{"id":10441,"date":"2019-12-07T13:00:27","date_gmt":"2019-12-07T07:30:27","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=10441"},"modified":"2021-07-02T15:43:23","modified_gmt":"2021-07-02T10:13:23","slug":"2nd-puc-chemistry-question-bank-chapter-7","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-chemistry-question-bank-chapter-7\/","title":{"rendered":"2nd PUC Chemistry Question Bank Chapter 7 The p-Block Elements"},"content":{"rendered":"

You can Download Chapter 7 The p-Block Elements Questions and Answers, Notes, 2nd PUC Chemistry\u00a0Question Bank with Answers<\/a> Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 2nd PUC Chemistry Question Bank Chapter 7 The p-Block Elements<\/h2>\n
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2nd PUC Chemistry The p-Block Elements NCERT Textbook Questions and Answers<\/h3>\n

Question 1.
\nDiscuss the general characteristics of Group 15 elements with reference to their electronic configuration, oxidation state, atomic size, ionisation enthalpy and electronegativity.
\nAnswer:
\n(i) Electronic configuration ‘All the elements in group 15 have 5 valence electrons. Their general electronic configuration is ns2<\/sup>np3<\/sup><\/p>\n

(ii) Oxidation states
\nAll these elements have 5 valence electrons and require three more electrons to complete 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 valance shell is relatively small. The remaining elements of this group show a formal oxidation state of -3 in their covalent compounds.<\/p>\n

In addition to – 3 state, N and P also show -1 and.-2 oxidation states. All the elements present in this group show +3 and +5 oxidation states. However stability of +5 oxidation state decreases down the group, whereas the stability of+3 oxidation state increases. This happens because of the inert pair effect.<\/p>\n

(iii) Ionization energy and electronegativity First ionization energy decreases on moving down a group. This is because of increasing atomic sizes. As we move down a group, electronegativity decreases, owing to an increase in size.<\/p>\n

(iv) Atomic size
\nOn moving down a group, the atomic size increases. This increase in the atomic size is attributed to an increase in number of shell.<\/p>\n

Question 2
\nWhy does the reactivity of nitrogen differ
\nfrom phosphorus? i
\nAnswer:
\nNitrogen is chemically less reactive. This is because of high stability of its molecule, N2<\/sub>. In N2<\/sub> the two nitrogen atoms form a triple bond. This triple bond has a very high bond strength, which is very difficult to break. It is because of nitrogens small size that it is able to form p\u03c0-p\u03c0 bonds with itself. This property is not exhibited by atoms such as phosphorus. Thus, phosphorus is more reactive than nitrogen.<\/p>\n

Question 3
\nDiscuss the trends in chemical reactivity of group 15 elements.
\nAnswer:
\nGeneral trends in chemical properties of group-15
\n(i) Reactivity towards hydrogen:
\nThe elements of group 15 react with hydrogen to form hydrides of type EH3<\/sub>, where E – N, P, As, Sb, or Bi. The stability of hydrides decreases on moving down from NH3 to BiH3<\/sub><\/p>\n

(ii) Reactivity towards oxygen:
\nThe elements of group 15 form two types of oxides: E2<\/sub>O3<\/sub> arid E2<\/sub>O5<\/sub>, where E = N, P, As, Sb or Bi. The oxide with the element in higher oxidation state is more acidic than the other. However, the acidic character decreases on moving down a group.<\/p>\n

(iii) Reactivity towards halogens:
\nThe group 15 element react with halogens to form two series of salts: EX3<\/sub> and EX5<\/sub>. However, nitrogen does not form NX5<\/sub> as H lacks the d-orbital. All trihalides (except NX3<\/sub>) are stable.<\/p>\n

(iv) Reactivity towards metals:
\nThe group 15 elements react with metals to form binary compounds in which these metals exhibit-3 oxidation states.<\/p>\n

Question 4
\nWhy does NH3<\/sub> form hydrogen bond but PH3<\/sub> does not?
\nAnswer:
\nNitrogen is highly electronegative as compared to phosphorus. This causes a greater attraction of electrons towards nitrogen in NH3<\/sub> than towards phosphorus in PH3<\/sub>. Hence, the extent of hydrogen bonding in PH3<\/sub> is very less as compared to NH3<\/sub>.<\/p>\n

\"KSEEB<\/p>\n

Question 5.
\nHow is nitrogen prepared in the laboratory? Write the chemical equations of the reactions involved.
\nAnswer:
\nAn aqueous solution of ammonium chloride is treated with sodium nitrite
\nNH4<\/sub>Cl + NaNO2<\/sub>(aq) \u2192 N2<\/sub>(g) + 2H2<\/sub>O(e) + NaCl(aq)
\nNO and HNO3<\/sub> are produced in small amounts. These are impurities that can be removed on passing nitrogen gas through aqueous H2<\/sub>SO4<\/sub> containing potassium dichromate<\/p>\n

Question 6
\nHow is ammonia manufactured industrially?
\nAnswer:
\nAmmonia is prepared on a large-scale by the Haber\u2019s proceed.
\n\"2nd<\/p>\n

Question 7
\nIllustrate how copper metal can give different products on reaction with HNO3<\/sub>.
\nAnswer:
\nOn heating with dil HNO3<\/sub>, copper gives copper nitrate and nitric oxide.
\n\"2nd<\/p>\n

Question 8
\nGive the resonating structures of NO2<\/sub> and N2<\/sub>O5<\/sub>.
\nAnswer:
\nResonating structures of NO2<\/sub> are:
\n\"2nd
\nResonating structures of N2<\/sub>O5<\/sub> are:
\n\"2nd<\/p>\n

Question 9.
\nThe HNH angle value is higher than HPH, HAsH and HSbH angles. Why?
\n[Hint: Can be explained on the basis of sp3<\/sub> hybridisation in NH3<\/sub> and only s-p bonding between hydrogen and other elements of the group].
\nAnswer:
\nHydride NH3<\/sub>, PH3<\/sub>, AsH3<\/sub>, SbH3<\/sub>
\nH-M-H angle 107\u00b0, 92\u00b0, 91\u00b0, 90\u00b0<\/p>\n

The above trend in H-M-H bond angle can be explained on the basis of the electronegativity tif central atom. Since nitrogen is highly electronegative, there is high electron density around nitrogen. This causes greater repulsion between the electron pairs around nitrogen, resulting in maximum bond angle. We know that electronegativity decreases on moving down a group consequently. The repulsive interactions between the electron pairs decreases, thereby decreasing the HMH bond angle.<\/p>\n

\"KSEEB<\/p>\n

Question 10.
\nWhy does R3<\/sub>P = O exist but R3<\/sub>N = O does not (R = alkyl group)?
\nAnswer:
\nN (unlike p) lacks d-orbital. This restricts nitrogen to expand its coordination number beyond four. Hence, R3<\/sub>N = O doesnot exits.<\/p>\n

Question 11.
\nExplain why NH3is basic while BiH3<\/sub> is only feebly basic.
\nAnswer:
\nNH3 is distinctly basic while BiH3<\/sub> is feebly basic.
\nNitrogen has a small size due to which the lone pair of electrons is concentrated in a small region. This means that the charge density per unit volume is high. On moving down a group, the size of the central atom increases and the charge gets distributed over a large area decreasing the electron density. Hence the electron donating capacity of group 15 elements hydrides decreases on moving down the group.<\/p>\n

Question 12.
\nNitrogen exists as diatomic molecule and phosphorus as P4<\/sub>. Why?
\nAnswer:
\nNitrogen owing to its small size has a tendency to form pn-pn multiple bonds with itself. Nitrogen thus forms a very stable diatomic molecule, N2<\/sub>. On moving down a group, the tendency to form p\u03c0-p\u03c0 bonds decreases (because of the large size of heavier elements). Therefore, phosphorus (like other heavier metals) exists in the P4<\/sub> state.<\/p>\n

Question 13
\nWrite main differences between the properties of white phosphorus and red phosphorus.
\nAnswer:<\/p>\n\n\n\n\n\n\n\n\n
White phosphorous<\/strong><\/td>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Red phosphorous<\/strong><\/td>\n<\/tr>\n
It is a soft and waxy solid. It possesses agarlic smell.<\/td>\nIt is hard and crystalline solid.<\/td>\n<\/tr>\n
It is poisonous<\/td>\nwithout any smell<\/td>\n<\/tr>\n
It is insoluble in water but soluble in carbon disulphide<\/td>\nIt is non-poisonous It is insoluble in both water and carbon<\/td>\n<\/tr>\n
It undergoes spontaneous combustion in air<\/td>\ndisulphide It is relatively<\/td>\n<\/tr>\n
In both solid and vapour states, it exists as a P4<\/sub> molecule<\/td>\nless reactive It exists as a chain of tetrahedral P4<\/sub>.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Question 14
\nWhy does nitrogen show catenation properties less than phosphorus?
\nAnswer:
\nThe property of catenation depends upon the strength of the element bond. Since the N- N (159 kJ mol-1<\/sup>) bond strength is much weaker than P-P (213 kJ mol-1<\/sup>) bond strength, therefore nitrogen show less catenation properties than phosphorus.<\/p>\n

Question 15
\nGive the disproportionation reaction of
\nH3<\/sub>PO3<\/sub>.
\nAnswer:
\nH3<\/sub>PO3<\/sub> on heating undergoes self oxidation reduction, i.e. disproportionation to form PH, in which P is reduced and H3<\/sub>PO4<\/sub> in which P is oxidised.
\n\"2nd<\/p>\n

Question 16.
\nCan PCl5<\/sub> act as an oxidising as well as a reducing agent? Justify.
\nAnswer:
\nPCl5<\/sub> can only act as an oxidizing agent. The highest oxidation state that P can show is +5. In PCl5<\/sub> , phosphorus is in its highest oxidation state (+5). However it can decrease its oxidation state and act as an oxidizing agent.<\/p>\n

Question 17.
\nJustify the placement of Q, S, Se, Te and Po in the same group of the periodic table in terms of electronic configuration, oxidation state and hydride formation.
\nAnswer:
\nThe elements of group 16 are collectively called chalcogens.
\n(i) Elements of group 16 have six valence electrons each. The general electronic configuration of these elements is ns2<\/sup> np4<\/sup>, where n varies from 2 to 6.
\n(ii) oxidation state:
\nAs these elements have six valence electrons (ns2<\/sup> np4<\/sup>). They should display an oxidation state of -2. However only oxygen predominantly shows the oxidation state of -2 owing to its high electronegativity. It also exhibits the oxidation . state of -1 (H2<\/sub>O2<\/sub>), Zero (O2<\/sub>), and +2 (OF2<\/sub>) However, 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 of the group show an oxidation state of +2, +4 and +6 due to the availability of d-orbitals.<\/p>\n

(iii) Formation of hydrides
\nThese elements form hydrides of formula H2<\/sub>E, where E = O, S, Se, Te, PO. Oxygen and sulphur also form hydrides of type H2<\/sup>E2<\/sup>. These hydrides are quite volatile in nature.<\/p>\n

\"KSEEB<\/p>\n

Question 18.
\nWhy is dioxygen a gas but sulphur a solid?
\nAnswer:
\nOxygen is smaller in size as compared to sulphur. Due to its smaller size, it can effectively form p\u03c0-p\u03c0 bonds and form O2<\/sub>\u00a0(O = O) molecule. Also, the intermolecular forces in oxygen are weak Van dar Wall\u2019s which cause
\nit to exist as gas. On the other hand, Sulphur does not form M2<\/sub> molecule but exists as a puckered structure held to gather by strong covalent bonds. Hence it is a solid.<\/p>\n

Question 19
\nKnowing the electron gain enthalpy . values for O \u2192 O–<\/sup> and O \u2192 O2-<\/sup> as -141 and 702 kj mol-1<\/sup> respectively, how can you account for the formation of a large number of oxides having O2-<\/sup> species and not O–<\/sup>?
\n(Hint: Consider lattice energy factor in the formation of compounds).
\nAnswer:
\nStability of an ionic compound depends on its lattice energy. More the lattice energy of a compound, more stable it will be. Lattice energy is directly proportional to the charge carried by an ion. When a metal combines with oxygen, the lattice energy of the oxide involving O2-<\/sup> ion is much more than the oxide involving O–<\/sup> ion. Hence, we can say that formation of O2-<\/sup> is energetically more favourable than formation of O–<\/sup><\/p>\n

Question 20
\nWhich aerosols deplete ozone?
\nAnswer:
\nFreons or chlorofluro carbons (CFCS<\/sub>) are aerosols that accelerate the depletion of ozone. In the presence UV radiations, molecules of CFC\u2019s break down to form chlorine – free radicals that combine with O3<\/sub> to form oxygen.<\/p>\n

Question 21.
\nDescribe the manufacture of H2<\/sub>SO4<\/sub> by contact process?
\nAnswer:
\nH2<\/sub>SO4<\/sub> is manufactured by contact process. It involves the following steps<\/p>\n

Step (1)
\nSulphur or sulphide ores are burnt in air to form SO2<\/sub>.<\/p>\n

Step (2)
\nBy a reaction with O2<\/sub>, SO2<\/sub> is converted to SO3<\/sub> in the presence of V2<\/sub>O5<\/sub> as a catalyst.
\n\"2nd<\/p>\n

Step (3)
\nSO3<\/sub> produced is absorbed on H2<\/sub>SO4<\/sub> to give H2<\/sub>S2<\/sub>O7<\/sub>(Oleum)<\/p>\n

SO3<\/sub>+ H2<\/sub>SO4<\/sub> \u2192 H2<\/sub>S2<\/sub>O7<\/sub><\/p>\n

This oleum is then diluted to obtain H2<\/sub>SO4<\/sub>, of the desired concentration. In practice, the plant is operated at 2 bar (pressure) and 720 K (temperature). The H2<\/sub>SO4<\/sub> thus obtained is 96 – 98% pure.<\/p>\n

Question 22.
\nHow is SO2<\/sub> an air pollutant?
\nAnswer:<\/p>\n

    \n
  • It combines with water vapour present in the atmosphere to form sulphuric acid. This causes acid rain. Acid rain damages soil, plants and buildings, especially those made of marble.<\/li>\n
  • Even in very low concentrations, SO2<\/sub> causes irritation in the respiratory tract. It causes throat
    \nand eye irritation and can also affect the larynx to cause breathlessness.<\/li>\n
  • It is extremely harmful to plants. Plants exposed to sulphur dioxide for a long time lose colour from their leaves. This condition is known as chlorosis. This happens because the formation of chlorophyll is affected by the presence of SO2<\/sub>.<\/li>\n<\/ul>\n

    Question 23.
    \nWhy are halogens strong oxidising agents?
    \nAnswer:
    \nThe general electronic configuration of halogens is np5<\/sup>, where n = 2-6. Thus, halogens need only one more e- to complete their octet and to attain the stable noble gas configuration. Also, halogens are highly electronegative with low dissociation energies and high negative electron gain enthalpies. Therefore, they have a high tendency to gain an election. Hence, they act as strong oxidizing agents.<\/p>\n

    \"KSEEB<\/p>\n

    Question 24.
    \nExplain why fluorine forms only one
    \nAnswer:
    \nFluorine forms only one oxoacid, i.e. HOF because of its high electronegativity and small size.<\/p>\n

    Question 25
    \nExplain why inspite of nearly the same electronegativity, oxygen forms hydrogen bonding while chlorine does not.
    \nAnswer:
    \nBoth chlorine and oxygen have almost the same electronegativity values, but chlorine rarely forms hydrogen bonding. This is because in comparison to chlorine, oxygen has a smaller size and as a result, a higher electron density per unit volume.<\/p>\n

    Question 26
    \nWrite two uses of ClO2<\/sub>.
    \nAnswer:<\/p>\n

      \n
    1. It is used for purifying water<\/li>\n
    2. It is a used as a bleaching agent.<\/li>\n<\/ol>\n

      Question 27
      \nWhy are halogens coloured?
      \nAnswer:
      \nAlmost all halogens are coloured. This is because halogens absorb radiations in the visible region. This results in the excitation of valence electrons to a higher energy region. Since the amount of energy required for excitation differ for each halogen, each halogen displays a colour.<\/p>\n

      Question 28
      \nWrite the reactions of F2<\/sub> and Cl2<\/sub> with water.
      \nAnswer:
      \n\"2nd<\/p>\n

      Question 29
      \nHow can you prepare Cl, from HCl and HCl from Cl2<\/sub>? Write reactions only.
      \nAnswer:
      \nCl2<\/sub> can be prepared from HC1 by Deacon\u2019s process
      \n\"2nd<\/p>\n

      (ii) HCl can be prepared from Cl by treating it with water
      \nCl2<\/sub> + H2<\/sub>O \u2192 HCl + HOCl<\/p>\n

      Question 30
      \nWhat inspired N. Bartlett for carrying out reaction between Xe and PtF6? ‘
      \nAnswer:
      \nNeil Bartlett initially carried out a reaction between oxygen and PtFg. This resulted in the formation of a red compound, O+<\/sup>2<\/sub> [Pt F6<\/sub> ].<\/p>\n

      Later he realised that the first ionization energy of oxygen (1175 kJ\/mol) and Xe (1170 kJ\/mol) is almost the same. Thus, he tried to prepare a compound with Xe and PtF6<\/sub>. He was successful and a red coloured compound, Xe+<\/sup> [PtF6<\/sub>]- was formed.<\/p>\n

      Question 31.
      \nWhat are the oxidation states of phosphorus in the following:
      \n(i) H3<\/sub>PO3<\/sub>
      \n(ii) PCl3<\/sub>
      \n(iii) Ca3<\/sub>P2<\/sub>
      \n(iv) Na3<\/sub>PO4<\/sub>
      \n(v) POF3<\/sub>?
      \nAnswer:
      \nLet oxidation state of P be x
      \n(i) H3<\/sub>PO3<\/sub>
      \n3 + x + 3 (-2) = 0
      \n3 + x – 6 = 0
      \nx = 3.<\/p>\n

      (ii) PCl3<\/sub>
      \nx + 3 (-1) = 0
      \nx – 3 = 0
      \nx = 3<\/p>\n

      (iii) Ca3<\/sub>P2<\/sub>
      \n3(2) + 2x = 0
      \n6 + 2x = 0
      \nx = -3<\/p>\n

      (iv) Na3<\/sub>PO4<\/sub>
      \n3 x 1 +x + 4(-2) = 0
      \n3 + x – 8 = 0
      \nx = 5<\/p>\n

      (v) POF3<\/sub>
      \nx + (-2) + 3x (-1) = 0
      \nx – 5 = 0
      \nx = 5.<\/p>\n

      Question 32.
      \nWrite balanced equations for the following:
      \n(i) NaCl is heated with sulphuric acid in the presence of MnO2<\/sub>.
      \n(ii) Chlorine gas is passed into a solution of Nal in water.
      \nAnswer:
      \n(i) 4NaCl + MnO2<\/sub> + 4H2<\/sub>SO4<\/sub> \u2192 MnCl2<\/sub> + 4NaHSO4<\/sub>+ 2H2<\/sub>O + Cl2<\/sub>
      \n(ii) Cl2<\/sub> + Nal \u2192 2NaCl + I2<\/sub><\/p>\n

      \"KSEEB<\/p>\n

      Question 33.
      \nHow are xenon fluorides XeF2<\/sub>, XeF4<\/sub> and XeF. obtained?
      \nAnswer:
      \nXeF2<\/sub>, X3<\/sub>F4<\/sub> and XeF6<\/sub> are obtained by a direct reaction between Xe and F2<\/sub>. The condition under which the reaction is carried out determines the product.
      \n\"2nd<\/p>\n

      Question 34
      \nWith what neutral molecule\u2019 is ClO–<\/sup> isoelectronic? Is that molecule a Lewis base?
      \nAnswer:
      \nClO–<\/sup> is isoelectronic to ClF. Also, both species contain 26e–<\/sup>s in all as shown.
      \nTotal electrons in ClO–<\/sup> = 17 + 8 + 1 = 26
      \nIn ClF =17 + 9 =26
      \nClF, acts like a lewis base as it accepts e-s from F to form ClF3<\/sub><\/p>\n

      Question 35
      \nHow are XeO3<\/sub> and XeOF4<\/sub> prepared?
      \nAnswer:
      \n(i) XeO3<\/sub> can be prepared in two ways as shown:
      \n6XeF4<\/sub> + 12H2<\/sub>O \u2192 4Xe + 2XeO3<\/sub> + 24HF + 3O2<\/sub> XeF6<\/sub> + 3H2<\/sub>O \u2192 XeO3<\/sub> + 6HF
      \n(ii) XeF4 can be prepared using XeF6
      \nXeF6<\/sub>+ H2<\/sub>O \u2192 XeOF4<\/sub> + 2HF<\/p>\n

      Question 36
      \nArrange the following in the order of property indicated for each set:
      \n(i) F2<\/sub>, Cl2<\/sub>, Br2<\/sub>, I2<\/sub> – increasing bond dissociation enthalpy.
      \nAnswer:
      \nBond dissociation energy usually decreases on moving down a group as the atomic size increases. However, the bond dissociation energy of F2<\/sub> is lower than that of Cl2<\/sub> and Br2<\/sub>. This is due to the small atomic size of fluorine, Thus, the increasing order for bond dissociation energy among halogens is as follows:<\/p>\n

      I2<\/sub> < F2<\/sub> < Br2<\/sub> < Cl2<\/sub><\/p>\n

      (ii) HF, HCl, HBr, HI – increasing acid strength.
      \nAnswer:
      \nHF < HCl < HBr < HI
      \nThe bond dissociation energy of HX molecules where X – F, Cl, Br, I, decreases with an increase in the atomic size. Since H – I bond is the weakest, HI is the strongest acid.<\/p>\n

      (iii) NH3<\/sub>, PH3<\/sub>, ASH3<\/sub>, SbH3<\/sub>, BiH3<\/sub> – increasing base strength.
      \nAnswer:
      \nBiH3<\/sub> < SbH3<\/sub> < AsH3<\/sub> < PH3<\/sub> < NH3<\/sub> On moving from nitrogen to bismuth, the size of the atom increases while the electron density on the atom decreases. Thus, the basic strength decreases.<\/p>\n

      \"KSEEB<\/p>\n

      Question 37.
      \nWhich one of the following does not exist?
      \n(i) XeOF4<\/sub>
      \n(ii) NeF2<\/sub>
      \n(iii) XeF2<\/sub>
      \n(iv) XeF
      \nAnswer:
      \n(ii) NeF2<\/sub><\/p>\n

      Question 38.
      \nGive the formula and describe the structure of a noble gas species which is isostructural with:
      \n(i) ICl4<\/sub>–<\/sup>
      \n(ii) IBr2<\/sub>–<\/sup>
      \n(iii) BrO3<\/sub>–<\/sup>
      \nAnswer:
      \n(i) XeF4<\/sub> is isoelectronic with ICl4<\/sub>–<\/sup> and has square planar geometry
      \n\"2nd<\/p>\n

      (ii) XeF2<\/sub> is isoelectronic to IBr2<\/sub>–<\/sup> and has a linear structure
      \n\"2nd<\/p>\n

      (iii) XeO3<\/sub> is isostructural to BrO3<\/sub>–<\/sup> and has a pyramidal molecular structure.
      \n\"2nd<\/p>\n

      Question 39.
      \nWhy do noble gases have comparatively large atomic sizes?
      \nAnswer:
      \nNoble gases do not form molecules. In case of noble gases. the atomic radii corresponds to Van der Waa1s radii. On the other hand, the atomic radii of other elements correspond to their covalent radii. By definition, Yan der WaaI\u2019s radii are larger than covalent radii. It is for this reason that noble gases are very large in size as compared to other atoms belonging to the same period.<\/p>\n

      \"KSEEB<\/p>\n

      Question 40
      \nList the uses of neon and argon gases.
      \nAnswer:
      \nUses of neon gas:
      \n(i) It is mixed with helium to protect electrical equipments from high voltage.
      \n(ii) It is filled in discharge tubes with characteristic colours.
      \n(iii) It is used in beacon lights.<\/p>\n

      Uses of argon gas:
      \n(i) Argon along with nitrogen is used in gas filled electric lamps. This is because Ar is more inert than N.
      \n(ii) It is usually used to provide an inert temperature in a high metallurgical process.
      \n(iii) It is also used in laboratories to handle air sensitive substances.<\/p>\n

      2nd PUC Chemistry The p-Block Elements Additional Questions and Answers<\/h3>\n

      Question 1.
      \nWhat is the electronic .configuration of p-block elements?
      \nAnswer:
      \nns2<\/sup>np0 – 5<\/sup><\/p>\n

      Question 2.
      \nWrite three uses of nitrogen?
      \nAnswer:<\/p>\n

        \n
      1. Used in the manufacture of nitric acid, ammonia, calcium cyanamide.<\/li>\n
      2. Inert atmosphere for iron and steel industry.<\/li>\n
      3. It is used as a refrigerant.<\/li>\n<\/ol>\n

        Question 3.
        \nWrite three physical properties of Ammonia?
        \nAnswer:<\/p>\n

          \n
        1. It is a colourless gas.<\/li>\n
        2. It is lighter than air.<\/li>\n
        3. Extremely soluble in water.<\/li>\n<\/ol>\n

          Question 4.
          \nWrite three physical properties of Nitric acid?
          \nAnswer:<\/p>\n

            \n
          1. Colourless fuming liquid.<\/li>\n
          2. Melting point is 231.4k.<\/li>\n
          3. Boiling point is 355.6 k.<\/li>\n<\/ol>\n

            Question 5.
            \nWhat are the three allotropic forms of phosphorous.
            \nAnswer:<\/p>\n

              \n
            1. \u00a0white phosphorous<\/li>\n
            2. Red phosphorous<\/li>\n
            3. Black phosphorous.<\/li>\n<\/ol>\n

              Question 6.
              \nWrite the chemical formula of Hypo phosphorus acid.
              \nAnswer:
              \nH3<\/sub>PO2<\/sub>.<\/p>\n

              \"KSEEB<\/p>\n

              Question 7.
              \nWrite the chemical formula of orthophosphorous acid.
              \nAnswer:
              \nH3<\/sub>PO3<\/sub><\/p>\n

              Question 8.
              \nWrite the chemical formula of pyrophosphorous acid.
              \nAnswer:
              \nH4<\/sub>P2<\/sub>O5<\/sub><\/p>\n

              Question 9.
              \nDraw the structure of PCl5.<\/sub>
              \nAnswer:
              \n\"2nd<\/p>\n

              Question 10.
              \nDraw the structure of SF6<\/sub>.
              \nAnswer:
              \n\"2nd<\/p>\n

              \"KSEEB<\/p>\n<\/div>\n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

              You can Download Chapter 7 The p-Block Elements Questions and Answers, Notes, 2nd PUC Chemistry\u00a0Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations. Karnataka 2nd PUC Chemistry Question Bank Chapter 7 The p-Block Elements 2nd PUC Chemistry The p-Block Elements NCERT Textbook Questions …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[82],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/10441"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=10441"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/10441\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=10441"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=10441"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=10441"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}