₃ (s) + 6HF (aq)

(b) (i) Because of enthalpy of dissociation of S- S bond is higher than O-O bond and the hydration energy of S^2- is less than that of O^2- ion.

(ii) Due to smaller size of F as compared to Cl, the N – F bond is much stronger than N-Cl bond while bond dissociation energy of F₂ is much lower than that of Cl₂. Therefore, energy released during the formation of NF₃ molecule is more than the energy needed to break N₂ and F₂ molecules into individual atoms. In other words, formation of NF₃ is an exothermic reaction.
The energy released during the formation of NCl₃ molecule is less than the energy needed to break N₂ and Cl₂ molecule into individual atoms. Thus formation of NCl₃ is an endothermic reaction.

(iii) The electronic configuration of Cl is 1s² 2s² 2p⁶ 3s² \(3 \mathrm{P}_{x}^{2} 3 \mathrm{P}_{y}^{2} 3 \mathrm{P}_{z}^{1}\). It has only one half filled orbital. But to form three Cl – F bonds, we need three half filled orbitals. To achieve this, one of the \(3 P_{y}^{2}\) electrons gets excited to 3d orbital. The resulting five orbitals of the third shell of Cl in the first excited state undergoes SP³d hybridization to give T-shaped structure to ClF³ molecule. Since Cl does not undergo SP² hybridization, therefore ClF³ does not have trigonal planar structure.

Question 136.
(a) Complete the following chemical reaction equations :
(i) P₄ + SO₂Cl₂ →
(ii) XeF₄ + H₂O →
(b) Explain the following observations giving appropriate reasons :
(i) The stability of +5 oxidation state decreases down the group in group 15 of the periodic table.
(ii) Solid phosphorus pentachloride behaves as an ionic compound.
(iii) Halogens are strong oxidizing agents. (Delhi 2010)
Answer:
(a)

(b) (i) Group 15 first elements are pentavalent, therefore they can show positive oxidation state +3 (due to P-electron) and +5 (due to P and S electrons). In a group the +5 oxidation state stability decreases but +3 oxidation state increases due to inter pair effect which results the 5- orbitals electrons to participate in bonding hence shows +3 oxidation state as their stable oxidation state.
(ii) PCl₅ conducts electricity in the molten state. This means that in solid state it exists as [PCl₄]⁺ [PCl₆]^– in which the cation is tetrahedral and the anion is octahedral.
2PCl₅ → [PCl₄]⁺ [PCl₆]^–
On melting, these ions become free to move and hence PCl₅ conducts electricity in the molten state.
(iii) As halogens are strong elctron acceptors and change to negative ions and thus undergo reduction, so they are strong oxidising agent.

Question 137.
(a) Explain the following :
(i) NF₃ is an exothermic compound ’ whereas NCl₃ is not.
(ii) F₂ is most reactive of all the four common halogens.
(b) Complete the following chemical equations :
(i) C + H₂SO₄ (cone) →
(ii) P₄ + NaOH + H₂O →
(iii) Cl₂ + F₂ (excess) → (Delhi 2010)
Answer:
(a) (i) F is more electronegative than Cl. The difference in the electronegativity between N and F is much more than the difference between electronegativity of N and Cl. So there is need of much more energy to break the N-F bond.
(ii) Because of the low bond dissociation energy F₂ readily dissociates into atoms and reacts with other substances readily.

(b) (i) C + 2H₂SO₄ (cone.) → CO₂ + 2SO₂ + 2H₂O
(ii) P₄ + 3NaOH + 3H₂O → PH₂ + 3NaH₂PO₂
(iii) Cl₂ + 3F₂ (excess) → 2ClF₃

Question 138.
(a) Account for the following :
(i) The acidic strength decreases in the order HCl > H₂S > PH₃
(ii) Tendency to form pentahalides
decreases down the group in group 15 of the
(b) Complete the following chemical equations :
(i) P₄ + SO₂Cl₂ →
(ii) XeF₂ + H₂O →
(iii) I₂ + HNO₃ (conc) → (Delhi 2010)
Answer:
(a) (i) Because of decrease in electronegativity from chlorine to phosphorous, the bond dissociation enthalpy from HCl to H-P increases and their tendency to release H⁺ decreases and thus acidic strength decreases.
(ii) Down the group, the tendency of next ‘s’ orbital’s electron to jump to previous’d’ orbital decreases very much due to inert pair effect.

Question 139.
(a) Draw the structures of the following molecules :
(i) (HPO₃)₃
(ii) BrF₃
(b) Complete the following chemical equations :
(i) HgCl₂ + PH₃ →
(ii) SO₃ + H₂SO₄ →
(iii) XeF₄ + H₂O → (All India 2010)
Answer:
(a) (i) (HPO₃)₃ :

Question 140.
(a) What happens when
(i) chlorine gas is passed through a hot concentrated solution of NaOH?
(ii) sulphur dioxide gas is passed through an aqueous solution of a Fe (III) salt?
(b) Answer the following :
(i) What is the basicity of H₃PO₃ and why?
(ii) Why does fluorine not play the role of a central atom in interhalogen compounds?
(iii) Why do noble gases have very low boiling points? (All India 2010)
Answer:

(b) (i) Basicity of H₃PO₃ = 2
Because basicity is the number of replaceable H⁺ ions in an acid and H₃PO₃ is a Dibasic acid.
(ii) Because F being smaller, it cannot accomodate larger sized other halogen atoms around it. Due to the absence of d-orbitals, F does not show positive oxidation state of +3, +5, +7 needed for the formation of polyatomic interhalogen compounds.
(iii) Because the atoms of these elements are held together by weak van der Waal’s forces of attraction.

Question 141.
(a) Complete the following chemical reaction equations :
(i) P₄ + SO₂Cl₂ →
(ii) XeF₆ + H₂O →
(b) Predict the shape and the asked angle (90° or more or less) in each of the following cases:
(i) \(\mathrm{SO}_{3}^{2-}\) and the angle O – S – O
(ii) ClF₃ and the angle F – Cl – F
(iii) XeF₂ and the angle F – Xe – F (Delhi 2012)
Answer:
(a) (i) P₄ + 10SO₂Cl₂ → 4PCl₅ + 10SO₂
or P₄ + 8SO₂Cl₂ → 4PCl₃ + 4SO₂ + 2S₂Cl₂
(ii) XeF₆ + H₂O → XeOF₄ + 2HF
or XeF₆ + 2H₂O → XeOF₄ + 4HF
or XeF₆ + 3H₂O → XeO₃ + 6HF

(b) (i) \(\mathrm{SO}_{3}^{2-}\)


Angle : The angle F-Cl-F is lesser than 90°
(iii) XeF₂; Structure :

Shape : Linear
Angle ; F-Xe – F > 90°

Question 142.
(a) Complete the following chemical equations :
(i) NaOH (hot and cone.) + Cl₂ →
(ii) XeF₄ + O₂F₂ →
(b) Draw the structures of the following molecules :
(i) H₃PO₂ (ii) H₂S₂O₇ (iii) XeOF₄ (Delhi 2012)
Answer:
(a) (i) 3Cl₂⁺ 6NaOH → 5 NaCl + NaClO₃ + 3H₂O
(ii) XeF₄ + O₂F₂ → XeF₆ + O₂
(b) (i) H₃PO₂:

(ii) H₂S₂O₇:

(iii) XeOF₂:

Question 143.
(a) Draw the molecular structure of following compounds :
(i) XeF₆ (ii) H₂S₂O₈
(b) Explain the following observations :
(i) The molecules NH₃ and NF₃ have dipole moments which are of opposite direction.
(ii) All the bonds in PCl₅ molecule are not equivalent.
(iii) Sulphur in vapour state exhibits paramagnetism. (Delhi 2012)
Answer:
(a) (i) XeF₆:

(b) (i) Because F is more electronegative than N in NF₃ whereas N is more electronegative than H in NH₃.
(ii) Because PCl₅ has a trigonal bipyramidal structure in which the three equatorial P-Cl bonds are equivalent while the two axial bonds are longer than equatorial bonds.
(iii) Because sulphur in vapour state has two unpaired electrons in the antibonding π* orbitals like O₂.

Question 144.
(a) Complete the following chemical equations
(i) XeF₄ + SbF₅ →
(ii) Cl₂ + F₂ (excess) →
(b) Explain each of the following :
(i) Nitrogen is much less reactive than phosphorus.
(ii) The stability of +5 oxidation state decreases down group 15.
(iii) The bond angles (O – N – O) are not of the same value in \(\mathrm{NO}_{2}^{-}\) and \(\mathrm{NO}_{2}^{+}\). (Delhi 2012)
Answer:
(a) (i) XeF₄ + SbF₅ → [XeF₃]⁺ [SbF₆]^–
(ii) Cl₂ + 3F₂ (excess) → 2ClF₃

(b) (i) Due to presence of triple bond between two nitrogen atoms/high bond dissociation enthalpy of nitrogen, whereas in phosphorus molecule there is single bond between four atoms of phosphorus so have low bond dissociation enthalpy.
(ii) Because the participation of outer s- electron pair goes on decreasing down the group due to inert pair effect.
(iii) Because \(\mathrm{NO}_{2}^{-}\) has sp² hybridisation whereas \(\mathrm{NO}_{2}^{+}\) has sp hybridisation.

Question 145.
(a) Draw the molecular structures of the following compounds :
(i) N₂O₅ (ii) XeOF₄
(b) Explain the following observations :
(i) Sulphur has a greater tendency for catenation than oxygen.
(ii) IC1 is more reactive than I₂.
(iii) Despite lower value of its electron gain
enthalpy with negative sign, fluorine (F₂) is a stronger oxidising agent than Cl₂. (All India 2012)
Answer:
(a)

(i) N₂O₅

(ii) XeOF₄

(b) (i) It is because S – S bond is stronger than 0-0 bonds as there is more interelectronic repulsion in O – O due to small size than in S – S.
(ii) ICl is more reactive than I₂ because I-Cl bond is polar and weaker while I-I bond is stronger and non-polar.
(iii) It is due to (a) low enthalpy of dissociation of F – F bond, (b) high hydration enthalpy of F.

Question 146.
(a) Complete the following chemical equations :
(i) Cu + HNO₃ (dilute) →
(ii) XeF₄ + O₂F₂ →
(b) Explain the following observations :
(i) Phosphorus has greater tendency for catenation than nitrogen.
(ii) Oxygen is a gas but sulphur a solid.
(iii) The halogens are coloured. Why? (All India 2012)
Answer:
(a) (i) 3Cu + 8HNO₃ (dil) → 3CU(NO₃)₂ + 2NO + 4H₂O
(ii) XeF₄ + O₂F₂ → XeF₆ + O₂

(b) (i) The self linking property or catenation property of nitrogen is less than that of phosphorus because N – N bond is weaker than P – P bond.

(ii) Oxygen forms a stable diatomic molecule. In 02 molecule two atoms of oxygen have joined together through double bond 0 = 0. The multiple bonding is possible due to small size of oxygen atoms. So oxygen is a gas. S has eight atoms arranged in the form of a puckered ring per molecule in which two sulphur atoms are joined by covalent bonds. So sulphur is a solid at room temperature.

(iii) All halogens are coloured. This is due to absorption of radiation in visible region which results in the excitation of outer electrons to higher energy level. So they display different colours.

Question 147.
(a) Draw the structures of the following molecules :
(i) H₃PO₂ (ii) ClF₃
(b) Explain the following observations :
(i) Nitrogen is much less reactive than phosphorus.
(ii) Despite having greater polarity, hydrogen fluoride boils at a lower temperature than water.
(iii) Sulphur has greater tendency for catenation than oxygen in the same group. (All India 2012)
Answer:
(a) (i) H₃PO₂ :

(ii) ClF₃ :

(b) (i) Because P-P bond is weaker than N ≡ N bond, N₂ is inert due to the presence of triple bond.
(ii) In water O is larger than F therefore Van- der-Waals forces of attraction increase with increase in size and hence the boiling point increases.
(iii) It is because S – S bond is stronger than 0-0 bond as there is more interelectronic repulsion in O – O than in S-S.

Question 148.
(a) Draw the structures of the following molecules :
(i) N₂O₅ (ii) HClO₄
(b) Explain the following observations :
(i) H₂S is more acidic than H₂O.
(ii) Fluorine does not exhibit any positive oxidation state.
(iii) Helium forms no real chemical compound. (All India 2012)
Answer:
(a) (i) N₂O₅ :

(ii) HClO₄ :

(b) (i) H₂S is more acidic than H₂O because bond dissociation enthalpy of H – S bond in H₂S is less than that of H – O bond in
H₂O.
(ii) F is the most electronegative element. It has no d-orbitals and therefore, there is no scope for any electron promotion. So it can only show oxidation state of -1 in its compounds.
(iii) Helium has very high ionisation enthalpy and no vacant d-orbitals, therefore no chemical compound of helium is known.

Question 149.
(a) Describe the conditions and the steps involved in the manufacture of sulphuric acid by contact process. Write the necessary reactions. (No diagram is required)
(b) Give reasons :
ii) Bond dissociation energy of F₂ is less than that of Cl₂.
(ii) Nitric oxide becomes brown when released in air. (Comptt. Delhi 2012)
Answer:
(a) Contact Process : Burning sulphur in an excess of air
S + O₂ → SO₂ (g)
or, By heating sulphide ores like pyrites in an excess of air :
4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂
In either case, an excess of air is used so that the SO₂ produced is already mixed with oxygen for the next stage. This is reversible reaction and the formation of SO₃ is exothermic in the presence of catalyst V₂O₅ at 720 K
2SO₂ + O₂ ⇌ 2SO₃ ΔH = -196 KJ/mol
This cannot be done by simply adding water to the S03. The reaction is so uncontrollable that it creates a fog of H2S04. Instead, the S03 is first dissolved in cone. H2S04.
H₂SO₄ + SO₃ → H₂S₂O₇
The product is known as fuming sulphuric acid or oleum to which water is added to get H₂SO₄
H₂S₂O₇ + H₂O → 2H₂SO₄

(b) (i) Bond dissociation energy of F₂ is less than that of Cl₂ because of large electron- electron repulsion among the lone pair in F₂ molecule.
(ii) Nitric oxide becomes brown when released in air because of the formation of NO₂ gas.

Question 150.
Account for the following :
(a) Thermal stability of water is much higher than that of H₂S.
(b) White phosphorus is more reactive than red phosphorus.
(c) Ammonia acts as a ligand.
(d) Bismuth is a strong oxidizing agent in pentavalent state.
(e) Concentrated sulphuric acid is a strong dehydrating agent. (Comptt. Delhi 2012)
Answer:
(a) The thermal stablility of hydrides decrease from H₂O to H₂S because as the size of the atom increases, the bond becomes weaker and thus breaks on heating.
(b) Red phosphorus is regarded as a polymer consisting of chains of P₄ tetrahedral linked together. This makes phosphorus denser and less reactive

But in white phosphorus (P₄), the four phosphorus atoms lie at the corners of a regular tetrahedron. Each phosphorus atom is linked to each of the other three atoms by covalent bond. The bond angle is equal to 60° which suggests that the molecule is under strain and hence active in nature.

(c) Due to the presence of a lone pair of electrons on nitrogen atom, it has a tendency to donate an electron pair, hence acts as a ligand.
(d) In Bismuth, the inert pair effect is very prominent. Thus +5 oxidation state is less stable in comparison to +3 oxidation state i.e
it readily accepts two electrons in pentavalent state and gets reduced to trivalent state. Therefore, it acts as a strong oxidising agent.
(e) Cone. H₂SO₄ has great affinity for water molecule i.e it acts as a dehydrating agent.

Question 151.
Account for the following :
(i) Chlorine water loses its yellow colour on standing.
(ii) BrCl₃ is more stable than BrCl₅.
(iii) Fluorine does not form oxoacids.
(iv) PCl₅ acts as an oxidising agent.
(v) SO₂ is an air pollutant. (Comptt. All India 2012)
Answer:
(i) Cl₂ water on standing loses its yellow colour due to the formation of HCl and HOCl. HOCl is unstable and decomposes to HCl. As a result, yellow colour disappears.

(ii) In BrCl₃, Br can accommodate three chlorine atoms around it and hence it is stable but in BrCl₅, five Cl atoms cannot be accommodated around Br and hence it is unstable.
(iii) Due to high electonegativity and small size F does not form oxoacids.
(iv) In PCl₅, the oxidation state of phosphorus is +5. Thus, it cannot increase its oxidation state beyond +5. So PCl₅ cannot act as a reducing agent. However, it can decrease its oxidation state from +5 to +3 or to some lower value. Thus, it acts as an oxidising agent.
(v) SO₂ is a pungent and irritating gas. It acts as an air pollutant due to the following reasons:

• It causes throat and eye irritation as it is absorbed readily by respiratory’ tract.
• It combines with moisture forming sulphurous acid. It is then converted into H₂SO₄. Both these acids cause acid rain and destroy the marble, corrode metals, deteriorate fabrics, paper, leather etc.

Question 152.
(a) With the help of chemical equations explain the principle of contact process in brief for the manufacture of sulphuric acid.
(No diagram)
(b) Account for the following :
(i) Bond dissociation energy of F₂ is less than that of Cl₂
(ii) Nitric oxide (NO) becomes brown when released in air. (Comptt. All India 2012)
Answer:
(a) Principle of contact process : The process involves the oxidation of sulphur dioxide by air in the presence of a catalyst.

Sulphur trioxide is dissolved in 98% H₂SO₄ acid and oleum is formed
H₂SO₂ + SO₃ → H₂S₂O₇ (oleum)
Oleum is diluted with water to form sulphuric acid
H₂S₂07 + H₂O → 2H₂S0₄

(b) (i) The low bond dissociation energy of F is due to high inter electronic repulsions between non-bonding electrons in 2p- orbitals as the size of F atoms is small. As a result, F-F bond is weaker than Cl-Cl bond.
(ii) It readily combines with 02 to form brown fumes of NO₂.
2NO + O₂→ 2NO₂ (brown fumes)

Question 153.
(a) Give reasons for the following:
(i) Bond enthalpy of F₂ is lower than that of Cl₂.
(ii) PH₃ has lower boiling point than NH₃,
(b) Draw the structures of the following molecules:
(i) ClF₃ (ii) (HPO₃)₃ (iii) XeF₄ (Delhi 2013)
Answer:
(a) (i) Due to smaller size of F than Cl as a result
of which electron-electron repulsions between the lone pairs of electrons are very large than that of Cl, hence bond dissociation enthalpy of F₂ is less than that of Cl₂
(ii) PH₃ has lower boiling point than NH₃ because PH₃ cannot form hydrogen bonds like NH₃.

(b) (i) BrF₃ :

(ii) (HPO₃)₃: Cyclotrimetaphosphoric acid

(iii) XeF4:

Question 154.
(a) Account for the following:
(i) Helium is used in diving apparatus.
(ii) Fluorine does not exhibit positive oxidation state.
(iii) Oxygen shows catenation behaviour less than sulphur.
(b) Draw the structures of the following molecules:
(i) XeF₂ (ii) H₂S₂O₈ (Delhi 2013)
Answer:
(i) Helium is used in diving apparatus because of its very low solubility in blood and therefore an oxygen-helium mixture is used for artificial respiration. (ii) Because it is most electronegative element and does not have d-orbitals for octet expansion, therefore it shows only a negative oxidation state of -1.
(ii) The greater catenation tendency of sulphur is due to two reasons :
(a) The lone pair of electrons feels more repulsion in 0-0 bond than S-S bond due to its small size and thus S-S forms strong bond.
(b) As the size of atom increases down the group from O – PO, the strength of bond increases and therefore catenation tendency also increases.

(b) (i) XeF₂ :

(ii) H₂S₂O₈ :

Question 155.
Give reasons for the following :
(i) Oxygen is a gas but sulphur is a solid.
(ii) O₃ acts as a powerful oxidising agent.
(iii) BiH₃ is the strongest reducing agent
amongst all the hydrides of Group 15 elements. (All India 2013)
Answer:
(i) Due to small size and high electronegativity, oxygen forms pπ – pπ multiple bonds and thus forms diatomic. O₂ molecule are held together by weak van der Waals forces of attraction which can be easily overcome by collisions at room temperature. Therefore O₂ is gas at room temperature. On the other hand due to higher tendency for catenation and lower tendency for pπ – pπ multiple bonds, intra-atomic S₈ has strong forces of attraction which cannot be overcome by collisions. Therefore S is solid at room temperature.

(ii) O₃ is a powerful oxidising agent due to its high energy content than oxygen and hence decomposes to give diatomic oxygen and atomic oxygen

(iii) As we move down the group, the E—H bond length increases and their strength decreases. Bi—H bond is the weakest. It can break easily and evolves H₂ gas which acts as the reducing agent.

Question 156.
(a) Write the balanced chemical equations for obtaining XeO₃ and XeOF₄ from XeF₆.
(b) Account for the following :
(i) H₂S is less acidic than H₂Te.
(ii) H<sub3PO₂ has reducing nature.
(iii) SO₂ is an air pollutant. (Comptt. Delhi 2013)
Answer:
(a) XeF₆ + 3H₂O → XeO₃ + 6HF
XeF₆ + H₂O → XeOF₄ + 2HF

(b) (i) In the group with increase in size of the element and increased bond distance, the bond dissociation energy decreases, therefore H-S bond dissociation energy is higher than H-Te and hence H-S bond breaks less easily than H-Te bond and H₂S is a weaker acid than H₂Te.
(ii) Because it contains two P-H bonds and thus reduces AgNO₃ to metallic silver
4AgNO₃ + H₃PO₂ + 2H₂O → 4Agi ↓ + H₃PO₄ + 4HNO₃
(iii) SO₂ is a pungent and irritating gas. It acts as an air pollutant due to the following reasons:

• It causes throat and eye irritation as it is absorbed readily by respiratory tract.
• It combines with moisture forming sulphurous acid. It is then converted into H₂SO₄. Both these acids cause acid rain and destroy the marble, corrode metals, deteriorate fabrics, paper, leather etc.

Question 157.
Account for the following :
(a) Phosphorus shows high tendency for catenation.
(b) F₂ is more reactive than ClF₃ but ClF₃ is more reactive than Cl₂.
(c) Nitrogen is found in gaseous state.
(d) Decomposition of ozone molecule is a spontaneous process.
(e) SF₆ is inert towards hydrolysis. (Comptt. Delhi 2013)
Answer:
(a) The bond strength of P-P is more tan N-N, therefore, phosphorous shows more tendency for catenation than nitrogen.
(b) Because the bond between Cl-F is weaker than the bond between Cl-Cl due to less effective overlapping of orbitals of Cl-F than Cl-Cl.
The F₂ is most reactive due to its high electrode potential.
(c) Due to small size and high electronegativity nitrogen forms pπ – pπ multiple bonds, which are held together by weak Van-der Waal’s forces of attraction which can be easily broken. Hence N₂ exists as a gas at room temperature.
(d) Because O₃ is thermodynamically unstable and its decomposition results in liberation of heat (ΔH is -ve) and increase in entropy (ΔS is +ve) so (ΔG becomes -ve and process becomes spontaneous.
(c) In SF₆, S is protected by 6F atoms and does not allow H₂O molecules to attach on it. So SF₆ inerts towards hydrolysis.

Question 158.
(a) Complete the following chemical equations:
(i) P₄ + NaOH + H₂O →
(ii) XeF₄ + O₂F₂ →
(b) How would you account for the following situations?
(i) The acidic strength of these
compounds increases in the following order :
PH₃ > H₂S > HCl
(ii) The oxidising power of oxoacids of chlorine follows the order :
HClO₄ > HClO₃ > HClO₂ > HClO
(iii) In vapour state sulphur exhibits paramagnetic behaviour. (Comptt. Delhi 2014)
Answer:
(a)

(b) (i) As the electronegativity increases in the same period from left to right so their electronegativity are in the increasing order, P < S < Cl.
In the same way the acid strength is also in the increasing order i.e. PH₃ < H₂S < HCl.

Acidic strength of oxoacids of the same halogen increases with increase in oxidation number of the halogen because of the relative stability of the anions left after removal of proton. Thus as the number of oxygen atoms in the anion increases, the dispersal of the negative charge through pn-pn back bonding also increases and hence stability and acidic strength increases

(iii) In vapour state sulphur partly exists as S₂ molecule which has two unpaired electrons in the antibonding π orbitals and hence exhibits paramagnetism.

Question 159.
(a) Using VSEPR theory predict the probable structures of the following :
(i) N₂O₃ (ii) BrF₃
(b) Arrange the following groups of substances in the order of the property indicated against each group :
(i) NH₃, PH₃, AsH₃, SbH₃ – increasing order of boiling points.
(ii) O, S, Se, Te – increasing order of electron gain enthalpy with negative sign.
(iii) F₂, Cl₂, Br₂, I₂ – increasing order of bond dissociation enthalpy. (Comptt. Delhi 2014)
Answer:
(a)

(b) (i) PH₃ < AsH₃ < NH₃ < SbH₃
(ii) O < Te < Se < S
(iii) The increasing order of bond dissociation enthalpy is :
Cl₂ > Br₂ > F₂ > I₂

Question 160.
(a) Write the formula and describe the structure of a noble gas species which is isostructural with
(i) IBr₂ (ii) BrO₃^–
(b) Assign reasons for the following :
(i) SF₆ is kinetically inert.
(ii) NF₃ is an exothermic compound whereas NCl₃is not.
(iii) HCl is a stronger acid than HF though flourine is more electronegative than chlorine. (Comptt. All India 2014)
Answer:
(a) (i) IBr₂^– : It has 2 bond pairs and 3 lone pairs.
Therefore according to VSEPR theory it should have Linear structure.
IBr2- has (7 + 2 × 7 + 1) i.e. 22 valence electrons.
No noble gas has 22 electrons

∴ The isoelectronic species for IBr₂^– is XeF₂
(ii) BrO₃^–: It has 3 bond pairs and 1 lone pair of electrons. Therefore according to VSEPR theory it has pyramidal structure.
It has 26 valence electrons i.e. (7 + 3 × 6 + Pyramidal shape 1 = 26).


A noble gas having 26 valence electrons is XeO₃ (8 + 3 × 6 = 26).
Thus XeO₃ also has pyramidal structure.
(b) (i) Because SF₆ is showing steric hindrance due to 6 (six) fluorine atoms which make it unable to react further with any other atom.
(ii)

• Because of enthalpy of dissociation of S- S bond is higher than 0-0 bond and the hydration energy of S^2- is less than that of O^2- ion.
• Due to smaller size of F as compared to Cl, the N – F bond is much stronger than N-Cl bond while bond dissociation energy of F₂ is much lower than that of Cl₂. Therefore, energy released during the formation of NF₃ molecule is more than the energy needed to break N₂ and F₂ molecules into individual atoms. In other words, formation of NF₃ is an exothermic reaction.
  The energy released during the formation of NCl₃ molecule is less than the energy needed to break N₂ and Cl₂ molecule into individual atoms. Thus formation of NCl₃ is an endothermic reaction.

(iii) Because HCl has large size and less bond strength than HF which makes easier liberation of H⁺.

Question 161.
(a) How is ammonia prepared on a large scale?
Name the process and mention the optimum conditions for the production of ammonia by this process.
(b) Assign reasons for the following :
(i) H₂S is more acidic than H₂O.
(ii) NH₃ is more basic than PH₃.
(iii) Sulphur has a greater tendency for catenation than oxygen. (Comptt. All India 2014)
Answer:
(a) According to Le Chatelier’s principle the favourable conditions for the production of NH₃ by Haber’s process are

• pressure of 200 atmosphere
• temperature of ~ 700 K
• use of catalyst like Fe₂O₃ with small amount of K₂O and Al₃O₃
  N₂ + 3H₂ ⇌ 2NH₃

(b) (i) H₂S is more acidic than H20 because bond dissociation enthalpy of H – S bond in H₂S is less than that of H – O bond in H₂O.
(ii) Since both P and N contain lone pairs of electrons but due to small size and high electronegativity of Nitrogen in NH₃, the electron density is much higher than PH₂ therefore it can easily donate electrons and acts as strong Lewis base than PH₂.
(iii) The greater catenation tendency of sulphur is due to two reasons :

• The lone pair of electrons feels more repulsion in O-O bond than S-S bond due to its small size and thus S – S forms strong bond.
• As the size of atom increases down the group from O-PO, the strength of bond increases and therefore catenation tendency also increases.

Question 162.
(a) Account for the following :
(i) Acidic character increases from HF to HI.
(ii) There is large difference between the melting and boiling points of oxygen and sulphur.
(iii) Nitrogen does not form pentahalide.
(b) Draw the structures of the following :
(i) ClF₃ (ii) XeF₄ (Delhi, All India 2015)
Answer:
(a) (i) Acidic character increases from HF to HI because bond dissociation enthalpy decreases from HF to HI.
(ii) Oxygen exists as diatomic O₂ molecule while sulphur exists as polyatomic S₈ molecule which has very high molecular mass therefore sulphur has much high melting and boiling points.
(iii) Nitrogen does not contain’d’ orbitals.

(b) (i) ClF₃

Question 163.
(i) Which allotrope of phosphorus is more reactive and why?
(ii) How the supersonic jet aeroplanes are responsible for the depletion of ozone layer?
(iii) F₂ has lower bond dissociation enthalpy than Cl₂. Why?
(iv) Which noble gas is used in filling balloons for meteorological observations?.
(v) Complete the equation :
XeF₂ + PF₅ → (Delhi, All India 2015)
Answer:
(i) White phosphorus is more reactive because it is less stable due to angular strain.
(ii) The oxides of nitrogen released by the exhausts of supersonic jetplanes are causing the depletion of ozone layer.
(iii) F₂ has lower dissociation enthalpy as due to small size of F there is strong repulsive forces between F-F electrons.Bond dissociation enthalpy of F₂ is less than that of Cl₂ because of large electron-electron repulsion among the lone pairs in small size F₂molecule.
(iv) Helium is filled in the balloons for meteorological observations.
(v) XeF₂ + PF₅ → [XeF]⁺ [PF]^–

Question 164.
Account for the following :
(i) Acidic character increases from HF to HI.
(ii) There is large difference between the melting and boiling points of oxygen and sulphur.
(iii) Nitrogen does not form pentahalide.
Answer:
(i) Acidic character increases from HF to HI bond because dissociation enthalpy decreases from HF to HI due to which H+ ion releases easily.
(ii) Oxygen exists as diatomic molecule O₂ while sulphur exists as S₈ molecule which has very high molecular mass therefore sulphur has high BP and MP.
(iii) Nitrogen does not contain’d’ orbitals.

Question 165.
(a) Complete the following chemical equations :
(i) Cu + HNO₃ (dilute) →
(ii) P₄ + NaOH + H₂O →
(b) (i) Why does R₃P = O exist but R₃N = O does not? (R = alkyl group)
(ii) Why is dioxygen a gas but sulphur a solid?
(iii) Why are halogens coloured? (Comptt. Delhi 2015)
Answer:
(a)

(b) (i) Nitrogen in R₃N = O cannot form pπ – dπ multiple bonds because it cannot expand its covalency beyond 4 due to absence of d-orbital, so it will not exist. But R₃P = O can do so due to presence of d-orbitals and formation of pπ – dπ multiple bonds which can expand its covalency up to 5.

(ii) Because of bigger size and the strong forces of attraction holding 8 atoms, their bonds cannot be broken easily and hence sulphur exists as solid while oxygen due to high electro-negativity and tendency to form pπ – dπ multiple bonds through Vander-waals forces of attraction can be broken easily and hence exists as gas.

(iii) All halogens are coloured due to absorption of light in the visible region as a result of which their electrons get excited to higher energy levels and while returning to lower level transmit energy of corresponding colour.

Question 166.
(a) Write balanced equations for the following reactions :
(i) Chlorine reacts with dry slaked lime.
(ii) Carbon reacts with concentrated H₂SO₄
(b) Describe the contact process for the manufacture of sulphuric acid with special reference to the reaction conditions, catalysts used and the yield in the process. (Comptt. Delhi 2015)
Answer:
(a)

(b) Contact process of sulphuric acid :
It involves the following steps :
(i) Formation of sulphur dioxide by burning either sulphur or iron pyrites in excess of air.
S + O₂ → SO₂
4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂
(ii) Catalytic oxidation of SO₂ into SO₃ by using V₂O₅ as catalyst

According to Le-Chatelier’s principle the reaction conditions are :
(a) High concentration of reactants
(b) Low temperature (623-723 K)
(c) High pressure (2 bar)
By obeying above conditions the yield of H₂SO₄ will be 96 – 98%
(iii) Absorption of S03 in 98% H2S04 to give Oleum (H2S207)

(iv) Dilution of Oleum to give sulphuric acid
H₂S₂O₇ + H₂O → 2H₂SO₄

Question 167.
(a) Elements of Gr. 16 generally show lower value of first ionization enthalpy compared to the corresponding periods of Gr. 15. Why?
(b) What happens when
(i) Concentrated H₂SO₄ is added to CaF₂?
(ii) Sulphur dioxide reacts with chlorine in the presence of charcoal?
(iii) Ammonium chloride is treated with Ca(OH)₂? (Comptt. All India 2015)
Answer:
(a) Elements of group 16, i.e., oxygen family have general electronic configuration of ns²np⁴ while elements of group 15, i.e., nitrogen family have general electronic configuration of ns²np³ which is a relatively stable half-filled configuration with high exchange energy and therefore require more ionization energy to release electrons from this stable configuration.
(b)

Question 168.
(a) Draw the structure of the following :
(i) BrF₃ (ii) XeO₃
(b) Answer the following :
(i) Why is NH₂ more basic than PH₃?
(ii) Why are halogens strong oxidising agents?
(iii) Draw the structure of XeOF₄. (Comptt. All India 2015)
Answer:
(a)

(b) (i) Both P and N contains lone pairs of electrons but due to small size and high electronegativity of nitrogen in NH₃, the electron density is much higher than PH₃, therefore it can easily donate electrons and acts as a strong Lewis base than PH₃.
(ii) Halogens have strong tendency to accept an electron due to high negative electron gain enthalpies. Hence halogens act as strong oxidising agents.
(iii) XeOF₄:

Question 169.
Account for the following:
(a) (i) Ozone is thermodynamically unstable.
(ii) Solid PCl₅ is ionic in nature.
(iii) Fluorine forms only one oxoacid HOF.
(b) Draw the structure of (i) BrF₅ (ii) XeF₄ (Delhi 2016)
Answer:
(a) (i) Ozone is thermodynamically very unstable because:
• The decomposition of ozone into oxygen is exothermic in nature. (ΔH = -ve)
• There is also increase in entropy which in turn makes ΔG -ve and reaction spontaneous.
(ii) In solid state, PCl₅ exists as [PCl₃]⁺ [PCl₆]^– in which the cation is tetrahedral and anion is octahedral. Because of the presence of strong attractive forces, it is a solid.
(iii) Due to absence of d-orbitals in fluorine, it can only form one oxoacid i.e., HOF.

(b) (i) Structure of BrF₅

Question 170.
(i) Compare the oxidizing action of F₂ and Cl₂ by considering parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy.
(ii) Write the conditions to maximize the yield of H²SO⁴ by contact process.
(iii) Arrange the following in the increasing order of property mentioned:
(a) H₃PO₃, H₃PO₄, H₃PO₂ (Reducing character)
(b) NH₃, PH₃, AsH₃, SbH₃, BiH₃ (Base strength) (Delhi 2016)
Answer:
(i) Since the standard reduction potential value of fluorine is more than that of chlorine, so fluorine is a stronger oxidising agent.

• Bond dissociation enthalpy of F₂ is less as compared to that of chlorine.
• The negative electron gain enthalpy of fluorine is slightly less than that of chlorine.
• The hydration enthalpy of flouride ion is much higher than that of Cl^– ion due to smaller size.

(ii) Favourable conditions for manufacturing sulphuric acid by contact process are:

• high pressure of about 2 bars
• low temperature 720 K
• presence of V₂O₅ catalyst

(iii) (a) H₃PO₄ < H₃PO₃ < H₃PO₂ (Reducing character)
(b) BiH₃ < SbH₃ < AsH₃ < PH₃ < NH₃ (Basic strength)

Question 171.
(a) What happens when :
(i) HCl reacts with finely powdered iron.
(ii) Cl₃ reacts with hot concentrated solution of NaOH.
(b) Give appropriate reason for each of the following :
(i) Sulphur vapour exhibits some paramagnetic character.
(ii) NH³ is more basic than PH³
(iii) Dioxygen is a gas but sulphur a solid. (Comptt. Delhi 2016)
Answer:
(a) (i) It forms FeCl₂ and H₂ gas

(b) (i) Sulphur above 1000 K exists partially as S₂ molecule which has two unpaired
electrons in its Anti Bonding orbitals.
(ii) Due to small size of N, NH₃ molecule has small surface area and high electron density in comparison to PH³. Hence it can donate lone pair of electron more readily.
(iii) In dioxygen there is P_π – P_π multiple bonding and no cation hence it is a gas.

Question 172.
(a) Complete the following equations :

(b) Explain giving reason in each case :
(i) Halogens are strong oxidising agents
(ii) Fluorine form only one oxoacid, HOF
(iii) Helium is used in diving apparatus (Comptt. Delhi 2016)
Answer:
(a)

(b) (i) Due to their high electron affinity. As they have seven electrons, they accept one electron readily.
(ii) Due to high electronegativity and small size of fluorine.
(iii) Due to low solubility in blood.

Question 173.
Draw the structure of
(i) BrF₃ (ii) XeOF₄
Explain giving reason in each case :
(i) Why H₂Te is more acidic than H₂S?
(ii) Why are halogens strong oxidising agents?
(iii) Why does nitrogen show catenation tendency less than phosphorus? (Comptt. All India 2016)
Answer:
(a)

(b) (i) Because of low bond dissociation enthalpy of H-Te bond as Te has larger size than S.
(ii) Because of their strong electron affinity they have seven valence electrons.
(iii0 Because N-N single bond is weaker than P-P single bond.

Question 174.
(i) Why PCl₅ gives fumes in moisture?
(ii) Why interhalogens are more reactive than pure halogens?
(b) Draw the structures of the following :
(i) PCl_5(s) (ii)H₂S₂O₈ (iii) XeF₄ (Comptt. All India 2016)
Answer:
(a) (i) Because it reacts with moisture present in the air giving HCl.
(ii) Because of low bond dissociation enthalpy of interhalogens.
(b) (i) PCl₅(s)

Question 175.
(a) When concentrated sulphuric acid was added to an unknown salt present in a test tube a brown gas (A) was evolved. This gas intensified when copper turnings were added to this test tube. On cooling the gas (A) changed into a colourless solid (B).
Identify (A) and (B). Write chemical reactions involved.
(b) Draw structure of XeOF₄. (Comptt. All India 2017)
Answer:
(a) The salt is sodium nitrate which on heating with cone. H₂SO₄ evolves a brown gas i.e. NO₂ which gets intensified when Cu turnings are added.

On cooling, NO₂ condenses as a brown liquid which turns paler on cooling and eventually becomes a colourless solid due to formation of dimerized NO₂ i.e. N₂O₄ (Dinitrogen tetraoxide).
A: Nitrogen dioxide (NO₂)
B : Dinitrogen tetraoxide (Na₂O₄)
(b) Structure of XeOF₄:

Question 176.
(a) Account for the following :
(i) Reducing character decreases from SO₂ to TeO₂.
(ii) HClO₃ is a stronger acid than HCIO.
(iii) Xenon forms compounds with fluorine and oxygen only.
(b) Complete the following equations :
(i) 4NaCl + MnO₂ + 4H₂SO₄ →
(ii) 6XeF₄ + 12H₂O → (Comptt. All India 2017)
Answer:
(a) (i) Reducing character decreases from SO₂ to TeO₂ because the pπ – pπ bonds in them become weaker with increase in size and bond length along the group.
(ii) HClO₃ is a stronger acid than HClO because with increase in oxidation state and oxidation number the acidic character increases i.e. HClO₃ (+5) and HCIO (+1)
(iii) Xenon forms compounds with fluorine and oxygen only due to their high electronegativity and reactivity. The first ionisation energy of it is fairly close to that of O₂ and F₂.

(b) (i) 4NaCl + MnO₂ + 4H₂SO₄ → 4NaHSO₄ + MnCl₂ + 2H₂O + Cl₂
(ii) 6XeF₄ + 12H₂O → 4Xe + 2XeO₃ + 24HF + 3O₂

Important Questions for Class 12 Chemistry