{"id":39541,"date":"2023-01-18T10:00:05","date_gmt":"2023-01-18T04:30:05","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=39541"},"modified":"2023-01-19T09:54:18","modified_gmt":"2023-01-19T04:24:18","slug":"plus-two-chemistry-chapter-wise-questions-answers-chapter-7","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/plus-two-chemistry-chapter-wise-questions-answers-chapter-7\/","title":{"rendered":"Plus Two Chemistry Chapter Wise Questions and Answers Chapter 7 The p Block Elements"},"content":{"rendered":"

Plus Two Chemistry Chapter Wise Questions and Answers Chapter 7 The p Block Elements are part of Plus Two Chemistry Chapter Wise Questions and Answers<\/a>. Here we have given Plus Two Chemistry Chapter Wise Questions and Answers Chapter 7 The p Block Elements.<\/p>\n

Kerala Plus Two Chemistry Chapter Wise Questions and Answers Chapter 7 The p Block Elements<\/h2>\n

Plus Two Chemistry The p Block Elements One Mark Questions and Answers<\/h3>\n

Question 1.
\nThe gas obtained by the thermal decomposition of barium azide is _________.
\nAnswer:
\nNitrogen<\/p>\n

Question 2.
\nThe number of replaceable hydrogen atoms in ortho phosphorous acid (H3<\/sub>PO3<\/sub>) is
\n(a) 3
\n(b) 2
\n(c) 1
\n(d) 0
\nAnswer:
\n(b) 2<\/p>\n

Question 3.
\nThe stable allotrope of sulphur at room temperature is ___________ .
\nAnswer:
\nRhombic sulphur<\/p>\n

Question 4.
\nWhich element would readily replace oxygen from an oxide?
\n(a) N
\n(b) S
\n(c) F
\n(d) Cl
\n(Answer:
\n(c) F<\/p>\n

Question 5.
\nThe geometry of XeF4<\/sub> is.
\nAnswer:
\nSquare planar<\/p>\n

Question 6.
\nThe shape and hybridisation of XeF4<\/sub> molecule
\nAnswer:
\nSquare planar and sp3<\/sup>d2<\/sup><\/p>\n

Question 7.
\nThe least stable hydride of 15th<\/sup> group elements is.
\nAnswer:
\n(BiH3<\/sub>)<\/p>\n

Question 8.
\nChoose the weak monobasic acid among the following.
\n(a) H3<\/sub>BO3<\/sub>
\n(b) H3<\/sub>PO2<\/sub>
\n(c) H3<\/sub>PO4<\/sub>
\n(d) HNO3<\/sub>
\nAnswer:
\n(a) H3<\/sub>BO3<\/sub><\/p>\n

Question 9.
\nThe bond enthalpy is highest for ________
\nAnswer:
\nH2<\/sub><\/p>\n

Question 10.
\nMagnetic moment of an atom with atomic no. 24 in aqeous solution is ____________.
\nAnswer:
\n4.90 B.M.<\/p>\n

Plus Two Chemistry The p Block Elements Two Mark Questions and Answers<\/h3>\n

Question 1.
\nIn the class, a student argued that \u201cheterogenous\u201d\u00a0catalysis and Le-Chatlier\u2019s principles are applied in contact process.”<\/p>\n

    \n
  1. Do you agree with this statement?<\/li>\n
  2. Justify.<\/li>\n<\/ol>\n

    Answer:
    \n1. Yes
    \n\"Plus
    \nThe reaction is exothermic, reversible and the forward reaction leads to a decrease in volume. Therefore, low temperature and high pressure are the favourable conditions for maximum yield according to Le Chatelier\u2019s principle.<\/p>\n

    2. Contact process is an example of heterogeneous catalysis since the catalyst V2<\/sub>O5<\/sub> is solid while the reactants are gases.<\/p>\n

    Question 2.
    \nClF3<\/sub> exists, but FCl3<\/sub> doesn\u2019t, why?
    \nAnswer:
    \nIn the valence shell of Cl vacant 3d orbitals are available. Hence it can form ClF3<\/sub>. But, F has no vacant d orbital to show higher oxidation state and FCl3<\/sub> is not possible.<\/p>\n

    Question 3.
    \nIn a classroom discussion, a student argued that carbon has the maximum catenation property.<\/p>\n

      \n
    1. Do you agree?<\/li>\n
    2. What do you understand by the term catenation?<\/li>\n
    3. List out some other elements which can show catenation?<\/li>\n<\/ol>\n

      Answer:<\/p>\n

        \n
      1. Yes.<\/li>\n
      2. Catenation is the self linking property of an atom. It is maximum for carbon, because C-C bond is very strong.<\/li>\n
      3. Sulphur and Phosphorus.<\/li>\n<\/ol>\n

        Question 4.
        \nSome elements and their ores are given in table. Arrange them correctly.<\/p>\n\n\n\n\n\n\n\n\n\n
        A<\/td>\nB<\/td>\n<\/tr>\n
        a) Calcium<\/td>\nGalena<\/td>\n<\/tr>\n
        b) Potassium<\/td>\nMagnetite<\/td>\n<\/tr>\n
        c) Lead<\/td>\nFluorapatite<\/td>\n<\/tr>\n
        d) Tin<\/td>\nGypsum<\/td>\n<\/tr>\n
        e) Phosphorus<\/td>\nCassiterite<\/td>\n<\/tr>\n
        <\/td>\nCarnallite<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

        Answer:<\/p>\n\n\n\n\n\n\n\n\n
        A<\/td>\nB<\/td>\n<\/tr>\n
        a) Calcium<\/td>\nGypsum<\/td>\n<\/tr>\n
        b) Potassium<\/td>\nCarnallite<\/td>\n<\/tr>\n
        c) Lead<\/td>\nGalena<\/td>\n<\/tr>\n
        d) Tin<\/td>\nCassiterite<\/td>\n<\/tr>\n
        e) Phosphorus<\/td>\nFluorapatite<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

        Question 5.
        \nAnalyse the statement: \u2018Ionisation enthalpy of halogens decreases with a decrease in atomic size\u2019. Is it true? Justify your answer.
        \nAnswer:
        \nNo. On moving from top to bottom in a given group, the size of the atom increases and hence ionisation enthalpy decreases.<\/p>\n

        Question 6.
        \nHydrides of group 15 are Lewis bases.<\/p>\n

          \n
        1. Give reason.<\/li>\n
        2. Arrange the group 15 hydrides in the decreasing order of basic strength.<\/li>\n<\/ol>\n

          Answer:<\/p>\n

            \n
          1. Due to the presence of lone pair on the central atom which is available for donation.<\/li>\n
          2. NH3<\/sub> > PH3<\/sub> > ASH3<\/sub> > SbH3<\/sub> \u2265 BiH3<\/sub>.<\/li>\n<\/ol>\n

            Question 7.
            \nAccording to VSEPR theory assign structure to XeOF4<\/sub>.
            \nAnswer:
            \nXeOF4<\/sub> possess square pyramidal shape.
            \n\"Plus<\/p>\n

            Question 8.
            \nWhy does the reactivity of nitrogen differ from phosphorus?
            \nAnswer:<\/p>\n

              \n
            1. Nitrogen has a small size, high electronegativity, high ionisation enthalpy as compared to phosphorus.<\/li>\n
            2. Nitrogen does not contain vacant d-orbitals in its valence shell whereas phosphorus contains vacant d-orbitals in its valence shell.<\/li>\n
            3. Nitrogen has the ability to form triple bond ( N \u2261 N) as a result of which its bond enthalpy (941.4 kJ mol-1<\/sup>) is very high making it less reactive.<\/li>\n<\/ol>\n

              Question 9.
              \nWhy does NH3<\/sub> form hydrogen bond but PH3<\/sub> does not?
              \nAnswer:
              \nIn NH3<\/sub>, the nitrogen atom forms hydrogen bond because of the following reasons:<\/p>\n

                \n
              • Small size of nitrogen<\/li>\n
              • High electronegativity (3.0) of nitrogen<\/li>\n<\/ul>\n

                N-H bond is polar forming hydrogen bond.
                \nP-H bond is almost purely covalent due to larger size and lesser electronegativity.<\/p>\n

                Question 10.
                \nThe HNH angle is higher than HPH, HAsH and HSbH angles. Why?
                \nAnswer:
                \nBecause in NH3<\/sub> is sp3<\/sup> hybridised. Due to lone pair of electrons the bond angle contracts from 109\u00b0 28\u2019 to 106.5\u00b0. The sp3<\/sup> hybridisation becomes less and less distinct with increasing size of the central atom. Thus, the bond angle of the hydrides of group 15 decreases as.
                \n\"Plus<\/p>\n

                Question 11.
                \nCan PCl3<\/sub> act as an oxidising as well as a reducing agent? Justify.
                \nAnswer:
                \nThis is because in PCl3<\/sub> phosphorus is in the intermediate oxidising state of -3.
                \n1. As reducing agent:
                \nThe following reactions support the reducing behaviour of PCl3<\/sub>.<\/p>\n

                  \n
                • PCl3<\/sub> + SO2<\/sub>Cl2<\/sub> \u2192 PCl5<\/sub> + SO2<\/sub><\/li>\n
                • PCl3<\/sub> + SO3<\/sub> \u2192 POCl3<\/sub> + SO2<\/sub><\/li>\n<\/ul>\n

                  2. As an oxidising agent:
                  \nIt oxidises metals to their respective chlorides.<\/p>\n

                    \n
                  • 12Ag + 4PCl3<\/sub> \u2192 12AgCl + P4<\/sub><\/li>\n
                  • 6Na + PCl3<\/sub> \u2192 3NaCl + Na3<\/sub>P<\/li>\n<\/ul>\n

                    Question 12.
                    \nExplain why inspite of nearly the same electronegativity, nitrogen forms hydrogen bonding while chlorine does not.
                    \nAnswer:
                    \nDue to its larger size (99 pm) as compared to oxygen (66 pm).<\/p>\n

                    Question 13.
                    \nOzone is a strong oxidising agent. Why?
                    \nAnswer:
                    \nO3<\/sub> undergoes dissociation to form nascent oxygen. This nascent oxygen is responsible for oxidising property of ozone.<\/p>\n

                    O3<\/sub> \u2192 O2<\/sub> + [O]<\/p>\n

                    Question 14.
                    \nWhite phosphorus is more reactive than other solid phases of phosphorus. Give reason.
                    \nAnswer:
                    \nThis is beacuse of angular strain in the P4<\/sub> molecule where the angles are only 60\u00b0.<\/p>\n

                    Question 15.
                    \nWhat happens when<\/p>\n

                      \n
                    1. Concentrated H2<\/sub>SO4<\/sub> is added to CaF2<\/sub>?<\/li>\n
                    2. SO3<\/sub> is passed through water?<\/li>\n<\/ol>\n

                      Answer:
                      \n1. It form hydrogen fluoride.
                      \nCaF2<\/sub> + H2<\/sub>SO4<\/sub> \u2192 CaSO4<\/sub> + 2HF<\/p>\n

                      2. It dissolves SO3<\/sub> to give H2<\/sub>SO4<\/sub>.
                      \nSO3<\/sub> + H2<\/sub>O \u2192 H2<\/sub>SO4<\/sub><\/p>\n

                      Question 16.
                      \nGive two important fluorides of Xenone. Predict their structure.
                      \nAnswer:<\/p>\n

                        \n
                      • XeF2<\/sub> – linear<\/li>\n
                      • XeF4<\/sub> – Square planar<\/li>\n
                      • XeF6<\/sub> – Distorted octahedral<\/li>\n<\/ul>\n

                        Question 17.
                        \nWhy is N2<\/sub> less reactive at room temperature?
                        \nAnswer:
                        \nDue to the presence of triple bond between N atoms N2<\/sub> has high bond dissociation energy and is less reactive.<\/p>\n

                        Question 18.
                        \nArrange the following in the increasing order of thermal stability.
                        \nASH3<\/sub>, NH3<\/sub>, PH3<\/sub>
                        \nAnswer:
                        \nThermal stability of hydrides, decreases from nitrogen to bismuth: NH3<\/sub> > PH3<\/sub> > AsH3<\/sub>.<\/p>\n

                        Question 19.
                        \nThe 3 allotropic forms of Pare white, red and black.<\/p>\n

                          \n
                        1. Name the thermodynamically most stable allotrope.<\/li>\n
                        2. Which allotrope of P is stored in water?<\/li>\n<\/ol>\n

                          Answer:<\/p>\n

                            \n
                          1. Black phosphorus<\/li>\n
                          2. White phosphorus<\/li>\n<\/ol>\n

                            Question 20.
                            \nNitrogen and Phosphorus are in the same group. PCI5<\/sub> is known but NCI5<\/sub> is not known. Why?
                            \nAnswer:
                            \nIn phosphorus vacant d-orbitals are present. It can form pentahalides. But due to absence of d-orbitals nitrogen cannot form NCI5<\/sub>.<\/p>\n

                            Question 21.
                            \nWhy is helium used in diving apparatus?
                            \nAnswer:
                            \nHelium is used in diving apparatus because of its very low solubility in blood.<\/p>\n

                            Question 22.
                            \nWhy has it been difficult to study the chemistry of radon?
                            \nAnswer:
                            \nRadon is radioactive with very short half-life which makes the study difficult.<\/p>\n

                            Question 23.
                            \nAccount for the following:
                            \nSO3<\/sub> is more covalent than SO2<\/sub>.
                            \nAnswer:
                            \nDue to high charge and small size of sulphur in +6 oxidation state in SO3<\/sub> it is more covalent than SO2<\/sub>, in which sulphur is in +4 oxidation state.<\/p>\n

                            Question 24.
                            \nI2<\/sub> is more soluble in KI than in water.
                            \nAnswer:
                            \nI2<\/sub> combines with KI to form the soluble complex, KI3<\/sub>.
                            \nKI + I2<\/sub> \u2192 KI3<\/sub><\/p>\n

                            Question 25.
                            \nH3<\/sub>PO3<\/sub> is dibasic while H3<\/sub>PO4<\/sub> is tribasic.
                            \nAnswer:
                            \nThere are three \u2018P-OH\u2019 bonds that are ionisable in H3<\/sub>PO4<\/sub>. So, it is tribasic. In H3<\/sub>PO3<\/sub> there are only 2 ionisable \u2018P-OH\u2019 bonds. Hence it is dibasic.
                            \n\"Plus<\/p>\n

                            Plus Two Chemistry The p Block Elements Three Mark Questions and Answers<\/h3>\n

                            Question 1.
                            \nConsider the given reaction:
                            \n2A + 6SiO2<\/sub> \u2192 6CaSiO3<\/sub> + B
                            \nB + 10C \u2192 D + 10CO<\/p>\n

                              \n
                            1. Identify A, B and D.<\/li>\n
                            2. D is stored underwater. Why?<\/li>\n
                            3. Give two examples of oxoacids of D and compounds of D.<\/li>\n<\/ol>\n

                              Answer:
                              \n1. A – Ca3<\/sub>(PO4<\/sub>)2<\/sub>
                              \nB – P4<\/sub>O10<\/sub>
                              \nD – P4<\/sub><\/p>\n

                              2. Because P4<\/sub> readily catches fire in air.<\/p>\n

                              3. Two examples of oxoacids of D and compounds of D<\/p>\n

                                \n
                              • Oxoacids of D:
                                \nOxoacids – Phosphorous acid(H3<\/sub>PO3<\/sub>), Phosphoric acid (H3<\/sub>PO4<\/sub>).<\/li>\n
                              • Compounds of D:
                                \nphosphine (PH3<\/sub>), PCI5<\/sub>, PCI4<\/sub>, P4<\/sub>O6<\/sub><\/li>\n<\/ul>\n

                                Question 2.<\/p>\n

                                  \n
                                1. Suggest a method for the preparation of PCI3<\/sub>.<\/li>\n
                                2. Why does PCI3<\/sub> fume in moist air?<\/li>\n<\/ol>\n

                                  Answer:
                                  \n1. By passing dry chlorine overheated white phosphorus.
                                  \nP4<\/sub> + 6CI2<\/sub> \u2192 4PCI3<\/sub><\/p>\n

                                  2. PCI3<\/sub> hydrolyses in the presence of moisture giving fumes of HCI.
                                  \nPCI3<\/sub> + 3H2<\/sub>O \u2192 H3<\/sub>PO3<\/sub> + 3HCI<\/p>\n

                                  Question 3.
                                  \nA student argued that electronegativity of p-block elements decrease along period and increases down the group.<\/p>\n

                                    \n
                                  1. Do you agree with this? Explain.<\/li>\n
                                  2. Write about the metallic character of p-block elements.<\/li>\n
                                  3. Arrange the following of p-block elements in the order of decreasing oxidising power.<\/li>\n<\/ol>\n

                                    F2<\/sub>\/F-( E\u00b0 =+2.85 V), Br\/Br–<\/sup>( E\u00b0 =+1.07 V), CI2<\/sub>\/CI–<\/sup> (E\u00b0 =+1.36V), I2<\/sub>\/I–<\/sup>( E\u00b0 =+0.57V).
                                    \nAnswer:<\/p>\n

                                      \n
                                    1. No. electronegativity of p-block elements increases along a period and decreases down a group.<\/li>\n
                                    2. Most of the p-block elements are non-metallic in nature. Among p-block elements the metallic character decreases along a period and increases down a group.<\/li>\n
                                    3. F2<\/sub> > CI2<\/sub> > Br2<\/sub> > I2<\/sub>, because the standard electrode potential values decreases in the same order.<\/li>\n<\/ol>\n

                                      Question 4.
                                      \nThe shape and hybridisation of some interhalogen compounds are given below in wrong order. Match them correctly.
                                      \n\"Plus
                                      \nAnswer:
                                      \n\"Plus<\/p>\n

                                      Question 5.
                                      \nAccount for the following:<\/p>\n

                                        \n
                                      1. H2<\/sub>O\u00a0is a liquid while H2<\/sub>S is a gas.<\/li>\n
                                      2. H2<\/sub>S is more acidic than H2<\/sub>O<\/li>\n
                                      3. SF6<\/sub> is known while SH6<\/sub> is not known.<\/li>\n<\/ol>\n

                                        Answer:<\/p>\n

                                          \n
                                        1. There is intermolecular hydrogen bonding in H2<\/sub>O\u00a0molecule but there is no hydrogen bonding in H2<\/sub>S .<\/li>\n
                                        2. The S-H bond is weaker than O-H bond because size of S atom is greater than that of O atom. Hence H2<\/sub>S can dissociate to give H+<\/sup> ions in aqueous solution.<\/li>\n
                                        3. In the higher oxidation state, S can combine only with highly electronegative elements like F.<\/li>\n<\/ol>\n

                                          Question 6.
                                          \nSulphur forms many allotropes such as a – sulphur, \u03b2 -sulphur etc.<\/p>\n

                                            \n
                                          1. What do you mean by allotropy?<\/li>\n
                                          2. How can you convert a – sulphur to b-sulphur?<\/li>\n<\/ol>\n

                                            Answer:<\/p>\n

                                              \n
                                            1. Certain elements can exist in different forms with different physical property and same chemical properties.<\/li>\n
                                            2. b -sulphur is prepared by melting a -sulphur in a dish and cooling, till crust is obtained. Holes are then pierced into the crust, and the liquid is taken out. On removing the crust, needle shaped \u03b2 -sulphur is obtained.<\/li>\n<\/ol>\n

                                              Question 7.
                                              \nAccount for the following:<\/p>\n

                                                \n
                                              1. PH3<\/sub> has lower boiling point than NH3<\/sub>.<\/li>\n
                                              2. Pentahalides are more covalent than trihalides.<\/li>\n
                                              3. ICI is more reactive than I2<\/sub>.<\/li>\n<\/ol>\n

                                                Answer:<\/p>\n

                                                  \n
                                                1. NH3<\/sub>, PH3<\/sub> molecules are not associated through intermolecular hydrogen bonding in liquid state.<\/li>\n
                                                2. Higher the positive oxidation state of central atom, more will be its polarising power which, in turn, increases the covalent character.<\/li>\n
                                                3. Bond energy of I-CI bond is less than that fo I-I bond.<\/li>\n<\/ol>\n

                                                  Plus Two Chemistry The p Block Elements Four Mark Questions and Answers<\/h3>\n

                                                  Question 1.
                                                  \nMy name is \u2018X\u2019. I am a poisonous colourless gas with the smell of rotten fish.<\/p>\n

                                                    \n
                                                  1. Identify \u2018X\u2019.<\/li>\n
                                                  2. Explain the laboratory preparation of X.<\/li>\n<\/ol>\n

                                                    Answer:<\/p>\n

                                                      \n
                                                    1. Phosphine(PH3<\/sub>).<\/li>\n
                                                    2. By heating white phosphorus with concentrated NaOH solution in an inert atmosphere of CO2<\/sub>.<\/li>\n<\/ol>\n

                                                      \"Plus<\/p>\n

                                                      Question 2.<\/p>\n

                                                        \n
                                                      1. How is bleaching powder prepared?<\/li>\n
                                                      2. Give the composition of bleaching powder.<\/li>\n<\/ol>\n

                                                        Answer:
                                                        \n1. By treating Cl2<\/sub> with dry slaked lime.
                                                        \n2Ca(OH)2<\/sub> + 2CI2<\/sub> \u2192 Ca(OCI)2<\/sub>\u00a0+ CaCI2<\/sub> + 2H2<\/sub>O<\/p>\n

                                                        2. Ca(OCI)2<\/sub> CaCI2<\/sub>.Ca(OH)2<\/sub>.2H2<\/sub>O.<\/p>\n

                                                        Question 3.<\/p>\n

                                                          \n
                                                        1. Why do noble gases form compounds with fluorine and oxygen only?<\/li>\n
                                                        2. Does the hydrolysis of XeF6<\/sub> lead to a redox reaction?<\/li>\n<\/ol>\n

                                                          Answer:
                                                          \n1. Fluorine and oxygen are small atoms with high value of electronegativity. Fluorine is also highly reactive in nature. It is for this reason they form compounds with noble gases.<\/p>\n

                                                          2. No, the products of hydrolysis are XeOF4<\/sub> and XeO2<\/sub>F2<\/sub> where the oxidation states of all the elements remain the same as it was in the reacting state.
                                                          \n\"Plus<\/p>\n

                                                          Question 4.<\/p>\n

                                                            \n
                                                          1. Interhalogen compounds are more reactive than halogens. Why?<\/li>\n
                                                          2. Is there any exception to the above generalisation. Explain.<\/li>\n<\/ol>\n

                                                            Answer:<\/p>\n

                                                              \n
                                                            1. This is because the bond in the interhalogen (X – X\u2019) is weaker than X – X bond in the halogens.<\/li>\n
                                                            2. Yes. F2<\/sub> is more reactive than interhalogen compounds. This is because the F – F bond is weaker than X – X\u2019 bond in interhalogen compounds.<\/li>\n<\/ol>\n

                                                              Question 5.
                                                              \nNitrogen cannot extend its covalency beyond four but it forms wide variety of oxides and also forms oxoacids.<\/p>\n

                                                                \n
                                                              1. Name the two oxoacids of nitrogen.<\/li>\n
                                                              2. Action of nitric acid with metals depends on many factors. Justify.<\/li>\n<\/ol>\n

                                                                Answer:
                                                                \n1. HNO2<\/sub>, HNO3<\/sub><\/p>\n

                                                                2. The product of oxidation depends upon the concentration of the acid, temperature and the nature of the material undergoing oxidation, e.g.<\/p>\n

                                                                a. 3Cu + 8HNO3<\/sub>(dilute) \u2192 3Cu(NO3<\/sub>)2<\/sub> + 2NO + 4H2<\/sub>O
                                                                \nCu + 4HNO3<\/sub>(conc.) \u2192 Cu(NO3<\/sub>)2<\/sub> + 2NO2<\/sub> + 2H2<\/sub>O<\/p>\n

                                                                b. Zinc reacts with dilute nitric acid to give N2<\/sub>O and with concentrated acid to give NO2<\/sub>.
                                                                \n4Zn + 4HNO3<\/sub>(dilute) \u2192 4Zn(NO3<\/sub>)2<\/sub> + 5H2<\/sub>O + N2<\/sub>O
                                                                \nZn + 10HNO3<\/sub>(conc.) \u2192 Zn(NO3<\/sub>)2<\/sub> + 2H2<\/sub>0 + 2NO2<\/sub><\/p>\n

                                                                Some metals (e.g., Cr, Al) do not dissolve in concentrated nitric acid due to the formation of a passive film of oxide on the surface.<\/p>\n

                                                                Question 6.<\/p>\n

                                                                  \n
                                                                1. Arrange hydrides of group 16 in the order of increasing acidic strength.<\/li>\n
                                                                2. Draw the structures of any three oxoacids of phosphorus and find out their basicity.<\/li>\n<\/ol>\n

                                                                  Answer:
                                                                  \n1. H2<\/sub>O < H2<\/sub>S < H2<\/sub>Se < H2<\/sub>Te
                                                                  \n2.
                                                                  \n\"Plus
                                                                  \n\"Plus<\/p>\n

                                                                  Question 7.<\/p>\n

                                                                    \n
                                                                  1. Noble gases have very low boiling points. Why?<\/li>\n
                                                                  2. XeF2<\/sub> and XeF4<\/sub> are important Xe compounds. How can we prepare them and what is the action of water on them?<\/li>\n<\/ol>\n

                                                                    Answer:
                                                                    \n1. Noble gases being monoatomic have no interatomic forces except weak dispersion forces and therefore, they are liquefied at very low temperatures. Hence, they have low boiling points.<\/p>\n

                                                                    2. XeF2<\/sub> and XeF4<\/sub> are formed by the direct reaction of elements under the specific conditions.
                                                                    \n\"Plus
                                                                    \nThey are readly hydrolysed even by traces of water,
                                                                    \ne.g. 2XeF2<\/sub>(s) + 2H2<\/sub>O(l) \u2192 2Xe(g) + 4HF(aq) + O2<\/sub>(g)<\/p>\n

                                                                    Question 8.
                                                                    \nIt is greenish yellow gas with an offensive smell used in water purification. It partially dissolves in water to give a solution which turns blue litmus to red. When it is passed through NaBr solution bromine is formed.<\/p>\n

                                                                      \n
                                                                    • Identify the gas.<\/li>\n
                                                                    • Identify the group to which it belongs.<\/li>\n
                                                                    • Write the electronic configuration.<\/li>\n
                                                                    • Write the equation showing its reaction with water.<\/li>\n<\/ul>\n

                                                                      Answer:<\/p>\n

                                                                        \n
                                                                      • Chlorine<\/li>\n
                                                                      • Halogen family (17th<\/sup> group)<\/li>\n
                                                                      • 1s2<\/sup> 2s2<\/sup> 2p6<\/sup> 3s2<\/sup> 3p5<\/sup><\/li>\n
                                                                      • CI2<\/sub> + H2<\/sub>O \u2192 HCI + HOCI<\/li>\n<\/ul>\n

                                                                        Question 9.
                                                                        \nBasic character of hydrides of group 15 is due to the presence of lone pair on the central atom.<\/p>\n

                                                                          \n
                                                                        1. NH3<\/sub> is strongly basic while BiH3<\/sub> is weakly basic. Why?<\/li>\n
                                                                        2. Differentiate between allotropic forms of phosphorus.<\/li>\n<\/ol>\n

                                                                          Answer:
                                                                          \n1. Due to the presence of lone pair of electrons on nitrogen atom of NH3<\/sub>. But in BiH3<\/sub>, due to the large size of Bi the electron density decreases and hence is less basic.<\/p>\n

                                                                          2. The allotropic forms of P are White, Red and Black phosphorus. White phosphorus consists of tetrahedral P4<\/sub> molecules. Red phosphorus is polymeric in structure consisting of chains of P4<\/sub> tetrahedra linked together. Black phosphorus has a layer type structure and has \u03b1 and \u03b2 forms.<\/p>\n

                                                                          Question 10.<\/p>\n

                                                                            \n
                                                                          1. Give a method for preparation of XeO3<\/sub>.<\/li>\n
                                                                          2. Deduce the molecular shape of BrF3 on the basis of VSEPR theory.<\/li>\n<\/ol>\n

                                                                            Answer:
                                                                            \n1. XeO3<\/sub> is prepared by hydrolysis of xenon hexa fluride in presence of water.
                                                                            \nXeF6<\/sub> + 3H2<\/sub>0 \u2192 XeO3<\/sub> + 6HF<\/p>\n

                                                                            2. In BrF3<\/sub> the central atom Br is sp3<\/sup>d hybridised with 3 bond pairs and 2 lone pairs. According to VSEPR theory the expected geometry is trigonal bipyramidal. But due to strong Ip – Ip and Ip – bp repulsions compared to weak bp – bp repulsion it Assumes a bend T – shape as shown below
                                                                            \n\"Plus<\/p>\n

                                                                            Question 11.
                                                                            \nAccount for the following:<\/p>\n

                                                                              \n
                                                                            1. BiH3<\/sub> is the strongest reducing agent among all the hydrides of group 15 elements.<\/li>\n
                                                                            2. Bleaching action of chlorine.<\/li>\n<\/ol>\n

                                                                              Answer:<\/p>\n

                                                                                \n
                                                                              1. It is because BiH3<\/sub> is least stable among the hydrides of group 15 elements.<\/li>\n
                                                                              2. Bleaching action of chlorine is due to oxidation. The nacent oxygen ([O]) produced is responsible for the bleaching action.<\/li>\n<\/ol>\n

                                                                                CI2<\/sub> + H2<\/sub>O \u2192 2HCI + [O]
                                                                                \nColoured substance + [O] \u2192 Colourless substance<\/p>\n

                                                                                Plus Two Chemistry The p Block Elements NCERT Questions and Answers<\/h3>\n

                                                                                Question 1.<\/p>\n

                                                                                  \n
                                                                                1. Noble gases have very low boiling points. Why?<\/li>\n
                                                                                2. XeF2<\/sub> and XeF4<\/sub> are important Xe compounds. How can we prepare them and what is the action of water on them?<\/li>\n<\/ol>\n

                                                                                  Answer:
                                                                                  \n1. Noble gases being monoatomic have no interatomic forces except weak dispersion forces and therefore, they are liquefied at very low temperatures. Hence, they have low boiling points.<\/p>\n

                                                                                  2. XeF2<\/sub> and XeF4<\/sub> are formed by the direct reaction of elements under specific conditions.
                                                                                  \n\"Plus
                                                                                  \nThey are readly hydrolysed even by traces of water,
                                                                                  \ne.g. 2XeF2<\/sub>(s) + 2H2<\/sub>O(l) \u2192 2Xe(g) + 4HF(aq) + O2<\/sub>(g)<\/p>\n

                                                                                  Question 2.
                                                                                  \nWhy does the reactivity of nitrogen differ from phosphorus?
                                                                                  \nAnswer:<\/p>\n

                                                                                    \n
                                                                                  1. Nitrogen has a small size, high electronegativity, high ionisation enthalpy as compared to phosphorus.<\/li>\n
                                                                                  2. Nitrogen does not contain vacant d-orbitals in its valence shell whereas phosphorus contains vacant d-orbitals in its valence shell.<\/li>\n
                                                                                  3. Nitrogen has the ability to form triple bond ( N \u2261 N) as a result of which its bond enthalpy (941.4 kJ mol-1<\/sup>) is very high making it less reactive.<\/li>\n<\/ol>\n

                                                                                    Question 3.
                                                                                    \nWhy does NH3<\/sub> form hydrogen bond but PH3<\/sub> does not?
                                                                                    \nAnswer:
                                                                                    \nIn NH3<\/sub>, the nitrogen atom forms hydrogen bond because of the following reasons:<\/p>\n

                                                                                      \n
                                                                                    • Small size of nitrogen<\/li>\n
                                                                                    • High electronegativity (3.0) of nitrogen<\/li>\n<\/ul>\n

                                                                                      Due to more difference of electronegativity between N and H atom the N-H bond is polar forming hydrogen bond. On the contrary, in PH3<\/sub> the P-H bond is almost purely covalent due to larger size and lesser electronegativity (2.11) of phosphorus and hence does not form hydrogen bond.<\/p>\n

                                                                                      Question 4.
                                                                                      \nThe HNH angle is higher than HPH, HAsH and HSbH angles. Why?
                                                                                      \nAnswer:
                                                                                      \nBecause in NH3<\/sub> is sp3<\/sup> hybridised. Due to lone pair of electrons the bond angle contracts from 109\u00b0 28\u2019 to 106.5\u00b0. The decreased bond angle in other hydrides is because of the fact that the sp3<\/sup> hybridisation becomes less and less distinct with increasing size of the central atom i.e., pure p-orbitals are utilised in M-H bonding or in simple words the s- orbital of H atom overlaps with orbital having almost pure p-character. Thus, the bond angle of the hydrides of group 15 decreases as
                                                                                      \n\"Plus<\/p>\n

                                                                                      Question 5.
                                                                                      \nCan PCI3<\/sub> act as an oxidising as well as a reducing agent? Justify.
                                                                                      \nAnswer:
                                                                                      \nYes, This is because in PCI3<\/sub> phosphorus is in the intermediate oxidising state of -3.
                                                                                      \n1. As an reducing agent:
                                                                                      \nThe following reactions support the reducing behaviour of PCI3<\/sub>.<\/p>\n

                                                                                        \n
                                                                                      • PCl3<\/sub> + SO2<\/sub>Cl2<\/sub> \u2192 PCl5<\/sub> + SO2<\/sub><\/li>\n
                                                                                      • PCl3<\/sub> + SO3<\/sub> \u2192 POCl3<\/sub> + SO2<\/sub><\/li>\n<\/ul>\n

                                                                                        2. As an oxidising agent:
                                                                                        \nIt oxidises metals to their respective chlorides.<\/p>\n

                                                                                          \n
                                                                                        • 12Ag + 4PCl3<\/sub> \u2192 12AgCl + P4<\/sub><\/li>\n
                                                                                        • 6Na + PCl3<\/sub> \u2192 3NaCl + Na3<\/sub>P<\/li>\n<\/ul>\n

                                                                                          Question 6.
                                                                                          \nExplain why inspite of nearly the same electronegativity, oxygen forms hydrogen bonding while chlorine does not.
                                                                                          \nAnswer:
                                                                                          \nOxygen atom can form hydrogen bond whereas chlorine does not. The tendency for hydrogen bonding depends upon<\/p>\n

                                                                                            \n
                                                                                          1. Small size and<\/li>\n
                                                                                          2. High electronegativity values<\/li>\n<\/ol>\n

                                                                                            Although the electronegativities of O and Cl are nearly the same yet chlorine does not form hydrogen bond due to its larger size (99 pm) as compared to oxygen (66 pm).<\/p>\n

                                                                                            We hope the given Plus Two Chemistry Chapter Wise Questions and Answers Chapter 7 The p Block Elements will help you. If you have any query regarding Plus Two Chemistry Chapter Wise Questions and Answers Chapter 7 The p Block Elements, drop a comment below and we will get back to you at the earliest.<\/p>\n","protected":false},"excerpt":{"rendered":"

                                                                                            Plus Two Chemistry Chapter Wise Questions and Answers Chapter 7 The p Block Elements are part of Plus Two Chemistry Chapter Wise Questions and Answers. Here we have given Plus Two Chemistry Chapter Wise Questions and Answers Chapter 7 The p Block Elements. Kerala Plus Two Chemistry Chapter Wise Questions and Answers Chapter 7 The […]<\/p>\n","protected":false},"author":7,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_genesis_hide_title":false,"_genesis_hide_breadcrumbs":false,"_genesis_hide_singular_image":false,"_genesis_hide_footer_widgets":false,"_genesis_custom_body_class":"","_genesis_custom_post_class":"","_genesis_layout":"","footnotes":""},"categories":[42728],"tags":[],"yoast_head":"\nPlus Two Chemistry Chapter Wise Questions and Answers Chapter 7 The p Block Elements - A Plus Topper<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.aplustopper.com\/plus-two-chemistry-chapter-wise-questions-answers-chapter-7\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Plus Two Chemistry Chapter Wise Questions and Answers Chapter 7 The p Block Elements\" \/>\n<meta property=\"og:description\" content=\"Plus Two Chemistry Chapter Wise Questions and Answers Chapter 7 The p Block Elements are part of Plus Two Chemistry Chapter Wise Questions and Answers. 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