{"id":9964,"date":"2019-12-05T11:55:00","date_gmt":"2019-12-05T06:25:00","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=9964"},"modified":"2021-07-02T15:43:57","modified_gmt":"2021-07-02T10:13:57","slug":"2nd-puc-chemistry-question-bank-chapter-1","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-chemistry-question-bank-chapter-1\/","title":{"rendered":"2nd PUC Chemistry Question Bank Chapter 1 The Solid State"},"content":{"rendered":"

You can Download Chapter 1 The Solid State 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 1 The Solid State<\/h2>\n
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2nd PUC Chemistry The Solid State NCERT Textbook Questions and Answers<\/h3>\n<\/div>\n<\/div>\n<\/div>\n

Question 1
\nDefine the term \u2018amorphous\u2019. Give a few examples of amorphous solids.
\nAnswer:
\nAmorphous solids are those solids in which the constituent particles may have short range order but do not have a long range order. They have irregular shapes and are isotropic in nature. They do not undergo a clean cleavage.They do not have sharp melting points or definite heat of fusion. E.g.: Glass, rubber and plastics.<\/p>\n

Question 2.
\nWhat makes a glass different from a solid such as quartz? Under what conditions could quartz be converted into glass?
\nAnswer:
\nGlass is an amorphous solid in which the constituent particles (SiO4<\/sub> tetrahedra) have only short range order and there is no long range order. On melting quartz and then cooling it rapidly, it is converted into glass.<\/p>\n

Question 3.
\nClassify each of the following solids as ionic, metallic, molecular, network (covalent) or amorphous.
\n(i) Tetra phosphorus deioxide (P4<\/sub>O10<\/sub>)
\n(ii) Ammonium phosphate (NH4<\/sub>)3<\/sub>PO4<\/sub>
\n(iii) SiC
\n(iv) I2<\/sub>
\n(v) P4<\/sub>
\n(vi) Plastic
\n(vii) Graphite
\n(viii) Brass
\n(ix) Rb
\n(x) LiBr
\n(xi) Si
\nAnswer:
\nIonic – (NH4<\/sub>)3<\/sub>PO4<\/sub> and LiBr
\nMetallic – Brass and Rb
\nMolecular – P4<\/sub>O10<\/sub>,I2<\/sub> P4<\/sub>
\nNetwork (Covalent) – Graphite, SiC, Si
\nAmorphous – Plastics<\/p>\n

\"KSEEB<\/p>\n

Question 4.
\n(i) What is meant by the term \u2018coordination number\u2019?
\n(ii) What is the coordination number of atoms:
\n(a) in a cubic close-packed structure?
\n(b) in a body-centred cubic structure?
\nAnswer:
\n(i) The coordination number of a constituent particles (atom, ion or molecule) in a crystal is the number of constituent particles which are the immediate neighbour of the particle in the crystal. In ionic crystals coordination number of an ion in the crystal is the number of oppositely charged ions surrounding that particular ion.
\n(ii) a] 12 b] 8.<\/p>\n

Question 5.
\nHow can you determine the atomic mass of an unknown metal if you know its density and the dimension of its unit cell? Explain.
\nAnswer:
\nThe atomic mass can be determined using the formula
\n\"2nd
\nwhere
\n\u03c1 is the density of the metal, a is the dimension of the unit cell, N0<\/sub> is Avogadro number and Z is the number of atoms in the unit cell.<\/p>\n

Question 6.
\n(a) \u2018Stability of a crystal is reflected in the magnitude of its melting points\u2019.
\n(b) The melting points of some compounds are given below:
\nWater = 273K, ethyl alcohol = 155.7 K, Diethyl ether = 156.8K, Methane = 90.5K.
\nWhat can you say about intermolecular forces between these molecules?
\nAnswer:
\n(a) Higher the melting point, greater are tire “forces holding the constituent particles together and hence greater is the stability.<\/p>\n

(b) The intermolecular forces in water and ethyl alcohol are mainly hydrogen bonding. Higher melting point of water than alcohol shows that hydrogen bonding in ethyl alcohol molecules is not as strong as in water molecules. Diethyl ether is a polar molecule. The intermolecular forces present in them are dipole -dipole attraction. Methane is a non-polar molecule. The only forces present in them are the weak Vander Waal\u2019s forces (London \/ dispersion forces).<\/p>\n

Question 7.
\nHow will you distinguish between the following pairs of terms:
\n(i) Hexagonal close-packing and cubic close-packing?
\n(ii) Crystal lattice and unit cell?
\n(iii) Tetrahedral void and octahedral void?
\nAnswer:
\n(i) When a third layer is placed over the second layer in such a way that the spheres cover the tetrahedral or \u2018C\u2019 voids, a three dimensional closest packing is obtained where the spheres in every third or alternate layers are vertically aligned (i.e. the third layer is directly above the first, the fourth above the second layer or so on) calling the first layer as layer A and the second layer as layer B, the arrangement is ABAB pattern or hexagonal close packing.<\/p>\n

When the third layer is placed over the second layer in such a way that spheres cover the octahedral or \u2018d\u2019 voids, a layer differentfrom layers A and B is produced. Let us call it as layer C, continuing further, a packing is obtained were the spheres in every fourth layer will be vertically aligned. This pattern of stocking spheres is called ABCABC… pattern or close cube packing (CCP).<\/p>\n

(ii) A regular three dimensional arrangement of points in space is called crystal lattice.
\nUnit cell is the smallest portion of a crystal lattice which, when repeated in different directions, generates the entire lattice.<\/p>\n

(iii) Wherever a sphere of the second layer is above the void of the first layer (or vice versa) a tetrahedral void is formed. These voids are called tetrahedral voids because a tetrahedron is formed when the center of these four spheres arejoined.
\nThe voids which are surrounded by six spheres is called octahedral voids.<\/p>\n

\"KSEEB<\/p>\n

Question 8.
\nHow many lattice points are there in one unit cell of each of the following lattice?
\n(i) Face-centred cubic
\n(ii) Face-centred tetragonal
\n(iii) Body-centred
\nAns:
\nLattice points in face centred cubic or face centred tetragonal
\n= 8 (at corners) + 6 (at face centres) = 14
\nHowever, particles per unit cell
\n8 \u00d7 \\(\\frac{1}{8}+\\) + 6 \u00d7 \\(\\frac{1}{2}+\\) =4
\nlattice points in body centred cube = 8 (at comers) + 1 (at body centre) = 9.
\nHowever, particles per unit cell = 8 \u00d7 \\(\\frac{1}{8}+\\) + 1 = 2<\/p>\n

Question 9.
\nExplain:
\n(i) The basis of similarities and differences between metallic and ionic crystals.
\n(ii) Ionic solids are hard and brittle.
\nAnswer:
\n(a) Similarities (i) Both ionic and metallic crystals have electrostatic forces of attraction. In ionic crystals, these are between the oppositely charged ions. In metals, these are among valence electrons and the kernels. That is why both have high melting point.<\/p>\n

(ii) in both cases the bond is non-directional Differences: (i) In ionic crystals, the ions are not free to move. Hence, they cannot conduct electricity in the solid state. They can do so only in the matter state or in aqueous solutions. In metals the electrons are free to flow. Hence they can conduct electricity in the solid state.<\/p>\n

(ii) Ionic bond is strong due to electrostatic forces of attraction.
\nMetalic bond may be weak or strong depending on the number of valance electrons and size of the kernels.<\/p>\n

Question 10.
\nCalculate the efficiency of packing in case of a metal crystal for
\n(i) simple cubic
\n(ii) body-centred cubic
\n(iii) face-centred cubic
\n(with the assumptions that atoms are touching each other).
\nAnswer:
\n(i)
\n\"2nd
\nTop\/Bottom\/side view of simple cubic unit cell As spheres are touching each other, evidently a = 2 r
\nNumber of spheres per unit cell =\\(\\frac { 1 }{ 2 }\\) \u00d7 8 = 1
\nvolume of the sphere = \\(\\frac { 4 }{ 3 }\\)\u03c0r3<\/sup>
\nvolume of the cube = a3<\/sup> = (2r)3<\/sup> = 8r3<\/sup>
\n\u2234 Fraction occupied, i.e. packing fraction
\n(\\(\\frac { 4 }{ 3 }\\)\u03c0r3<\/sup>) \/ 8r3<\/sup>
\nor % occupied = 52.4%.<\/p>\n

\"KSEEB<\/p>\n

(ii)
\n\"\"
\nspheres at the corners, body diagonal, AD = 4r further, face diagonal,
\n\"2nd
\nNumber of spheres per unit cell = 8 \u00d7 \\(\\frac { 1 }{ 8 }\\) + 1 = 2
\nvolume of 2 spheres = 2 \u00d7 \\(\\frac { 4 }{ 3 }\\) \u03c0r3<\/sup> = \\(\\frac { 8 }{ 3 }\\) \u03c0r3<\/sup>
\n\u2234 Fraction occupied i.e. packing fraction
\n\"2nd<\/p>\n

(iii)
\n\"2nd
\nA,B,C are the centres of the sphere . As sphere on the face centre is touching the spheres at the corners evidently AC = 4r
\nBut from right angled triangle ABC
\n\"2nd
\n\u2234 volume of the unit cell
\n\"2nd
\nNumber of spheres in the unit cell
\n= 8 \u00d7 \\(\\frac { 1 }{ 8 }\\) + 6 \u00d7 \\(\\frac { 1 }{ 2 }\\) = 4
\nvolume of 4 spheres = = 4 \u00d7 \\(\\frac { 4 }{ 3 }\\) \u03c0r3<\/sup> = \\(\\frac { 16 }{ 3 }\\) \u03c0r3<\/sup>
\n\u2234 fraction occupied, i.e. packing fraction
\n\"2nd
\nor % occupied = 74%.<\/p>\n

Question 11.
\nSilver crystallises in fee lattice. If edge length of the cell is 4.07 x 10-8<\/sup> cm and density is 10.5 g cm-3<\/sup>, calculate the atomic mass of silver.
\nAnswer:
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 12.
\nA cubic solid \u00a1s made of two elements P and Q. Atoms of Q are at the corners of the cube and P at the bodycentre. What is the formula of the compound? What are the coordination numbers of P and Q?
\nAnswer:
\nAs atoms Q are present at the 8 corners of the cube, therefore, number of atoms of in the unit cell = \\(\\frac { 1 }{ 8 }\\) \u00d7 8 = 1
\nAs atoms P are present at the body centre, therefore, number of atoms of x in the unit cell = 1
\n\u2234Ratio of atoms x:y =1:1
\nHence the fonnula of the compound is PQ coordination number of each of P and Q = 8.<\/p>\n

Question 13.
\nNiobium crystallises in body-centred cubic structure. If density is 8.55 g cm-3<\/sup>, calculate atomic radius of niobium using its atomic mass 93 u.
\n\"2nd
\nor x = antilog0.519 = 3.304 for body – centred cubic,
\nr = \\(\\frac{\\sqrt{3}}{4}\\) a = 0.433a = 0.433 x 330.4 pm = 143.1 pm<\/p>\n

Question 14.
\nIf the radius of the octahedral void is r and radius of the atoms in close packing is R, derive relation between r and R.
\nAnswer:
\n\"2nd
\nA sphere fitting into the octahedral void is shown by shaded circle. A sphere above and a sphere below this small (shaded) sphere have not been shown in the figure. Obviously, ABC is a right angled triangle. Applying Pythagoras theorem,
\nBC2<\/sup> = AB2<\/sup> + BC2<\/sup>
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 15.
\nCopper crystallises into a fee lattice with edge length 3.61 x 10-8<\/sup> cm. Show that the calculated density is in agreement with its measured value of 8.92 g cm-3<\/sup>.
\nAnswer:
\n\"2nd
\nwhich is in close agreement with the measured value.<\/p>\n

Question 16.
\nAnalysis shows that nickel oxide has the formula Ni098<\/sub> O1.00<\/sub> What fractions of nickel exist as Ni2+<\/sup> and Ni3+<\/sup> ions?
\nAnswer:
\nThe formula Ni098<\/sub> O1.00<\/sub> shows that Ni : O= 0.98 : 1.00 = 98 : 100.
\nThus, if there are 100 O-atoms, then Ni atoms = 98
\ncharge on 100 O2′ ions = 100 x (-2) = – 200
\nsuppose Ni atoms as Ni2+<\/sup> = x
\nThen Ni atoms as Ni3+<\/sup> = 98 – x
\nTotal charge on Ni2+ and Ni3+<\/sup>
\n= (+2) x + (+3) (98 – x) = 294 – x
\nAs metal oxide is neutral, total charge on
\ncations = total charge on anions
\n294 – x = 200
\nx = 94
\n\u2234 % of Ni as Ni2+<\/sup> =
\ngxlOO
\n98
\n= 96% % of Ni as Ni3+<\/sup> = 1000 – 96 = 4.<\/p>\n

Question 17.
\nWhat is a semiconductor? Describe the two main types of semiconductors and contrast their conduction mechanism.
\nAnswer:
\nSubstances whose conductance lies in between that of metals (conductors) and insulators are called semiconductors. Two main types of semiconductors are n-type and p-type. n-type semiconductor: When a silicon crystal is doped with atoms of group-15 elements, such as P, As, Sb or Bi, then only four of the five valance electrons of each impurity atom participate in forming covalent bonds and fifth electron is almost free to conduct electricity. Silicon that has been doped with group-15 element is called n-type semiconductor.
\n\"2nd<\/p>\n

(ii) p-type semiconductor: when a silicon crystal is doped with atoms of group-13 elements, such as B, Al, Ga or In, each impurity atom forms only three covalent bonds with the host atoms. The place where the fourth electron is missing t is called a hole which moves through the crystal like a positive charge and hence increases its conductivity. Silicon that has been doped with group-13 element is p-type semiconductor.
\n\"2nd<\/p>\n

Question 18.
\nNon-stoichiometric cuprous oxide, Cu2<\/sub>O can be prepared in laboratory. In this oxide, copper to oxygen ratio is slightly less than 2:1. Can you account for the fact that this substance is a p-type semiconductor?
\nAnswer:
\nThe ratio less than 2:1 in Cu2<\/sub>O shows that some cuprous (Cu+<\/sup>) ions have been replaced by cupric (Cu2+<\/sup>) ions. To maintain electrical neutrality, every two Cu+<\/sup> ions will be replaced by one Cu2+<\/sup> ion thereby creating a hole. As conduction will be due to presence of these positive holes, hence it is a p-type semiconductor.<\/p>\n

Question 19.
\nFerric oxide crystallises in a hexagonal close-packed array of oxide ions with two out of every three octahedral holes occupied by ferric ions. Derive the formula of the ferric oxide.
\nAnswer:
\nO2-<\/sup> ions form cubic close packed lattice
\nNumber of O2-<\/sup> = 8 x \\(\\frac { 1 }{ 8 }\\) = 1
\nNumber of Fe3+<\/sup> = \\(\\frac { 2 }{ 3 }\\) octahedral voids = \\(\\frac { 2 }{ 3 }\\)
\nFormula of the compound =
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 20.
\nClassify each of the following as being either a p-type or a n-type semiconductor: (i) Ge. doped with In (ii) B doped with Si.
\nAnswer:
\n(i) Ge is group 14 element and In is group 13 element. Hence, an electron deficit hole is created and therefore, it is p-type.
\n(ii) B is group 13 element and Si is group 14 element, there will be free electron. Hence it is p-type.<\/p>\n

Question 21.
\nGold (atomic radius = 0.144 nm) crystallises in a face-centred unit cell. What is the length of a side of the cell?
\nAnswer:
\nFor fee lattice, edge length,
\na = 2 \u221a2 x 0.144 nm = 0.407 nm<\/p>\n

Question 22.
\nIn terms of band theory, what is the difference
\n(i) between a conductor and an insulator
\n(ii) between a conductor and a semi\u00acconductor?
\nAnswer:
\n(i) The energy gap between the valence band and conduction band in an insulator is very large whereas in a conductor, the energy gap is very small or there is overlapping between valance band and conduction band.
\n(ii) In a conductor, there is very small energy gap or there is overlapping between valence band and conduction band but in a semiconductor there is always a small energy gap between them. \u2019<\/p>\n

Question 23.
\nExplain the following terms with suitable examples:
\n(i) Schottky defect
\n(ii) Frenkel defect
\n(iii) Interstitials and
\n(iv) F-centres.
\nAnswer:
\n(i) Schottky defect:
\nIt arises due to the missing of equal number of cations and anions from their normal positions leaving behind a pair of holes. This defect is usually observed in ionic compounds having high coordination number and ions of almost similar size. The density of the crystal is lowered due to the presence of vacancies in the crystal lattice. Schottky defect increases slightly the electrical conductance of the crystal. E.g. NaCl, KC1, CsCl.<\/p>\n

(ii) Frenkel defect:
\nIt arises w’hen an ion, usually cation, leaves its normal site and occupies an interstitial site. This defect is usually observed in ionic \u2018 compounds having low co-ordination number , and crystals with anions are much larger in size than the cations. Since no ions are missing from the crystals as whole, it doesn\u2019t affect the density of the crystal.
\nE.g.: AgCl, ZnS, AgBr<\/p>\n

(iii) Interstitials:
\nAtoms or ions which occupy the normally vacant interstitial positions in a crystal are called interstitials.<\/p>\n

(iv) F-centres:
\nThe electron occupying holes, created by missing of anions from the lattice sites are called F-centres. These F-centres are responsible for colour of compound.<\/p>\n

Question 24.
\nAluminium crystallises in a cubic close- packed structure. Its metallic radius is 125 pm.
\n(i) What is the length of the side of the unit cell?
\n(ii) How many unit cells are there in 1.00 cm3 of aluminium?
\nAnswer:
\n(i) For CCP structure, edge length, a = 2\u221a2r = 2\u221a2 x 125pm = 354pm
\n(ii) volume of unit cell = a3<\/sup> = (354 \u00d7 10-8<\/sup>m)3<\/sup>
\n\u2234 Number of unit cells in 1.00 cm3<\/sup>
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 25.
\nIf NaCl is doped with 10-3<\/sup> mol% of SrCl2<\/sub>, what is the concentration of cation vacancies?
\nAnswer:
\nIntroduction of one Sr2+<\/sup> introduces one cation vacancy because Sr2+<\/sup> replaces two Na+<\/sup> ions. Introduction of 10-3<\/sup> moles of SrCl2<\/sub> per 100 moles of NaCl would introduce 10’3 mole cation vacancies
\n\u2234 Number of vacancies per mole of NaCl
\n= \\(\\frac{10^{-3}}{100}\\) = 10-5<\/sup> mole = 10-5<\/sup> x 6.02 x 1023<\/sup> 1 (F = 6.02 x 1018<\/sup> vacancies.<\/p>\n

Question 26.
\nExplain the following with suitable examples:
\n(i) Ferromagnetism
\n(ii) Paramagnetism
\n(iii) Ferrimagnetism
\n(iv) Antiferromagnetism
\n(v) 12 – 16 and 13 – 15 group compounds.
\nAnswer:
\n(i) Ferromagnetism is considered as an extreme case of paramagnetism. Those substances with a strong attraction to an external magnetic field and those retain a permanent magnetism even when the field is removed are known as ferromagnetic substances. Ferrimagnetism is caused by spontaneous alignment of magnetic moment of atoms or ions in the same direction. Examples of ferromagnetic substances are Fe, CO, Ni, CrO2<\/sub> alloys of Fe, Co and Ni etc. Once such a material is, magnetised, it remains permanently magnetised Ferromagnetic
\n\"2nd<\/p>\n

(ii) Paramagnetic substances are those which are attracted by magnetic field and have impaired electrons. They lose magnetism in the absence of magnetic field. The greater the number of unpaired electrons, the greater the magnetic moment of the substance and hence greater the paramagnetism. E.g.: O2<\/sub>, CuO, Fe3+<\/sup>, Al, Mn, Cu2+<\/sup>\u00a0etc.<\/p>\n

(iii) These are substances which are expected to have large magnetism on the basis of unpaired electrons but actually have small net magnetic moment. This is due to the alignment of magnetic moment in opposite direction in unequal numbers resulting in a net magnetic moment.
\n\"2nd<\/p>\n

(iv) These are substances which are expected to possess paramagnetism or ferromagnetism on the basis of unpaired electrons but they actually possess zero net magnetic moment. It arises due to the alignment of magnetic moments in opposite direction in a compensatory manner resulting a zero magnetic moment.
\n\"2nd<\/p>\n

(v) Germanium and silicon are group 14 elements and have a characteristic valance of four and form four bonds as in diamond. A large variety of solid state materials have been prepared by combination of groups 13 and 15 or 12 and 16 to stimulate average valence of four as Ge or Si. Typical compounds of group 13 -15 are InSb, A1P and GaAs. Gallium arsenide (GaAs) semiconductors have very fast response and have revelutionised semi – conductor devices. ZnS, CdS, CdSe and HgTe are examples of group 12 – 16 compounds. In. these compounds, the bonds are not perfectly covalent and the ionic character depends on the electronegativities of the two elements.<\/p>\n

2nd PUC Chemistry The Solid State One Mark Questions and Answers<\/span><\/h3>\n

Question 1.
\nHow many atoms are present in the unit
\ncell of
\n(a) simple cubic lattice
\n(b) BCC
\n(c) FCC
\nAnswer:
\n(a) 1
\n(b) 2 and
\n(c) 4.<\/p>\n

Question 2.
\nHow can you convert NaCl structure into CsCl structure and vice versa.
\nAnswer:
\nNaCl structure can be converted into CsCl. structure by application of pressure while the reverse can be done by heating to 760 k.<\/p>\n

Question 3.
\nHow does amorphous silica differ from quartz?
\nAnswer:
\nIn amorphous Silica, SiO4<\/sub> tetrahedra are randomly joined to each other whereas in quartz, they are linked in a regular manner.<\/p>\n

Question 4.
\nWhat are stoichiometric defects or intrisnic defects in ionic crystals?
\nAnswer:
\nStoichiometric defects are those defects in which the ratio of cations to anions remains the same are represented by the molecular formula<\/p>\n

\"KSEEB<\/p>\n

Question 5.
\nWhat is the effect of the presence of Schottky defects on the density of a crystal?
\nAnswer:
\nThe density of the crystal decreases.<\/p>\n

Question 6.
\nName one Solid which has both Schottky and Frenkel defects. !
\nAnswer:
\nSilver bromide (AgBr).<\/p>\n

Question 7.
\nWhat is Frenkel, defect?
\nAnswer:
\nWhen some ions (usually cations) are missing from the lattice sites and they occupy the interstitial sites so that electrical neutrality as well as stoichiometric ratio is maintained, it is called Frenkel defect.<\/p>\n

Question 8.
\nWhy does the conductivity of metals decrease with rise of temperature?
\nAnswer:
\nThe kernals begin to vibrate in the path of the flow of electrons creating hinderance in the flow.<\/p>\n

Question 9.
\nWhat are the type of lattice imperfections found in crystals?
\nAnswer:
\n(a) Stiochiometric defects, viz, Schottky defect and Frenkel defect and
\n(b) Non- Stiochiometric defects, viz., metal excess, metal deficiency and
\n(c) Impurity defects.<\/p>\n

Question 10.
\nDefine coordination number of a metal ion in an ionic crystal.
\nAnswer:
\nIn an ionic crystal, coordination number of a metal ion (+ve ion) is the number of negative ions surrounding the metal ion, i.e. which are present as its nearest neighbours.<\/p>\n

Question 11.
\nWhy are amorphous solids to be considered as supercooled liquids?
\nAnswer:<\/p>\n

    \n
  • Like liquids, amorphous solids are isotropic<\/li>\n
  • Like liquids, they posses fluidity.<\/li>\n<\/ul>\n

    \"KSEEB<\/p>\n

    2nd PUC Chemistry The Solid State Two Marks Questions and Answers<\/span><\/h3>\n

    Question 1.
    \nRefractive index of a solid is observed to have the same value along all directions. Comment on the nature of solid. Would it show cleavage property?
    \nAnswer:
    \nAs the solid has same value of refractive index along all directions. This means that it is isotropic and hence amorphous. Being an amorphous solid, it would not show a clean cleavage when cut with a knife. Instead it would
    \nbreak into pieces with irregular surfaces.<\/p>\n

    Question 2.
    \nIonic solids conduct electricity in molten state but not in solid state. Explain.
    \nAnswer:
    \nIn molten state, ionic solids dissociate to give free ions and hence can conduct electricity.<\/p>\n

    However, in the solid state, as the ions are not free but remain held together by strong electrostatic forces of attraction, they cannot
    \nconduct electricity in the solid state.<\/p>\n

    Question 3.
    \nName the parameters that characterize a unit cell.
    \nAnswer:
    \nA unit cell is characterized by following six parameters: (i) The dimensions of the unit cell along the three edges. These are represented by a, b and c. The edges may or may not be mutually perpendicular. (ii) The angles between the edges. These are represented by \u03b1, \u03b2\u00a0 and \u03b3. The angle \u03b1\u00a0 is. between b and c, \u03b2 is between a and c and \u03b3
    \nis between a and b.<\/p>\n

    Question 4.
    \nWhat type of defect can arise when a solid is heated? Which physical property is effected and in what way?
    \nAnswer:
    \nWhen a solid is heated, vacancy defect is produced in the crystal. This is because of heating, some atoms or ions leave the lattice site completely, i.e. some lattice site becomes vacant. As a result of this defect, the density of the substance decreases because some atoms\/ions leave the crystal completely.<\/p>\n

    Question 5.
    \nA group 14 element is to be converted into n-type semiconductor by doping it with a suitable impurity. To which group should the impurity belong?
    \nAnswer:
    \nn-type semiconductor means conduction due to presence of excess of negatively charged electrons. Hence to convert group 14 element into n-type semiconductor, it should be doped with group 15 element.<\/p>\n

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

    You can Download Chapter 1 The Solid State 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 1 The Solid State 2nd PUC Chemistry The Solid State 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\/9964"}],"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=9964"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/9964\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=9964"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=9964"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=9964"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}