{"id":8764,"date":"2019-11-27T11:06:21","date_gmt":"2019-11-27T05:36:21","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=8764"},"modified":"2021-07-02T15:44:53","modified_gmt":"2021-07-02T10:14:53","slug":"1st-puc-chemistry-question-bank-chapter-3","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-chemistry-question-bank-chapter-3\/","title":{"rendered":"1st PUC Chemistry Question Bank Chapter 3 Classification of Elements and Periodicity in Properties"},"content":{"rendered":"

You can Download Chapter 3 Classification of Elements and Periodicity in Properties Questions and Answers, Notes, 1st PUC Chemistry Question Bank with Answers<\/a> Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 1st PUC Chemistry Question Bank Chapter 3 Classification of Elements and Periodicity in Properties<\/h2>\n

1st PUC Chemistry Classification of Elements and Periodicity in Properties One Mark Questions and Answers<\/h3>\n

Question 1.
\nDefine atomic radius.
\nAnswer:
\nAtomic radius is the distance from the center of the nucleus to the point where the electron density is effectively zero.<\/p>\n

Question 2.
\nDefine Vander Waal\u2019s radius.
\nAnswer:
\nVander Waal\u2019s radius is one half of the distance between the nuclei of two non bonded adjacent atoms belonging to two neighboring molecules of an elements in the solid state.<\/p>\n

Question 3.
\nDefine ionic radius.
\nAnswer:
\nIonic radius is the distance from the nucleus of an ion to the point up to which the nucleus has influence on its electron cloud.
\nOR
\nIonic radius is the distance from the nucleus of an ion to the outer most orbital containing electrons.<\/p>\n

\"KSEEB<\/p>\n

Question 4.
\nWhat is ionization energy?
\nAnswer:
\nIonisation energy is the amount of energy required to remove the most loosely bound electron from an isolated neutral gaseous atom.<\/p>\n

Question 5.
\nWhy ionisation potential of inert gases are comparatively higher?
\nAnswer:
\nInert gases have completely filled stable electronic configuration. A lot of energy is required to disturb that stable electronic configuration and remove an electron. Hence inert gases have a high ionsation potential.<\/p>\n

Question 6.
\nAmong Na+<\/sup>, Ca+2<\/sup>, Al+3<\/sup> which is having smallest size?
\nAnswer:
\nAl+3<\/sup>.<\/p>\n

Question 7.
\nWhich has got the smallest size among Fe, Fe+2<\/sup> & Fe3+<\/sup>?
\nAnswer:
\nFe2+<\/sup> is smallest. ,<\/p>\n

Question 8.
\nThe electron affinity of Nitrogen is more than that of oxygen, why?
\nAnswer:
\nBecause Nitrogen contains halfly filled two orbitals, which is an extra stable state.<\/p>\n

Question 9.
\nAlkali metals have low ionization energy why?
\nAnswer:
\nAlkali metals are present in the periodic table after inert gases. By loosing one electron they gets electronic configuration of inert gas.<\/p>\n

Question 10.
\nWhat are iso electronic ions?
\nAnswer:
\nIons having same number of electrons but differ in the atomic number.<\/p>\n

Question 11.
\nHow does ionization energy varies along a period and down the group?
\nAnswer:
\nIonisation energy increases along the period and decreases down the group.<\/p>\n

\"KSEEB<\/p>\n

Question 12.
\nDefine covalent radius.
\nAnswer:
\nCovalent radius is one half of the distance between nuclei of two covalently bound atoms of the same element in a molecule.<\/p>\n

Question 13.
\nName the element having highest electron affinity.
\nAnswer:
\nChlorine.<\/p>\n

Question 14.
\nArrange F, Cl, Br and I in the order of increasing electron affinity.
\nAnswer:
\nI, Br, F, Cl.<\/p>\n

Question 15.
\nWhat is meant by electro negativity of an atom?
\nAnswer:
\nIt is tendency of an atom in a molecule to attract the shared pair of electron to itself.<\/p>\n

Question 16.
\nGroup the following species that are isoelectronic.
\nBe2+<\/sup>, F–<\/sup>, Fe2+<\/sup>, N3-<\/sup>, He, S2-<\/sup> , CO3+<\/sup>, Ar
\nAnswer:
\n(Be2+<\/sup>,He); (F–<\/sup>,N3-<\/sup>); (Fe3+<\/sup>,CO3+<\/sup>); (S2-<\/sup>,Ar)<\/p>\n

Question 17.
\nWhich one has the larger size : Fe2+<\/sup> or Fe3+<\/sup> ?
\nAnswer:
\nFe2+<\/sup><\/p>\n

Question 18.
\nState the modern periodic law.
\nAnswer:
\nProperties of elements are periodic functions of their atomic numbers.<\/p>\n

\"KSEEB<\/p>\n

Question 19.
\nName the element which is most electronegative and the element which is least electronegative in the periodic chart.
\nAnswer:
\nFluorine is the most electronegative (EN – 4.0) element
\nCeasium is the least electronegative (EN = 0.7) element<\/p>\n

Question 20.
\nWrite the general outer electronic configurations of the following elements,
\na) alkali metals
\nb) alkaline earth metals
\nc) halogens
\nd) nobel gases
\nAnswer:
\n(a) alkali metals – ns-1<\/sup>
\n(b) alkaline earth metals – ns2<\/sup>
\n(c) halogens – ns2<\/sup>np5<\/sup>
\n(d) nobel gases – ns2<\/sup>np6<\/sup><\/p>\n

Question 21.
\nWhat is the decreasing order of shielding effect of orbitals s, p, d and f.
\nAnswer:
\nDecreasing order : s > p > d > f.<\/p>\n

Question 22.
\nWhy do alkali metals have lowest ionization energy?
\nAnswer:
\nThey have largest atomic size, therefore, there is less force of attraction between valence electrons, and nucleus.<\/p>\n

Question 23.
\nWhich is smallest among Na+<\/sup>, Mg2+<\/sup>, Al3+<\/sup>, and why?
\nAnswer:
\nAl3+<\/sup> is smallest because it has highest number of protons (13) among Na+<\/sup>, Mg2+<\/sup>, and Al3+<\/sup> ions, due to which effective nuclear charge is maximum.<\/p>\n

\"KSEEB<\/p>\n

Question 24.
\nWhich has largest ionic radius among Ca2+<\/sup>, Mg2+<\/sup>, Ba2+<\/sup>?
\nAnswer:
\nBa2+<\/sup>.<\/p>\n

Question 25.
\nDefine (i) metallic radius, (ii) van der Waal\u2019s radius.
\nAnswer:
\n(i) Metallic radius is half the distance between centres of nuclei of two atoms of metal held together by metallic bond.
\n(ii) Van der Waals\u2019 radius is half of the distance between centres of nuclei of two atoms held by weak van der Waal\u2019s forces of attraction.<\/p>\n

Question 26.
\nHow does electronegativity vary (i) down the group, (ii) across the period from left to right?
\nAnswer:
\n(i) Electronegativity goes on decreasing down the group.
\n(ii) It goes on increasing along the period from left to right.<\/p>\n

Question 27.
\nWhat is the nature of oxides formed by most of p-block elements?
\nAnswer:
\nThey form mostly acidic oxides. Some of them form amphoteric and neutral oxides also.<\/p>\n

Question 28.
\nWhich of the following pairs of elements would you expect to have lower first ionization energy? (i) Cl or F, (ii) Cl or S, (iii) K or Ar, (iv) Kr or Xe.
\nAnswer:
\n(i) Cl, (ii) S, (iii) K, (iv) Xe.<\/p>\n

\"KSEEB<\/p>\n

1st PUC Chemistry Classification of Elements and Periodicity in Properties Two Marks Questions and Answers<\/h3>\n

Question 1.
\nAmong C, N, B and O which element has the highest ionization potential and which element has the lowest ionization potential. Give reason.
\nAnswer:
\nElement having highest ionization potential is N. Due to extra stability of half filled 2p orbital. Element having lowest ionization potential is B. Due to shielding effect of completely filled 2s orbital.<\/p>\n

Question 2.
\nWhich is the most electronegative element and the most electropositive element in the modern periodic table?
\nAnswer:
\nMost electronegative element is fluorine. Most electropositive element is Cesium<\/p>\n

Question 3.
\nWhich of the following two elements belong to i) the same group ii) same period in the periodic table 4Be <\/sup>; 3Li<\/sup>; 12Ms<\/sup>, 35Br<\/sup>
\nAnswer:
\n\"1st
\nPrincipal quantum number of outermost electron in Be and Li is 2. Hence both Be and Li belong to the same period i.e., 2nd<\/sup> period. Be and Mg have the same number of outermost electrons, i.e., 2. Hence both Be and Mg belong to the same group in the periodic table.<\/p>\n

Question 4.
\nWhat is ionization energy? How does it change in a period as well as in a group?
\nAnswer:
\nThe amount of energy required to remove the most loosely bound electron from an isolated gaseous atom. Ionization energy increases along the period and decreases down the group.<\/p>\n

Question 5.
\nWhat is electro negativity? How does it change in a period as well as in a group?
\nAnswer:
\nThe ability of an atom to attract the shared electron pair (of a covalent bond) in a molecule towards itself is called electro negativity. In a period from left to right the electronegativity increases. Down a group electronegativity value decreases.<\/p>\n

\"KSEEB<\/p>\n

Question 6.
\nWhat is electron affinity? How does it vary along the period and group?
\nAnswer:
\nThe energy released when an electron is added to the outer most orbit of an isolated
\nneutral gaseous atom. It increases along the period and decreases down the group.<\/p>\n

Question 7.
\nTo which blocks do the elements with following atomic number belong ? 7, 13, 25, 42
\nAnswer:
\nAt No. Electronic configuration Block of the element
\n\"1st<\/p>\n

Question 8.
\nHow does covalent radii vary in a period as well as in a group in the periodic table? What is the reason?
\nAnswer:
\nIn a period as we move from left to right covalent radius decreases in a period. As we move’ from left to right atomic number or nuclear charge increases. The pull of the electron cloud by the nucleus increases, the electron cloud shrinks and as a result the covalent radius decreases.<\/p>\n

In a group, covalent radius increases from top to bottom. In a group as we move from top to bottom one by one new orbitals are added up thereby the size of the atom goes on increasing. The atomic radius also goes on increasing.<\/p>\n

Question 9.
\nWhat are s, p, d and f block elements?
\nAnswer:
\nThe elements for which the last electron has entered in s- orbital are called s- block\u00a0elements, p-block elements are those for which the last electron has entered in the p – orbital, d-block elements are those in which the last electron has entered the d- orbital. The elements in which the last electron entered into the f – orbital of their atoms are called f – block elements.<\/p>\n

\"KSEEB<\/p>\n

Question 10.
\nGive four characteristics of s-block elements.
\nAnswer:<\/p>\n

    \n
  1. They are soft metals<\/li>\n
  2. They are highly electropositive<\/li>\n
  3. They are good . reducing agents<\/li>\n
  4. They have low melting and boiling points<\/li>\n
  5. They impart specific colour to the flame.<\/li>\n<\/ol>\n

    Question 11.
    \nGive four defects of Mendeleev\u2019s periodic table.
    \nAnswer:<\/p>\n

      \n
    1. Isotopes should be given separate place because they have different atomic mass and character<\/li>\n
    2. The increasing order of atomic weight is not maintained<\/li>\n
    3. Some elements in the same group differ in their properties<\/li>\n
    4. The position of hydrogen is not justified.<\/li>\n<\/ol>\n

      Question 12.
      \nGive two reasons, why the number of elements in first period is only 2?
      \nAnswer:
      \nIt is because 1st<\/sup>\u00a0energy level can have only Is orbital which can have two electrons. When n = 1, then 1 = 0.<\/p>\n

      Question 13.
      \nOn the basis of their electronic configurations, explain why alkali metals are highly reactive.
      \nAnswer:
      \nAlkali metals have general electronic configuration ns1. They can lose 1 electron to acquire stable electronic configuration. They have large atomic size, therefore, they can lose electron easily and they are most reactive.<\/p>\n

      Question 14.
      \nGive the order in which the melting points of halides of sodium decrease and why?
      \nAnswer:
      \nNaF > NaCl > NaBr > Nal. Greater the difference in electronegativity more will be ionic character, higher will be melting point due to high lattice energy.<\/p>\n

      Question 15.
      \nWhy are group I elements called alkali metals and group 17 are called halogens?
      \nAnswer:
      \nGroup I elements are called alkali metals because their hydroxides form soluble bases called alkalies and their ashes are alkaline in nature. Group 17 are called halogens because they are salt-producer.<\/p>\n

      Question 16.
      \nGive four characteristics of d-block elements.
      \nAnswer:<\/p>\n

        \n
      1. They show variable oxidation state<\/li>\n
      2. They form coloured ions<\/li>\n
      3. They are used as catalyst<\/li>\n
      4. They form alloys.<\/li>\n<\/ol>\n

        \"KSEEB<\/p>\n

        Question 17.
        \nGive any two features of Mendeleev\u2019s periodic table.
        \nAnswer:<\/p>\n

          \n
        1. It was based on atomic mass<\/li>\n
        2. It has places for undiscovered elements.<\/li>\n<\/ol>\n

          Question 18.
          \nHow do the solubilities of alkaline earth metal sulphate and carbonates vary down the group and why?
          \nAnswer:
          \nSolubilities of alkali earth metal sulphates and carbonate decrease down the group because lattice energy dominates over hydration energy.<\/p>\n

          Question 19.
          \nWhy is melting point of LiCl lower than NaCl?
          \nAnswer:
          \nLiCl has lower melting point than NaCl because it is covalent whereas NaCl is ionic.<\/p>\n

          Question 20.
          \nArrange the following in increasing order: (i) BeCO3<\/sub>, BaCO3<\/sub>, CaCO3<\/sub>, MgCO3<\/sub> of Thermal stability; (ii) BeCl2<\/sub>, BaCl2<\/sub>, SrCl2<\/sub>, CaCl2<\/sub> Ionic character.
          \nAnswer:
          \n(i) BeCO3<\/sub> < MgCO3<\/sub> < CaCO3<\/sub> < BaCO3<\/sub>. (ii) BeCl2<\/sub> < CaCl2<\/sub> < SrCl2<\/sub> < BaCl2<\/sub>.<\/p>\n

          Question 21.
          \nWhich alkali metal carbonate is thermally unstable and why?
          \nAnswer:
          \nLi<sub>2<\/sub>CO<sub>3<\/sub> is thermally unstable because it is covalent.<\/p>\n

          \"KSEEB<\/p>\n

          Question 22.
          \nOut of O and S which has higher negative electron gain enthalpy and why?
          \nAnswer:
          \nS has high electron gain enthalpy because in oxygen there is more inter-electronic
          \nrepulsion than sulphur, therefore, more energy is released in case of sulphur on gaining electrons.<\/p>\n

          Question 23.
          \nPredict which atom in each of the following pairs has the highest first ionization energy, (a) B and C, (b) N and O, (c) F and Ne.
          \nAnswer:
          \n(a) C, (b) N, (c) Ne.<\/p>\n

          Question 24.
          \nAmong the elements Li, K, Ca, S and Kr, which one is expected to have the lowest first ionization enthalpy and which one has the highest first ionization enthalpy? ,
          \nAnswer:
          \nK has lowest first ionization enthalpy whereas Kr has highest first ionization enthalpy.<\/p>\n

          Question 25.
          \nAmong thp elements of the third period Na to Ar pick out the element:
          \n(i) with highest first ionization enthalpy, (ii) with largest atomic radius,
          \n(iii) that is most reactive non-metal, (iv) that is most reactive metal.
          \nAnswer:
          \n(i) Ar, (ii) Na, (iii) Cl, (iv) Na.<\/p>\n

          Question 26.
          \nName a species that will be isoelectronic with each of the following atoms or ions: (i) Ne, (ii) Cl–<\/sup> (iii) Ca2+<\/sup>, (iv) Rb.
          \nAnswer:
          \n(i) Na+<\/sup>, (ii) Ar, (iii) S2+<\/sup>, (iv) Y2+<\/sup>.<\/p>\n

          Question 27.
          \nArrange the following ions in the order of increasing size: Be3+<\/sup>, Cl–<\/sup>, S2-<\/sup>, Na+<\/sup>, Mg+<\/sup>, Br–<\/sup>.
          \nAnswer:
          \nBe2+<\/sup> < Mg2+<\/sup> < Na+<\/sup> < Cl–<\/sup>< S2<\/sup> < Br–<\/sup>.<\/p>\n

          Question 28.
          \nWrite the characteristics of p-block elements.
          \nAnswer:
          \n(i) p-block elements consists of metals, non-metals and metalloids, (ii) p-block consists of solids, liquids and gases.<\/p>\n

          \"KSEEB<\/p>\n

          Question 29.
          \nDescribe the characteristic properties of d and f-block elements.
          \nAnswer:
          \n(i) All of them are metals, (ii) All of them are solids except Hg. (iii) They show variable oxidation state (valency), (iv) They are good conductor of heat and electricity, (v) Most of them are meltable and ductile, (vi) They form alloys, (vii) They form coloured ions.<\/p>\n

          Question 30.
          \nArrange the following elements in increasing order of metallic character: B, Al, Mg. K.
          \nAnswer:
          \nB, Al, Mg, K is increasing order of metallic character. .<\/p>\n

          Question 31.
          \nArrange the following elements in increasing order of non-metallic character: B, C, Si, N, F.
          \nAnswer:
          \nSi, B, C, N, F is increasing order of non-metallic character.<\/p>\n

          Question 32.
          \nA, B, C, D and E have the following electronic configuration:
          \nA\u00a0 = 1s2<\/sup> 2s2<\/sup> 2p1<\/sup>; B = 1s2<\/sup> 2s2<\/sup> 2p5<\/sup> 3s2<\/sup> 3P1<\/sup>; C = 1s2<\/sup> 2s2<\/sup> 2p6<\/sup> 3s2<\/sup> 3P3<\/sup>; D = 1s2<\/sup> 2s2<\/sup> 2p6<\/sup> 3s2<\/sup> 3P5<\/sup>; E = 1s2<\/sup> 2s2<\/sup> 2p6<\/sup> 3s2<\/sup> 3P6<\/sup>4s2<\/sup> Which among these belong to the same group in the periodic table?
          \nAnswer:
          \nA and B belong to same group of periodic table because they have same number of valence electrons.<\/p>\n

          Question 33.
          \nPredict the formulae of the stable binary compounds that would be formed by the following pairs of elements: (a) Silicon and oxygen, (b) Aluminium and bromine, (c)-Calcium and iodine, (d) element with atomic number 114 and fluorine, (e) element with atomic number 120 and oxygen.
          \nAnswer:
          \n(a) SiO2<\/sub> (b) AlBr3<\/sub>, (c) CaI2<\/sub>, (d) UnqF4<\/sub>, .(e) XO where X is element with atomic number 120.<\/p>\n

          Question 34.
          \nAmongst the elements B, Al, C and Si: (a) Which has the highest first ionization enthalpy? (b) Which has most negative electron gain enthalpy? (c) Which has the largest atomic radius? (d) Which has the most metallic character?
          \nAnswer:
          \n(a) Carbon, (b) Carbon, (c) Al. (d) Al.<\/p>\n

          Question 35.
          \nWhich of the elements Na, Mg, Si and P would have greatest difference between first and second ionization enthalpies?
          \nAnswer:
          \nNa has greatest difference between first and second ionization enthalpies because Na+ has stable electronic configuration, i.e., 1s2<\/sup> 2s2<\/sup> 2p6<\/sup>, therefore, it has very high second ionization energy.<\/p>\n

          Question 36.
          \nDiscuss significance of atomic number as the basis of classification of elements over mass number.
          \nAnswer:
          \nProperties of elements depend upon number of valence electrons which depend upon electronic configuration. Atomic number is needed to write electronic configuration of an element. It shows that atomic number is more important to determine chemical properties of elements than atomic mass.<\/p>\n

          Question 37.
          \n(a) Why do group I metals have lower ionization enthalpy than corresponding group II metals? (b) Why is an anion larger in size than its neutral atoms?
          \nAnswer:
          \n(a) Group I elements are larger in size than alkli earth metals (Group II elements), therefore, there is less force of attraction between nucleus and valence electron, that is why their ionization energy is lower, (b) Anions are larger than neutral atom because electrons are more than protons, therefore, effective nuclear charge is less, therefore, distance between centre of nucleus and valence electrons is more.<\/p>\n

          Question 38.
          \nDefine electro negativity. How does it differ from electron affinity?
          \nAnswer:
          \nElectronegativity is defined as measure of tendency to attract shared pair of electrons towards itself in a covalently bound molecule. It has arbitrary value whereas electron affinity has absolute value. Electro negativity is property of covalently bound atom whereas electron affinity is property of isolated atom.<\/p>\n

          \"KSEEB<\/p>\n

          1st PUC Chemistry Classification of Elements and Periodicity in Properties Three Mark Questions and Answers<\/h3>\n

          Question 1.
          \nExplain the features that influence\/affect the ionization energy.
          \nAnswer:<\/p>\n

            \n
          • Effective Nuclear charge:- If effective nuclear charge increases Ionization energy decreases.<\/li>\n
          • Atomic Size:- If atomic size increases Ionization energy increases.<\/li>\n
          • Half or completely filled orbitais:- From Half or completely filled orbitais, starting high lE. . . .<\/li>\n
          • Orbital of same energy: If orbitais having the same principle quantum number n, .<\/li>\n
          • Shielding (Screening) Effect:- Inner electrons repel the outer valence electrons. They reduce the nuclear force.<\/li>\n<\/ul>\n

            Question 2.
            \nExplain classification of elements into different blocks in the periodic table.
            \nAnswer:
            \n(a) s-block elements:- An element in which the outermost (differentiating) electron of its atom belongs to s-orbital of valence shell is called s-block elements. Kept in left hand side of the periodic table, group I & II (or IA & lIA).<\/p>\n

            (b) p-block elements:- An element in which the outermost (differentiating) electron of its atom belongs to p-orbital of valence shell is called p-block elements. Kept in right hand side of the periodic table, group 13 to 18.
            \nElements in group 18 are called Aerogens or Noble gases.
            \nGeneral electronic configuration of p-block elements is ns2<\/sup> np1-6<\/sup>.
            \nBoth s and p block together is called representative or normal elements.<\/p>\n

            (c) d .block elements:- The atom of an element in which outermost (differentiating)
            \nelectron enters to d- sub shell of pen ultimate (n-1) shell is called d-block elements.
            \nKept at middle (between s and p block elements) portion of the periodic table, group 3 to 12.
            \nd) f-block elements: An element in which outermost (differentiating) electron of its atom enters to f-sub shell of anti-penultimate (n-2) shell is called f -block elements. Placed separately at the bottom portion of the periodic table.
            \nGeneral electronic configuration of f-block elements is (n – 2) f1-14<\/sup> (n-l)d1<\/sup>ns2<\/sup>.<\/p>\n

            Question 3.
            \nArrange the following ions in the order of increasing size? Be2+<\/sup>, Cl–<\/sup>, S2-<\/sup>, Na+<\/sup>, Mg2+<\/sup>, Br–<\/sup>.
            \nAnswer:
            \nArranging the given ions into different groups and periods in order of increasing atomic numbers of their respective elements, we have,
            \n\"1st<\/p>\n

            \"KSEEB<\/p>\n

            Question 4.
            \nConsider the elements N, P, O, S and arrange them in order of (a) increasing first ionization enthalpy, (b) increasing negative electron gain enthalpy, (c) increasing non-metallic character.
            \nAnswer:
            \nArranging all the given elements into different groups and periods in order of their increasing atomic numbers, we have,
            \n(a) Since \u0394i<\/sub>H1<\/sub> decreases down a group, therefore, \u0394i<\/sub>H1<\/sub> of N and O are higher than those of P and S. Further since N has more stable exactly half-filled electronic configuration in the 2p-subshell, therefore, it is more difficult to remove out an electron from N than from O even though O has higher nuclear charge. Similarly, P has exactly half-filled electronic configuration in the 3p-subshell. Therefore, \u0394i<\/sub>H1<\/sub> of P is higher than that of S. Thus, the overall increasing order of first ionization enthalpy of these elements follows the order: S < P < O < N.<\/p>\n

            (b) Since adding an electron to smaller size 2p-orbital causes greater repulsion than adding an electron to larger 3p-orbital, therefore, electron gain enthalpies of P and S are more negative than those of N and O respectively. Further since S has higher nuclear charge but P has more stable exactly half-filled electronic configuration in the 3p-subshell, therefore, it is easier to add an electron to S than to P. In other words, electron gain enthalpy of S is more negative than that of P. Similarly, N has exactly half-filled electronic configuration in the 2p-subshell but O has higher nuclear charge.<\/p>\n

            But the addition of an electron to N causes repulsions to such an extent that electron gain enthalpy of N is actually positive while that of O as expected in negative. Combining all the above results, the increasing order of negative electron gain enthalpy of these elements follows the order: N < P < O < S.<\/p>\n

            (c) Since non-metallic character decreases down a group but increases along a period, therefore, O is the most non-metallic element while P is the least non-metallic element. The actual order of increasing non-metallic character is : P < S < N < O.<\/p>\n

            Question 5.
            \nThe first ionization energy of carbon atom is greater than that of boron whereas the reverse is true for the second ionization energy. Explain.
            \nAnswer:
            \nE.C. of C-atom is 1s2<\/sup> 2s2<\/sup> 2p2<\/sup> and E.C. of B-atom is 1s2<\/sup> 2s2<\/sup> 2p1<\/sup>. The first electron to be removed in both cases is from a 2p-orbital but nuclear charge of C is more than that of – B. Therefore, the \u0394i<\/sub>H1<\/sub> of C is greater than that of B. After the removal of first electron, the second electron to be removed from C-atom is from a 2p-orbital whereas that from B-atom is from a 2s orbital. Since a s-orbital is more penetrating and hence is more strongly attracted by the nucleus then a p-orbital, therefore, \u0394i<\/sub>H2<\/sub> of B is higher than that of C.<\/p>\n

            Question 6.
            \nArrange the following ions in order of their increasing ionic radii: Li+<\/sup>, Mg2+<\/sup>, K+<\/sup>, Al3+<\/sup>.
            \nAnswer:
            \n(i) The ionic radius of any cation increases as the number of energy shells increases and decreases as the magnitude of the positive charge increases.<\/p>\n

            (ii) Mg2+<\/sup> (1s2<\/sup> 2s2<\/sup> sp6<\/sup>) and Al3+<\/sup> (1s2<\/sup> 2s2<\/sup> 2p6<\/sup>) are isoelectronic ions and each one of these has two energy shells. Since the positive charge on Al3+ is higher than that on Mg2+<\/sup>, therefore ionic radius of Al3+<\/sup>is lower than that of Mg2+<\/sup>.<\/p>\n

            (iii) Since, K+ (1s2<\/sup> 2s2<\/sup> 2p6<\/sup> 3s2<\/sup> 3p6<\/sup>) has three shells and Mg2+<\/sup> and Al3+<\/sup> have two shells each, therefore, ionic radius of K+ is the largest followed by Mg2+<\/sup> and then Al3+<\/sup>.<\/p>\n

            (iv) Now Li+<\/sup> (1s2<\/sup>) has one shell and +1 charge but Al3+ (Is2 2s2 2p6) has two shells and +3 charge. Since the increase in the ionic radius of Al3+<\/sup> due to the presence of two shells is more than counter balanced by the decrease in its size due to an increase in charge from +1 in Li+ to +3 in Al3+<\/sup>, therefore, the ionic radius of Al3+<\/sup> is lower than that of Li+<\/sup>.
            \nThus, the ionic radii of these four ions increase in the order : Al3+<\/sup> < Li+<\/sup> < Mg2+<\/sup> < K+<\/sup>.<\/p>\n

            \"KSEEB<\/p>\n

            Question 7.
            \nArrange the elements of second period in order of increasing second ionization enthalpies.
            \nAnswer:
            \nThe electronic configuration of the ions obtained after removal of first electron from the elements of 2nd<\/sup> period from left to right are: Li+<\/sup> (1s2<\/sup>), Be+<\/sup> (Is2<\/sup> 2s1<\/sup>), B+<\/sup> (1s2<\/sup> 2s2<\/sup>), O (1s2<\/sup> 2s2<\/sup> 2p1<\/sup>), N+<\/sup> (1s2<\/sup> 2s2<\/sup> 2p2<\/sup>), O–<\/sup> (1s2<\/sup> 2s2<\/sup> 2p3<\/sup>), F (1s2<\/sup> 2s2<\/sup> 2p4<\/sup>), Ne+<\/sup> (1s2<\/sup> 2s2<\/sup> 2p3<\/sup>).
            \nThe following conclusions can be drawn from the above configurations:<\/p>\n

            (i) Li+ has noble gas, i.e., He gas configuration, therefore, AJH2 of Li is the highest in the second period.<\/p>\n

            (ii) Since in B+, the electron has to be removed from a more stable fully filled 28- orbital while in Be+, it has to be lost from the less stable half-filled 2s-orbital and furthermore, the loss of an electron from Be+ gives more stable Be2+ ion with noble gas configuration, therefore, AjH2 of Be is lower than that of B.<\/p>\n

            (iii) Since more energy is required to remove an s-electron than a p-electron of the same energy level, therefore, more energy is required to remove a 2s-electron from B+ (1s2<\/sup> 2s2<\/sup>) than a 2p-electron from C+ (Is2<\/sup> 2s2<\/sup> 2P1<\/sup>). In other words, \u0394i<\/sub>H2<\/sub> of C is lower than that of B.<\/p>\n

            (iv) As we move from C to N to O, the nuclear charge increases by one unit at a time, therefore, their \u0394i<\/sub>H2<\/sub> also increase accordingly. In other words, \u0394i<\/sub>H2<\/sub> of O is higher than that of N which, in turn, is higher than that of C.<\/p>\n

            (v) In case of O+<\/sup> (1s2<\/sup> 2s2<\/sup> 2p3<\/sup>) an electron is to be lost from an exactly half-filled 2p-orbital but in case of F+<\/sup>( 1s2<\/sup> 2s2<\/sup> 2p4 <\/sup>) this is not so. However, loss of an electron from F+<\/sup> gives an exactly half-filled 2p-orbital (i.e., F2+<\/sup> ( 1s2 <\/sup>2s2<\/sup> 2p3<\/sup>), therefore, \u0394i<\/sub>H2 <\/sub>of F should be lower than that of O.<\/p>\n

            (vi) Like O+<\/sup> (1s2<\/sup> 2s2<\/sup> 2p3<\/sup>) and F+ ( 1s2<\/sup> 2s2<\/sup> 2p4<\/sup>), in case of Ne+ (1s2<\/sup> 2s2<\/sup> 2p5<\/sup>) also an electron is to be removed from a 2p-orbital. Since Ne has the highest nuclear charge in 2nd period, \u0394i<\/sub>H2 <\/sub>of Ne is expected to be much higher than that of O or F.
            \nFrom the above discussion, it follows that \u0394i<\/sub>H2<\/sub> of the elements of 2nd<\/sup> period increase in the order: Be < C < B < N < F < O < Ne < Li.<\/p>\n

            Question 8.
            \nClassify the elements having atomic numbers as given below into three separate pairs on the basis of similar chemical properties. Give brief electronic explanation: 9, 12, 16, 34, 53, 56.
            \nAnswer:
            \nThe second period ends at atomic number 10 while the third period ends at atomic number 18. Therefore, 9, 12 and 16 are the first elements in their respective groups. The atomic numbers of the other elements of the same group can be deduced by adding magic numbers of 8, 18, 18 and 32 to elements of 2nd<\/sup> period and by adding magic numbers of 18,18 and 32 to the elements of 3rd<\/sup> period. Thus,
            \n9 + 8 + 18 + 18 = 53
            \n12 + 8 + 18 + 18 = 56
            \n16 + 18 = 34
            \nelements with atomic numbers 9 (F) and 53 (I) belong to halogen family (group 17); elements with atomic numbers 12 (Mg) and 56 (Ba) belong to alkaline earth metals (Group 2) while elements with atomic numbers 16 (S) and 34 (Se) belong to oxygen family (Group 16).<\/p>\n

            Question 9.
            \nGive the name and atomic number of the inert gas atom in which the total number of d-electrons is equal to the difference in numbers of total p and s-electrons.
            \nAnswer:
            \nThe first inert gas which contains d-electrons is krypton. Its atomic number is 36 and its electronic configuration is: 1s2<\/sup> 2s2<\/sup> 2p6<\/sup> 3s2<\/sup> 3p6<\/sup> 3d10<\/sup> 4s2<\/sup> 4p6<\/sup>.
            \nTotal number of d-electrons = 10
            \nTotal number of p-electrons = 6 + 6 + 6 = 18.
            \nTotal number of s-electrons = 2 + 2 + 2 + 2 = 8.
            \n\u2234Difference in total number of p- and s-electrons = 18 – 8 = 10.
            \nThus, the inert gas is krypton.<\/p>\n

            \"KSEEB<\/p>\n

            Question 10.
            \nGive four characteristics of f-block elements. Why are they called inner transition metals?
            \nAnswer:
            \n(a) They form coloured ions, (b) They are paramagnetic in nature, (iii) They form basic , oxides and hydroxides, (iv) They get transmitted in air. They are called inner transition metals because inner 4f-orbital is progressively filled.<\/p>\n

            Question 11.
            \nArrange the species in each group in order of increasing ionization energy and give reason: (a) K+<\/sup>, Cl–<\/sup>, Ar, (b) Na, Mg, Al, (c) C, N,).
            \nAnswer:
            \n(a) Cl–<\/sup>< Ar < < K+<\/sup> became nuclear change goes on increasing, (b) Na < Al < Mg because Mg has stable electronic configuration and Na has large atomic size due to which it has lowest ionization energy, (c) C < O < N because N has half filled p-orbital which is more stable whereas carbon is larger in size.<\/p>\n

            Question 12.
            \nWhat are the factors that affect electron affinity?
            \nAnswer:
            \n(i) Atomic size
            \n(ii) Stability of electronic configuration
            \n(iii) Inter-electronic repulsion
            \n(iv) Screening effect.<\/p>\n

            Question 13.
            \nExplain the terms (i) screening effect, (ii) penetration effect, (iii) metallic character.
            \nAnswer:
            \n(i) Screening effect: The inner electrons between valence electron and nucleus shield’s the valence electron from nucleus; it is called shielding effect.
            \n(ii) Penetration effect: Due to shape of the orbital, s-electron penetrates nearer to the nucleus than p, d or ^electrons and more tightly held.
            \n(iii) Metallic character: Lower the ionization energy, more will be tendency to lose electron, higher will be metallic character.<\/p>\n

            Question 14.
            \n(a) Explain why the second ionization energy of B is significantly higher
            \nthan the second ionization energy of C, even though the first ionization energy of B is less than C. ‘
            \n(b) Which has higher 1st ionization energy B or Be and why?
            \nAnswer:
            \n(a) B(5) 1s2<\/sup> 2s2<\/sup> 2p1 <\/sup>C(6) 1s2<\/sup> 2s2<\/sup> 2p2<\/sup>; B after losing one electron, has completely filled s-orbital from which removal of second electron is more difficult than carbon.
            \n(b) Be has higher first ionization energy due to completely filled valence s-orbital.<\/p>\n

            Question 15.
            \nGive the reasons of the following: (a) Fluorine has less negative electron gain enthalpy than chlorine
            \n(b) Noble gases tend to be less reactive
            \n(c) First ionization enthalpy of Mg is more than that of Na but second ionizationb enthalpy of Mg is less than that of Na.
            \nAnswer:
            \n(a) It is due to more inter electronic repulsion in fluorine atom due to smaller atomic size than chlorine.
            \n(b) It is because they have stable electronic configuration, i.e., their octet is complete except He.
            \n(c) First ionization energy of Mg is more than that of Na because it has smaller atomic size but 2nd ionization of Mg is less because after losing one electron sodium acquires nearest noble gas configuration.<\/p>\n

            \"KSEEB<\/p>\n

            Question 16.
            \n(a) Arrange F, Cl, Br and I in increasing order of electron affinity.
            \n(b) Predict the position of the element with atomic number 26 in the periodic table.
            \n(c) Why I.E. of oxygen is less than that of nitrogen?
            \nAnswer:
            \n(a) I < Br < F < Cl is increasing order of electron affinity.
            \n(b) It belongs to group 8 of periodic table because its electronic configuration is [Ar] . 4s2<\/sup> 3d6<\/sup>.
            \n(c) It is because nitrogen has stable electronic configuration, i.e., half filled p-orbitals. Therefore, its I.E. is more than that of oxygen.<\/p>\n

            Question 17.
            \n(a) Account for the following: (i) Mg+2<\/sup> ion is smaller than O-2<\/sup> ion although both have same electronic structure, (ii) Ionisation enthalpy of nitrogen is more than that of oxygen, (b) Write the IUPAC name and the symbol for the element with at. no. 118.
            \nAnswer:
            \n(a) (i) It is due to greater nuclear charge and greater effective nuclear charge in Mg2+<\/sup> and O2-<\/sup>.
            \n(ii) It is due to stability of electronic configuration of nitrogen. .
            \n(b) Uuo (Ununoctium).<\/p>\n

            Question 18.
            \n(a) Write the general electronic configuration for f-block elements
            \n(b) Which of the following atoms and ions will have the largest and smallest size? Al, Mg, Al+3<\/sup>, Mg+2<\/sup>.
            \nAnswer:
            \n(a) (n-2)f1-14<\/sup>(n-l)d0-1<\/sup>ns2<\/sup>.
            \n(b) Mg will be largest and Al3+<\/sup> will be smallest.
            \n(c) III set because non-metal cannot lose electrons easily therefore, it will have high first and second ionization energies.<\/p>\n

            Question 19.
            \n(a) Predict the position of the element in periodic table having valency shell electronic configuration of (n-l)d1<\/sup>ns2<\/sup>; n = 4. (b) Why noble gases have bigger atomic size than halogens? Why electron gains enthalpy of noble gases are positive?
            \nAnswer:
            \n(a) (n -l)d1<\/sup>ns2<\/sup>, with n = 4, 3d1<\/sup>4s2<\/sup> is Scandium belongs to 4th<\/sup> period and group 3.
            \n(b) It is because we can measure van der Waal\u2019s radii in noble gases which are bigger than covalent radii. It is because electron has to enter the next higher energy level leading to very unstable electronic configuration.<\/p>\n

            Question 20.
            \n(a) Predict the position of the element in the periodic table satisfying the electronic configuration (n-l)d1<\/sup>ns2<\/sup> when n = 4. (b) Name the species which is isoelectronic with Cl–<\/sup>, (c) Why are f-block elements are placed in a separate row at the bottom of periodic table.
            \nAnswer:
            \n(a) It belongs to group 3 and fourth period.
            \n(b) Ar is isoelectronic with Cl+<\/sup>.
            \n(c) It is because they resemble each other but do not resemble any other group elements.<\/p>\n

            Question 21.
            \nSome elements are wrongly placed in the decreasing order of the property mentioned. Rectifying the fault, place them in correct order of the property. Also, furnish reason for the correction done
            \n(a) F > O > N > C (second ionization potential
            \n(b) N > Si > C > P (electronegativity of the elements)
            \n(c) Na > Mg > A1 > Si (First ionization potential).
            \nAnswer:
            \n(a) O > F > N > C, second ionization potential of oxygen is highest since electron is to be removed from half filled configuration for rest follow the order of size.<\/p>\n

            \"KSEEB<\/p>\n

            (b) N > C > P > Si, nitrogen has smallest size and so has higher tendency to attract shared pair of electrons.<\/p>\n

            (c) Si > Al > Mg > Na, silicon has highest first ionization potential due to smallest size, Mg is exception i.e., has higher ionization potential than aluminium because of ns2 configuration which is stable.<\/p>\n

            Question 22.
            \nWhich of the second, third or fourth ionization energy values for calcium shows a sudden increase? Why?
            \nAnswer:
            \n3rd ionization energy shows a sudden increase because electron is pulled from 3p orbital which is completely filled orbital instead of 4s.<\/p>\n

            Question 23.
            \nIdentify the element out of the choices for which hints are given:
            \n(a) Hint (This metal is extracted from sea water), choices (Mg, Be, Ca, Sr)
            \n(b) Hint (Material used in solar cells contain the metal); choices (Cs, Si, K, Rb).
            \n(c) Hint (most electropositive element amongst alkaline earth metals); choices (Be, Ba, Ca, Mg).
            \nAnswer:
            \n(a) Mg because Mg and …. a are present in large amount in sea water.
            \n(b) Si is used in solar cells.
            \n(c) Ba is most electropositive because electrospositive character increases down the group.<\/p>\n

            Question 24.
            \nVariation of first ionization enthalpies with Z = 1 to 60.
            \n\"1st
            \nStudy the figure given above and answer the following:
            \n(a) Compounds of Xe are known but other noble gases do not form compounds
            \n(b) Giving reason arrange alkali metals in increasing order of first ionization enthalpies
            \n(c) Estimate the first ionization enthalpy values of Mg and Al.
            \nAnswer:
            \n(a) Ionisation energy of Xe is minimum, hence electron can be taken out from it stable noble gas configuration allowing it to make some compounds,
            \n(b) Li > Na > K > Rb > Cs. Ionisation energy decreases as size increases. As size increases, outermost elements are away from nuclear pull, so electron can be easily removed.
            \n(c) Mg = 700 kJ\/mol. Al = 600 kJ\/mol.<\/p>\n

            Question 25.
            \nValues of two types of radius of sodium are 186 and 102 pm. Which value indicates metallic and which indicates ionic radius of sodium?
            \nAnswer:
            \n186 pm represents metallic radius, 102 pm represents ionic radius. Reason: Ionic radius is smaller due to removal of an electron, cation is formed which has smaller radius than its parent atom.<\/p>\n

            Question 26.
            \nFill the arrows (I), (II) and (III) in the following diagram choosing appropriately from the options as electronegativity, atomic radius and non metallic character.
            \n\"1st
            \nAnswer:
            \nI. Electronegativity, II. Non-metallic character, III. Atomic radius.<\/p>\n

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

            You can Download Chapter 3 Classification of Elements and Periodicity in Properties Questions and Answers, Notes, 1st PUC Chemistry Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations. Karnataka 1st PUC Chemistry Question Bank Chapter 3 Classification of Elements and Periodicity in Properties …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[81],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/8764"}],"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=8764"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/8764\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=8764"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=8764"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=8764"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}