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CLASS-IX MPC BRIDGE COURSE Note: A molecule splits into atoms first before taking part in a chemical reaction. CHEMISTRY Steps DAY-1 : SYNOPSIS Atomic mass unit (a.m.u.): It is the smallest unit of mass and is used to measure the masses of atoms and subatomic particles. The mass of one a.m.u. is equal to the mass of 1 th 12 1 th 12 of mass of C- 12 isotopes atom. For example, the atomic weight of calcium is 40. This means that an atom of calcium is on average is 40 times the mass of 1/ 12 the mass of C- 12 isotope’s atom. Atomic weights of many elements are not whole numbers due to the presence of stable isotopes. The number of atoms of a particular isotope present in 100 atoms of a natural sample of that element is called its relative abundance which always remains constant for a given element. Natural chlorine is a mixture of two isotopes with relative abundances 75% (Cl-35) and 25% (Cl-37) approximately. Then, the atomic weight of chlorine is 75 35 25 37 35.5 100 Mass of one atom of an element = Relative atomic mass mass of mass of C- 12 Molecu le: 1 th 12 the The term molecule was introduced by Avogadro. Molecule is the smallest particle of matter that exists independently and is formed by the combination of atoms. Molecule is also defined as the smallest particle of matter that can exist and retains all the properties of that substance. NARAYANA GROUP OF SCHOOLS 2. the mass of C-12 atom. The other names of a.m.u. are Aston, Dalton and Avogram. Note:1 a.m.u.=1.66 10–24 g or 1.66 0–27 kg. Atomic weight has no units. The relative atomic mass of an element indicates the number of times one atom of that element is heavier than 1. 3. 4. to calculate the molecular weight: Write the formula of the compound or the molecule. Identify the different types of elements present in it and write their symbols along with the number of atoms. Now multiply the number of atoms with the atomic weights of the respective elements Finally add them to get molecular weight. DAY-1: WORKSHEET 1. 1 amu is equal to the mass of 1) 1 th of C - 12 atom 12 1 th of O-16 atom 14 3) 1g of H2 4) 1.66 × 10–23 kg 2. 1 atomic mass unit = 2) 1 th mass of a carbon - 12 atom 12 2) 1.66 × 10–24g 3) 6.023 × 10–23g 4) 6.023 × 1023g 3. The ratio of weight of one atom of an element to its atomic weight is equal to 1) 1 amu 1) 1 th of C – 12 isotopic atom 12 3) 12 amu 4) None 4. The mass of one atom of an element is 40 × 1.66 × 10–24g. The number of protons in its nucleus is 1) 40 2) 20 3) 10 4) 5 5. The weight of Helium atom in grams is 1) 2 2) 4 –24 3) 6.64 × 10 4) 1.66 × 10–24 6. The symbol of carbon is C. It means that 1) ‘C’ represents one atom of carbon. 2) ‘C’ also represents 1 mole of carbon atoms. 3) ‘C’ also represents 12g of C. 4) All 2) mass of 74 CLASS-IX 7. Atomic weight of an element is x. It means that weight of one atom of that element is 1 xg 12 3) 12 × x g 4) 1.66x × 10–24g 8. The mass of an atom of an element ‘x’ is 39. The number of atoms of it present in gram atomic weight of it is_______. 1) 1 2) 1.66 × 1024 23 3) 6.023 × 10 4) 96500 9. The total mass of 100 atoms of silicon is 1) 2800 2) 2800 amu 3) 28 × 1.66 × 10–22g 4) 280 kg 10. The approximate number of electrons that are required to make 1 smallest unit of mass is 1) 6.023 × 1023 2) 1.66 × 1024 3) 1852 4) 2500 1) ‘x’ g 2) DAY-2 : SYNOPSIS Gram Atomic Weight (GAW): (a) Atomic weight of an element expressed in grams is known as its gram atomic weight. For example, the atomic weight of hydrogen is 1.008. So, the gram-atomic weight of hydrogen is 1.008 g. (b)Gram atomic weight of any substance is also called its gram atom. For example, 1 gram atom of carbon weighs 12 gram and 1 gram atom of nitrogen weighs 14 grams. (c)Number of gram atoms Given weight = Gram atomic weight . For example, the number of gram atoms in 5 g of hydrogen =5/1 = 5. (d)Weight of x gram atoms = x Gram atomic weight. (e) 1 gram atom or gram atomic weight of an element contain = 6.023 10 23 atoms. (f) Number of atoms in a given substance ( given element) = Number of gram atoms 6.023 1023. NARAYANA GROUP OF SCHOOLS MPC BRIDGE COURSE (g)Number of atoms in 1 gram of an element = Gram 6.023×1023 . Atomic weight Molecular Weight (GMW): (a) It is the molecular weight of an element or compound expressed in grams. For example, the molecular weight of hydrogen gas is 2. So, the gram molecular weight of hydrogen is 2 g. (b)Gram molecular weight of a substance is also called its gram molecule or mole molecule. For example, the weight of 1 gram molecule or mole molecule of H2O is 18 grams and the weight of 1 gram molecule of N2O is 44 grams. (c)Number of moles = Given weight . Gram Molecular weight (d) Weight of x moles of any compound = x Gram molecular weight. (e)Number of molecules in a given substance= Number of gram molecules 6.023 1023. (f) Weight of substance in grams = Number of gram molecules GMW. DAY-2: WORKSHEET 1. Gram atom of any element contains 1) 6.023 × 1023 atoms 2) 3.0115 × 1023 atoms 3) 1.505 × 1023 atoms 4) 12.0 × 1023 atoms 2. Which of the following is correct? 1) Molecular mass of oxygen is 32. 2) Gram molecular mass of sulphur (S8) is 252 g. 3) The weight of one molecule of O3 is 48 amu. 4) All 3. The ratio of number of molecules present in 1 gram mole of O2 to one gram mole of SO2 is 1) 1 : 4 2) 1 : 2 3) 1 : 8 4) 1 : 1 4. The ratio of weights of hydrogen and helium is 1 : 2. Find the ratio of number of gram atoms. 1) 2 : 1 2) 1 : 1 3) 1 : 4 4) 4 : 1 75 CLASS-IX 5. Among all the naturally occuring elements, which one can generate the maximum number of gram atoms from a given amount? 1) Hydrogen 2) Uranium 3) Calcium 4) Mercury 6. How many gram atoms of the lightest element weigh same as 1 gram atom of the heaviest element? 1) 1 2) 235 3) 238 4) 100 7. Among all the naturally occuring elements, one gram atom of which element contains the maximum amount of it? 1) Hydrogen 2) Uranium 3) Calcium 4) Mercury 8. Identify the element whose 2 gram atoms weigh 8g. 1) Hydrogen 2) Helium 3) Oxygen 4) Sulphur 9. Find the number of gram molecules present in the following: i) 5g of Neonii) 7 g of nitrogen (i) (ii) 1) 0.25 0.25 2) 0.25 0.5 3) 0.5 0.25 4) 1 2 10.1 x = 6.023 × 10 23 molecules = gram molecular mass of a substance. Then, x = _____ 1) gram atom 2) mole 3) molecular weight 4) gram DAY-3 : SYNOPSIS Important relations related to mole: a) 1 mole of particles = 6.023 1023 particles (atoms/ molecules/ions/electrons/ protons/neutrons/nucleons). b) The weight of 1 mole atoms of an element = gram atomic weight of the element. c) The weight of 6.023 1023 atoms of an element = gram atomic weight of the element. d) The weight of 1 mole molecules of a compound = gram molecular weight of a compound. NARAYANA GROUP OF SCHOOLS MPC BRIDGE COURSE e ) The weight of 6.023 1023 molecules of a compound = gram molecular weight of the compound. f) The weight of 1 mole of formula units of a salt = gram formula weight of the salt. Some more important relations: No of gram atoms or mole atoms = Given weight Gram atomic weight . Number of moles (n) Given weight = Gram Molecular weight Weight of x gram atoms = x Gram atomic weight Weight of x moles of any compound = x Gram molecular weight DAY-3: WORKSHEET 1. The number of atoms in 8g of Sulphur is 1) 6.02 × 1023 2) 3. 01 × 1023 24 3) 12.04 × 10 4) 1.505 × 1023 2. 12 g of Carbon contains equal number of atoms as 1) 12 grams of Mg 2)40 grams of Calcium 3) 32 grams of Oxygen 4) 7 grams of nitrogen 3. 6.02 × 1022 particles present in 32 g of oxygen is 1) 0.1 mole 2) 1 mole 3) 10 moles 4) 100 moles 4. Number of molecules present in 32g of oxygen is 1) 3.2 × 1010 2) 6.02 × 1023 23 3) 3.2 × 10 4) 6.02 × 1010 5. Which of the following has more number of molecules? 1) 1 g O2 2) 1 g N2 3) 1 g F2 4) 1 g CO2 6. Which of the following has more number of atoms? 1) 1 g Ca 2) 1 g C 3) 1 g Cu 4) 1 g Cl2 7. Which of the following pairs of gases contain equal number of particles? 1) 1 g He, 1 g H2 2) 1 g He, 2 g H2 3) 4 g He, 2 g H2 4) 4 g He, 4 g H2 76 CLASS-IX 8. If equal mass of N2 and O2 are taken, the ratio of number of molecules in these gases would be 1) 1 : 1 2) 7 : 8 3) 8 : 7 4)28:32 9. Which of the following contains largest number of atoms? 1) 4 g of H2 2) 1 g of O2 3) 28 g of N2 4) 18 g of H2O 10. Which one of the following pairs of gases contain the same number of molecules? 1) 16 g of O2 and 14 g of N2 2) 8 g of O2 and 22 g of CO2 3) 28 g of N2 and 22 g of CO2 4) 32 g of O2 and 32 g of N2 DAY-4 : SYNOPSIS Dalton’s atomic theory Matter consists of small indivisible particles called atoms. Atoms of same element are alike in all respects. Atoms of different elements are different in all respects. Atoms combine in small whole numbers to form compound atoms (molecules). Atom is the smallest unit of matter which takes part in a chemical reaction. All the points put forward in Dalton’s atomic theory have been contradicted by modern research, except that atom is the smallest unit of matter, which takes part in a chemical reaction. These particles were affected by the electric and magnetic fields but in the direction, opposite to the cathode rays. Amongst the positively charged particles formed by the discharge in gases, it was found that the particles formed during the discharge through hydrogen were lightest. Further on, the magnitude of charge on these particles was same as on electron, but positive in nature. The lightest positively charged particle of hydrogen was named proton. Characteristics of a Proton The electric charge on a proton is +1.6 × 10–19 C. The mass of proton is 1.67 × 10–24 g. It is estimated that a proton is 1837 times as heavy as an electron. NARAYANA GROUP OF SCHOOLS MPC BRIDGE COURSE DAY-4: WORKSHEET 1. The term ‘atom’ was given by 1) Democritus 2) John Dalton 3) William Crookes 4) Maharishi Kanada 2. Which of the following is true according to Dalton’s atomic theory? 1) Matter consists of small indivisible particles called atoms. 2) Atoms of same element are alike in all respects 3) Atoms combine in small whole numbers to form compound atoms (molecules) 4) Atom is the smallest unit of matter which takes part in a chemical reaction. 3. The first atomic theory was given by 1) Democritus 2) John Dalton c) William Crookes 4) Maharishi Kanada 4. Which is not correct about electrons? 1) Discovered by Chadwick 2) Named by J.L. Stoney 3) Present inside the nucleus 4) It has maximum e/m ratio 5. Which of the following is never true for cathode rays? 1) They possess kinetic energy 2) They are electromagnetic waves 3) They produce heat 4) They produce mechanical pressure 6. The discharge tube experiment in which cathode rays are emitted has shown that 1) All nuclei contain positive charge 2) All forms of matter contain electrons 3) Protons are positively charged 4) Mass of proton and that of neutron are almost equal 7. Assertion : The charge to mass ratio of the particles in anode rays dependson nature of the gas taken in the discharge tube. Reason: The particles in anode rays carry positive charge. 1) Assertion is correct and reason is the correct explanation of assertion. 2) Reason is correct but assertion is incorrect. 77 CLASS-IX 3) Assertion is correct and reason is not the correct explanation of assertion. 4) Reason and assertion both are incorrect. 8. Positive rays are 1) Electromagnetic waves 2) Electrons 3) Positively charged gaseous ions 4) Neutrons 9. The ratio of e/m for p+ and - particle is 1) 1:2 2) 2:1 3) 1:3 4) 1:1 10.Which is not correctly matched: 1) Particle nature of e– was given by Bohr. 2) The heaviest sub-atomic particle Neutron among e, p, n. 3) Electron is non fundamental particle. 4) Outside the nucleus neutron is unstable. DAY-5 : SYNOPSIS Thomson’s Atomic model: • His model of atom has been given following different names: Water Melon Model (or) Plum Pudding Model (or) Raisin Pudding Model. • He proposed that atom consists of positive charge in which the negatively charged electrons are embedded. • He could not explain how the electrons are protected from the effect of positive charge. Hence model is considered to be a failure. Rutherford’s Planetary Model: • This is the first atomic model which has successfully explained the structure of atom. • His explanation is based on the gold foil experiment. • His experiment contain (i) source of alpha particles (ii) lead block (iii) golden foil of thickness 0.0004mm (iv) ZnS screen. • He made alpha rays to pass though the golden foil of very less thickness (0.0004mm). • Defects of Rutherford’s atomic model: (i) According to the Classical laws of mechanics or dynamics of physics, any NARAYANA GROUP OF SCHOOLS MPC BRIDGE COURSE charged particle revolving around another charged particle should lose energy continuously. Hence electron revolving round the nucleus should lose energy and fall inside the nucleus. But nucleus is found to be stable. Thus Rutherford’s atomic model does not explain the stability of an atom. (ii) It could not explain the distribution of electrons around the nucleus and does not tell us anything about their energies. (iii) If the electron loses energy continuously, then the atomic spectra should be continuous but it is discontinuous. Hence It could not explain the line spectrum. • Quantum theory (i) Energy is emitted or absorbed not continuously but discontinuously in the form of small packets of energy called quanta. • The quantum in case of light is called photon. (ii)Each quantum is associated with definite amount of energy. (iii)The amount of energy associated with a quantum of radiation is proportional to the frequency of radiation. E E h , where h = Planck’s constant and is equal to 6.625 10–34 Jsec. • The energy of a photon of light in terms of wavelength and velocity of light ‘c’, is given by E hc . • A body can emit or absorb energy only in terms of the integral multiples of quantum i.e.,E = nh where n=1,2,3,4 .. • Main features of Bohr’ atomic model In order to overcome the drawbacks of Rutherford’s atomic model, Neil’s Bohr proposed a new atomic model based upon 78 CLASS-IX quantum theory of radiations. (i) The electrons in an atom revolve around the nucleus only in certain selected circular orbits. These orbits are associated with definite energies and are called shells or energy levels. (ii) These are numbered as 1, 2, 3, 4 .... etc., or designated as K, L, M, N ...etc. shells (iii) Only those orbits are permitted in which the angular momentum of an electron is a whole number multiple of nh , where h is Planck’s constant. That 2 is, Angular momentum of an electron , mvr nh , 2 Where n= 1, 2, 3, ......n; m is the mass of the electron, v is the velocity of the electron and r is the radius of the orbit. • n is also known as principal quantum number. (iv) As long as the electron revolves in a particular orbit, its energy remains constant. Hence the orbits are known as stationary orbits. (v) When an electron gains energy, it jumps to higher energy level by absorbing a definite amount of energy (equal to the difference in energy between the two energy levels). When the electron jumps back to the lower energy level it radiates same amount of energy in the form of a photon. E E 2 E1 h , where the is frequency of the radiation emitted or absorbed when an electron jumps from one orbit to another orbit. • Success of Bohr’s atomic model: (i) Bohr’s model could explain the stability of atom. (ii) Bohr’s theory helped in calculating the radius and energy of an orbit. (iii) It could explain the atomic spectrum of hydrogen. • Spectrum: The group of wavelengths or frequencies is known as a spectrum. NARAYANA GROUP OF SCHOOLS MPC BRIDGE COURSE • Electromagnetic spectrum: This is a spectrum in which all the electromagnetic radiations are arranged in the increasing order of their wavelengths. • Drawbacks: (i) Bohr’s model could not explain the spectra of atoms containing more than one electron. (ii) It was observed that in the presence of a magnetic field, each spectral line gets split up into closely spaced lines. This phenomenon, known as Zeeman effect, could not be explained by Bohr’s model. Similarly, the splitting of spectral lines under the effect of applied electric field (Stark effect), could not be explained. (iii)Bohr’s model could not justify the quantization of angular momentum i.e., why the angular momentum of an electron is mvr nh . 2 DAY-5: WORKSHEET 1. Rutherford’s experiment on scattering of alpha particles showed for the first time that atom has 1) Nucleus 2) Electrons 3) Protons 4) Neutrons 2. Rutherford’s model is related to explain 1) Discovery of nucleus 2) Spectrum of Hydrogenic species 3) Planetary motion of electrons around nucleus 4) All of these 3. Rutherford’s - particle dispersion experiment concludes? 1) All –Ve ions are deposited at small part. 2) Proton moves around the nucleus. 3) All +Ve ions are deposited at small part. 4) Neutrons are charged particles. 4. When alpha particles are sent through a thin metal foil, most of them go straight through the foil because 1) Alpha particles are much heavier than electrons 2) Alpha particles are positively charged 3) Most part of the atom is empty space 4) Alpha particles move with high velocity 79 CLASS-IX 5. Rutherford’s alpha particles scattering experiment eventually led to the conclusion that 1) Mass and energy are related 2) Electrons occupy space around the nucleus 3) Neutrons are deep in the nucleus 4) The point of impact with matter can be precisely determined. 6. The space between nucleus and electrons in an atom is 1) full of electromagnetic radiation 2) full of air 3) empty 4) none of the above 7. Rutherford’s experiment which established the nuclear model of the atom used a beam of 1) -particles which impinged on a metal foil and got absorbed. 2) -rays which impinged on a metal foil and ejected electrons. 3) Helium atoms, which impinged on a metal foil and got scattered. 4) Helium nuclei which impinged on a metal foil and got scattered. 8. Assertion A : Size of the nucleus is very small as compared with size of the atom. Reason (R) : Almost all the mass of the atom is concentrated in the nucleus. 1) Both Assertion and Reason are correct and Reason is the correct explanation of Assertion 2) Both Assertion and Reason are correct but Reason is not the correct explanation of Assertion 3) Assertion is correct and Reason is incorrect 4) Assertion is incorrect and Reason is correct 9. Very few (1 in 10000 or 20000) - particles are deviated through an angle of 180 o C. This has lead to the discovery of 1) Proton 2) Neutron 3) Both 4) Nucleus NARAYANA GROUP OF SCHOOLS MPC BRIDGE COURSE 10. The electrons of Rutherford’s model of the atom are expected to lose energy because they 1) are attracted by the nucleus 2) strike each other 3) are accelerated 4) are in motion DAY-6 : SYNOPSIS • Sommerfeld’s atomic model: (i) Electrons revolve in elliptical orbit. Each ellipse has a major axis and minor axis. (ii) He proposed that the angular momentum of an electron moving in an elliptical orbit is equal to the integral multiples of h/2 i.e., mvr kh , where 2 k is called azimuthal quantum number. (iii) The relation between principal quantum number (n) and azimuthal quantum number(k) is as follows: n length of the major axis = k length of the minor axis (iv) According to him, as the value of k decreases, the ellipticity of the orbit increases. Thus, when n=k, the orbit will be circular. (v) For a given value of n, the values of k are 1 to n. In other words for a given stationary orbit, a set of substationary orbits are present. (vi) The substationary orbits are also known as sublevels or subshells or subenergy levels. The subshells of a given shell have slight energy difference. n=4, k=4 n=4, k=3 n=4, k=2 n=4, k=1 • Success of sommerfeld’s atomic model: It could explain the reason for further splitting of spectral lines in electric and magnetic fields. This is due to the presence of sub stationary orbits or subshells. 80 CLASS-IX Quantum numbers A When each orbital in an atom is specified by a set of three quantum numbers (n, l, m) and each electron is designated by a set of four quantum numbers (n, l, m and s). (1) Principal quantum number (n) (i) It was proposed by Bohr and denoted by ‘n’. (ii) It determines the average distance between electron and nucleus, means it denotes the size of atom. (iii) It determines the energy of the electron in an orbit where electron is present. (iv) The maximum number of electrons in an orbit represented by this quantum number is 2n2. No energy shell in atoms of known elements possess more than 32 electrons. (v) It gives the information of orbits like K, L, M, N----------. (vi) Angular momentum can also be calculated by using principal quantum number. (2) Azimuthal quantum number (l) (i) Azimuthal quantum number is also known as ‘ angular quantum number’. Proposed by Sommerfeld and denoted by ‘l’. (ii) It determines the number of subshells or sub-levels to which the electron belongs. (iii) It tells about the shape of subshells. (iv) It also expresses the energies of subshells s < p < d < f (increasing energy). (v) The value of l = (n – 1) always. Where ‘n’ is the number of principle shells, Value of l Name of subshell Shape of subshell = = 0 s = Spherical 1 p 2 d Dumbbell Double dumbbell 3……(n-1) f Complex MPC BRIDGE COURSE s – sub-shell 2 electrons, d – sub-shell 10 electrons p – sub-shell 6 electrons, f– sub-shell 14 electrons. (viii) For a given value of ‘n’ the total values of ‘l’ is always equal to the value of ‘n’. (3) Magnetic quantum number (m) (i) It was proposed by Zeeman and denoted by ‘m’. (ii) It gives the number of permitted orientations of sub-shells. (iii) The value of ‘m’ varies from –l to +l through zero. (iv) It tells about the splitting of spectral lines in the magnetic field, i.e., this quantum number proves the Zeeman effect. (v) For a given value of ‘n’ the total value of ‘m’ is equal to n2. (vi) For a given value of ‘l’ the total value of ‘m’ is equal to(2l + l). (vii) Degenerate orbitals : Orbitals having the same energy are known as degenerate orbitals. e.g. for ‘p’ subshell px py pz. (viii) The number of degenerate orbitals of ‘s’ subshell = 0. (4) Spin quantum number ( s) (i) It was proposed by Goldschmidt and Uhlenback and denoted by the symbol of ‘s’. (ii) The value of ‘s’ is +½ and –½, which signifies the spin or rotation or direction of electron on it’s axis during the movement. (iii) The spin may be clockwise or anticlockwise. (iv) It represents the value of spin angular momentum, and is equal h s s 1 . 2 (vi) It represents the orbital angular momentum, which is equal to to h l l 1 . 2 (v) Maximum spin of an atom = ½ × number of unpaired electrons. (vi) This quantum number is not the result of solution of Schrodinger equation as solved for H–atom. (vii) The maximum number of in subshell = 2(2l +1). NARAYANA GROUP OF SCHOOLS electrons 81 CLASS-IX MPC BRIDGE COURSE DAY-6: WORKSHEET 3) n = 4, l = 0, m = 0, s = + 1. The ratio of energies of two radiations, one with a wavelength of 400 nm and the other with 800 nm. 1) 1 : 4 2) 2 : 1 3) 8 : 1 4) 1 : 6 2. Find the number of photons of light with wavelength 4000pm that provide 1J of energy. 1) 4.01×1015 2) 2.01×1016 3) 3.01× 1015 4) 1.01× 1015 3.3 × 10 18 photons of a certain light radiation are found to produce 1.5 J of energy. Calculate the wavelength of light radiations. (h = 6.63 × 10–34 Js). 1) 4527A0 2) 9870A0 3) 3978A0 4) 7956A0 4. If the principal quantum number is 3, the azimuthal quantum number can have values 1) 1, 2, 3 2) 3, 2, 1, 0, –1, –2, –3 3) 0, 1, 2 4) 1 1 , 2 2 5. Which one of the following sets of quantum numbers represent an impossible arrangement ? n l ml ms 1) 3 2 – 2 1/2 2) 4 0 0 1/2 3) 3 2 – 3 1/2 4) 5 3 0 – 1/2 6. Which of the following represents the correct set of the four quantum numbers for 4d-electrons ? 1) 4, 3, 2, + 1/2 2) 4, 2, 1, 0 3) 4, 3, –2, + 1 2 4) 4, 2, 1, – 1 2 7. In an atom, the maximum number of electrons having the quantum numbers n = 6, l = 3, s = + 1 are 2 1) 14 2) 7 3) 6 4) 2 8. Which of the following sets of the quantum numbers is permitted ? 1) n = 4, l = 2, m = + 3, s = + 1 2 2) n = 3, l = 3, m = + 3, s = + 1 2 NARAYANA GROUP OF SCHOOLS 1 2 4) n = 4, l = 3, m = + 1, s = 0. 9. Which of the following sets of quantum numbers is correct? 1) n = 4, l = 3, m = + 4, s = + 1 2 2) n = 3, l = 2, m = + 3, s = – 1 2 3) n = 2, l = 2, m = + 2, s = + 1 2 4) n = 1, l = 0, m = 0, s = – 1 2 10.For n = 4, 1) The total possible values of ‘l’ are 3. 2) The highest value of ‘l’ is 4. 3) The total number of possible values of ‘m’ is 7. 4) The highest value of ‘m’ is + 3. DAY-7 : SYNOPSIS Shapes of orbitals: (1) Shape of ‘s’ orbital: (i) For ‘s’ orbital l = 0 & m = 0.So ‘s’ orbital have only one unidirectional orientation i.e., the probability of finding the electrons is same in all directions. (ii) The size and energy of ‘s’ orbital with increasing ‘n’ will be 1s < 2s < 3s < 4s. (iii) s-orbitals are known as radial node (or) nodal surface. But there is no radial node for ‘1s’ orbital since it is starting from the nucleus. (2) Shape of ‘p’ orbitals: (i) For ‘p’ orbital, l = 1 and m=+l,0,–1 means there are three ‘p’ orbitals, which is symbolised as px, py, pz. (ii) Shape of ‘p’ orbital is dumbbell, in which the two lobes on opposite side are separated by the nodal plane. (iii) p-orbital has directional properties. 82 CLASS-IX (3) Shape of ‘d’ orbitals: (i) For the ‘d’ orbital l =2, then the values of ‘m’ are –2, –1, 0, +1, +2. It shows that the ‘d’ orbital has five orbitals. (ii) Each ‘d’ orbital is identical in shape, size and energy. (iii) The shape of ‘d’ orbital is double dumbbell . (iv) It has directional properties. (4) Shape of ‘f’ orbital: (i) For the ‘ f ’ orbital l = 3 then the values of ‘m’ are –3. –2, –1, 0, + 1, + 2, + 3. It shows that the ‘ f ’ orbitals have seven orientations as, f x x2 y2 , f y x2 y2 , f z x2 y2 , f xyz, f z3 , f 2 , f 2 . yz xz A (ii) The ‘ f ’ orbital is complicated in shape. Note: The surface at which the probability of finding an electron is zero is called a node or nodal plane. The spherical ‘s’ orbitals do not have nodal planes but have nodal regions equal to n -1 which are present between spherical ‘s’ orbitals. p-orbitals have both nodal regions equal to n - 2 and nodal plane equal to the value of l ie 1. The nodal plane for px orbital is yz. The nodal plane for py orbital is xz. The nodal plane for pz orbital is xy. Each d - orbital has nodal regions equal to n – 3 and nodal planes equal to the value of l ie 2. For any orbital, the number of nodal regions is equal to ‘n – l – 1’ and nodal planes is equal to ‘l’. For any orbital, the total number of nodal regions and nodal planes is equal to n–1 The nodal regions are known as radial nodes and the nodal planes are known as angular nodes. Relative energies of atomic orbitals: The atomic spectra provide information regarding the energies of the atomic orbitals. The order of the energies may be wirtten as, NARAYANA GROUP OF SCHOOLS MPC BRIDGE COURSE 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4s < 3d < 4p < 5s <4d < 5p < 6s < 4f < 5d < 6p < 7s. A The electrons occupy orbitals with lower energy. Moeller has given a simple procedure representing the orbitals of increasing energies in the above order. According to this procedure the orbitals of increasing energies may be shown by a diagram. A As per this diagram, the electrons occupy ‘1s’ orbital, next the ‘2s’ orbital will be occupied. The sequence continues as shown by the arrows in this diagram. This procedure is fully valid for lighter elements. For heavier elements the order may change. Thus, for example, the electronic configuration of Lanthanum, La (Z = 57) should be [Xe] 6s2 4f1. DAY-7: WORKSHEET 1. When an atom is in a magnetic field, the possible number of orientations for an orbital of azimuthal quantum number 3, is a) Three b) One c) Five d) Seven 2. Which of the d-orbitals lie/s in the xyplane? a) dxz only b) dxy only c) d x 2 y 2 only d) d xy and d x 2 y 2 3. Which of the following orbitals is cylindrically symmetrical about z-axis? a) dxz b) dyz c) d z2 d) dxy 4. Which of the following orbitals has/have two nodal planes? a) dxy b) d x2 y 2 c) dyz d) All of these 83 CLASS-IX 5. Which of the following orbitals is not cylindrically symmetrical about z-axis? a) 3 d2z b) 4pz c) 6s d) 3dyz 6. The number of spherical nodes in ‘6s’ orbital is a) 6 b) 5 c) 3 d) 0 7. Which of the following orbitals has/have two lobes along y-axis? a) dxy b)dyz MPC BRIDGE COURSE A If one electron in an atom has the quantum numbers n = 1, l = 0 , m = 0 and s = +l/2, no other electron can have the same four quantum numbers. In other words, we cannot place two electrons with the same value of ‘s’ in ‘1s’ orbital. A The orbital diagram c) d x2 y 2 d) All of these 8. When ‘3d’ orbital is complete, the new electron will enter into the a) 4p – orbital b) 4f – orbital c) 4s – orbital d) 4d – orbital 9. In a potassium atom, which of the following order of electronic energy levels is correct? a) 4s > 3d b) 4s > 4p c) 4s < 3d d) 4s < 3p 10.Which of the following sets of orbitals may degenerate? a) 2s, 2px, 2py b) 3s, 3px, 3dxy c) 1s, 2s, 3s d) 2px, 2py, 2pz A A A DAY-8 : SYNOPSIS A The atom is built up by filling electrons in various orbitals according to the following rules. (1) Aufbau’s principle A This principle states that the electrons are added one by one to the various orbitals in order of their increasing energy starting with the orbital of lowest energy. The increasing order of energy of various orbitals is, Is < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s <5f < 6d < 7p......... (2) (n+l) Rule A In neutral isolated atom, the lower the value of (n + l ) for an orbital, lower is its energy. However, if the two different types of orbitals have the same value of (n + l ), the orbitals with lower value of ‘n’ has lower energy. (3) Pauli’s exclusion principle A According to this principle, “no two electrons in an atom will have same values of all the four quantum numbers”. NARAYANA GROUP OF SCHOOLS A A does not represent a possible arrangement of electrons . Because there are only two possible values of ‘s’. An orbital can hold not more than two electrons. (4) H und’s Ru le of ma x imum multiplicity This rule deals with the filling of electrons in the orbitals having equal energy (degenerate orbitals). According to this rule, “Electron pairing in p, d and f orbitals cannot occur until each orbital of a given subshell contains one electron each or is singly occupied”. This is due to the fact that electrons being identical in charge, repel each other when present in the same orbital. This repulsion can however be minimised if two electrons move as far apart as possible by occupying different degenerate orbitals. All the unpaired electrons in a degenerate set of orbitals will have same spin. As we know the Hund’s rule, let us see how the three electrons are arranged in ‘p’ orbitals. The important point to be remembered is that, all the singly occupied orbitals should have electrons with parallel spins i.e., in the same direction eitherclockwise or anti-clockwise. 2px 2py 2pz or 2px 2py 2pz Electronic configuration of elements:– A On the basis of the electronic configuration principles, the electronic configuration of various elements are given in the following table : 84 CLASS-IX MPC BRIDGE COURSE A The above method of writing the electronic configurations is quite cumbersome. Hence, usually the electronic configuration of the atom of any element is simply represented by the notation. Number of Electrons present x nl Number of Principal Principle shell shells Symbol of subshell Some Unexpected Electronic Configuration A Some of the exceptions are important. Because they occur with common elements, notably chromium and copper. A ‘Cu’ has 29 electrons. Its expected electronic configuration is Is2 2s2 2p6 3s2 3p6 4s2 3d9. But in reality the configuration is 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 3d 10 (as this configuration is more stable). Similarly ‘Cr’ has the configuration of 1s2 2s2 2p 6 3s2 3p5 4s1 3d6 instead of 1s2 2s2 2p6 3s2 3p6 4s2 3d4. A Factors responsible for the ex tra stability of half-filled and completely filled subshells: (i) Symmetrical distribution: It is well known fact that symmetry leads to stability. Thus the electronic configuration in which all the orbitals of the same subshell are either completely filled or exactly half filled are more stable because of symmetrical distribution of electrons. (ii) Exchange energy: The electrons with parallel spins present in the degenerate orbitals tend to exchange their position. The energy released during this exchange is called exchange energy. The number of exchanges that can take place is maximum when the degenerate orbtials (orbitals of same subshell having equal energy) are exactly half-filled or completely filled. As a result, the exchange energy is maximum and so it is stable. NARAYANA GROUP OF SCHOOLS DAY-8: WORKSHEET 1. No two electrons in an atom can have a) The same principal quantum numbers. b) The azimuthal quantum numbers. c) The same magnetic quantum numbers. d) An identical set of four quantum numbers. 2. In the electronic configuration given below, which rule is violated ? 1s 2s 2P a) Aufbau rule b)Pauli’s exclusion principle c) Hund’s rule d) The configuration is correct 3. According to Aufbau principle, the 19 th electron in an atom goes into the a) 4s-orbital. b) 3d-orbital. c) 4p-orbital. d) 3p-orbital. 4. The orbital diagram, in which both Pauli’s exclusion principle and Hund’s rule are violated? 2s 2p a) b) c) d) 5. Nitrogen has the electronic configuration 1s 2 , 2s 2 , 2p1x , 2p1y , 2p1z . Which is determined by a) Aufbau’s principle. b) Pauli’s exclusion principle. c) Hund’s rule. d) Uncertainty principle. 6. Pauli’s exclusion principle states that, b a) Nucleus of an atom contains no negative charge. b) Electrons move in circular orbits around the nucleus. c) Electrons occupy orbitals of lowest energy. d) All the four quantum numbers of two electrons in an atom cannot be equal. 85 CLASS-IX 7. The statements (i) In filling a group of orbitals of equal energy, it is energetically preferable to assign electrons to empty orbitals rather than pair them into a particular orbital. (ii) When two electrons are placed in two different orbitals, energy is lower if the spins are parallel are valid for a) Aufbau principle b) Hund’s rule c) Pauli’s exclusion principle d) Uncertainty principle 8. The orbital diagram in which the Aufbau’s principle is violated? 2s 2p a) 2p 2s 2p b) 2s c) 2s 2p d) 9. The electrons would go to lower energy levels first and then to higher energy levels, according to which of the following? a) Aufbau principle b) Pauli’s exclusion principle c) Hund’s rule of maximum multiplicity d) Heisenberg’s uncertainty principle 10.The number of unpaired electrons in the ground state of vanadium (Z = 23) is a) 1 b) 2 c) 5 d) 3 DAY-9 : SYNOPSIS 1. Nece ssity fo r class ificatio n of elements: (a) The classification may help to study them better. (b)The classification may lead to correlate the properties of the elements with some fundamental property that is characteristic of all the elements. (c)The classification may further reveal relationship between the different elements. 2. Early attempts for classification: (i) Clas sificat ion into Metals and Nonmetals: Lavoisier was the first (a) Such a classification hardly serves any purpose, as the elements are divided into two major groups only. NARAYANA GROUP OF SCHOOLS MPC BRIDGE COURSE (b)There is no justification for more active metals or more active nonmetals. (c)Certain elements exhibited metallic as well as nonmetallic character and, hence, were named as metalloids. (ii) Classification on the basis of valency: (i) Many elements have variable valencies. This makes the position of such elements uncertain. (ii) It has not explained the diverse nature of elements having the same valency. (iii) Doberiner’s Triads: (a) This was given by Johann Doberiner, a German chemist. (b)He classified elements into sets of three chemically similar elements, called triads. (c)It states that, if the elements of triad are arranged in the increasing order of their atomic weights, then the atomic weight of the middle element is approximately equal to the average of the atomic weights of the other two. This is known as Doberiner’s law of triads. • Reason for rejection of Doberiner Triads: (a) The Doberiner’s method of classification could arrange only a limited number of elements out of those known at that time in the form of triads. Therefore, the idea of triads could not be applied to all the elements then known. (b)Elements with dissimilar properties can also be arranged in the form of triads. This is against the rule of classification. (iv) Newland’s law of octaves: (a) This was given by John Newland’s, an English chemist and a musician in 1864 (b)It states that, when the elements are arranged in the order of their increasing atomic weights, then the properties of the elements were repeated at every eighth element like the eighth note of an octave in music. 86 CLASS-IX Example: Thus, the properties of sodium (Na) and potassium (K) are similar to those of lithium (Li). Similarly, chlorine (Cl) resembles fluorine (F). • Reason for rejection of Newland’s law of octaves: (a)Newland’s classification failed badly while dealing with the heavier elements beyond calcium (Ca). (b) When the noble gases were discovered, the idea of octaves could not be held. For example, with the discovery of neon (Ne) between F and Na, and argon (Ar) between Cl and K, it becomes the ninth element and not the eighth, which has similar properties. (v) Lother Meyer’s classification: (a) This was given by Lothar Meyer, a German chemist in 1869. (b)He plotted a graph of atomic volume (atomic mass/density) versus atomic mass for various elements. He noticed that the elements with similar properties occupied similar positions on the curve. (c)For example, • The most electropositive elements like Lithium (Li), Sodium (Na), potassium (K), Rubidium (Rb), Cesium (Cs), etc,. occupy the peak positions. • The moderate electropositive elements like magnesium (Mg), calcium (Ca), strontium (Sr) and barium (Ba) are placed on the descending curve. NARAYANA GROUP OF SCHOOLS MPC BRIDGE COURSE • The most electronegative elements like fluorine (F), chlorine (Cl), bromine (Br), Iodine (I) are placed on the ascending curve. Thus, Lothar Meyer observed a periodicity in the properties of the elements with atomic mass. • Main features of Mendeleev’s periodic table (i) In Mendeleev’s table, the elements were arranged in vertical columns, called groups. (ii) There were in all eight groups: Group I to VIII. The group numbers were indicated by Roman numerals. i.e., I, II, III, IV, V, VI, VII & VIII. (iii) Except VIII, every group is further divided into subgroups i.e., A and B. Groups VIII occupy three triads of three elements each, i.e., in all nine elements (iv) The properties of the elements in same group or subgroup are similar. (v) There is no resemblance in the elements of subgroups A and B of same group except valency. (vi) The horizontal rows of the periodic table are known as periods. (vii) There were seven pe riods, represented by Arabic numerals 1 to 7. To accommodate more elements, the periods 4, 5, 6 and 7 were divided into two halves. The first half of the elements are placed in the upper left corner and second half in the lower right corner. For example, the elements occupying the box corresponding to group I and period 4 are potassium (K) and copper (Cu), K is written in the top left corner, while Cu is written in the lower right corner. (viii) A period comprises the entire range of elements after which the properties repeat themselves. (ix) In a period, the properties of the elements gradually change from metallic to nonmetallic while moving from left to right. (x)There were gaps left in the periodic table. Mendeleev left these gaps knowingly, as these elements were not discovered at that time. 87 CLASS-IX • Merits of the Mendeleev’s Periodic Table: Mendeleev’s classification was considered superior to the others proposed earlier because of the following reasons: (i) It is based on the more fundamental property of atomic weight of an element. Thus it is better than earlier classification. (ii) It helped in systematic study of the elements. Mendeleev’s classification condensed the study of about 90 elements (only 65 were known at that time, but he left a provision for many more) to the study of only 8 groups of elements. (iii) Some gaps were left knowing my Mendeleev for undiscovered elements. This accelerated the process of discovering these elements, as their properties were predicted by Mendeleev on the basis of other elements present in the same group. (iv) By placing elements strictly according to the similarity in their properties, he was also able to correct certain atomic weights . • For example, he corrected the atomic masses of beryllium (Be), gold (Au) and platinum (Pt). 4. 5. 6. DAY-9: WORKSHEET 1. G, O, D are the correct symbols of right elements of the periodic table arranged in the increasing order of their atomic weights. The atomic weight of ‘G’ is 40 and that of D is 137. If G, O, D are the elements of a Dobereiner triad, then find the atomic weight of ‘O’. a) 88.5 b) 120.5 c) 99.5 d) 77.5 2. Which of the following is not a Dobereiner triad? a) Cl, Br, I b) Ca, Sr, Ba c) Li, Na, K d) Fe, Co, Ni 3. Statement A : Classification of elements is not useful to reveal the relationship between the different elements. NARAYANA GROUP OF SCHOOLS 7. 8. MPC BRIDGE COURSE Statement B : Classification of elements may lead to correlate the properties of the elements with some fundamental property that is characteristic of all the elements. a) Statement ‘A’ is correct but ‘B’ is incorrect. b) Statement ‘B’ is correct but ‘A’ is incorrect. c) Statements ‘A’ and ‘B’ are incorrect. d) Both ‘A’ and ‘B’ statements are correct. (i) Elements with both metallic and nonmetallic characters are called_____. (ii) Arrangement of elements into groups of three is called ________. (i) (ii) a) Active metals Octaves b) Metallic elements Metals c) Triads Metals d) Metalloids Triads Select the following pair of elements in which their arithmetic mean of atomic weights is equal to the atomic weight of strontium. a) Lithium, Barium b) Sodium, Calcium c) Calcium, Barium d) Sodium, Barium Which of the following is wrong triad? a) Chlorine, bromine, iodine b) Lithium, sodium, potassium c) Carbon, nitrogen, oxygen d) Calcium, strontium, barium Which of the following is an achievement of the triads classification? a) Relation between all properties of an element. b) Relation between only atomic weights of an element. c) Relation between the properties of same elements. d) Relation between the atomic mass of all elements. Dobereiner’s law is rejected due to Statement A : Quite large number of elements cannot be grouped into tirads. Statement B : It was possible to group quite dissimilar elements into triads. a) Statement ‘A’ is correct but ‘B’ is incorrect 88 CLASS-IX b) Statement ‘B’ is correct but ‘A’ is incorrect c) Statement ‘A’ and ‘B’ are incorrect d) Both ‘A’ and ‘B’ statements are correct 9. The _______ was the basis of the classification proposed by Dobereiner, Newlands and Mendeleev. a) Atomic number b) Atomic weights c) Atomic mass d) None 10. The discovery of which of the following group of elements gave a death blow to the Newlands law of octaves. a) Inert gases b) Alkaline earths c) Rare earths d) Actinides DAY-10 : SYNOPSIS Defects (Limitations) of Mendeleev’s Periodic table Anoma lous pai rs : In Mendeleev’s periodic table, the elements are arranged on the basis of their atomic weights. However, there are few such pairs in which atomic weights of preceeding elements is higher than that of the following elements. Preceeding Elements Following Elements Argon (40) Potassium (39) Cobalt (58.9) Nickel (58.6) Tellurium (128.0) Iodine (127.0) The above pairs go against Mendeleev’s periodic law. Position of hydrogen: Hydrogen is not given a definite position. It is placed in group ‘Ib’ and group ‘VIIb’ of Mendeleev’s original periodic table. Position of rare earth elements and actinides : The position of rare earth elements and actinides cannot be justified on the basis of atomic weight. Positi on of isot opes: Mendeleev’s periodic table is silent about isotopes. The position of various isotopes of the elements cannot be justified on the basis of atomic weight. Positi on of trans ition elem ents : Mendeleev’s concept of transition elements was defective. He regarded elements of group VIII as transition elements. NARAYANA GROUP OF SCHOOLS MPC BRIDGE COURSE Over looking chemical similarities : In Mendeleev’s periodic table, there are certain relationships which are excessive. Examples : There was hardly any relationship between alkali metals and copper, silver and gold in group I. There was hardly any relationship between fluorine and manganese in group ‘VIIa’. Some obvious similarities between copper and nickel, platinum and gold were overlooked. Modern Periodic Table - Long form of Periodic Table: In 1913, H.G. J. Moseley showed by Xray analysis that the atomic number is more fundamental property of an element than its atomic weight. Therefore, he slightly modified Mendeleev’s periodic law and replaced the word atomic weight by atomic number. Modern Periodic Law : It states that physical and chemical properties of all elements are periodic function of their atomic numbers. On the basis of above law, Moseley prepared long form of periodic table which consists of 7 periods and 18 groups. Description of long form of (Extended form) of Periodic Table In long form of the periodic table, the elements are arranged in the order of increasing atomic numbers in horizontal rows called periods, such that all elements having same number of valence electrons come under the same vertical column called group. This not only ensures periodicity in el ectronic configuration, but periodicity in chemical properties also. Characteristics of long form of periodic table : The subgroups ‘A’ and ‘B’ are separated in this table. In a group elements, the electrons in outermost shell participates in chemical reactions, whereas in B group elements, the electrons from outermost and inner shells participates. 89 CLASS-IX transition elements are The accommodated in the middle of the table in three series. The strongly metallic elements (alkali metals and alkaline earth metals) occupy groups IA and ‘IIA’ respectively on the left hand of transition elements. The non-metallic elements are placed on the right hand of transition elements. The rare gases (noble gases) are placed in zero group at the end (last column) of periodic table. The elements occupying left and right wing vertical columns (groups) are called normal representative elements. The rare earths (Lanthanides) and Actinides are called inner transition elements. They are kept outside the periodic table to mark their peculiar properties. The horizontal rows in the periodic table are called periods. There are seven periods in all, such that each period has consecutive (or continuous) atomic number. The number of elements in a period corresponds to maximum number of electrons which can be accommodated in its one shell. The number of period to which an element belongs is given by its quantum number(n), i.e., the number of outermost shell as counted from nucleus. The vertical columns in periodic table are called groups. There are 18 groups in long form of periodic table. The elements in a group do not have consecutive atomic numbers. However, each element in a group has same number of electrons in its outermost shell and hence, all elements of a group have same chemical properties. The elements in zero group are called rare gases or noble gases. All the elements in this group (with exception of Helium which has 2 electrons) have eight electrons in their outermost shell. The elements in the group IA are called alkali metals. NARAYANA GROUP OF SCHOOLS MPC BRIDGE COURSE The elements in group IIA are called alkaline earth metals. The elements in group VII A are called halogens. The elements in groups IIIB, IVB, VB, VIB, VIIB, VIII, IB and IIB are called transition elements. In transition elements the outermost shell (valence shell) and penultimate shell (the shell before outermost shell) are incomplete. Common names of the elements: Merits of long form of periodic table: The classification of elements is based on fundamental property of elements, i.e., atomic number. It relates position of an element to its electronic configuration in the valence shell and hence, it posses similar chemical properties. It explains variations and similarities in the properties of elements in terms of electronic configuration. Inert gases in which valence shells are completely filled have been placed at the end of each period. Such a location of inert gases shows logical completion of each period. The elements of a subgroup in a given group have been placed separately. Thus, dissimilar elements do not fall together. It provides clear demarcation of different kinds of elements such as : (i) active metals (ii) non-metals (iii) metalloids (iv) transition elements (v) inert gases (vi) lanthanides (vii) actinides. Defects (De-merits) of long form of periodic table: Position of hydrogen is unresolved. It fails to accommodate lanthanides and actinides in the main body of the table. The arrangement is unable to reflect electron configuration of many elements in transition group, lanthanides and actinides. 90 CLASS-IX DAY-10: WORKSHEET 1. Which of the following elements position cannot be justified on the basis of atomic weight? a) Rare earth elements b) Alkali metals c) Transition elements d) Actinides 2. What are the indefinite positions of hydrogen given in Mendeleev’s periodic table? a) 1 b, III b b) I a, II b c) I b, VII b d) VII a, III b 3. An element ‘E’ has atomic number 14. To which period this element belongs? How many maximum number of elements are present in the period to which element ‘E’ belongs? a) 1st period and 6 elements. b) 3rd period and 8 elements. c) 4th period and 8 elements. d) 5th period and 13 elements. 4. Find the period number and the group number in which the element with atomic number 24 is present. a) 2, VB b) 4, VIB c) 5, VIIB d) 3, VB 5. Outer electronic configuration of the atoms of the four elements are given below: (i) 3d0 4s1 (ii) 3s2 3p5 (iii) 4s2 4p6 (iv) 3d8 4s2 In which group and period of the periodic table are they situated respectively? (i) (ii) (iii) (iv) a) 4 and IA 3 and VIIA 4 and 0 4 and VIII b) 4 and 0 3 and VII A 4 and IA 4 and VIII c) 2 and IIA 4 and II A 4 and VIIA 4and 0 d) 3 and VIIA 4 and IA 4 and 0 4 and I A 6. Electronic configuration of three elements are given below: (i) 1s2 2s1 (ii) 1s2 2s2 2p5 2 2 (iii) 1s 2s 2p6 3s2 3p6 4s2 Choose the right option: a) b) c) d) ( i) A lk a line ea rt h m e ta l H a lo g en A lk a li m e ta l A lk a li m e ta l (ii ) A lk a li m e ta l A lk a line ea rth m e ta l H alo g en A lk a line ea rth m e ta l NARAYANA GROUP OF SCHOOLS (i ii) H alog e n A lk ali m e tal A lk alin e e a rth m et al H alog e n MPC BRIDGE COURSE 7. Three elements A, B and C have atomic number Z, Z + 2 and Z + 3 respectively. Among these, ‘C’ is an alkali metal. To which groups of the periodic table do the elements ‘A’ and ‘B’ belong respectively? a) 16, 18 b) 14, 16 c) 15, 17 d) 12, 14 8. Match the following: Column - I Column - II 1) Pnicogens p) Fluorine family 2) Chalcogens q) Nitrogen family 3) Halogens r) Helium family 4) Aerogens s) Oxygen family a) 1 p, 2 q, 3 r, 4 s b) 1 q, 2 r, 3 p, 4 s c) 1 q, 2 s, 3 p, 4 r d) 1 r, 2 p, 3 q, 4 s 9. How many elements are present in, ( i ) Fifth period (ii) Sixth period a) (i) 16 (ii) 32 b) (i) 18 (ii) 32 c) (i) 14 (ii) 28 d) (i) 14 (ii) 32 10.Which of the following pair is against to Mendeleev’s periodic law? a) Chromium, Manganese b) Sodium, Magnesium c) Copper, Zinc d) Tellurium, Iodine DAY-11 : SYNOPSIS 1 Classification of elements into blocks (i) s - block: (a)The elements in which the last electron enters the s-subshell of their outermost energy level are called s- block elements. (b)This block is situated at the extreme left of the periodic table. (c)It contains elements of groups IA and IIA. (d)The general electronic configuration of these elements is ns 1- 2 , where ‘n’ represents the outermost shell. (e)The elements IA group elements are called Alkali metals and that of IIA group elements are known as Alkaline earth metals. General characteristics of s-block elements are: • They are soft metals and have low melting points. 91 CLASS-IX • They are highly electropositive and have low ionisation energies . • They are highly reactive and form ionic compounds. • They show oxidation states of +1 and +2 and are good reducing agents. (ii) p - block: (a) The elements in which the last electron enters the p-subshell of their outermost energy level are called p- block elements. (b)This block is situated at the extreme right side of the periodic table. (c)It contains elements of groups IIIA, IVA, VA, VIA, VIIA and VIIIA (exceptionhelium). (d) The general electronic configuration of these elements is ns2 np1-6, where ‘n’ represents the outermost shell. (e) It includes metals, non metals , metalloids and inert gases. General characteristics of p-block elements are: • They form ionic as well as covalent compounds. • They have relatively high values of ionisation energies. • Most of them are non metals and are highly electronegative. • They show variable oxidation states and form acidic oxides. (iii) d - block: (a) The elements in which the last electron enters the d-subshell of the penultimate energy level are called dblock elements. (b)This block is situated in between ‘s’ and ‘p’ blocks of the periodic table. (c)It contains elements of groups IB, IIB, IIIB, IVB, VB, VIB, VIIB and VIII groups. (d) The general electronic configuration of these elements is (n-1) d 1-10 ns 1- 2 , where (n-1) represents the penultimate shell and ‘n’ represents the outermost shell. (e) It includes weak metals. NARAYANA GROUP OF SCHOOLS MPC BRIDGE COURSE General characteristics of d-block elements are: • They are hard, high melting metals showing variable oxidation states. • They form coloured complexes and form ionic as well as covalent compounds. • Most of them exhibit paramagnetism and possess catalytic properties. • They form alloys and are good conductors of heat and electricity. (iv) f - block: (a) The elements in which the last electron enters the f-subshell of the antipenultimate (third to the outer most shell ) shell are called f- block elements. (b)This block is placed separately at the bottom of the main periodic table. (c)It consists of two series of elements placed at the bottom of the periodic table. The elements of first series follow lanthanum and are called Lanthanides and the elements of second series follow actinium and are called actinides. (d) The general electronic configuration of these elements is (n-2)f1-14 (n-1) d 0-1 ns2, where ‘n’ represents the outermost shell, (n-1) represents the penultimate shell and (n - 2) represents the anti penultimate shell. General characteristics of f-block elements are: • They are hard, high melting metals showing variable oxidation states. • They form coloured complexes and have high densities. • Most of the elements of actinide series are radioactive. (v) Inert gases: (a) The elements of VIII A group, the last column of p-block are known as inert gases or noble gases or aerogens. (b)The outermost shell of these elements is completely filled and are hence nonreactive. (vi) Representative elements: (a) The elements of ‘s’ and ‘p’ block elements, except inert gases are known as representative elements. 92 CLASS-IX (b)The outermost shell of these elements are incompletely filled and hence are highly reactive. (c)These are also known as main group elements or normal elements. (vii) Transition elements: (a) The elements of d- block in which the outermost shell and penultimate shell are incompletely filled are known as transition elements. (b)The name is derived from the fact that they represent transition (change) in character from reactive metals (elements from IA and IIA) on one side and less active metals of groups IB and IIB on the other side. (c)These elements are less reactive than representative elements. (viii) Inner transition elements: (a) The elements in which the last three principal shells are incomplete are known as inner transition elements. (b)The lanthanides and actinides are included in this category. (c)These are less reactive than transition elements. DAY-11: WORKSHEET 1. Match the correct pairs: Electronic configurations Type of element i) n s 2np 5 1) Alkali metal 1 ii) ns 2) Alkaline earth metal 2 6 iii) ns np 3) Transition Metal iv) ns 2 4) Inert gas 1 – 10 v) (n – 1) d ns1 – 2 5) Non metal a) i 3, ii 1, iii 4, iv 5, v 2 b) i 1, ii 5, iii 4, iv 3, v 2 c) i 5, ii 1, iii 4, iv 2, v 3 d) i 5, ii 4, iii 2, iv 1, v 3 2. The electronic configuration of halogen is: a) ns2np 6 b) ns2np3 c) ns2np 5 d) ns2 3. Hydrogen by containing one electron forms H +. In this property, it resembles with: a) Transitional metals b) Alkali earth metals c) Alkali metals d) Halogens NARAYANA GROUP OF SCHOOLS MPC BRIDGE COURSE 4. The electronic structure (n – 1) d 1 – 10 ns0 –2 is characteristic of: a) Transition elements b) Lanthanides c) Actinides d) Rare – earths 5. The elements having the electronic configuration, [Kr] 4d 10 f 14, 5s 2 p 6d 2 , 6s 2 belongs to: a) s-block b) p-block c) d-block d) f-block 6. An element has the electronic configuration 1s2, 2s22p 6, 3s23p 63d 5, 4s1. It is: a) Li b) Cr c) B d) Be 7. An element has electronic configuration 1s 2 2s 2 2p 63s 2 3p 4 . Predict their period, group and block. a) Period = 3rd, block = p, group = 16 b) Period = 5th, block = s, group = 1 c) Period = 3rd, block = p, group = 10 d) Period = 4th, block = d, group = 12 8. In the long form of the periodic table, all the non-metals are placed under a) s-block b) p-block c) d-block d) f-block 9. In the modern periodic table, the place of the element with atomic number 31 is in: a) s-block b) d-block c) p-block d) f-block 10. Elements of ‘d’ group are called: a) Transition elements b) Transuranic elements c) Metals d) Metalloids DAY-12 : SYNOPSIS Periodic Properties: Properties which are directly or indirectly related to the electronic configuration of the elements and show a regular gradation when we move from left to right in a period or from top to bottom in a group are called periodic properties. Some important periodic properties are atomic size, ionization energy, electron affinity, electronegativity, valency, density, atomic volume, melting and boiling points etc. ATOMIC SIZE Atomic size: It refers to the distance between the centre of the nucleus of the atom to the outermost shell containing electrons. 93 CLASS-IX Since, absolute value of the atomic size cannot be determined, it is usually expressed in terms of the following operational definitions. Units: Atomic size is expressed in terms of angstrom (1A° = 10–10 m). In a period, on moving from left to right in a period, the size of the elements decreases due to a gradual increase in the nuclear pull. In a group atomic size increases from top to bottom due to increase in the number of shells. Ionic size: An atom can be changed to a cation by loss of electrons and to an anion by gain of electrons. A cation is always smaller than the parent atom because, during its formation effective nuclear charge increases and sometimes a shell may also decrease. On the other hand, the size of an anion is always larger than the parent atom because, during its formation effective nuclear charge decreases. Isoelectronic ions or species are the neutral atoms, cations or anions of different elements which have the same number of electrons but different nuclear charge. The size of the isoelectronic species depends upon their nuclear charge. Greater the nuclear charge, smaller the size. IONISATION ENERGY (IE) The energy required to remove the outermost electron from an isolated gaseous atom of the element is called Ionisation energy. A g IE A + g + e . Ionisation energy is expressed in eV/ atom or kJ/mole 1 eV/atom = 96.45 kJ/mole. Ionisation energy depends on the following factors : (i) Size of the atom: Greater the size of atom, smaller is the IE. (ii) Nuclear charge: Greater the nuclear charge, greater is the IE. (iii) Screening effect: Greater the screening effect of the inner electrons, smaller is the IE. NARAYANA GROUP OF SCHOOLS MPC BRIDGE COURSE (iv) Penetration effect: Greater the penetrating effect of the outermost occupied orbital, greater is the IE. For a particular energy level, the penetration effect is in the order : s > p > d > f. (v) Electronic configuration: If the electronic configuration of the atom is stable, it would have relatively higher IE. Ionisation energy in general increases on moving along the period and decreases on going down the group. Be, Mg, N, P and noble gases have relatively higher values of IE due to their stable electronic configuration. Alkali metals have the least and noble gases have the highest ionisation energies in their respective periods. Helium (He) has the highest IE among all the elements. Cesium (Cs) has the least IE among all the elements (except Fr which is radioactive). The amount of energy required to remove the outermost electron from an isolated gaseous atom of the element is called first ionisation energy (IE1). The amount of energy required to remove the outermost electron from an uni positive ion of the element is called second ionisation energy (IE2). Successive ionisation energies are always greater than the first ionisation energy. IE3 > IE2 > IE1 As we move from top to bottom in the periodic table, the atomic size increases and the nuclear pull decreases. The weaker nuclear pull results in the outer electron being held loosely, thereby requiring less energy to remove the outer electron and hence, ionisation energy decreases.Thus, down the group the ionisation energy decreases due to increase in size. As we move from left to right in the periodic table, the atomic size decreases and the nuclear pull increases. This results in the outer electron being held more tightly, increasing the ionisation energy. Thus, as we move from left to right, the ionisation energy increases due to the decrease in size. 94 CLASS-IX DAY-12: WORKSHEET 1. The atomic radii in case of inert gases is: a) Ionic radii b) Covalent radii c) vander Waals’ radii d) None 2. The covalent and van der Waal’s radii of hydrogen respectively are: o o a) 0.37 A , 1.2 A o o o b) 0.37 A , 0.37 A o o o c) 1.2 A , 1.2 A d) 1.2 A , 0.37 A 3. The size of the species, Pb, Pb 2+, Pb 4+ decreases as: a) Pb4+ > Pb2+ > Pb b) Pb > Pb2+ > Pb4+ c) Pb > Pb4+ > Pb2+ d) Pb4+ > Pb > Pb2+ 4. Which of the following has largest radius? a) 1s2, 2s2, 2p6, 3s2 b) 1s2, 2s2, 2p6, 3s2 3p1 NARAYANA GROUP OF SCHOOLS 5. 6. 7. 8. 2 6 FIRST I.P. Some trends among the elements of second and third periods are given below: IE1 : Li < B < Be < C < O < N < F < Ne. IE1 : Na < Al < Mg < Si < S < P < Cl < Ar. IE2 : Li > C > B > Be. IE2 : O > F > N > C. IE2 : Na > Al > Mg. IE3 : Mg > Na > Al. While comparing IE 2 , consider the electronic configuration of A+ ion, keeping in mind the stability of electronic configuration with half filled and fully filled subshells. Electronegativity can be defined as the tendency of an atom in a molecule to attract the shared pair of electrons towards itself. As we move from top to bottom in a periodic table, the atomic size increases and the nuclear attraction decrease. This decreases the electronegativity. As we move from left to right in a period the atomic size decreases and the nuclear attraction increases. This increases the electronegativity. Fluorine is the most electronegative element. In a period, the highest electronegativity is of halogens and the lowest is of alkali metals. MPC BRIDGE COURSE c) 1s , 2s , 2p , 3s2 3p3 d) 1s2, 2s2, 2p6, 3s2 3p5 Which of the following combinations contains only isoelectronic ions? a) N3–, O2–, Cl–, Ne b) F–, Ar, S2–, Cl– c) P3–, S2–, Cr, Ar d) N3–, F–, O2–, Ar Arrange the elements S, P, As in the order of their increasing ionization energies. a) S < P < As b) P < S < As c) As < S < P d) As < P < S The factor that is not affecting the ionisation energy is a) Size of atom b) Charge in the nucleus c) Type of bonding in the crystalline lattice d) Type of electron involved The graph of first ionisation enthalpy versus atomic number is as follows: 2 ATOMIC NUMBER Which of the following statements is correct ? a) Alkali metals are at the maxima and noble gases at the minima. b) Noble gases are at the maxima and alkali metals at the minima. c) Transition elements are at the maxima. d) Minima and maxima do not show any regular behaviour. 9. Atoms which have high first ionisation energy have a) High nuclear charge. b) Small atomic size. c) Metallic properties. d) Strongly bound valence electrons. 10. Arrange the following atoms in order of increasing first ionisation energies. K, Cs, Rb, Ca a) Cs, Rb, K, Ca b) Rb, Cs, K, Ca c) Cs, Rb, Ca, K d) Rb, Cs, Ca, K 95 CLASS-IX DAY-13 : SYNOPSIS ELECTRO POSITIVITY, METALLIC/NONMETALLIC CHARACTER Metals have the tendency to form cations by loss of electrons and this property makes the elements as electropositive elements or metals. M g M g e The tendency of an element to lose electrons is closely connected to the (IE) of the element. The smaller the Ionisation energy (IE) of an element, the greater will be its tendency to lose electrons and thus greater will be its metallic character. Tendency to oxidise itself provides reducing property to the elements thus, smaller the (IE), greater the metallic character hence, greater the reducing nature : (IE) increases on moving along a period left to right and decreases down the group, hence metallic and reducing nature decreases along the period and increases down the group. The most reactive metals are on the left of the periodic table, whereas the least reactive metals are in the transition metal groups closer to the right side of the table. Variation of metallic character and non-metallic character In a group, as we move from top of bottom, the size of atoms increases, resulting in an increase in electropositive character. Thus, the metallic character increases down the group. As we move from top to bottom, the size of atoms increases resulting in the decrease in ionisation energy. Thus, the non-metallic character decreases down the group. So, as we move down the group, the metallic character increases and the non-metallic character decreases. This is understood from the following example. NARAYANA GROUP OF SCHOOLS MPC BRIDGE COURSE In a period as we move from left to right, the size of atom decreases, resulting in a decrease in electropositivity. Thus, metallic character decreases as we move from left to right in a period. As we move from left to right, the size of atoms increases, resulting in an increase in ionisation energy or electronegativity. Thus non-metallic character increases, as we move from left to right in a period. Thus metallic character decreases and non-metallic character increases from left to right in a period. ELECTRON AFFINITY (EA] The energy released, when an isolated atom of the element in gaseous state accepts an electron to form univalent negative ion is called electron affinity. It is measured in eV/atom or kJ/mole. X g e X g + EA It is also called electron gain enthalpy. The energy change in the process of addition of an electron to monovalent negative ion of the element in gaseous state is called second electron affinity (EA2). X g e X2 g + EA2 . ‘EA 1 ’ is generally positive (energy is released), whereas, ‘EA 2 ’ is always negative (energy is absorbed). In other words, energy is generally released when the electron is added to atom of the element in gaseous state, whereas energy is always required when an electron is added to monovalent negative ion of the element in gaseous state. Electron affinity is numerically equal to ionisation energy but are opposite to each other Electron affinity depends on the following factors : (i) Atomic size: Smaller the size of the atom, greater is the EA. (ii) Nuclear charge: Greater the nuclear charge, greater is the EA. 96 CLASS-IX MPC BRIDGE COURSE Reducing Agents: Electropositive elements can lose electrons easily and hence, can act as good reducing agents. The elements of IA and IIA groups are electropositive and are good reducing agents. Down the group, the electropositivity increases and also the reducing character. The Best Reducing AgentAn exception: Lithium is the strongest reducing agent due to high hydration energy of Li+ ion. (iii) Elect ronic co nfigurat ion: If electronic configuration of the element is stable, its EA would be exceptionally low. ‘EA’ (electron affinity) in general, increases on moving across the period and decreases on going down the group. Be, Mg, N and P have exceptionally low values of ‘EA’ due to their stable electronic configurations. Noble gases (He, Ne, Ar, Kr, Xe, Rn) have negative values of ‘EA’ due to their stable electronic configuration. Halogens have the highest ‘EA’ in their respective periods. Among halogens, fluorine has lower ‘EA’ than chlorine due to greater interelectronic repulsions in small sized ‘2p’ subshell. Chlorine has the highest ‘EA’ among all the elements. Some trends in the values of electron affinities : EA1 : Cl > F > Br > I EA1 : S > O > Se > Te EA1 : C > B > Li > Be EA1 : Si > Al > Na > Mg EA1 : F > O > N > Ne. NARAYANA GROUP OF SCHOOLS Oxidising Agents: Elements which can gain electrons easily act as good oxidising agents. The elements of VI A and VII A groups gain electrons easily and are good oxidising agents. From left to right in a period, the electron affinity and electronegativity increases, and also the tendency to gain electrons. So the oxidising character increases from left to right in a table. The Best Oxidising Agent: Fluorine is the strongest oxidising agent due to its highest electronegativity. REDUCING, OXIDISING CHARACTERS AND NATURE OF OXIDES Redox reactions are common for almost every element in the periodic table except for the noble gas elements of group VIIIA. In general, metals act as reducing agents, and reactive non-metals, such as O and the halogens act as oxidising agents, Variation of acidic and basic character Metals are characterised by basic character and non-metal characterised by acidic character. Down the group, the metallic character increases and the non-metallic character decreases. Thus, the basic character increases and acidic character decreases down the group. From left to right in a period, the metallic character decreases and the nonmetallic character increases. Thus, the basic character decreases from left to right, whereas, the acidic character increases. 97 CLASS-IX MPC BRIDGE COURSE DAY-13: WORKSHEET DAY-14 : SYNOPSIS 1. The electronic configurations of the elements X, Y, Z and J are given below. Which element has the highest metallic character ? Why do atoms combine? Atoms combine to attain stability (i) by attaining octet configuration. (ii) by reducing the energy. How do atoms acquire stable octet configuration? Atoms can complete the valence shell by acquiring octet configuration in two ways. 1. By transfer of one or more electrons, from one atom to another. Generally, electropositive elements lose electrons and electronegative elements gain electrons. 2. By sharing one or more electrons between two or more atoms. Thus, we can conclude that, atoms tend to acquire 8 electrons in their outermost shell (except hydrogen, lithium and beryllium which tend to acquire 2 electrons), in order to attain stable state. This is called ‘octet rule’. What is a Chemical Bond ? During a chemical reaction, atoms come closer and are held together by a force of attraction to form molecules. This force of attraction is called a chemical bond. Chemical bonds are responsible for the existence of molecules. Electron theory of valency - Kossel and Lewis’ approach of bonding A number of attempts were made to explain formation of chemical bonds in terms of electrons, but it was only in 1916, Kossel and Lewis’ succeeded independently in giving a satisfactory explanation. They proposed a theory, based on electronic concept of atoms, known as electron theory of valency. The main postulates of this theory are: 1. The secret of stability of atoms: Atoms with eight electrons in the outermost shell (two in the case of Hydrogen, Helium, Lithium and Beryllium) are chemically more stable. 2. Cause of chemical reaction: The cause for chemical reaction is to attain stability. An atom achieves this by acquiring the octet configuration (inert gas configuration) in its outermost shell. a) X = 2, 8, 4 b) Y = 2, 8, 8 c) Z = 2, 8, 8, 1 d) J = 2, 8, 8,7 2. Which of the following is arranged in the order of decreasing electropositive character ? a) Fe, Mg, Cu b) Mg, Cu, Fe c) Mg, Fe, Cu d) Cu, Fe, Mg. 3. Which of the following sets of elements has the strongest tendency to form positive ions in gaseous state? a) Li, Na, K b) Be, Mg, Ca c) F, Cl, Br d) O, S, Se 4. Which of the following elements is most electropositive? a) C b) N c) Be d) O 5. Out of C, Si, Ge, Sn, Pb metallic nature is in a) Ge, Sn, Pb b) Sn, Pb c) Ge, Pb d) Ge, Sn 6. Which of the following elements is most metallic? a) Al b) Mg c) P d) S 7. A metal has an atomic number of : a) 9 b) 18 c) 35 d) 37 8. Ionisation energy of F– is 320 kJ mol –1. The electron affinity of fluorine would be: a) ––320 kJ mol–1 b) ––160 kJ mol–1 c) 320 kJ mol–1 d) 160 kJ mol–1 9. The element having very high ionisation energy but zero electron affinity is: a) H b) F c) He d) Be 10. The electron affinities of B, C, N and O are in the order: a) B < C < N < O b) B < C < O > N c) B < C > O > N d) B > C < O < N NARAYANA GROUP OF SCHOOLS 98 CLASS-IX 3. Type of electrons taking part in a chem ical rea ction ( or) chem ical bonding: The electrons present in the outermost shell of an atom are responsible for chemical reaction. The outermost shell is called valence shell and hence, the electrons present in it are called valence electrons. The number of electrons taking part in a chemical reaction is called valency of that element. 4. Atta inment o f neare st inert gas configuration: The atoms of various elements achieve the nearest inert gas configuration, either by transfer (losing or gaining) or by sharing of electrons with another atom. This transfer or sharing of electrons results in the development of an attractive force between the atoms, which holds the atoms together by a bond. Electron Dot structure of atoms - Lewis’ Symbols In the formation of any molecule or formula unit, only the electrons present in the outermost shell are shown. The reason for not showing the inner shell electrons is that, they are well protected and do not involve in chemical reaction. Therefore, valence electrons are considered for the formation of the chemical bonds. G.N. Lewis’ introduced simple symbols called Lewis’ symbols to denote the valence electrons in an atom. Lewis’ symbols: The symbol of the element, surrounded by the valence electrons of its atom, represented in the form of dots around it, is known as Lewis’ symbol or electron dot symbol. Significance of Lewis’ Symbols (i) The number of dots present around the symbol, gives the number of electrons present in the outermost shell i.e., number of valence electrons. (ii) The number of electrons present in the outermost shell is the common valency of the element. NARAYANA GROUP OF SCHOOLS MPC BRIDGE COURSE The common valency of the element is equal to the number of dots around the symbol (if the dots are < 4, then the valency is equal to the number of dots and if the number of dots > 5, then the valency is 8 – number of dots.) For example: Li, Be, B and C have valencies 1, 2, 3 and 4 respectively and N, O and F have 3, 2, 1 respectively (i.e. 8 – number of dots). DAY-14: WORKSHEET 1. It was found that atoms having atomic numbers of 2, 10, 18, 36, 54, 86 are very stable and do not show any chemical reactivity, these elements were found to be gases and are called: a) Inert gases b) Diatomic gases c) Monoatomic gases d) Noble gases 2. Which inert gas has a duplet configuration in its valence shell? a) Heliumb) Neon c) Argon d) Krypton 3. Assertion : Duplet configuration implies that a given atom has 2 electrons in its valence shell. Reason : Elements with octet configuration in their valence shell are stable. a) Both assertion and reason are correct and reason is the correct explanation of assertion. b) Both assertion and reason are correct but reason is not the correct explanation of assertion. c) Assertion is correct and reason is incorrect. d) Assertion is incorrect and reason is correct. 4. “The duplet and octet configuration of electrons in the valence shell is most stable and any atom having this configuration will be in minimum state of energy”. This statement was given by a) Kossel and Lewis b) Lewis and Debye c) Kossel and London d) Lewis and London 99 CLASS-IX 5. Chemical reactivity of an elements depends on: a) Outer shell electronic configuration. b) Reactivity of the nucleus. c) Core electrons. d) None. 6. To attain a ______ participating atoms redistribute their electrons to get a electronic configuration of the nearest noble gas either octet or duplet. a) State of maximum energy b) State of minimum energy c) Stability d) None 7. In each of the following sets, identify the more stable species. i) Al or Al3+, ii) N or N3–, iii) H or H2 i ii iii 3– a) Al N H b) Al3+ N 3– H2 3+ c) Al N H d) Al N H 8. Statement A : Pure metal is more stable than its ore. Statement B : Ore of metal is more stable than pure metals. a) ‘A’ is correct, ‘B’ is incorrect. b) ‘B’ is correct, ‘A’ is incorrect. c) ‘A’ and ‘B’, both are correct. d) ‘A’ and ‘B’, both are incorrect. 9. Valency of the metal atom with respect to oxygen is maximum in: a) Mn2O7 b) OsO4 c) MnO2 d) CrO3 10. An element A is tetravalent and another element B is divalent. The formula of the compound formed by the combination of these elements is: a) A2B b) AB c) AB2 d) A2B3 DAY-15 : SYNOPSIS Ionic bond and its formation The bond that is formed by transfer of electrons from one atom to another atom is called an ionic or electrovalent bond. The cations and anions formed as a result of electron transfer are drawn towards each other due to the electrostatic force (coulomb force) of attraction. They form an ionic bond or an electrovalent bond. NARAYANA GROUP OF SCHOOLS MPC BRIDGE COURSE Note: The bond between two elements is ionic if the EN difference between them is greater than 1.7 The number of electrons transferred during an ionic bond formation is known as an electrovalency. Compounds containing ionic bonds are called ionic compounds. Examples of ionic compounds are NaCl(Na + Cl – ), 2+ 2 2+ CaO(Ca O ), MgO(Mg O 2– ) and MgCl 2 (Cl –Mg++ Cl–). Features of donor atoms: Donor atoms lose electrons with greater ease if the following conditions are satisfied: 1) Less ionisation potential (I.P.) 2) More size 3) Less cation charge Features of acceptor atoms Acceptor atoms gain electrons more easily in the following conditions: 1) High electron affinity and electronegativity 2) Less size 3) Less anion charge a) For donor atoms to lose electrons easily, they should possess less I.P., big size and less charge on cations. IA group has the above mentioned characters and therefore is the best donor group. b) For acceptor atoms to gain electrons easily, they should possess more EA, less size and less negative charge on anions. Group VII A has these features and so is regarded as the best acceptor group. Therefore, IA and VII A get along well with each other, forming a strong ionic bond. Properties of ionic compounds 1. Physical state:Generally, ionic solids are relatively hard. It is because of the close packing due to strong inter-ionic force of attraction present between oppositely charged ions. 2. Structure of ionic solids: Unit cell: There is a basic unit in an ionic crystal, which when repeated three-dimensionally, gives complete crystal. This basic unit is called the unit cell. 100 CLASS-IX 3. Melting and boiling points: Ionic compounds possess high melting and boiling points. Reason: Melting and boiling points of ionic compounds involves breaking of the lattice structure and setting the ions free. In a lattice, there are strong electrostatic forces between oppositely charged ions. To break these strong electrostatic forces, considerable amount of energy is required. Hence, the melting points and boiling points of ionic compounds are high. 4. Solubility Ionic compounds are soluble in water. Reason: Dissolving an ionic solid involves the setting of opposite ions free from the lattice into the solvent. This can happen when the strong electrostatic force of attraction between the opposite ions is weakened. Therefore, solvents having oppositely charged ions, called polar solvents should be used. The best polar solvent is water. Therefore, all ionic compounds are dissolved in water. 5. Electrical conductivity: Even though ionic solids consist of opposite ions, they are bad conductors of electricity. In ionic solids, a strong electrostatic force of attraction, making the ions immobile, holds the oppositely charged ions together. Hence, conductivity is not possible. However, in their fused or aqueous state, ionic compounds are good conductors of electricity owing to the presence of mobile ions. For instance, NaCl in its fused state or in its aqueous solution, has free Na+ and Cl– ions. The mobility of Na+ and Cl– results in conduction. 6. High reactivity Ionic compounds react instantaneously in fused state. This is because of easy formation of free ions, rapid union of these ions in solutions, form new compounds. NARAYANA GROUP OF SCHOOLS MPC BRIDGE COURSE For example, the reaction between NaCl and AgNO3 is very rapid in solution state, resulting in the formation of AgCl and a precipitate of NaNO3. 7. Directional properties A given ion in the ionic crystal is surrounded by a uniform electric field around it. Therefore, the electrostatic bonding force acts equally on the ion in all the directions. As there is no specific direction for the electrostatic bonding force, the ionic bond is a non-directional bond. 8. Isomorphism Crystals of different ionic compounds having similar arrangement of ions as well as geometry are known as isomorphs, and the phenomenon is known as isomorphism. Eg: ZnSO4. 7H2O & FeSO4. 7H2O A crystal of an isomorph, if placed in a saturated solution of other isomorphs, grows in size. The valency of elements forming isomorphous compounds is same. DAY-15: WORKSHEET 1. Number of electrons transferred from one atom to another during bond formation in Aluminium Nitride: a) 1 b) 2 c) 3 d) 4 2. The electrovalency of N in magnesium nitride: a) one b) two c) three d) four 3. Which one of the following has an electrovalent linkage? a) CH4 b) MgCl2 c) SiCl4 d) BF3 4. The electronic structure of four elements a, b, c and d are: a) 1s2 b) 1s2 2s2 2p2 c) 1s2 2s2 2p5 d) 1s2 2s2 2p6 101 CLASS-IX MPC BRIDGE COURSE The tendency to form electrovalent bond is greatest in a) 1 b) 2 c) 3 d) 4 5. Which of the following is easily formed? a) Calcium chloride b) Calcium bromide c) Potassium chloride d) Potassium bromide 6. Which of the following is least ionic? a) CaF2 b) CaBr2 c) CaCl2 d) CaI2 7. Which of the following is more ionic? a) Si3N4 b) AlN c) BN d) Ca3N2 8. Arrange the bonds in order of increasing ionic character in the molecules: LiF, K2O, N2, SO2 and ClF3. a) N2 < ClF3 < SO2 < LiF < K2O b) N2 < SO2 < ClF3 < K2O < LiF DAY-16 : SYNOPSIS Covalent bond and its formation A bond formed by the equal contribution and equal sharing of electrons between two atoms or more atoms is known as covalent bond (co-sharing, valence valence electron). Since, the formation of a covalent bond results in the formation of a molecule, it is also called molecular bond. G.N. Lewis did the study of covalent bond. He explained covalent bond formation by the electron dot structure called Lewis Structure. When is the bond between two atoms covalent? a) Both assertion and reason are correct and reason is the correct explanation of assertion. When non-metals come together, the tendency to donate or accept the electrons is not possible due to the less electronegativity (EN) difference. Thus, in order to acquire stable configuration (an octet or duplet) of a noble gas, sharing takes place between them, resulting in formation of covalent bond. Generally, if the EN difference between two nonmetals is less than 1.7, a covalent bond is formed between them due to their combination. b) Both assertion and reason are correct but reason is not the correct explanation of assertion. Representation of covalent bond: The covalent bond between a pair of two atoms is represented by a small line[ – ]. c) Assertion is correct and reason is incorrect. For example, H 2 can be represented as H–H. d) Assertion is incorrect and reason is correct. Covalency: The number of electron pairs shared between two atoms of the same element or different elements during the formation of a molecule is known as covalency. c) SO2 < N2 < ClF3 < LiF < K2O d) ClF3 < N2 < SO2 < K2O < LiF 9. Assertion (A) : non-volatile. Ionic compound tend to be Reason (R) : Inter molecular forces in these compounds are weak. The correct answer is: 10.Which of the following true for ionic compounds? a) They are hard solids b) They can be broken down into pieces very easily c) They are soluble in non-polar solvents d) all the above NARAYANA GROUP OF SCHOOLS Eg: Covalency of hydrogen molecule is equal to 1 and that oxygen molecule is 2. 102 CLASS-IX Bond pairs and lone pairs Bond pair of electrons: The shared pair of electrons, which result in the formation of a bond, is called the “bonded pair”. Lone pair of electrons: The pair of electrons, present in the valence shell but not involved in the bonding is called the “non-bonded pair” or “lone pair.” Conditions favourable for covalent bond formation: Covalent bonding is all about sharing of electrons. The ideal conditions necessary for atoms to share electrons are: MPC BRIDGE COURSE Covalent bonds based on the type of atoms involved in bonding Based on the types of atoms involved in bonding, covalent bonds are classified into homogeneous and heterogeneous covalent bonds. 1. Homogeneous covalent bond: It is a covalent bond formed between the atoms of similar type. Examples: (a) Formation of hydrogen molecule: 1) Atoms should be of small size. 2) They should have high electronegativity and ionisation potential. (or) H – H (hydrogen molecule) (b) Formation of chlorine molecule: 3) The electronegativity and the ionisation potential of the combining atoms should be almost the same. Favourable conditions for the covalent bond formation are satisfied by the elements of VA, VIA, and VIIA groups. (or) Cl – Cl (chlorine molecule) The electrons in the outermost shell (valence electrons) in the elements of VA, VIA and VIIA groups are 5, 6 and 7 respectively, and they can have stable octet configuration by sharing 3, 2 and 1 electron respectively. The molecules formed among the elements of VA, VIA and VIIA are mostly covalent molecules. Eg: SO2, PCl3, PBr3, SF6, SCl2, IF7, O2, N2, F2, Cl2, Br2, I2, etc. (or) O = O (oxygen molecule) Note: Number of electron pairs shared between two atoms of the same element is equal to the number of electrons short of octet. (c) Formation of oxygen molecule: 2. Heterogeneous covalent bond: It is a covalent bond formed between the atoms of different types. Examples: (a)Formation of Hydrogen Chloride HCl: Examples of covalent compounds: (or) F2, Cl2, I2, O2, N2, H2, HCl, H2O, NH3 etc. Note: The force of attraction present between the molecules of inert gases is the Vander Waal’s forces. NARAYANA GROUP OF SCHOOLS H – Cl (Hydrogen chloride) After formation of a covalent bond, hydrogen has stable duplet 103 CLASS-IX S.No. 1. 2 3. S.No. 1. 2 3. MPC BRIDGE COURSE Type of Definition covalent bond Single The covalent covalent bond bond formed by sharing of 1 pair of electrons Double The covalent covalent bond bond formed by sharing of 2 pairs of electrons Triple covalent The covalent bond bond formed by sharing of 3 pairs of electrons Representation Examples Type of Definition covalent bond Single The covalent covalent bond bond formed by sharing of 1 pair of electrons Double The covalent covalent bond bond formed by sharing of 2 pairs of electrons Triple covalent The covalent bond bond formed by sharing of 3 pairs of electrons Representation H × H H2 , Cl2 , F2 etc (f) Formation of Carbondioxide molecule – CO2: H-H O2 × ×× O O × ×× II. Covalent bonds based on the number of electron pairs shared. O=O N2 × × × (or) O = C = O (Carbondioxide) N N × × Summary: DAY-16: WORKSHEET N N H H × Examples H2 , Cl2, F2 etc H-H O2 × ×× O O × ×× O=O N2 × × × N N × × N N configuration, and chlorine has stable octet configuration. (b) Formation of water molecule – H2O: 1. When two atoms of chlorine combine to form one molecule of chlorine gas, the energy of the molecule is: a) Greater than that of separate atoms b) Equal to that of separate atoms c) Lower than that of separate atoms d) None of these 2. In the formation of covalent bond, a) Transfer of electrons take place b) Electrons are shared by only one atom c) Sharing of electrons take place d) None of these 3. Silicon has 4 electrons in the outermost orbit. In forming the bonds: a) It gains electrons b) It loses electrons c) It shares electrons d) None of the above (c) Formation of Ammonia molecule – NH3: 4. Which of the following is a covalent compound? a) H2 b) CaO c) KCl d) Na2S 5. Which of the following substance has covalent bonding? (d) Formation of methane molecule – a) Sodium chloride b) Solid neon c) Copper d) BeCl2 6. Which is a covalent compound? :4CH a) RbF (e) Formation of carbon tetrachloride – CCl 4 : b) MgCl2 c) CaC2 d) NH3 7. An element ‘Y’ has the ground state electronic configuration 2, 8, 8. The type of bond that exists between the atoms of ‘Y’ is: a) Ionic b) Covalent c) Metallic d) Vander Waal’s (Carbontetrachloride) NARAYANA GROUP OF SCHOOLS 104 CLASS-IX MPC BRIDGE COURSE 8. The bond between two identical nonmetal atoms has a pair of electrons: a) Unequally shared between the two b) Transferred fully from one atom to another c) With identical spin d) Equally shared between them 9. The maximum number of covalent bonds by which the two atoms can be bonded to each other is: a) Four b) Two c) Three d) No fixed number 10. The molecule, which contains maximum number of electrons, is: a) CH4 b) CO2 c) NH3 d) BCl3 DAY-17 : SYNOPSIS Properties of covalent compounds i) Physical state. Generally, covalent compounds are gases, liquids or soft solids at room temperature. Reaso n : Covalent compounds are composed of molecules. There exists a intermolecular force of attraction known as Vander waal’s force of attraction between these molecules’. These Vander waal’s forces are weak and hence covalent compounds exist as soft solids, liquids and gases. ii) Melting and boiling Points: Covalent compounds possess low melting and boiling points. Reason: Melting and boiling involves the breaking of intermolecular force of attraction between the molecules. As the intermolecular forces are weak between the molecules, less energy is required to break them. Hence, melting and boiling points of covalent compounds are low. Exception: Daimond iii) Solubility: Covalent compounds are soluble in non-polar solvents. NARAYANA GROUP OF SCHOOLS Reason: Like dissolves like. Covalent compounds have molecules without the opposite polarity and hence they are nonpolar compounds. Therefore, covalent compounds are soluble in non-polar solvents like kerosene, benzene, ether, carbon disulphide, carbon tetrachloride, etc. Exception: Covalent compounds, being non-polar, are insoluble in polar solvents like water. But some covalent compounds like alcohol, urea and sugar are soluble in water. Because, this is due to the attraction between covalent molecules like alcohol, urea, sugar and the water molecule. This attraction is called “Hydrogen bonding”. The attractive force of Hydrogen bond pulls out-the urea molecule from its solid phase into the solvent (water) and the urea gets dissolved.[Note: We shall learn about Hbonding in detail, in future classes]. iv) Conductivity: Charge carriers are required for substances to conduct electricity. Ionic solutions have charge carriers in the form of mobile ions. Metals have charge carriers in the form of electrons. But covalent compounds do not have any charge carriers and hence they are bad conductors of electricity. Exception: Covalent compounds do not conduct electricity. But graphite is a good conductor of electricity even though it is a covalent compound. Graphite has a structure comprising layers. Each layer consists of a network of ‘hexagonal carbon rings’. Each carbon atom in the ring is covalently bonded to three adjacent carbon atoms. Therefore, out of the four valence electrons of carbon, three get trapped in the covalent bond, leaving the fourth electron free. This free electron in graphite is responsible for the conductivity of graphite. 105 CLASS-IX v) Speed of reaction: Reactions of covalent compounds are slow. Reas on: The reactions of covalent compounds involve, firstly, the breaking of the existing bonds in the molecules and then the formation of new bonds. Breaking of bonds requires extra energy, and until sufficient energy is available, the reaction is not initiated. Further, to initiate the reaction, the reacting molecules should collide. And among all the collisions, only a small fraction is fruitful. Since a covalent reaction involves a number of operations, it is slow. MPC BRIDGE COURSE During the unequal sharing, the atom which exercises more influence on the electron pair gets the partial negative ( ) charge and the other atom gets the partial positive ( ) charge. For example, in H 2 O molecule the electronegativity of “O” is more, and its influence is greater on the shared pair. This results in a partial negative charge on oxygen and partial positive charge on hydrogen and electron pair is unequally shared. H vi) Directional property: Covalent bonds are directional. The direction is along the inter-nuclear axis (line joining the nuclei of two reacting atoms). This directional property leads to different arrangement of the atoms along the line of direction. The result being that different molecules are formed with the same structure. This phenomenon is called Isomerism and the molecules are called Isomers. Comparison between ionic and covalent compounds Polar covalent bond and its formation Polar covalent bond: We have seen that covalent bond is all about sharing of electrons. Consider a covalent compound AB formed from A and B. If both A and B exercise the same amount of influence on the shared electrons, then A and B are said to share the electron pair equally. This is seen in H2, where both atoms A and B belong to the same element. Such compounds are called non-polar covalent compounds. It is interesting to see what happens when one of the atoms in the covalent bond influences the shared electron pair, more than the other. If this happens, then the electron pair is said to be shared unequally between A and B. NARAYANA GROUP OF SCHOOLS O H The covalent bond formed due to unequal sharing of electrons is called a Polar covalent bond and the molecule is called a Polar molecule. Examples of polar molecules are HF, NH3, HCl, etc. Co-ordinate covalent bond: The bond in which the shared pair of electrons is contributed by only one atom but, is shared by both atoms or molecules is known as co-ordinate covalent or dative bond. Sidgwick and Powell proposed the coordinate covalent bond. Eg : NH3 BF3, NH4, NH4 , H3O+ H F H N B H F F H F H N B H F F The atom which contributes the shared pair is called the donor atom, and the other atom which makes use of it is known as the acceptor atom. The coordinate covalent bond is represented by an arrow ( ) directed from the donor to the acceptor. 106 CLASS-IX According to Lewis, a base is one, which donates a lone pair of electrons and an acid is one which accepts the lone pair of electrons. So, in a co-ordinate covalent bond donor atom called as Lewis base and accepted atom called as Lewis acid. DAY-17: WORKSHEET 1. Substance X has a melting point of 1500°C. It conducts current at room temperature. Substance Y is gas at room temperature. It does not conduct current. Substance Z is soluble in water and the solution conducts current. However, in solid state Z does not conducts current but melts at 1020°C to form a conducting liquid. Deduce whether X, Y, Z. i) have giant or molecular structure. ii) are bonded covalently or ionically. 2. CCl4 is insoluble in water because, a) H2O is non-polar. b) CCl4 is non-polar. c) They do not form inter molecular H– bonding. d) They do not form intra molecular H– bonding. 3. Which of the following when dissolved in water forms a non conducting solution? a) Green vitriol b) Chile or Indian salt petre c) Alcohol d) Potash alum 4. Which of the following properties would suggest that a compound under investigation is covalent? MPC BRIDGE COURSE 5. Assertion (A) : Graphite is a good electrical conductor. Reason (R) : The free electrons in graphite conducts electricity. a) Both assertion and reason are correct and reason is the correct explanation of assertion. b) Both assertion and reason are correct but reason is not the correct explanation of assertion. c) Assertion is correct and reason is incorrect. d) Assertion is incorrect and reason is correct. 6. Which of the following statements is incorrect? i) Covalent bond is highly directional. ii) Ionic compounds do not exhibit space is isomerism. iii) Molecules react quickly than ions. a) Both ii and iii c) Both i and iii b) Both i and ii d) Only iii 7. The electronegativity values of C, H, O and N are 2.5, 1, 3.0 and 2.5 respectively. The most polar bond is a) S - H b) O - H c) N - H d) C - H 8. Which of the following molecules is non polar but it contains polar bonds in it? a) BCl3 b) H2 c) NH3 d) CHCl3 9. The two compounds that are covalent when taken pure but produce ions when dissolved in water. a) NaCl, NaBr b) H2O, Cl2 c) HCl, H2O d) HF, NH3 a) It conducts electricity on melting. b) It is a non-electrolyte. c) It has a high melting point. d) It is a compound of a metal and a nonmetal. NARAYANA GROUP OF SCHOOLS 107 CLASS-IX MPC BRIDGE COURSE MATHEMATICS KEY DAY-5 KEY DAY-1 KEY 1) 3 2) 2 3) 3 4) 1 5) 2 6) 1 7) 2 8) 1 9) 2 10) 1 11) 2 12) 3 13) 2 14) 1 15) 1 16) 3 17) 1 18) 1 19) 1 DAY-2 KEY 1) 1 2) 4 3) 4 4) 1 5) 4 6) 3 7) 2 8) 2 9) 1 10) 1 11) 3 12) 1 13) 2 14) 1 15) 1 16) 3 17) 3 18) 4 19) 1 20) 1 21) 1 22) 4 1) 2 2) 1 3) 3 4) 1 5) 4 6) 3 7) 1 8) 2 9) 4 10) 3 11) 3 12) 1 13) 4 14) 3 15) 1 16) 4 17) 1,2,3,4 18) 1 19) 3 20) 3 21) 4 22) 1 23) 3 24) 2 25) a-s; b-r; c-q; d-r 26) 0 27) 9/7 28) 3 DAY-6 KEY DAY-3 KEY 1) 3 2) 4 3) 3 4) 1 5) 2 6) 2 7) 1 8) 3 9) 4 10) 2 11) 3 12) 1 13) 4 14) 1 15) 1 16) 1 17) 1 18) 4 19) 1 20) 3 21) a-s; b-s; c-q; d-p,r 1) 1 2) 3 3) 4 4) 1 5) 2 6) 2 7) 3 8) 3 9) 4 10) 3 11) 1 12) 1 13) 1 14) 4 15) 4 16) 1 17) 2 18) 4 19) 1 DAY-7 KEY 1) 3 2) 1 3) 4 4) 2 5) 2 6) 1 7) 1 8) 1 9) 2 10) 2 11) 3 12) 1 13) 1 14) 1 15) 2 DAY-4 KEY DAY-8 KEY 1) 3 2) 4 3) 2 4) 3 5) 3 6) 1 7) 1 8) 3 9) 1 10) 4 11) 2 12) 2 13) 1 14) 3 15) 3 16) 2 NARAYANA GROUP OF SCHOOLS 1) 1 2) 2 3) 1 4) 3 5) 2 6) 2 7) 3 8) 1 9) 2 10) 1 11) 2 12) 2,3 13) 1 14) 2 15) 3 108 CLASS-IX MPC BRIDGE COURSE DAY-9 KEY DAY-13 KEY 1) 4 2) 3 3) 2 4) 3 1) 4 5) 2 6) 3 7) 1 8) 3 10) 1,2 9) 4 10) 1 11) 1 12) 2 13) 4 14) 4 15) 1 2) 4 3) to 9) Graphs DAY-14 KEY 16) 1,2,3 17) 1 18) 3 19) 1 20) a-p,q,r,t; b-q; c-q,r; d-p,q,r,t 21) 1,2 DAY-10 KEY 1) 2 2) 3 3) 3 4) 3 5) 4 6) 1 7) 2 8) 1 9) 3 10) 2 11) 2 12) 1 13) 2 14) 2 15) 3 16) 2 17) 3 21) 2 18) 3 19) 2 20) 1 1) 2 2) 2 3) 3 4) 1 5) 3 6) 4 7) 2 8) 1 9) 2 10) 2 11) 1 12) 2 13) 1 14) 1 15) 1 16) 1 DAY-15 KEY 1) 2 2) 4 3) 4 4) 1 5) 2 6) 1 7) 3 8) 2 9) 3 10) 2 11) 3 12) 2 13) 2 14) 1 15) 4 16) 1 DAY-11 KEY DAY-16 KEY 1) 1 2) 3 3) 1 4) 1 1) 2 2) 1 3) 4 4) 3 5) 1 6) 2 7) 3 8) 2 5) 1 6) 4 7) 3 8) 3 9) 1 10) 2 11) 2 12) 1 9) 4 10) 2 11) 1 12) 1 13) 4 14) 1 15) 3 16) 1 13) 1 14) 4 15) 1 17) 1 18) 1 19) a-t; b-q; c-s; d-p DAY-17 KEY DAY-12 KEY 1) 3 2) 3 3) 2 4) 2 1) 3 2) 1 3) 1 4) 3 5) 4 6) 3 7) 1 8) 1 5) 1 6) 1 7) 3 8) 1 9) 3 10) 2 11) 1 12) 4 9) 1 10) 1 11) 1 12) 2 13) 3 14) a-s; b-r; c-p; d-q 13) 2 14) 3 15) 1 16) 3 17) 1 18) 1 19) 3,4 NARAYANA GROUP OF SCHOOLS 109 CLASS-IX MPC BRIDGE COURSE PHYSICS KEY DAY-7 KEY DAY-1 KEY 1) 1 5) 3 2) 2 6) 1 3) 4 7) 2 1) 4 2) 2 3) 3 4) 1 5) 3 6) 1 7) 1 8) 4 9) 1 10) 2 11) 1 12) 3 4) 2 8) 1 DAY-8 KEY DAY-2 KEY 1) 3 2) 1 3) 4 4) 3 1) 2 2) 2 3) 3 4) 3 5) 3 6) 1 7) 1 8) 2 5) 2 6) 1 7) 4 8) 1 9) 3 10) 2 11) 1 12) 3 13) 2 9) 1 DAY-9 KEY DAY-4 KEY 1) 3 2) 2 3) 2 5) 1 6) 1 7) 3 4) 2 1) 2 2) 4 3) 2 4) 2 5) 3 6) 1 7) 3 8) 3 9) 4 10) 2 11) 2 12) 3 DAY-4 KEY DAY-10 KEY 1) 3 2) 3 3) 4 4) 3 5) 1 6) 2 7) 2 8) 4 9) 3 10) 2 1) 1 2) 2 3) 1 4) 2 5) 3 6) 4 7) 2 8) 1 9) 1 10) 1 11) 3 12) 1 DAY-5 KEY DAY-11 KEY 1) 2 2) 1 3) 3 4) 1 1) 2 2) 1 3) 2 4) 3 5) 1 6) 2 7) 3 8) 2 5) 1 6) 1 7) 2 8) 3 9) 2 10) 2 11) 2 13) 4 9) 2 10) 2 11) 3 12) 4 13) 3 14) 4 DAY-12 KEY DAY-6 KEY 1) 2 2) 3 3) 1 5) 1 6) 4 7) 1 NARAYANA GROUP OF SCHOOLS 4) 2 1) 2 2) 1 3) 1 4) 1 5) 3 6) 2 7) 2 8) 3 9) 1 10) 1 11) 1 110 CLASS-IX MPC BRIDGE COURSE DAY-13 KEY DAY-18 KEY 1) 1 2) 3 3) 2 4) 1 1) 1 2) 1 3) 1 4) 1 5) 4 6) 2 7) 2 8) 2 5) 2 6) 1 7) 1 8) 1 9) 2 10) 2 11) 2 12) 2 9) 2 10) 1 11) 1 12) 4 13) 4 CHEMISTRY KEY DAY-14 KEY DAY-1 KEY 1) 1 2) 2 3) 2 4) 3 1) 1 2) 1,2,3,4 5) 3 6) 1 7) 4 8) 3 4) 1 5) 2 6) 2 9) 4 10) 4 8) 3 9) 2 10) 3 DAY-15 KEY 3) 2 7) 3 DAY-2 KEY 1) 1 2) 2 3) 3 4) 4 1) 1 2) 4 3) 3 4) 3 5) 1 6) 2 7) 2 8) 3 5) 2 6) 3 7) 4 8) 2 9) 4 10) 3 11) 1 12) 1 9) 4 10) 4 13) 1 14) 1 DAY-3 KEY DAY-16 KEY 1) 1 2) 1,2 3) 1,2 4) 2 7) 4 8) 3 1) 2 2) 1 3) 2 4) 3 5) 3 6) 4 5) 3 6) 2 7) 3 8) 1 9) 2,3 10) 3 9) 1 10) 2 11) 1 DAY-4 KEY DAY-17 KEY 1) 2 2) 2 3) 4 4) 1 5) 1 6) 4 7) 3 8) 3 9) 2 10) 4 11) 2 12) 2 NARAYANA GROUP OF SCHOOLS 1) 1 2) 1,2,3,4 4) 1 8) 3 5) 2 9) 2 3) 2 6) 2 10) 3 7) 3 DAY-5 KEY 1) 1 2) 4 3) 3 4) 3 5) 2 6) 3 7) 4 8) 2 9) 4 10) 4 111 CLASS-IX MPC BRIDGE COURSE DAY-6 KEY DAY-12 KEY 1) 2 2) 2 3) 3 4) 3 1) 3 2) 1 3) 2 4) 1 5) 3 6) 4 7) 2 8) 3 5) 3 6) 3 7) 3 8) 2 9) 4 10) 3,4 9) 1,3,4 10) 1 DAY-7 KEY DAY-13 KEY 1) 4 2) 4 3) 3 4) 4 1) 3 2) 3 3) 1 4) 3 5) 4 6) 2 7) 3 8) 1 5) 2 6) 2 7) 4 8) 1 9) 3 10) 4 9) 3 10) 2 DAY-8 KEY DAY-14 KEY 1) 4 2) 3 3) 1 4) 1 1) 1,3,4 2) 1 3) 2 4) 1 5) 3 6) 4 7) 2 8) 2 5) 1 6) 2,3 7) 2 8) 2 9) 1 10) 4 9) 2 10) 3 DAY-9 KEY DAY-15 KEY 1) 1 2) 4 3) 2 4) 4 1) 3 2) 3 3) 2 4) 3 5) 3 6) 3 7) 1 8) 4 5) 3 6) 4 7) 4 8) 1 9) 2,3 10) 1 9) 3 10) 1,2 DAY-10 KEY DAY-16 KEY 1) 1,2 2) 3 3) 2 4) 2 1) 3 2) 3 3) 3 4) 1 5) 1 6) 3 7) 1 8) 3 5) 4 6) 4 7) 4 8) 4 9) 2 10) 4 9) 3 10) 4 DAY-11 KEY DAY-17 KEY 1) 3 2) 3 3) 3 4) 1 1) 1 2) 2 3) 3 4) 2 5) 3 6) 2 7) 1 8) 2 5) 1 6) 4 7) 2 8) 1 9) 3 10) 1 NARAYANA GROUP OF SCHOOLS 9) 4 112