The p – Block Elements  Class 12 Chemistry Important Questions

Please refer to The p – Block Elements  Class 12 Chemistry Important Questions with solutions provided below. These questions and answers have been provided for Class 12 Chemistry based on the latest syllabus and examination guidelines issued by CBSE, NCERT, and KVS. Students should learn these problem solutions as it will help them to gain more marks in examinations. We have provided Important Questions for Class 12 Chemistry for all chapters in your book. These Board exam questions have been designed by expert teachers of Standard 12.

Class 12 Chemistry Important Questions The p – Block Elements 

Very Short Answer Questions

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Question. Why is dinitrogen very unreactive as compared to phosphorus?
Answer. Bond dissociation enthalpy of N₂ molecule is very high (941.4 kJ mol^–1) and so is least reactive at normal temperature and remains inert even in atmosphere.
Yellow phosphorus exists as P₄ molecule with a tetrahedral shape. The P – P – P bond angle is about 60° because of which the molecule is under tremendous strain and so phosphorus is reactive.

Question. Give reasons for the following :
N₂ is less reactive at room temperature.
Answer. In N₂ molecule N atoms are held by triple bonds.
It has very high bond dissociation energy (941.4 kJ mol_–1). Therefore N₂ is inert at room temperature.

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

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

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

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

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

Question. Complete the following chemical reaction equations ; 
Cu + HNO₃(dilute)
Answer. 3 Cu + 8HNO3 _{(dilute) →} 3Cu(NO₃)₂ + 2NO + 4H₂O

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

Question. Explain the following situations :
In the structure of HNO₃ molecule, the N O bond (121 pm) is shorter than N OH bond (140 pm).
Answer. HNO₃ is supposed to exist in two resonating forms. D ue to two resonating structures, N — O bond is shorter than N—OH bond

Question. Which allotrope of phosphorus is more reactive and why? 
Answer. White phosphorus is most reactive of all the allotropes because it is unstable due to the angular strain on P₄ molecule with bond angle of 60°.

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

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

Question. Draw the structure of the following :
Red P₄
Answer. 

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

Question. Account for the following :
There is large difference between the melting and boiling points of oxygen and sulphur.
Answer. Oxygen molecules are held together by weak van der Waals’ forces because of the small size and high electronegativity of oxygen. Sulphur shows catenation and the molecule is made up of 8 atoms with strong intermolecular forces. Hence, there is large difference in the melting and boiling points of oxygen and sulphur

Question. Give reasons for the following :
H₂Te is the strongest reducing agent amongst all the hydrides of group 16 elements.
Answer. Bond dissociation enthalpy and hence thermal stability of hydrides decreases from H₂O to H₂Te. Therefore, H_2Te releases hydrogen readily. Hence, H₂Te is the strongest reducing agent amongst all the hydrides of group 16.

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

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

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

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

Question. Arrange the following group of substances in the order of the property indicated against the group :
O, S, Se, Te – increasing order of electron gain enthalpy with negative sign.
Answer. Electron gain enthalpy of oxygen is less negative than sulphur due to compact size of oxygen atom (inter-electronic repulsion is more in O). From sulphur onwards enthalpy again becomes less negative upto
Po. O < S > Se > Te > Po

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

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

Question. Account for the following :
Boiling point of water is much higher than that of hydrogen sulphide.
Answer.     H2O               H2S
Boiling point 373 K   >    213 K
The abnormally high boiling point of H2O is due to strong intermolecular H-bonding. Since, all other elements have much lower electronegativity than oxygen, they do not undergo H-bonding.

Question. Account for the following :
Thermal stability of water is much higher than that of H₂S.
Answer. The thermal stability of the hydrides decrease from H₂O to H₂Te. This is because as the size of central element increases, the bond E—H become weaker and thus breaks on heating.

Question. Explain the following situations :
SF₄ is easily hydrolysed whereas SF₆ is not easily hydrolysed.
Answer. In SF₆, six F atoms protect the sulphur atom from attack by the reagent to such an extent that even thermodynamically most favourable reactions like hydrolysis do not occur. But in S_F4, S is not sterically protected as it is surrounded by only four F atoms hence, SF4 is reactive.

Question. Give reasons for the following :
OF₆ compound is not known.
Answer. OF₆ compound is not known because oxygen cannot expand its octet due to unavailability of d-orbital.

Question. Draw the structures of the following molecule :
SF₄
Answer. 

SF₄ is a gas with sp³d-hybridisation and have trigonal bipyramidal geometry and see-saw structure due to the presence of lone pair of electrons in equatorial positions.

Question. Draw the structure of the following :
H₂S₂O₈
Answer. 

Question. Draw the structure of the following molecule :
H₂S₂O₇ 
Answer.

Question. Write the structure of the following species :
H₂SO₅.
Answer. 

Question. Write the conditions to maximize the yield of H₂SO₄ by Contact process.
Answer. In Contact process, the rate determining step is

This reaction is reversible and exothermic i.e., ΔH is negative. Thus, according to Le-Chatelier’s principle, the conditions to maximise the yield are as follows :
(a) At lower temperature : As heat is evolved in the reaction so, at lower temperature the reaction proceeds more in forward direction.
(b) At higher pressure : As three moles of gaseous reactants give two moles of gaseous products thus, at higher pressure reaction moves in forward direction

Question. Complete the following equation :
Cu + conc. H₂SO₄4 →
Answer.

Question. Complete the following equation :
CaF₂ + H₂SO₄ →
Answer. CaF₂ + H₂SO₄ → CaSO₄ + 2HF

Question. Complete the following equation :
C + conc. H₂SO₄ →
Answer. C + 2H₂SO₄ (conc.) → CO_2(g) + 2H2O_(l) + 2SO_2(g)

Question. Why is K_a2 << K_a1 for H₂SO₄ in water?
Answer. H₂SO_4(aq) + H2O_(l) → H₃O⁺_(aq) + HSO^–_4(aq) K_a1 > 10, very large
HSO^–_4(aq) + H₂O_(l) → H₃O⁺_(aq) + SO^2–_4(aq)K_a2 = 1.2 × 10^–2
K_a2 is smaller thanKa1 because dissociation of HSO^–₄ is less probable due to presence of negative charge on the ion

Question. Account for the following :
Concentrated sulphuric acid has charring action on carbohydrates.
Answer. Concentrated H₂SO₄ removes water from organic compounds hence, it has charring action on carbohydrates.

Question. Complete the following equation :
SO₃ + H₂SO₄
Answer. SO₃ + H₂SO₄ → H₂S₂O₇

Question. Account for the following :
Acidic character increases from HF to HI.
Answer. The acidic strength of the hydrohalic acids in the order :
HF < HCl < HBr < HI
This order is a result of bond dissociation enthalpies of H — X bond decreases from H — F to H — I as the size of halogen atom increases

Question. F₂ has lower bond dissociation enthalpy than Cl₂. Why? 
Answer. F₂ has lower bond dissociation enthalpy than Cl₂ because F atom is very small and hence the electron-electron repulsions between the lone pairs of electrons are very large.

175. Why are halogens coloured?
Answer. Halogens absorb radiations in visible region which results in excitation of outer electrons to higher level resulting in different colours.

Question. Answer the following :
Why are halogens strong oxidising agents?
Answer. General electronic configuration of halogens is ns²np⁵. They easily accept one electron to complete their octet. This makes them a good oxidising agent.

Question. Arrange the following in the order of property indicated against each set :
HF, HCl, HBr, HI – increasing bond dissociation enthalpy 
Answer. HI < HBr < HCl < HF

Question. Arrange the following groups of substances in the order of the property indicated against each group :
F₂, Cl₂, Br₂, I₂ – increasing order of bond dissociation enthalpy.
Answer. Increasing bond dissociation enthalpy order is I₂ < F₂ < Br₂ < Cl₂ Bond dissociation enthalpy of F₂ is less than that of Br₂ and Cl₂ due to the lone pair – lone pair repulsions.

Question. Why is F₂ a stronger oxidising agent than Cl₂?
Answer. Fluorine is the strongest oxidising agent as it accept electron easily. It oxidise other halide ions in solution or even in solid phase.
F₂ + 2X– → 2F– + X₂ (X Cl, Br or I)
Cl₂ + 2X^– → 2Cl– + X₂ (X Br or I)
Br₂ + 2I^– → 2Br– + I₂

Question. Assign reasons for the following :
HCl is a stronger acid than HF though fluorine is more electronegative than chlorine.
Answer. HF is the weakest acid because of its high bond dissociation energy due to small size of fluorine atom.

Question. Account for the following :
Fluorine does not exhibit positive oxidation state.
Answer. Since, fluorine is the most electronegative element, it shows only a negative oxidation state of –1, and does not show any positive oxidation state.

Question. Fluorine exhibits only –1 oxidation state whereas other halogens exhibit +1, +3, +5 and +7 oxidation state also. Why is it so?
Answer. This is due to non-availability of d-orbitals in valence shell of fluorine

Question. Account for the following :
HF is not stored in glass bottles but is kept in wax-coated bottles.
Answer. HF acid attacks glass with the formation of fluoro silicate ions. Thus it is stored in wax-coated glass bottles to prevent the reaction.

Question. How are interhalogen compounds formulated and how are they prepared?
Answer. Interhalogen compounds are formulated as XX′, XX′₃, XX′₅ and XX′₇ where X is halogen of larger size and X′ of smaller size. The interhalogen compounds can be prepared by direct combination or by the action of halogen on lower interhalogen compounds. The product formed depends upon some specific conditions, for example,

Question. Draw the structure of XeF₄.
Answer. 

Question. What happens when XeF₄ reacts with SbF₅?
Answer. XeF₄ + SbF₅ → [XeF₃]+ [SbF₆]^–

Question. Write the structures of the following molecule :
XeOF₄. 
Answer. XeOF₄ is square pyramidal

Question. Account for the following :
Unlike xenon, no distinct chemical compound of helium is known.
Answer. Extremely small size and fully filled outer orbital makes helium very stable and reistant to chemical reaction and hence, it does not form compounds unlike bigger atoms of other elements of noble gas family.

Question. Write the balanced chemical equation for the reaction of Cl₂ with hot and conc. NaOH solution. Justify that this reaction is a disproportionation reaction.
Answer. 3Cl₂ + 6NaOH → 5NaCl + NaClO₃ + 3H₂O This reaction is a disproportionation reaction as chlorine from zero oxidation state is changed to – 1 and + 5 oxidation states.

Question. Iron dissolves in HCl to form FeCl₂ and not FeCl₃
Answer. Its reaction with iron produces H₂.
Fe + 2HCl → FeCl₂ + H₂
Liberation of hydrogen prevents the formation of ferric chloride.

Question. Account for the following :
Fluorine forms only one oxoacid HOF.
Answer. Fluorine forms only one oxoacid HOF. Because for the formation of other oxoacids d orbitals are required for the multiple p𝝅 – d𝝅 bonding between extra oxygen atoms and fluorine. High electronegativity and small size of fluorine also favours only the formation of one oxoacid.

Question. Complete the following chemical reaction equations :
I^–_(aq) + H₂O_(l) + O_3(g) →
Answer. 2I^–_(aq) + H2O_(l) + O_3(g) → 2OH^–_(aq) + I2(s) + O_2(g)

Question. Which allotrope of sulphur is thermally stable at room temperature? 
Answer. Rhombic sulphur

Question. Account for the following :
Sulphur in vapour form exhibits paramagnetic behaviour.
Answer. At elevated temperature, sulphur vapour exists as S₂ molecules which are paramagnetic like O₂.

Question. Draw the structure of the following : S₈
Answer. 

Question. What inspired N. Bartlett for carrying out reaction between Xe and PtF₆?
Answer. Neil Bartlett first prepared a red compound which is formulated as O₂⁺PtF₆^–. He then realised that the first ionisation enthalpy of molecular oxygen (1175 kJ/mol) is almost identical with Xe (1170 kJ/mol). He made efforts to prepare same type of compound with Xe and was successful in preparing another red compound Xe⁺PtF₆^–

Question. Despite having greater polarity, hydrogen fluoride boils at a lower temperature than water.
Answer. Higher boiling point of H₂O is due to the extensive H-bonding than HF.

Question. Explain giving reason for the following situation.
In aqueous medium HCl is stronger acid than HF.
Answer. In aqueous medium HCl is stronger acid than HF because bond dissociation enthalpy of H—Cl is lower than that of HF

Question. Answer the following :
Why does fluorine not play the role of a central atom in interhalogen compounds?
Answer. Fluorine does not have d-orbitals and its cannot show higher oxidation state. Therefore it does not play the role of a central atom in inter halogen compounds.

Question. Assign reasons for the following :
Of the noble gases only xenon is known to form well-established chemical compounds.
Answer. Except radon which is radioactive, Xe has least ionisation energy among noble gases and hence it readily forms chemical compounds particularly with oxygen and fluorine.

Question. Complete the following chemical reaction equation :
XeF₄ + H₂O →
Answer. 6XeF₄ + 12H₂O → 2XeO₃ + 24HF +3O₂ + 4Xe

Question. Draw the structure of : BrF₅
Answer. Geometry – Octahedral
Shape – Square pyramidal
Hybridisation – sp^3d2

Question. Give reasons : ICl is more reactive than I₂.
Answer. Interhalogen compounds are more reactive than halogens (except flourine) because X — X′ bond
(I—Cl bond) in interhalogens is weaker than
X—X bond (I—I bond) in halogens except
F — F bond. In other words I —Cl bond is weaker than
I — I bond. That’s why ICl is more reactive than I₂.

Question. Draw the structure of the following :
ClF₃
Answer. Hybridisation – sp³d
Structure – Trigonal bipyramidal
Shape – Bent (T-shaped)
Angle F–Cl–F : less than 90°

Question. Draw the structures of the following :
BrF₃
Answer. Hybridisation – sp³d
Structure – Trigonal bipyramidal
Shape – Bent -T

Question. Write the formula and describe the structure of noble gas species which is isostructural with BrO^–₃ 
Answer. The central atom Br has seven electrons. Four of these electrons form two double bonds or coordinate bonds with two oxygen atoms while the fifth electron forms a single bond with O^– ion. The remaining two electrons form one lone pair. Hence, in all there are three bond pairs and one lone pair around Br atom in BrO₃^–. Therefore, according to VSEPR theory, BrO₃^– should be pyramidal. Here, BrO₃^– has 26(7 + 3 × 6 + 1 = 26) valence electrons. A noble gas species having 26 valence electrons is XeO₃(8 + 3 × 6 = 26). Thus, like BrO₃^– ,XeO₃ is also pyramidal.

Question. Complete the following chemical equation :

Answer. 

Question. F₂ is more reactive than ClF₃ but ClF₃ is more reactive than Cl₂.
Answer. Interhalogen compounds are more reactive than halogen because the X′ — X bond in interhalogens is weaker than X—X bond in halogens. But in case of flourine, the F—F bond is weaker. This is because of the small size of fluorine atoms, the F—F bond distance is very small due to which there is appreciable inter-electronic repulsion. This repulsion weakens the bond between two fluorine atoms. Hence, F₂ is more reactive than ClF₃ but ClF₃ is more reacitve than Chlorine.

Question. Account for the following :
BrCl₃ is more stable than BrCl₅.
Answer. O.S. of Br in BrCl5 is +5, whereas in case of BrCl₃ is +3. As Br is more stable in +3 oxidation state than +5, due to inert pair effect. Therefore, it is unstable
and readily reduces from +5 to +3 oxidation state

Question. Why do some noble gases form compounds with fluorine and oxygen only?
Answer. Fluorine and oxygen are the most electronegative elements and hence are very reactive. Therefore, they form compounds with noble gases particularly with xenon.

Question. List the uses of neon and argon gases.
Answer. Neon is used in discharge tubes and fluorescent bulbs for advertisement display purposes. Argon is used to provide an inert atmosphere in high temperature metallurgical processes and for filling electric bulbs.

Question. Account for the following :
Electron gain enthalpy with negative sign for fluorine is less than that for chlorine.
Answer. The electron gain enthalpy of fluorine is less negative than that of chlorine due to the small size of fluorine atom.

Question. Explain the following observation :
Hydrogen fluoride has a much higher boiling point than hydrogen chloride.
Answer. Hydrogen uoride has much higher boiling point (b.p. 293 K) than hydrogen chloride (b.p. 189 K) due to strong hydrogen bonding.

Question. Suggest a possible reason for the following observations :
Fluorine forms the largest number of interhalogen compounds amongst the halogens.
Answer. Fuorine is the most electronegative element among halogens and it cannot exhibit any positive oxidation state. Therefore, it form largest number of interhalogen compounds.

Question. Complete the following chemical equation :
F_2(g) + H₂O_(l) → ……….
Answer. 2F_2(g) + 2H₂O_(l) → 4H⁺_(aq.) + 4F^–_(aq.) + O_2(g)

Short Answer Questions

Question. Compare the oxidizing action of F₂ and Cl₂ by considering parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy.
Answer. Oxidising power of a substance depends on the factors like bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy. Due to small size of fluorine, its electron gain enthalpy is less than that of chlorine. However, its low bond dissociation enthalpy and high hydration enthalpy compensate the low electron gain enthalpy. Fluorine because of its small size has higher hydration enthalpy than chlorine. Also, due to repulsion between electrons it has lower bond dissociation energy. Thus, fluorine has better oxidising action than chlorine

Question. Draw the structure of
(i) Hypochlorous acid.
(ii) Chlorous acid.
Answer. 

Question. How is XeO₃ obtained? Write the related chemical equations. Draw the structure of XeO₃.
Answer. XeO₃ can be obtained by hydrolysis of XeF₄ and XeF₆.
6XeF₄ + 12H₂O → 4Xe + 2XeO₃ + 24 HF + 3O₂.
XeF₆ + 3H₂O → XeO₃ + 6HF

Question. Write the formulae and the structure of noble gas species which are isostructural with
(i) ICl^–₄ (ii) BrO^–₃
Answer. 

(i) Structure of ICl₄^– : I in ICl₄^– has four bond pairs and two lone pairs. Therefore, according to VSEPR theory, it should be square planar as shown. Here, ICl₄^– has (7 + 4 × 7 + 1) = 36 valence electrons. A noble gas species having 36 valence electrons is XeF₄ (8 + 4 × 7 = 36). Therefore, like ICl₄^–, XeF₄ is
also square planar.

(ii) Structure of BrO₃^– : The central atom Br has seven electrons. Four of these electrons form two double bonds or coordinate bonds with two oxygen atoms while the fifth electron forms a single bond with O^– ion. The remaining two electrons form one lone pair. Hence, in all there are three bond pairs and one lone pair around Br atom in BrO₃^–.Therefore, according to VSEPR theory, BrO₃^– should be pyramidal.

Here, BrO₃^– has 26 (7 + 3 × 6 + 1 = 26) valence electrons. A noble gas species having 26 valence electrons is XeO₃ (8 + 3 × 6 = 26). Thus, like Br_O3^–, XeO₃ is also pyramidal

Long Answer Questions

Question. (a) How is sulphur dioxide prepared in :
(i) Laboratory
(ii) Industrially?
(b) What happens when sulphur dioxide is passed through water and reacts with sodium hydroxide? Write balanced equation.
(c) Write its any two uses.
Answer. (a) (i) It is prepared in laboratory by treating a sulphite with dil. H₂SO₄.
SO^2–_3(aq) + 2H⁺_(aq) → H2O_(l) + SO_2(g)
(ii) It is produced industrially as a by – product of the roasting of sulphide ores.
4FeS_2(s) + 11O_2(g) → 2Fe₂O_3(s) + 8SO_2(g)
(b) When sulphur dioxide is passed through water, it forms a solution of sulphurous acid.

When sulphur dioxide reacts with sodium hydroxide solution sodium, sulphite is formed.
2NaOH + SO₂ → Na₂SO₃ + H₂O
                        Sod. sulphite
(c) Uses : (i) In rening sugar and petroleum.
(ii) In bleaching wool and silk.
(iii) As an anti – chlor, disinfectant and preservative.
(iv) In the manufacture of sulphuric acid, sodium hydrogen sulphite and calcium hydrogen sulphite.