NCERT Solutions for Class 12 Chemistry Chapter 7 The p-block elements

Question 7.2 Why is $Bi{H}_3$ the strongest reducing agent amongst all the hydrides of Group 15 elements ?

Answer :

We see that the stability of hydrides becomes lesser as we go from $N{H}_3$ to $Bi{H}_3$ .this can be seen from dissociation enthalpy of their bond. due to that, the reducing character of the hydrides will be more. Ammonia is only a very mild reducing agent while $Bi{H}_3$ is the strongest reducing agent amongst all of the hydrides.

Question 7.3 Why is $N_2$ less reactive at room temperature?

Answer :

$N_2$ reacts poorly at room temperature. high bond enthalpy of N≡N bond is the reason behind this. Reactivity, however, increases rapidly with increase in temperature.

Question 7.4 Mention the conditions required to maximise the yield of ammonia.

Answer :

Ammonia is produced by Haber's process-

$N_2+3H_2\stackrel{700k}{\leftrightharpoons }2N{H}_3\mathrm{\Delta }{H}^o=-46.1KJmo{l}^{-1}$

The maximum yield conditions for the production of ammonia are-

1. pressure = $200atmor200\times {10}^{5}Pa$ ,
2. Temperature of around 700 K
3. catalyst = Iron oxide
4. Promotor= small amounts of $K_{2}O$ and $A{l}_2{O}_3$ to increase the rate of attainment of equilibrium.

Question 7.5 How does ammonia react with a solution of $C{u}^{2+}$ ?

Answer :

Ammonia is a Lewis base due to The presence of a lone pair of electrons on the nitrogen atom of the ammonia molecule. It forms a linkage with metal ions by donating the electron pair.

$C{u}^{2+}(aq)(blue)+4N{H}_3\to [Cu(N{H}_3{)}_4{]}^{2+}(aq)(deepblue)$

Question 7.6 What is the covalence of nitrogen in $N_2{O}_5$ ?

Answer :

We can see From the structure of $N_2{O}_5$ that covalence of nitrogen is four.

Question 7.7 Bond angle in $P{H}_4^{+}$ is higher than that in $P{H}_3$ . Why?

Answer :

As we can see Both are $s{p}^3$ hybridised. In $P{H}_4^{+}$ all of the 4 orbitals are bonded whereas in $P{H}_3$ there is a lone pair of electrons on P.this lone pair is responsible for lone pair-bond pair repulsion in $P{H}_3$ , which results in reducing the bond angle to less than 109° 28′.

Question 7.7 What is formed when $P{H}_3$ reacts with an acid?

Answer :

$P{H}_3$ reacts with acids like $HI$ to form $P{H}_{4}I$ which shows that it is basic in nature. Because of lone pair on phosphorus atom, $P{H}_3$ is acting as a Lewis base in the above reaction

$P{H}_3+HI\to P{H}_{4}I$

$P{H}_3+Acid\to salt$

Question 7.8 What happens when white phosphorus is heated with concentrated $NaOH$ solution in an inert atmosphere of $C{O}_2$ ?

Answer :

When white phosphorus is heated with the concentrated $NaOH$ solution in an inert atmosphere of $C{O}_2$ we see that phosphine and sodium hypophosphite is formed.

$P_4+3NaOH+H_{2}O\to 3Na{H}_{2}P{O}_2+P{H}_3$

Question 7.9 What happens when $PC{I}_5$ is heated?

Answer :

When we heat $PC{I}_5$ , it sublimes but decomposes on stronger heating and phosphorus trichloride is formed.

$PC{I}_5$ + Heat $\to$ $PC{I}_3$ + $C{l}_2$

Question 7.10 Write a balanced equation for the reaction of $PC{I}_5$ with heavy water.

Answer :

$$\begin{gathered} StepI:PC{l}_5+D_{2}O\to POC{l}_3+2DCl \\ StepII:POC{l}_3+3D_{2}O\to D_{3}P{O}_4+3DCl \end{gathered}$$

$OverallreactionPC{l}_5+4D_{2}O\to D_{3}P{O}_4+5DCl$

Question 7.11 What is the basicity of $H_{3}P{O}_4$ ?

Answer :

There are three P–OH bonds present in the molecule of $H_{3}P{O}_4$ . Hence, its basicity is three.

$H_{3}P{O}_4\stackrel{H_{2}O}{\rightleftharpoons }3H^{+}+P{O}_4^{3-}$

Question 7.12 What happens when $H_{3}P{O}_3$ is heated?

Answer :

$H_{3}P{O}_4$ when heated, it will disproportionates and give orthophosphoric acid (or phosphoric acid) and phosphine.

$4H_{3}P{O}_3\to 3H_{3}P{O}_4+P{H}_3$

Question 7.13 List the important sources of sulphur.

Answer :

• The presence of sulphur in the earth’s crust is only about 0.03-0.1%.
• mixed sulphur exists primarily as sulphates such as gypsum $CaS{O}_4.2H_{2}O$ , Epsom salt $MgS{O}_4.7H_{2}O$ , baryte $BaS{O}_4$ .
• another source is by sulphides such as galena $PbS$ , zinc blende $ZnS$ , copper pyrites $CuFe{S}_2$ . some sulphur also occurs as hydrogen sulphide in volcanoes. eggs, proteins, garlic, onion, mustard, hair and wool also contain sulphur.

Question 7.14 Write the order of thermal stability of the hydrides of Group 16 elements.

Answer :

Hydrides of group 16 elements become less thermally stable as we go down the group, i.e., $H_{2}O$ > $H_{2}S$ > $H_{2}Se$ > $H_{2}Te$ > $H_{2}Po$ . This is due to M-H bond dissociation energy decreases down the group as we increase in the size of the atom.

Question 7.15 Why is $H_{2}O$ a liquid and $H_{2}S$ a gas ?

Answer :

$H_{2}O$ has oxygen as the centre atom. oxygen is small in size as well as high electronegative when we compare it with sulpher.molecules of water are highly associated through hydrogen bonding which is not present in sulpher.molecules of $H_{2}S$ are connected to each other through weak van der Wal force only. Hence a $H_{2}O$ liquid and $H_{2}S$ a gas.

Question 7.16 Which of the following does not react with oxygen directly? Zn, Ti, Pt, Fe

Answer :

Since Platinum(Pt) is a noble metal .it will not react with oxygen directly

Question 7.17 Complete the following reactions:

(i) $C_2{H}_4+O_2\to$

Answer :

The reaction is:

$C_2{H}_4+3O_2\to 2C{O}_2+2H_{2}O$

Question 7.17 Complete the following reactions:

(ii) $4AI+3O_2\to$

Answer :

The complete reaction is:

$4AI+3O_2\to$ $2A{l}_2{O}_3$

Question 7.18 Why does $O_3$ act as a powerful oxidising agent?

Answer :

$O_3$ act as a powerful oxidising agent .This is because of the ease with which it frees atoms of nascent oxygen.i.e.

$O_3\to O_2+O$

Question 7.19 How is $O_3$ estimated quantitatively?

Answer :

A quantitative method for estimating O3 gas is:

When we reacts $O_3$ with an excess of potassium iodide solution which is buffered with a borate buffer (pH 9.2), iodine is released which can be then titrated against a standard solution of sodium thiosulphate.

$2I^{-}+H_{2}O+O_3\to 2O{H}^{-}+I_2+O_2$

we use starch as an indicator when $I_2$ liberated is titrated against a standard solution of sodium thiosulphate.

Question 7.20 What happens when sulphur dioxide is passed through an aqueous solution of Fe(III) salt?

Answer :

When we pass sulphur dioxide through an aqueous solution of Fe(III) salt, it converts iron(III) ions to iron(II) ions.

$2F{e}^{3+}+S{O}_2+2H_{2}0\to 2F{e}^{2+}+S{O}_4^{2-}+4H^{+}$

Question 7.21 Comment on the nature of two $S-O$ bonds formed in $S{O}_2$ molecule. Are the two $S-O$ bonds in this molecule equal ?

Answer :

The two $S-O$ bonds in $S{O}_2$ molecule are covalent and have equal strength because of having resonating structures.

Question 7.22 How is the presence of $S{O}_2$ detected ?

Answer :

Presence of sulphur dioxide is measured by the following reaction. it decolourises acidified potassium permanganate(VII) solution.

$5S{O}_2+2Mn{O}_4^{-}+2H_{2}O\to 5S{O}_4^{2-}+4H^{+}+2M{n}^{2+}$ .

it decolourises acidified potassium permanganate(VII) solution.

Hence This can be used to detect the presence of $S{O}_2$ .

Question 7.23 Mention three areas in which $H_{2}S{O}_4$ plays an important role.

Answer :

1. Manufacture of fertilisers (e.g., ammonium sulphate, superphosphate) from $H_{2}S{O}_4$ .
2. Use is petroleum refining
3. Manufacture of pigments, paints and dyestuff intermediates and detergent industry.

Question 7.24 Write the conditions to maximise the yield of $H_{2}S{O}_4$ by Contact process.

Answer :

Contact process which we use to create sulphuric acid is exothermic, reversible and the forward reaction which leads to a decrease in volume. hence, low temperature and high pressure are the optimum conditions for maximum yield. But if the temperature will be very low then the rate of reaction will become slow. Also, the presence of catalyst $V_2{O}_5$ fastens the reaction.

Question 7.25 Why is for $H_{2}S{O}_4$ in water ?

Answer :

$H_{2}S{O}_4$ is a very strong acid in water mostly due to its first ionisation to $H_3{O}^{+}$ and $HS{O}_4^{-}$ .The ionisation of $HS{O}_4^{-}$ to $H_3{O}^{+}$ and $S{O}_4^{2-}$ is minuscule. That is the reason why Ka₂ << Ka₁.

Question 7.26 Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidising power of $F_2$ and $C{I}_2$ .

Answer :

$F_2$ is much more powerful in oxidising, than $C{l}_2$ .The reason being, hydration enthalpy of F– ions (515 kJ mol^–1) is much higher than that of Cl– ion (381 kJ mol^–1). the dissociation energy of bond F-F is less than Cl-Cl bond but The former factor more than compensate the less negative electron gain enthalpy of F2. Hence it is a much stronger oxidising agent.

Question 7.27 Give two examples to show the anomalous behaviour of fluorine.

Answer :

We see anomalous behaviour of fluorine and this is because of its small size, highest electronegativity, very low F-F bond dissociation enthalpy, and non-availability of d orbitals in the valence shell.

1. ionisation enthalpy, electronegativity, and electrode potentials are all higher for fluorine than expected from the trends set by other halogens

2.ionic and covalent radii, melting point and boiling point., enthalpy for bond dissociation and electron gain enthalpy are very much lower than expected

3. Fluorine shows only an oxidation state of –1 due to non-availability of d-orbitals in its valence shell.

Question 7.28 Sea is the greatest source of some halogens. Comment.

Answer :

The water of the sea contains bromides, chlorides, and iodides of sodium, magnesium, potassium and calcium, but mainly solution of sodium chloride (2.5% by mass). The deposits of dried up seas have these compounds in it, e.g., sodium chloride and carnallite, KCl.MgCl2 .6H2O. iodine is also formed in Certain forms of marine life in their systems; many seaweeds, for example, contain up to 0.5% of iodine. Chile saltpeter contains up to 0.2% of sodium iodate. That's why the sea is the greatest source of halogens.

Question 7.29 Give the reason for bleaching action of $C{I}_2$ .

Answer :

Chlorine is a powerful bleaching agent and its bleaching action happens due to oxidation.

$C{l}_2+H_{2}O\to 2HCl+\left[O\right]$

Chlorine + Water $\to$ Hydrochloric acid + nascent Oxygen

Coloured substance + nascent Oxygen $\left[O\right]$ $\to$ Colourless substance

Question 7.30 Name two poisonous gases which can be prepared from chlorine gas.

Answer :

The poisonous gases which we can be prepared from chlorine are

1. phosgene ( $COC{l}_2$ )
2. mustard gas $ClC{H}_{2}C{H}_{2}SC{H}_{2}C{H}_{2}CL$ .

Question 7.31 Why is ICl more reactive than $I_2$ ?

Answer :

ICl is more reactive than $I_2$ . this is because interhalogen compounds are more reactive than halogens (except fluorine). This is because X–X′ bond in interhalogens is weaker than X–X bond in halogens except F–F bond.

Question 7.32 Why is helium used in diving apparatus?

Answer :

Helium is used in diving apparatus because it is very low soluble in blood.

Question 7.33 Balance the following equation:

$Xe{F}_6+H_{2}O\to Xe{O}_2{F}_2+HF$

Answer :

The balanced reaction is:

$Xe{F}_6+2H_{2}O\to Xe{O}_2{F}_2+4HF$

Question 7.34 Why has it been difficult to study the chemistry of radon?

Answer :

It has been difficult to study the chemistry of radon because radon is a radioactive element and it has a short half-life.

NCERT Solutions for Class 12 Chemistry Chapter 7 The p-block elements- Exercise Questions

Question 7.1 Discuss the general characteristics of Group 15 elements with reference to their electronic configuration, oxidation state, atomic size, ionisation enthalpy and electronegativity

Answer :

Since all the elements in group 15 have 5 valence electrons, Electronic configuration of group 15 element is ns2np3 where n= 2 to 6. All element requires three more electrons to complete their octets. However, gaining electrons is very difficult as the nucleus will have to attract three more electrons. This can take place only with nitrogen as it is the smallest in size and the distance between the nucleus and the valence shell is relatively small. The rest elements of this group show a formal oxidation state of -3 in their covalent compounds. N and P also show -1 and -2 oxidation states In addition to the -3 state. every element which is present in this group shows +3 and +5 oxidation states. whereas, the stability of the +5 oxidation state decreases as we go down a group, whereas the stability of +3 oxidation state increases. This happens because of the inert pair effect.

First ionization decreases on moving down a group. This is because of increasing atomic sizes. As we move down a group, electronegativity decreases, due to an increase in size. As we go down in the group, the atomic size increases. This increase in the atomic size is credited to an increase in the number of shells.

Question 7.2 Why does the reactivity of nitrogen differ from phosphorus?

Answer :

Nitrogen is a diatomic molecule $N\equiv N$ . The two atoms of nitrogen form a triple bond which makes it highly stable. The triple bond present is very strong and difficult to break due to the small size of the nitrogen atom, this is not the case in phosphorus atom and phosphorus exists in a tetra-atomic molecule. Since $P-P$ single bond (213KJ/mol) is weaker than $N\equiv N$ triple bond (941KJ/mol) hence they both react differently..

Question 7.3 Discuss the trends in chemical reactivity of group 15 elements.

Answer :

The element of group 15 :

React with hydrogen in order to form hydrides of type $E{H}_3$ , where E = N, P, As, Sb, or Bi.

React with oxygen in order to form two types of oxides: $E_2{O}_{3}and{E}_2{O}_5$ where E = N, P, As, Sb, or Bi.

React with halogens in order to form two series of salts: $E{X}_3$ and $E{X}_5$ . Except $NB{r}_3,N{I}_{3}andN{X}_5$ because it lacks the d -orbital.

React with metals for forming binary compounds in which metals exhibit -3 oxidation states.

Question 7.4 Why does $N{H}_3$ form hydrogen bond but $P{H}_3$ does not?

Answer :

$N{H}_3$ form hydrogen bond but $P{H}_3$ does not because Nitrogen has the massive attraction of the electron to the nucleus due to its higher electronegativity in comparison to the phosphorus. hence H-bonding in PH3 is very less as compared to NH3.

Note: Conditions for the formation of H-bond are-

• high electronegativity
• small size

Question 7.5 How is nitrogen prepared in the laboratory? Write the chemical equations of the reactions involved.

Answer :

We prepare nitrogen by the following method,

when An aqueous solution of ammonium chloride is reacted with sodium nitrite.

$N{H}_{4}Cl(aq)+NaN{O}_2(aq)\to N_2(g)+2H_{2}O(l)+NaCl$

here, NO and HNO ₃ are produced in small amounts. These are counted in impurities that we can remove by passing nitrogen gas through aqueous sulphuric acid, containing potassium dichromate.

Question 7.6 How is ammonia manufactured industrially?

Answer :

Ammonia is produced by Haber's process-

$N_2+3H_2\stackrel{700k}{\leftrightharpoons }2N{H}_3\mathrm{\Delta }{H}^o=-46.1KJmo{l}^{-1}$

According to Le-Chatelier's principle, High pressure would favour the production. The maximum yield conditions for the production of ammonia are-

1. pressure = $200atmor200\times {10}^{5}Pa$ ,
2. Temperature of around 700 K
3. catalyst = Iron oxide
4. Promotor= small amounts of $K_{2}O$ and $A{l}_2{O}_3$ to increase the rate of attainment of equilibrium.

Question 7.7 Illustrate how copper metal can give different products on reaction with $HN{O}_3$ .

Answer :

Concentrated nitric acid has a strong oxidizing property. It is used for oxidizing most metals. The concentration of the acid and temperature decides the products of oxidation.

(i) Cu reacts with dilute $HN{O}_3$

$3Cu+8HN{O}_3(dilute)\to 3Cu(N{O}_3{)}_2+2NO+4H_{2}O$

(i) Cu reacts with conc. $HN{O}_3$

$Cu+4HN{O}_3(conc)\to Cu(N{O}_3{)}_2+2N{O}_2+2H_{2}O$

Question 7.8 Give the resonating structures of $N{O}_2$ and $N_2{O}_5$ .

Answer :

Resonance structure of $N{O}_2$ and $N_2{O}_5$ are

Question 7.9 The HNH angle value is higher than HPH, HAsH and HSbH angles. Why?

Answer :

The angle value of HNH is higher than HPH, HAsH and HSbH angles. This is due to the higher electronegativity of the electron. Since nitrogen is highly electronegative, there is high electron density around the atom of nitrogen. This causes greater repulsion between the electron pairs which are around nitrogen, resulting in maximum bond angle.

Question 7.10 Why does $R_{3}P=0$ exist but $R_{3}N=0$ does not $(R=alkylgroup)$ ?

Answer :

$N$ does not have any $d$ -orbitals but phosphorus( $P$ ) does. This is the restriction which comes in nitrogen( $N$ ) to expand its coordination number beyond four. Hence, $R_{3}N=O$ does not exist.

Question 7.11 Explain why $N{H}_3$ is basic while $Bi{H}_3$ is only feebly basic.

Answer :

Nitrogen has a small size because of which the lone pair of electrons are concentrated in a small region. As we go down a group, the size of the central atom increases and the charge gets distributed over a large area which results in decreasing the electron density. Hence, the electron donating capacity(Basicity) of group 15 element hydrides decreases on moving down the group. And that's why electron releasing tendency(basicity) of $Bi{H}_3$ is less than ammonia.

Question 7.12 Nitrogen exists as diatomic molecule and phosphorus as $P_4$ . Why?

Answer :

The nitrogen atom has small size and high electronegativity due to this nitrogen form $p\pi -p\pi$ multiple bonds with itself and with other elements which have small size and high electronegativity (e.g., C, O). The elements which are heavier of this group do not form $p\pi -p\pi$ bonds because their atomic orbitals are so large and diffuse that they cannot have effective overlapping. Thus, nitrogen exists as a diatomic molecule with a triple bond (one s and two p) between the two atoms. On the contrary, phosphorus has less the tendency to form pπ-pπ bonds and hence it exists in the form $P_4$ .

Question 7.13 Write main differences between the properties of white phosphorus and red phosphorus.

Answer :

Question 7.14 Why does nitrogen show catenation properties less than phosphorus?

Answer :

The single $N-N$ bond in nitrogen is weaker than the single $P-P$ bond because of high interelectronic repulsion of the non-bonding electrons in $N_2$ , due to the small bond length. Therefor, the catenation tendency is weaker in nitrogen.

Question 7.15 Give the disproportionation reaction of $H_{3}P{O}_3$ .

Answer :

When we heat, orthophosphorus acid (H ₃ PO ₃ ) disproportionates into orthophosphoric acid (H ₃ PO ₄ ) and phosphine (PH ₃ ).

$\text{dpi}100\stackrel{+3}{4H_{3}P{O}_3}\to \stackrel{-5}{3H_{3}P{O}_4}+\stackrel{-3}{P{H}_3}$

Question 7.16 Can $PC{I}_5$ act as an oxidising as well as a reducing agent? Justify.

Answer :

No $PC{I}_5$ can not act as reducing agent but it can act as an oxidising. In $PC{I}_5$ , phosphorus have its highest oxidation state (+5) which cannot be increased further but it can decrease its oxidation state and act as an oxidizing agent. For example-

$Sn+\stackrel{+5}{2PC{l}_5}\to SnC{l}_4+\stackrel{+3}{2PC{l}_3}$

Question 7.17 Justify the placement of $O$ , $S$ , $Se$ , $Te$ and $Po$ in the same group of the periodic table in terms of electronic configuration, oxidation state and hydride formation.

Answer :

• Electronic Configuration-
  $O$ , $S$ , $Se$ , $Te$ and $Po$ , all have six valance electron each. The general electronic configuration of these elements is $n{s}^2,n{p}^4$ , where $n$ varies from 2 to 6.
• Oxidation state-
  As all of these elements have six valence electrons, they should display an oxidation state of -2. The stability of the -2 oxidation state decreases on moving down a group due to a decrease in the electronegativity of the elements. The heavier elements show +2, +4 and +6 oxidation state due to availability of $d$ -orbitals. It also exhibits the oxidation state of -1 ( $H_2{O}_2$ ), zero ( $O_2$ ), and +2 ( $O{F}_2$ )
• Hydrides-
  They all form hydrides of formula $H_{2}E$ , where $E=O,S,Se,Te,Po.$ Oxygen and sulphur also form hydrides of type $H_2{E}_2$ . These hydrides are volatile in nature.

Question 7.18 Why is dioxygen a gas but sulphur a solid?

Answer :

Oxygen is smaller in size as compared to the sulphur. Thus it can effectively form $p\pi -p\pi$ bond and form $O_2(O=O)$ molecules. The intermolecular forces in oxygen are weak van der Wall's, which cause it to exist as gas.whereas sulphur exists as a puckered structure held together by strong covalent bonds. Hence, it is solid.

Question 7.19 Knowing the electron gain enthalpy values for $O\to O^{-}$ and $O\to O^{2-}$ as $-141$ and $702kJmo{l}^{-1}$ respectively, how can you account for the formation of a large number of oxides having $O^{2-}$ species and not $O^{-}$ ?

Answer :

Lattice energy directly depends on the charge carried by an ion. More the lattice energy, more stable the compound will be. When metal and oxygen combine, the lattice energy of the oxide, which involves $O^{2-}$ ion is much more than the oxide which involves $O^{-}$ ion. Ionic compound stability depends on the lattice energy of the compound. Thus the oxides of $O^{2-}$ is more stable than oxides having $O^{-}$ .

Question 7.20 Which aerosols deplete ozone?

Answer :

Freons which are used in aerosol sprays and as refrigerants is accountable for the depletion of the ozone layer, freons are also called chlorofluorocarbons.\

Question 7.21 Describe the manufacture of $H_{2}S{O}_4$ by contact process?

Answer :

Sulphuric acid is manufactured by the Contact Process that involves three steps:

(i) burning of sulphide ores or sulphur in the air to generate. $S{O}_2$

(ii) conversion of $S{O}_2$ to $S{O}_3$ by the reaction with oxygen in the presence of a catalyst ( $V_2{O}_5$ ), and

(iii) absorption of $S{O}_3$ in $H_{2}S{O}_4$ to give Oleum ( $H_2{S}_2{O}_7$ )

Diluting the oleum with water gives $H_{2}S{O}_4$ of the desired concentration.

$2S{O}_2+O_2\to 2S{O}_3$

$S{O}_3+H_{2}S{O}_4\to H_2{S}_2{O}_7(olium)$

Question 7.22 How is $S{O}_2$ an air pollutant?

Answer:

• Sulphur dioxide $S{O}_2$ is considered an air pollutant because it readily undergoes oxidation in the atmosphere to form Sulphur trioxide $S{O}_3$ which then reacts with water vapour to form sulphuric acid $H_{2}S{O}_4$ . Which comes down in the form of acid rain. acid rain causes deforestation which is also not good for the environment.
• Even in low concentration of $S{O}_2$ causes the irritation in the eyes, respiratory problem and due to this it affects the larynx to cause breathlessness.
• Harmful for plants also, long exposure to $S{O}_2$ can reduce the colour of the leaves. It is because the formation of chlorophyll is affected by sulphur dioxide.

Question 7.23 Why are halogens strong oxidising agents?

Answer :

Halogens have 7 electrons in their valance shell and they need only one more electron to complete their octet and to attain the stable noble gas configuration. So they have a high tendency to gain an electron. Also, halogens are highly electronegative with low dissociation energies and high negative electron gain enthalpies which just increase the tendency to gain an electron. Hence they are strong oxidising agent.

Question 7.24 Explain why fluorine forms only one oxoacid, $HOF$ .

Answer :

In fluorine d-orbitals are absent and also it has very high electronegativity and small size.o it shows only +1 oxidation state in oxo-acid, but not + 3, + 5 or + 7. Hence It forms only one oxoacid $HOF$ and doesn't form oxoacid having other oxidation states than +1 like $HOFO,HOF{O}_{2}andHOF{O}_3.$

Question 7.25 Explain why inspite of nearly the same electronegativity, nitrogen forms hydrogen bonding while chlorine does not.

Answer :

Inspite of nearly the same electronegativity, nitrogen forms hydrogen bonding while chlorine does not, the reason behind this is the small size of nitrogen atom as compared to the chlorine atom. The small size makes electron density per volume higher.

Question 7.26 Write two uses of $CI{O}_2$ .

Answer :

Uses of $CI{O}_2$ .

1. $Cl{O}_2$ is used as a bleaching agent for paper pulp and in textiles
2. $Cl{O}_2$ is used as a germicide in water treatment.

Question 7.27 Why are halogens coloured?

Answer :

All halogens are coloured due to the absorption of radiations which comes under visible region, which results in the excitation of outer electrons to higher energy level. The different quanta of radiation absorb by different halogens and they display different colours, for example is, $F_2$ , has yellow, $C{l}_2$ , greenish yellow, $B{r}_2$ , red and $I_2$ , violet colour.

Question 7.28 Write the reactions of $F_2$ and $C{I}_2$ with water.

Answer :

$C{l}_2$ with water

$C{l}_2+H_{2}O\to HCl+HOCl$

$F_2$ with water

$2F_2+2H_{2}O\to 4H^{+}+4F^{-}+O_2+4HF$

Question 7.29 How can you prepare $C{I}_2$ from $HCL$ and $HCL$ from $C{I}_2$ ? Write reactions only.

Answer :

Chlorine has a great affinity for hydrogen. Chlorine reacts with hydrogen-containing compounds to form HCl.

$H_2+C{l}_2\to 2HCl$

$H_{2}S+C{l}_2\to 2HCl+S$

$C_{10}H16+8C{l}_2\to 16HCl+10C$

HCL to Chlorine

$4HCL+O_2\to 2C{l}_2+2H_{2}O$

Question 7.30 What inspired $N$ . Bartlett for carrying out reaction between $Xe$ and $Pt{F}_6$ ?

Answer :

Initially, he prepared a red compound of formula $O_2^{+}Pt{F}_6^{-}$ with the help of oxygen and $Pt{F}_6$ . Later, he realised that the first ionisation enthalpy of molecular oxygen (1175 kJ/mol) and that of xenon (1170 kJ/mol) are almost identical and then he tried to prepare the same type of compound with Xe and $Pt{F}_6$ . He was successful in preparing another red colour compound. $X{e}^{+}$ $Pt{F}_6^{-}$

Question 7.31 What are the oxidation states of phosphorus in the following:

$H_{3}P{O}_{{}_3}$

Answer :

It is known that the oxidation state of H = 1 and O is -2.
Let the oxidation state of $P$ be x
$$\begin{gathered} 3+x+3(-2)=0 \\ 3+x-6=0 \\ x-3=0 \\ x=3 \end{gathered}$$

hence oxidation state of $H_{3}P{O}_{{}_3}$ is 3

Question 7.31 What are the oxidation states of phosphorus in the following:

$PC{l}_3$

Answer:

It is known that the oxidation state of chlorine is -1
let oxidation state $P$ be $x$
$$\begin{gathered} x+3(-1)=0 \\ x-3=0 \\ x=3 \end{gathered}$$

hence oxidation state phosphorus in $PC{l}_3$ is +3

Question 7.31 What are the oxidation states of phosphorus in the following:

$C{a}_3{P}_2$

Answer :

We know that the oxidation state of calcium is +2
let oxidation state $P$ be $x$
$$\begin{gathered} 3(+2)+2(x)=0 \\ 6+2x=0 \\ 2x=-6 \\ x=-6/2 \\ x=-3 \end{gathered}$$

Hence the oxidation state of the phosphorus is -3

Question 7.31 What are the oxidation states of phosphorus in the following:

$N{a}_{3}P{O}_4$

Answer :

we know the oxidation state of sodium( $Na$ ) is +1 and oxygen( $O_2$ ) is -2
Let Oxidation state = x

$$\begin{gathered} 3(+1)+x+4(-2)=0 \\ 3+x-8=0 \\ x-5=0 \\ x=5 \end{gathered}$$

Thus the oxidation state of phosphorus in $N{a}_{3}P{O}_4$ is +5

Question 7.31 What are the oxidation states of phosphorus in the following:

$PO{F}_3$

Answer :

It is known that the oxidation state of the oxygen and fluorine are -2 and -1 respectively
Let oxidation state be x
$$\begin{gathered} x+(-2)+3(-1)=0 \\ x-2-3=0 \\ x-5=0 \\ x=5 \end{gathered}$$
So, the oxidation state of the phosphorus in $PO{F}_3$ is +5

Question 7.32 Write balanced equations for the following:

(i) $NaCl$ is heated with sulphuric acid in the presence of $Mn{O}_2$ .

Answer :

Nacl is heated with sulphuric acid in presence of Kmno4

$4NaCl+Mn{O}_2+4H_{2}S{O}_4\to MnC{l}_2+4NaHS{O}_4+2H_{2}O+C{l}_2$

Question 7.32 Write balanced equations for the following:

(ii) Chlorine gas is passed into a solution of $NaI$ in water.

Answer :

Chlorine gas is passed into a solution of water

$C{l}_2+2NaI\to 2NaCl+I_2$

Question 7.33 How are xenon fluorides $Xe{F}_2$ , $Xe{F}_4$ and $Xe{F}_6$ obtained?

Answer :

Under different concentration of Xenon, it forms , XeF2 , XeF4 and XeF6 by the direct reaction.

$F_2+Xe(in-excess)\to Xe{F}_2;under-673k,1bar$

$2F_2+Xe(1:5ratio)\to Xe{F}_4;under-873k,7bar$

$3F_2+Xe(1:20ratio)\to Xe{F}_6;under-573k,65bar$

XeF6 can also be made by the interacting $Xe{F}_4$ and $O_2{F}_2$ at 143K.

$Xe{F}_4+O_2{F}_2\to Xe{F}_6+O_2$

Question 7.34 With what neutral molecule is $Cl{O}^{-}$ isoelectronic? Is that molecule a Lewis base?

Answer :

Total electrons in $Cl{O}^{-}=17+8+1=26$

$Cl{O}^{-}$ is isoelectronic with two neutral molecules. And these two are $ClF$ and $O{F}_2$

In $ClF=17+9=26$

In $O{F}_2=8+9\times 2=26$

both species also contain 26 electrons.

Question 7.35 How are $Xe{O}_3$ and $XeO{F}_4$ prepared?

Answer :

• When we do Hydrolysis of $Xe{F}_4$ and $Xe{F}_6$ with water we get $XeO{F}_3$

$6Xe{F}_4+12H_{2}O\to 4Xe+2Xe{O}_3+24HF+3O_2$

$Xe{F}_6+3H_{2}O\to Xe{O}_3+6HF$

• And Partial hydrolysis of $Xe{F}_6$ gives us, $XeO{F}_4$

$Xe{F}_6+H_{2}O\to XeO{F}_4+2HF$

Question 7.36 Arrange the following in the order of property indicated for each set:

(i) $F_2$ , $C{I}_2$ , $B{r}_2$ , $I_2$ - increasing bond dissociation enthalpy.

Answer :

Bond dissociation energy usually decreases as we move down in a group, Bond dissociation energy usually decreases as the atomic size increases. whereas, the bond dissociation energy of $F_2$ is lower than that of $C{l}_2$ and $B{r}_2$ . This is due to the small atomic size of fluorine. hence,

_{$I_2<F_2<B{r}_2<C{l}_2$}

Question 7.36 Arrange the following in the order of property indicated for each set:

(ii) $HF$ , $HCl$ , $HBr$ , $HI$ - increasing acid strength.

Answer :

The dissociation energy of bond of H-X molecules where X = F, Cl, Br, I, decreases as we increase the atomic size.

HI is the strongest acid Since H-I bond is the weakest

$HF<HCl<HBr<HI$

Question 7.36 Arrange the following in the order of property indicated for each set:

(iii) $N{H}_3$ , $P{H}_3$ , $As{H}_3$ , $Sb{H}_3$ , $Bi{H}_3$ – increasing base strength.

Answer :

As we move from nitrogen to bismuth, the size of the atom increases and the electron density on the atom decreases. hence, the basic strength will decrease.

$Bi{H}_3<Sb{H}_3<As{H}_3<P{H}_3<N{H}_3$

Question 7.37 Which one of the following does not exist?

$(i)XeO{F}_4$

$(ii)Ne{F}_2$

$(iii)Xe{F}_2$

$(iv)Xe{F}_6$

Answer :

$Ne{F}_2$ does not exist because neon has very high ionization enthalpy. But ionization enthalpy of xenon is low.

Question 7.37 Which one of the following does not exist?

$Ne{F}_2$

Answer :

NeF2 Does not Exists.

Question 7.37 Which one of the following does not exist?

$Xe{F}_2$

Answer :

XeF2 Exists.

Question 7.37 Which one of the following does not exist?

$Xe{F}_6$

Answer :

XeF6 Exists.

Question 7.38 Give the formula and describe the structure of a noble gas species which is isostructural with:

$IC{I}_4^{-}$

Answer :

XeF ₄ has square planar geometry and is isoelectronic with ICl ^- _{4 .}

Question 7.38 Give the formula and describe the structure of a noble gas species which is isostructural with:

$IB{r}_2^{-}$

Answer :

XeF ₂ has a linear structure and is isoelectronic to IBr ^- ₂ and

Question 7.38 Give the formula and describe the structure of a noble gas species which is isostructural with:

$Br{O}_3^{-}$

Answer :

XeO ₃ has a pyramidal molecular structure and is isostructural to BrO ^- ₃ and.

Question 7.39 Why do noble gases have comparatively large atomic sizes?

Answer :

The atomic radius of an element corresponds to the covalent radius. but noble gases do not form any molecule, so for them, the radius is Vander walls radius. Vander wall radius is larger than the covalent radius.

Question 7.40 List the uses of neon and argon gases.

Answer :

We use Neon in discharge tubes and fluorescent bulbs for advertisement display purposes. Neon bulbs are used in botanical gardens and in green houses. We use Argon mainly to provide an inert atmosphere during metallurgical processes involving high temperature(arc welding of metals or alloys) and for filling electric bulbs. We use It in the laboratory too for handling substances that are air-sensitive