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

 

NCERT solutions for class 12 chemistry chapter 7 The p-block elements - In class 11th, you must have learnt that the p-block elements are placed from 13 to 18 groups of the periodic table but in class 11 you have studied only two groups 13 and 14, so in solutions of NCERT class 12 chemistry chapter 7 the p-block elements, you are going to study and get questions from groups from 15 to 18 of the periodic elements and their answers. The properties of p-block elements in comparison to others are greatly influenced by atomic sizes, ionisation enthalpy, electron gain enthalpy, and electronegativity. Also, this group has all the three types of elements, metals, non-metals, and metalloids. As this is one of the most important chapter of inorganic chemistry hence it is necessary for you to clear your doubts before moving further to other chapters. The CBSE NCERT solutions for class 12 chemistry chapter 7 The p-block elements will help you to score well in your exams.

If you are appearing for boards exam then NCERT solutions for class 12 chemistry chapter 7 The p-Block elements can be of a great help because this chapter carries a huge weightage of 8 marks out of 70 marks and are also important for competitive exams like JEE, SRMJEEE, VITEEE, BITSAT, etc. In chapter 7 the p-block elements, there are 34 intext questions and 40 questions are given in the exercise. The step-by-step NCERT solutions are arranged in a sequential manner which are prepared by subject experts. In this chapter, you will learn the preparation, properties, and uses of dinitrogen, phosphorous, dioxygen, ozone, simple oxides, chlorine and hydrochloric acid and also study the uses of noble gases.

 

After completing class 12 chemistry chapter 7 the p-block elements students will be able to explain the general trends in the chemistry of elements of groups 15, 16, 17 and 18 and also able to explain the importance of these elements and their compounds in our daily life. The p-block elements are given in the following table-

13

14 15 16 17

18

          He
B C N O F Ne
Al Si P S Cl Ar
Ga Ge As Se Br Kr
In Sn Sb Te I Xe
Ti Pb Bi Po At Rn

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

Solutions to In Text Questions Ex 7.1  to 7.34

Answer:

Pentahalides are more covalent than trihalides. This is due to the fact that in pentahalides +5 oxidation state exists while in the case of trihalides +3 oxidation state exists. So, Higher the +ve O.S of the central atom more will be the polarising power and more will be the covalent character in the bond between the central atom and a halogen atom.  Since elements in +5 oxidation state will have more polarising power than in +3 oxidation state, the covalent character of bonds is more in pentahalides.

Question 7.2     Why is BiH_{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 NH_3  to BiH_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 BiH_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} \overset{700k}{\leftrightharpoons} 2NH_{3}\:\:\:\:\: \Delta H^{o}=-46.1KJmol^{-1}

The maximum yield conditions for the production of ammonia are-

  1. pressure = 200 \:atm \:or\:200\times 10^5 Pa,

  2. Temperature of around 700 K 

  3. catalyst = Iron oxide

  4. Promotor= small amounts of K_2O and Al_2O_3 to increase the rate of attainment of equilibrium.

 

Question 7.5     How does ammonia react with a solution of Cu^{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.

Cu^{2+} (aq)(blue) +4NH_3\rightarrow [Cu(NH_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_2O_5  that covalence of nitrogen is four.

Question 7.7      Bond angle in PH_{4}^{+} is higher than that in PH_{3}. Why?

Answer:

As we can see Both are  sp^3  hybridised. In PH_4^+ all of the 4 orbitals are bonded whereas in PH_3 there is a lone pair of electrons on P.this  lone pair is responsible for lone pair-bond pair repulsion in PH_3 , which results in reducing the bond angle to less than 109° 28′.

 

Question 7.7    What is formed when PH_{3}  reacts with an acid?

Answer:

PH_3 reacts with acids like HI to form PH_4I which shows that it is basic in nature. Because of  lone pair on phosphorus atom, PH_3 is acting as a Lewis base in the above reaction

PH_3 + HI\rightarrow PH_4I

PH_3 + Acid\rightarrow salt

 

Question 7.8     What happens when white phosphorus is heated with concentrated NaOH solution in an inert atmosphere of CO_{2} ?

Answer:

When white phosphorus is heated with the concentrated NaOH solution in an inert atmosphere of CO_{2}  we see that phosphine and sodium hypophosphite is formed.

 

P_{4}+\:3NaOH\:+H_{2}O\rightarrow 3NaH_{2}PO_{2}\:+ \:PH_{3}

 

Question 7.9     What happens when PCI_{5}  is heated?

Answer:

When we heat  PCI_{5} , it  sublimes but decomposes on stronger heating and phosphorus trichloride is formed.

PCI_{5}  +  Heat    \rightarrow PCI_{3}Cl_{2}

 

Question 7.11     What is the basicity of H_{3}PO_{4} ?

Answer:

There are three P–OH bonds present in the molecule of H_{3}PO_{4}. Hence, its basicity is three.

H_{3}PO_{4}\overset{H_{2}O}{\rightleftharpoons } 3H^{+}\:+\:PO_{4}^{3-}

 

Question 7.12     What happens when  H_{3}PO_{3}  is heated?

Answer:

H_3PO_4 when heated, it will disproportionates and give orthophosphoric acid (or phosphoric acid) and phosphine.

4H_3PO_3 \rightarrow 3H_3PO_4 + PH_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 CaSO_4.2H_2O , Epsom salt MgSO_4.7H_2O, baryte BaSO_4.

  • another source is by sulphides such as galena PbS, zinc blende ZnS, copper pyrites CuFeS_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_2O> H_2S> H_2Se> H_2Te> H_2Po. 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 aH_{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}\rightarrow

Answer:

The reaction is:

C_2H_4 + 3O_2\rightarrow 2CO_2+2H_2O

Question 7.17     Complete the following reactions:

            (ii)  4AI+3 O_{2}\rightarrow

Answer:

The complete reaction is:

4AI+3 O_{2}\rightarrow 2Al_2O_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\rightarrow 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_2O+O_3\rightarrow 2OH^-+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.

2Fe^{3+}+SO_2+2H_20\rightarrow 2Fe^{2+}+SO_4^{2-}+4H^+

Question 7.21     Comment on the nature of two S-O bonds formed in SO_{2} molecule. Are the two S-O  bonds in this molecule equal ?

Answer:

The two S-O  bonds  in SO_{2}molecule are covalent and have equal strength because of having resonating structures.

 

Question 7.22     How is the presence of  SO_{2}  detected ?

Answer:

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

5SO_2+2MnO_4^-+2H_2O\rightarrow 5SO_4^{2-}+4H^++2Mn^{2+}.

it decolourises acidified potassium permanganate(VII) solution.

Hence This can be used to detect the presence of SO_{2}.

Question 7.23     Mention three areas in which  H_{2}SO_{4} plays an important role.

Answer:

  1. Manufacture of fertilisers (e.g., ammonium sulphate, superphosphate) from H_2SO_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}SO_{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_2O_5 fastens the reaction.

Question 7.25     Why is K_{a_{2}}<<K_{a}_{_{1}}  for  H_{2}SO_{4}  in water ?

Answer:

H_2SO_4 is a very strong acid in water mostly due to its first ionisation to H_3O^+ and HSO_4^-.The ionisation of HSO_4^- to H_3O^+ and SO_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 CI_{2}.

Answer:

F_2 is much more powerful in oxidising, than Cl_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 CI_{2}.

Answer:

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

Cl_{2}\:+\:H_{2}O\rightarrow 2HCl + \left [ O \right ]

Chlorine + Water  \rightarrow  Hydrochloric acid + nascent Oxygen

Coloured substance + nascent Oxygen\left [ O \right ]  \rightarrowColourless 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 (COCl_2

  2. mustard gas ClCH_2CH_2SCH_2CH_2CL.

 

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:

              XeF_{6}+H_{2}O\rightarrow XeO_{2}F_{2}+HF

Answer:

The balanced reaction is:

XeF_{6}+2H_{2}O\rightarrow XeO_{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 moleculeN\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 EH_3, where E = N, P, As, Sb, or Bi.

 React with oxygen in order to form two types of oxides: E_2O_3\: and\: E_2O_5  where E = N, P, As, Sb, or Bi.

 React with halogens in order to form two series of salts: EX_3 and EX_5. Except NBr_{3},\:NI_{3} \:and \:NX_{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 NH_{3}  form hydrogen bond but PH_{3} does not?

Answer:

NH_{3}  form hydrogen bond but PH_{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.

NH_4Cl(aq)+NaNO_2(aq)\rightarrow N_2(g)+2H_2O(l)+NaCl

here, NO and HNO3 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} \overset{700k}{\leftrightharpoons} 2NH_{3}\:\:\:\:\: \Delta H^{o}=-46.1KJmol^{-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 = 200 \:atm \:or\:200\times 10^5 Pa,

  2. Temperature of around 700 K 

  3. catalyst = Iron oxide

  4. Promotor= small amounts of K_2O and Al_2O_3 to increase the rate of attainment of equilibrium.

 

Question 7.7     Illustrate how copper metal can give different products on reaction with HNO_{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 HNO_3

              3Cu + 8HNO_3(dilute)\rightarrow 3Cu(NO_3)_2+2NO +4H_2O

(i) Cu reacts with conc. HNO_3

Cu + 4HNO_3(conc)\rightarrow Cu(NO_3)_2+2NO_{2} +2H_2O

Question 7.8     Give the resonating structures of NO_{2}and N_{2}O_{5}.

Answer:

Resonance structure of NO_{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 = alkyl\: group)?

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_3N=O does not exist.

Question 7.11     Explain why NH_{3} is basic while  BiH_{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 BiH_{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:

White phosphorus

Red phosphorus

It is a translucent white waxy solid

It is crystalline solid.

It is insoluble in water but soluble in carbon disulphide

 It is insoluble in water as well as in carbon disulphide 

poisonous

non-poisonous

It consists of discrete tetrahedral P4 molecule

red phosphorus is polymeric, consisting of chains of P4 tetrahedra linked together

 

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}PO_{3} .

Answer:

When we heat, orthophosphorus acid (H3PO3) disproportionates into orthophosphoric acid (H3PO4) and phosphine (PH3).

\dpi{100} \overset{\:\:\:+3}{4H_3PO_3}\rightarrow \overset{\:\:\:-5}{3H_3PO_4}+\overset{-3}{\:\:\:\:PH_3}

Question 7.16   Can PCI_{5} act as an oxidising as well as a reducing agent? Justify.

Answer:

No PCI_{5} can not  act as reducing agent but it can act as an oxidising. In PCI_{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 \:+\:\overset{+5}{\:\:2PCl_{5}}\rightarrow SnCl_{4}+\:\overset{+3}{\:2PCl_{3}}

Question 7.17  Justify the placement of O, S, Se, Teand 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 ns^2,\ np^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 (OF_{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\rightarrow O^{-} and O\rightarrow O^{2-} as -141 and 702 kJ mol^{-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}SO_{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.SO_2

(ii) conversion of SO_2to SO_3 by the reaction with oxygen in the presence of a catalyst (V_2O_5 ), and

(iii) absorption of SO_3in H_2SO_4to give Oleum (H_2S_2O_7)

Diluting the oleum with water gives H_2SO_4 of the desired concentration.

2SO_2+O_2\rightarrow 2SO_3

SO_3+H_2SO_4\rightarrow H_2S_2O_7(olium)

Question 7.22  How is SO_{2} an air pollutant?

Answer:

  •  Sulphur dioxide SO_2 is considered an air pollutant because it readily undergoes oxidation in the atmosphere to form Sulphur trioxide SO_3 which then reacts with water vapour to form sulphuric acid H_2SO_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 SO_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 SO_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 likeHOFO,HOFO_2 \:and \:HOFO_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 CIO_{2}.

Answer:

Uses of CIO_{2}.

  1.  ClO_2  is used as a bleaching agent for paper pulp and in textiles

  2. ClO_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, Cl_2, greenish yellow, Br_2, red and I_2, violet colour.

 

Question 7.28     Write the reactions of F_{2} and CI_{2} with water.

Answer:

Cl_2 with water

Cl_2+H_2O\rightarrow HCl+HOCl

F_2 with water

2F_2+2H_2O\rightarrow 4H^++4F^-+O_2+4HF

Question 7.29 How can you prepare CI_{2}  from HCL  and HCL  from CI_{2} ? Write reactions only.

Answer:

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

H_2+Cl_2\rightarrow 2HCl

H_2S+Cl_2\rightarrow 2HCl+S

C_{10}H{16}+8Cl_2\rightarrow 16HCl+10C

HCL to Chlorine

4HCL+O_2\rightarrow 2Cl_2+2H_2O

Question 7.30     What inspired N. Bartlett for carrying out reaction between Xe and PtF_{6} ?

Answer:

Initially, he prepared a red compound of formula  O_2^+ PtF_6^- with the help of oxygen and PtF_{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 PtF_{6}. He was successful in preparing another red colour compound.Xe^+PtF_6^-

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

              H_{3}PO_{_{3}}

Answer:

It is known that the oxidation state of H = 1 and O is -2.
Let the oxidation state of P be x


\\3 + x + 3(-2) = 0\\ 3 + x - 6 = 0\\ x - 3 = 0\\ x = 3

hence oxidation state of H_{3}PO_{_{3}} is  3

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

              PCl_{3}

Answer:

It is known that the oxidation state of chlorine is -1
let oxidation state P be x
\\x + 3(-1) = 0\\ x - 3 = 0\\ x = 3

hence oxidation state phosphorus in  PCl_{3} is +3

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

              Ca_{3}P_{2}

Answer:

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

Hence the oxidation state of the phosphorus is -3

 

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

               Na_{3}PO_{4}

Answer:

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

\\3(+1) + x + 4(-2) = 0\\ 3 + x - 8 = 0\\ x - 5 = 0\\ x = 5

Thus the oxidation state of phosphorus in Na_{3}PO_{4} is +5

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

              POF_{3}

Answer:

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

 

Question 7.32     Write balanced equations for the following:

            (i)  NaCl  is heated with sulphuric acid in the presence of  MnO_{2}.

Answer:

Nacl is heated with sulphuric acid in presence of Kmno4

4NaCl + MnO_2+4H_2SO_4\rightarrow MnCl_2+4NaHSO_4+2H_2O+Cl_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

Cl_2+2NaI\rightarrow 2NaCl+I_2

 

Question 7.33     How are xenon fluorides XeF_{2}XeF_{4}  and XeF_{6}  obtained?

Answer:

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

F_2+Xe(in-excess)\rightarrow XeF_2;under-673k,1bar

2F_2+Xe(1:5ratio)\rightarrow XeF_4;under-873k,7bar

3F_2+Xe(1:20ratio)\rightarrow XeF_6;under-573k,65bar

XeF6 can also be made by the interacting XeF_4and O_2F_2at 143K.

XeF_4+O_2F_2\rightarrow XeF_6+O_2

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

Answer:

Total electrons in  ClO^- = 17 + 8 + 1 = 26

ClO^- is isoelectronic with two neutral molecules. And these two are ClF and OF_2

In ClF= 17 + 9 = 26

In OF_2=8+ \:9\times 2=26

both species also contain 26 electrons.

 

Question 7.35     How are XeO_{3} and XeOF_{4}  prepared?

Answer:

  • When we do Hydrolysis of XeF_4and XeF_6 with water we get XeOF_3                               

                 6XeF_4+12H_2O\rightarrow 4Xe+2XeO_3+24HF+3O_2

                     XeF_6+3H_2O\rightarrow XeO_3+6HF

  • And  Partial hydrolysis of XeF_6 gives us, XeOF_4

                  XeF_6+H_2O\rightarrow XeOF_4+2HF

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

        (i)     F_{2}, CI_{2}, Br_{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 Cl_2and Br_2. This is due to the small atomic size of fluorine. hence,

I_2< F_2<Br_2<Cl_2

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

           (ii)  HF, HCl, HBrHI - 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)  NH_{3}, PH_{3},AsH_{3} , SbH_{3}BiH_{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.

BiH_3<SbH_3<AsH_3<PH_3<NH_3

Question 7.37    Which one of the following does not exist?

              (i)\:XeOF_{4}

               (ii)\:NeF_{2}

                (iii)\:XeF_{2}

               (iv)\:XeF_{6}              

Answer:

NeF_{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?

             NeF_{2}

Answer:

NeF2 Does not Exists.

Question 7.37     Which one of the following does not exist?

              XeF_{2}

Answer:

XeF2 Exists.

Question 7.37     Which one of the following does not exist?

               XeF_{6}

Answer:

XeF6 Exists.

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

              ICI_{4}^{-}

Answer:

XeF4  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:

            IBr_{2}^{-}

Answer:

XeF2 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:

               BrO_{3}^{-}

Answer:

XeOhas a pyramidal molecular structure and is isostructural to BrO-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

 

NCERT Solutions Class 12 Chemistry

Chapter 1

CBSE NCERT solutions for class 12 chapter 1 The Solid State

Chapter 2

NCERT solutions for class 12 chemistry chapter 2 Solutions

Chapter 3

Solutions of NCERT class 12 chemistry chapter 3 Electrochemistry

Chapter 4

CBSE NCERT solutions for class 12 chemistry chapter 4 Chemical Kinetics

Chapter 5

Solutions of NCERT class 12 chemistry chapter 5 Surface chemistry

Chapter 6

NCERT solutions for class 12 chemistry General Principles and Processes of isolation of elements

Chapter 7

CBSE NCERT solutions for class 12 chemistry chapter 7 The P-block elements

Chapter 8

Solutions of NCERT class 12 chemistry chapter 8 The d and f block elements

Chapter 9

NCERT solutions for class 12 chemistry chapter 9 Coordination compounds

Chapter 10

Solutions of NCERT class 12 chemistry chapter 10 Haloalkanes and Haloarenes

Chapter 11

CBSE NCERT solutions for class 12 chemistry Alcohols, Phenols, and Ethers

Chapter 12

Solutions of NCERT class 12 chemistry chapter 12 Aldehydes, Ketones and Carboxylic Acids

Chapter 13

NCERT solutions for class 12 chemistry chapter 13 Amines

Chapter 14

CBSE NCERT solutions for class 12 chemistry chapter 14 Biomolecules

Chapter 15

Solutions of NCERT class 12 chemistry chapter 15 Polymers

Chapter 16

NCERT solutions for class 12 chemistry chapter 16 Chemistry in Everyday life

NCERT Solutions for Class 12 Subject wise

Solutions of NCERT class 12 biology

NCERT solutions for class 12 maths

CBSE NCERT solutions for class 12 chemistry

Solutions of NCERT class 12 physics

Benefits of NCERT solutions for class 12 chemistry chapter 7 The p-block elements

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