Lakhmir Singh Chemistry Class 9 Solutions Structure of The Atom
Text Book Solutions – Lakhmir Singh and Manjit Kaur – Chemistry
CBSE 9 – Chemistry Chapter 3
Lakhmir Singh Class 9 CBSE Chemistry Text Book PAGE No 191 Answers
Neutron is not present in ordinary hydrogen atom.
J. J. Thomson
Maximum of 2 e- can be accommodated in K shell of an atom.
Maximum of 8 e- can be accommodated in L shell of an atom.
Maximum of 18 e- can be accommodated in M shell of an atom.
Maximum of 32 e- can be accommodated in N shell of an atom.
(a). Maximum of 2 e- can be accommodated in innermost shell of an atom.
(b). Maximum of 8 e- can be accommodated in outermost shell of an atom.
Three subatomic particles present in atom are electrons, protons and neutrons.
Electron is negatively charged particle present in atoms of all the elements.
J. J. Thomson discovered electron.
Lakhmir Singh Class 9 CBSE Chemistry Text Book Page No 192 Answers
Nucleus is the central part of an atom in which protons and neutrons are held together.
K, L, M, N were the letters used by Bohr to represent electron shells in an atom.
Protons and neutrons actually determine the mass of atom.
Proton is the positively charged particle present in atoms of all the elements.
Electronic configuration of hydrogen : 1
Proton is 1840 times heavier than electron.
Hydrogen gas produces anode rays consisting of protons in the discharge tube experiment.
Nucleus was discovered by Rutherford in the alpha particle scattering experiment.
Positive charge on the nucleus is due to presence of protons.
(a). 8 electrons are present in outermost shell of Neon.
(b). 7 electrons are present in outermost shell of Chlorine.
(a). L shell can accommodate maximum of 8 e-.
(b). N shell can accommodate maximum of 32 e-.
(a). K shell can accommodate maximum of 2 e-.
(b). M shell can accommodate maximum of 18 e-.
(i). Chadwick discovered ‘neutron’.
(ii). Thomson discovered ‘electron’.
(iii). Goldstein discovered ‘proton’.
(a). Proton has relative charge of +1.
(b). Electron has relative charge of -1.
(c). Neutron has relative charge of 0.
(a) Atomic number
(b) Mass number
(l) Negative; Positive; No
Electron is a negatively charged particle found in the atoms of all elements.
The relative mass of an electron is 1/1840 u.
A charge of -1 is carried by an electron.
The deflection of fast moving alpha- particles through small and large angles in Rutherford’s scattering experiment is the evidence for the presence of nucleus in an atom.
Important information furnished about nucleus in Rutherford’s alpha- particle scattering experiment is:
(i). Nucleus of an atom is positively charged.
(ii). Nucleus of an atom is very hard and dense.
(iii). Nucleus of an atom is very small as compared to the size of an atom as a whole.
Most of the alpha- particles passed straight through the gold foil without any deflection in Rutherford’s alpha- particle scattering experiment, this shows that most of the space in an atom is empty.
There are equal no. of positive and negative particles in an atom, so it is neutral as a whole.
(a). Proton is present in same fixed no. in the atoms of any particular element.
(b). Atomic no. is characteristic for any particular element.
Lakhmir Singh Class 9 CBSE Chemistry Text Book Page No 193 Answers
Protons are positively charged particle found in the atoms of all the elements.
Relative mass of proton is 1u.
Relative charge of proton is +1 C.
Difference between proton and neutron-
(1). Proton is positively charged while electron is negatively charged.
(2). Proton is much heavier than electron.
Two observations which shows that atom is not indivisible are-
(1). In J. J. Thomson’s experiment, the stream of cathode rays in the gas discharge tube shows the presence of negatively charged subatomic particles called electrons.
(2). In Goldstein’s experiment, the faint red glow in the gas discharge tube shows the presence of positively charged subatomic particles called protons.
(i). Formation of cathode rays tells about the presence of negatively charged electrons in all the atoms.
(ii). Formation of anode rays tells about the presence of positively charged protons in all the atoms.
The arrangement of electrons in various shells of an atom of the element is known as electronic configuration of the element.
Electronic configuration of oxygen (atomic no. = 8) is (2,6)
Electronic configuration of element with atomic no. 12- (2,8,2)
So, K-2 ; L-8 ; M-2
(a). Nucleus is a small positively charged part at the center of an atom. Nucleus is positively charged.
(b). Rutherford discovered nucleus of an atom.
Alpha – particles were used by Rutherford in his experiment on the discovery of nucleus.
Alpha – particles have +2 units of charge.
(a). There are 13 e- in each atom of the element.
(b). Electronic configuration of given element- (2,8,3)
K-2 ; L-8 ; M-3
Atomic No. – 18
Electronic configuration – (2,8,8)
The special thing about the outermost shell is that it is completely filled with the electrons.
The neutron is a neutral particle found in the nucleus of an atom. Its relative mass is 1 u. It has no charge.
Electron has relative mass of 1/1840 u, proton has 1u and neutron also has 1u.
Electron has relative charge of -1u, proton has +1u and neutron has 0 relative charge.
Protons are positively charged particle found in the atoms of all the elements whereas neutron is a neutral particle found in the nucleus of an atom.
Electron has relative mass of 1/1840 u and proton has relative mass of 1u.
Electron has relative charge of -1u while proton has +1u of relative charge.
Proton has relative mass of 1u and neutron also has relative mass of 1u.
Proton has relative charge of +1u and neutron has no relative charge.
Electron has relative charge of -1 u whereas neutron has no relative charge.
Also, electron has relative mass of 1/1840 u and neutron has relative mass of 1 u.
Protons and neutrons are collectively present in the nucleus at the center while electrons revolve rapidly round the nucleus in fixed circular orbits called energy levels.
(a). The stream of particles coming out from cathode (negative electrode) are called cathode rays. Cathode rays are negatively charged.
(b). When electricity at high voltage is passed through a gas at very low pressure taken in discharge tube, stream of minute particles are given out by the cathode. These stream of particles are called cathode rays.
(c). The conclusion is that all the atoms contain negatively charged particles called electrons.
(a). According to Thomson model of atom- An atom consists of a sphere of positive charge with negatively charged electrons embedded in it. The positive and negative charges in an atom are equal in magnitude.
Neutron was not present in the Thomson model of atom.
(b). When mass no. is 18 and no. of electrons is 7 then
(i). No. of protons = 7
(ii). No. of neutrons = 18 – 7 = 11
(iii). Atomic no. = 7
(a).Rutherford’s model of atom-
1. An atom consists of positively charged, dense and very small nucleus containing all the protons and neutrons. Almost all the mass of atom is concentrated in the nucleus.
2. The nucleus is surrounded by negatively charged electrons. The electrons are revolving at very high speed round the nucleus in fixed circular orbits.
3. The electrostatic attraction between the positively charged nucleus and negatively charged electrons keep the atom held together.
4. An atom is electrically neutral.
5. Most of the space in an atom is empty.
The major drawback of Rutherford model of atom is that it does not explain the stability of the atom.
(b).Given: Mass no. = 23
No. of electrons = 11
Then, no. of protons = 11
No. of neutrons = 23 – 11 = 12
Atomic no. = 11
(a). Bohr’s model of atom-
1. An atom is made up of three particles, namely electrons, protons and neutrons.
2.The protons and neutrons are located in the small nucleus at the center of atom.
3. Electrons revolve round the nucleus in fixed circular orbits.
4. Maximum no. of electrons for any given shell is fixed. Any shell cannot exceed that maximum value.
5. Each given shell is associated with fixed amount of energy.
6. There is no change in energy of electrons as long as they keep revolving in the same energy level, and the atom remains stable.
(b).Given: Atomic no. = 11
Mass no. = 23
Then, electronic configuration – (2,8,1)
Nuclear composition is – 11 protons and 12 neutrons
Lakhmir Singh Class 9 CBSE Chemistry Text Book Page No 194 Answers
(a). (i). Atomic no. is the number of protons in one atom of an element.
(ii). Mass no. is the total number of protons and neutrons present in one atom of the element.
Example- The total no. of protons in a carbon atom is 6, so its atomic no. is 6.
Also, one atom of Na contains 11 protons and 12 neutrons, so its mass no. is 23.
(b). Mass No. = Atomic no. + No. of neutrons
(c). No. of neutrons = Mass No. – Atomic no.
= 24 – 12 = 12
(i). Mass no. = 31
(ii). Atomic no. = 15
(iii). E.C. = (2,8,5)
(a). E.C. – (2,8,7)
(b). Atomic No. = 17
(d). Anion; X-
(e). X must be Chlorine
(a). Atomic no. = 3
(b). Mass no. = 3 + 4 = 7
(c). E.C. – (2, 1)
(e). Cation will be formed; because outermost single electron can be easily donated.
Lakhmir Singh Class 9 CBSE Chemistry Text Book Page No 195 Answers
(a). Mass number
(b). Atomic number
(c). No. of protons = 4
(d). No. of neutrons = 9 – 4 = 5
(e). No. of electrons = 4
(f). Electrons in outermost orbit = 2
(a). Atomic no. = 18
(b). Element Z is non-metal
(c). As the outermost shell of element Z is completely filled so, it will not form any ion.
(d). Outermost electronic shell is completely filled with electrons.
(e). Name of element ‘Z’ = Argon
Symbol is Ar
(f). Z belongs to the group ‘Noble gases’.
Lakhmir Singh Class 9 CBSE Chemistry Text Book Page NO 210 Answers
E.C of Nitrogen = 2, 5
So, no. of valence electrons in Nitrogen atom = 5
Helium has less than 8 electrons in the valence shell of an atom. Its atomic no. is 2
Radioactive isotopes are used in the treatment of cancer.
One such isotope is Cobalt-60.
Uranium-235 isotope is used as a fuel in the reactors of nuclear power plants for generating electricity.
Cobalt-60 radioisotope is used in the treatment of cancer.
Iodine-131 radioisotope is used to determine the activity of thyroid gland.
Radioactive isotopes are used in industry to detect the leakage in underground oil pipelines, gas pipelines and water pipes.
The given statement is false.
Atoms containing same number of protons and electrons but different number of neutrons are called ISOTOPES.
The given pair are isotopes.
Radioactive isotopes have unstable nuclei and emit various types of radiations.
(a). Atomic no. = 5
(b). Mass no. = 6 + 5 = 11
(c). No. of electrons = 5
(d). No. of valence electrons, per atom = 3
Atomic No. = 17
E.C. = (2, 8, 7)
Valency = 8 – no. of valence electrons = 8 – 7 = 1
Atomic No. of X = 16
E.C. of X = (2, 8, 6)
Valency of X = 8 – no. of valence electrons = 8 – 6 = 2
Lakhmir Singh Class 9 CBSE Chemistry Text Book Page No 211 Answers
Valency shown by A (atomic no. 2) – 0
Valency shown by B (atomic no. 4) – 2
Valency shown by C (atomic no. 8) – 2
Valency shown by D (atomic no. 10) – 0
Valency shown by E (atomic no. 13) – 3
Uranium-235 : This isotope is used as a fuel in the reactors of nuclear power plants for generating electricity.
Cobalt-60 : This is used in the treatment of cancer.
The difference in the masses of isotopes of an element is due to different number of neutrons in their nuclei.
Because all the isotopes of an element have identical atomic configuration containing same number of valence electrons, therefore, all the isotopes of an element show identical chemical properties.
For example- Cl-35 and Cl-37, show identical chemical properties as they have same no. of 7 valence electrons.
Due to slight difference in the masses of the isotopes of an element, the physical properties of the isotopes are slightly different.
The fractional atomic masses of elements are due to the existence of their isotopes having different masses.
Deuterium, Protium and Tritium are isotopes.
Argon and Calcium are isobars.
(i). Due to identical electronic configuration containing the same no. of valence electrons, these isotopes have almost identical chemical properties.
(ii). All of them have 1 electron and 1 proton, so, they are electrically neutral.
Isobars are the atoms of different elements having different atomic numbers but the same mass number (or same atomic mass).
H – 1 proton, 1 electron and no neutron.
D – 1 proton, 1 electron and 1 neutron.
T – 1 proton, 1 electron and 2 neutrons.
Atomic No. = 7
E.C = 2, 5
Valency of given element = 3
Given element is NITROGEN.
(a). The number of electrons present in the valence shell are called valence electrons.
Valence electrons are situated in the outermost shell.
(b). There are 3 valence electrons present in the element with atomic no. 13.
Valence shell of this atom is M.
(a).The isotopes which are unstable due to presence of extra neutrons in their nuclei and emit various types of radiations, are called radioactive isotopes or radioisotopes.
For example: Carbon – 14 , Arsenic – 74
(b). Uses of isotopes-
(i). They are used in the treatment of cancer.
(ii). Radioactive isotopes are used as ‘tracers’ in medicine to detect the presence of tumors and blood clots in human body.
(a). The capacity of an atom of an element to form chemical bonds is known as its valency.
Valency of an atom with atomic no. 14 is 4.
(b). The valency of an element is either equal to the number of valence electrons in its atom or equal to the number of electrons required to complete eight electrons in the valence shell.
Valency of metal = No. of valence electron in its atom
Valency of a non-metal = 8 – No. of valence electron in its atom
For example- Valency of sodium (metal) is 1 and that of chlorine (non-metal) is also 1.
Lakhmir Singh Class Page No 212
Lakhmir Singh Class 9 CBSE Chemistry Text Book Page No 213 Answers
A and B are the example of isobars. This is because they have same number of nucleons.
Mass no. of A and B is 40.
The two species are isobars.
A represents Argon (Atomic no. = 18) while B represents Calcium (Atomic no. = 20).
A and D are isotopes as they have the same number of protons.
(i). Mass number of X = 8 + 8 = 16
(ii). Mass number of Y = 8 + 9 = 17
(iii). X and Y are isotopes.
(iv). X and Y represent Oxygen.