NCERT solutions for class 11 chemistry chapter 2 Structure of Atom In this chapter, we will study about the discovery of fundamental particles like electron, proton and neutrons. Solutions of NCERT class 11 chemistry chapter 2 structure of atom deals with the questions based on various advancements being made to explain different atomic models which gave the basic concept of quantization of energy and gave a modern structure of atoms. In this chapter, there are 67 question in the exercise. The NCERT solutions for class 11 chemistry chapter 2 structure of atom are prepared by our subject experts which provide all the answers. These NCERT solutions help students to clear their doubts in other subjects and other classes as well.
The rich diversity of chemical behavior of various elements can be discovered to the differences in the internal structure of atoms of these elements. This chapter starts with some experimental observations made by scientists towards the end of 19th and beginning of the 20th century. According to these observations, atoms are made of subatomic particles, such that protons, electrons and neutrons. This concept is very different from Dalton's atomic theory. NCERT solutions for class 11 chemistry chapter 2 structure of atom has covered all the topic wise answers as well as the exercise answers.
After completing CBSE NCERT solutions for class 11 chemistry chapter 2 Structure of Atom, students will be able to know about the discovery of proton, electron and neutron; describe Rutherford, Thomson and Bohr atomic models; understand nature of Planck's quantum theory and electromagnetic radiation; explain the photoelectric effect; state the de Broglie relation and Heisenberg uncertainty principle. This chapter also explains atomic orbitals in terms of quantum numbers, Pauli exclusion principle, Aufbau principle and Hund's rule of maximum multiplicity and at the end of chapter students will learn how to write the electronic configurations of atoms. Please scroll down to get free NCERT solutions for class 11 chemistry chapter 2 Structure of Atom.
Topics of NCERT class 11 chemistry chapter 2 Structure of Atom
2.1 Discovery of Subatomic Particles
2.2 Atomic Models
2.3 Developments Leading to the Bohr’s Model of Atom
2.4 Bohr’s Model for Hydrogen Atom
2.5 Towards Quantum Mechanical Model of the Atom
2.6 Quantum Mechanical Model of Atom
As the mass of one electron we know is .
Therefore,
.
As the mass of one electron is equal to
Therefore, Mass of 1 mole or electrons .
Charge on one electron is .
Therefore, the charge on 1 mole of electrons will be:
.
1 molecule of methane contains electrons.
Therefore, 1 mole of methane will contain:
electrons.
As 1 atom of ^{14}C contains .
and the number of atoms in ^{14}C in 1 mole is atoms.
Therefore, the number of neutrons in 14g of ^{14}C in 1 mole .
The number of neutrons in :
.
As the mass of one neutron is .
Then the mass of total neutrons in 7grams of ^{14}C:
.
1 mole of ammonia .
and 1 atom of contains .
Therefore, the number of protons in 1 mole of .
Number of protons in 3mg of :
No, there will be no effect of temperature and pressure.
Will the answer change if the temperature and pressure are changed?
As the mass of one proton is
Therefore, the mass of will be:
No, there will be no effect of temperature and pressure.
Given the nucleus of carbon:
Atomic number (Z) = 6
Mass number (A) = 13
Number of protons (Z) = 6
Number of neutrons (AZ) = 136 = 7
Given the nucleus of oxygen:
Atomic number (Z) = 8
Mass number (A) = 16
Number of protons (Z) = 8
Number of neutrons (AZ) = 168 = 8
2.3How many neutrons and protons are there in the following nuclei?
Edit Q
Answer:
Given the nucleus of Magesium:
Atomic number (Z) = 12
Mass number (A) = 24
Number of protons (Z) = 12
Number of neutrons (AZ) = 2412 = 12
Given the nucleus of Iron:
Atomic number (Z) = 26
Mass number (A) = 56
Number of protons (Z) = 26
Number of neutrons (AZ) = 5626 = 30
=
Given the nucleus of Strontium:
Atomic number (Z) = 38
Mass number (A) = 88
Number of protons (Z) = 38
Number of neutrons (AZ) = 8838 = 50
For the given atomic number Z=17 and mass number A=35;
Atom is .
For the given atomic number Z=92 and mass number A=233;
Atom is .
For the given atomic number Z=4 and mass number A=9;
Atom is .
Given the wavelength of the yellow light emitted from a sodium, lamp is .
And the frequency will be:
Therefore the wavenumber,
If a photon has a frequency of .
Then, the energy of each of the photons will be:
(ii) have wavelength of 0.50 Å.
For the wavelength .
The energy of each of the photons will be:
Given frequency, wavelength, and the wave number of a light wave:
Given the wavelength of light .
and Energy is 1J of energy:
Therefore, the number of photons of light with a wavelength of 4000 pm that provides 1J of energy is:
The photon is having a wavelength of strikes on a metal surface, where the work function of the metal being is .
So, Energy of the photon:
The photon is having a wavelength of strikes on a metal surface, where the work function of the metal being is .
The kinetic energy of the emission will be:
The photon is having a wavelength of strikes on a metal surface, where the work function of the metal being is .
From the previous part, we have the Kinetic Energy (K.E.):
Given the wavelength of the electromagnetic radiation is which is just sufficient to ionize the sodium atom.
So, the ionization energy required will be:
Given that the light is monochromatic yellow of wavelength .
Hence the energy emitted by the bulb will be:
Therefore, the number of photons emitted per second:
Given the wavelength of radiation is .
But the electrons are emitted with zero velocity from a metal surface when it is exposed to radiation. That means the kinetic energy will be zero.
So, the Threshold frequency will be:
and the Work function will be:
When an electron in a hydrogen atom undergoes a transition from an energy level with to an energy level , there will be an emission of energy whose wavelength can be found by:
and wavelength will be equal to:
The energy which is required to ionize an H atom if the electron occupies n=5 orbit is:
For ionization from 5th orbit,
Therefore,
For ionization from 1st orbit,
Therefore,
Hence, 25 times less energy is required to ionize an electron in the 5th orbital of the hydrogen atom as compared to that in the ground state.
The number of lines produced when an electron from shell drops to the ground state:
.
According to the question, the maximum number of emission lines when the excited electron of an H atom in drops to the ground state will be:
These are produced because of the following transitions:
The energy associated with the first orbit in the hydrogen atom is
The energy of an electron in shell is given by:
So, the energy associated with the fifth orbit would be:
The radius of Bohr's orbit for the hydrogen atom is given by,
So, for we have
Balmer formula:
As we can note from the formula that the wavenumber is inversely proportional to the wavelength.
Hence, for the longest wavelength transition in the Balmer series of atomic hydrogen wavenumber has to be the smallest or should be minimum i.e., .
For the Balmer series,
Thus, the expression of wavenumber is given by,
The ground state energy:
The energy required to shift the electron from the first Bohr orbit to the fifth Bohr orbit is:
And the expression for the energy of an electron is given by:
where m is mass of an electron, Z is the atomic mass of an atom, e is a charge of an electron, and h is the Planck's constant.
Now, substituting the values in the equation, we get
Hence, the wavelength of the emitted light will be:
The expression for the energy of an electron in hydrogen is:
Where m is mass of electrons, Z is the atomic mass of an atom, e is the charge of an electron, and h is the Planck's constant.
and electron energy in the hydrogen atom is given by,
The electron energy in orbit is:
Therefore, the energy required for the ionization from is
Now, the longest wavelength of light that can be used to cause this transition will be:
The wavelength of an electron is given by the de Broglie's equation:
Where,
is the wavelength of moving particle,
m is the mass of the particle, i.e.,
v is the velocity of the particle,i.e., (Given)
and h is the Planck's constant value, i.e.,
Now, substituting the values in the equation, we get
Hence, the wavelength of the electron moving with a velocity of is .
The wavelength of an electron can be found by de Broglie's equation:
Given the K.E. of electron which is equal to .
Hence we get,
Hence the wavelength is given by,
Calculating the number of electrons for each species.
has 11 electrons then, will have electrons.
has 19 electrons then, will have electrons.
has 12 electrons then, will have electrons.
has 20 electrons then, will have electrons.
has 16 electrons then, will have electrons.
has 18 electrons.
Hence, the following are isoelectronic species:
having 10 electrons each.
having 18 electrons each.
(a)
The electronic configuration of is :
Now, the electronic configuration of will be .
(b)
The electronic configuration of having is :
Now, the electronic configuration of will be .
(c)
The electronic configuration of having is :
Now, the electronic configuration of will be .
With given outermost electrons ,
The complete electronic configuration is
Hence the number of electrons present in the atom of the element is:
.
Therefore, the atomic number of the element is which is Sodium (Na).
With given outermost electrons ,
The complete electronic configuration is
Hence the number of electrons present in the atom of the element is:
.
Therefore, the atomic number of the element is 7 which is NItrogen (N).
With given outermost electrons ,
The complete electronic configuration is
Hence the number of electrons present in the atom of the element is:
.
Therefore, the atomic number of the element is 17 which is Chlorine (Cl).
The electronic configuration of the element is or
Therefore, the atomic number of the element is 3 which is Lithium, a pblock element.
The electronic configuration of the element is or
Therefore, the atomic number of the element is 15 which is Phosphorus, a pblock element.
The electronic configuration of the element is or
Therefore, the atomic number of the element is 21 which is Scandium, a dblock element.
For gorbital, the value of Azimuthal quantum number (l) will be 4.
As for any value 'n' of the principal quantum number, the Azimuthal quantum number (l) can have a value from zero to (n1).
Therefore, for l =4, the minimum value of n should be 5.
For dorbital, the value of Azimuthal quantum number (l) = 2.
When l =2, the values of m are:
Now, for the 3d orbital:
The value of Principal quantum number,
Azimuthal quantum number,
Magnetic quantum number, =
Given an atom of an element contains 29 electrons and 35 neutrons.
Now, for an atom to be neutral, the number of protons is equal to the number of electrons.
Therefore, the number of protons in the atom of the given element will be 29.
Given an atom of an element contains 29 electrons and 35 neutrons.
The electronic configuration of the atom will be:
which is the electronic configuration of copper.
The number of electrons in molecule is .
The number of electrons in molecule will be one less than the number of electrons in molecule. i.e, .
The number of electrons present in molecule will be one less than the number of electrons present in molecule. i.e.,
For a given value of the principal quantum number , the azimuthal quantum number can have values from to .
Therefore, for given atomic orbital ,
The value of can take values from to , i.e., .
And for a given value of , the Magnetic quantum number can have values.
When value of then, ,
or then,
or then,
or then,
For 3dorbital, the values of Principal quantum number is and Azimuthal quantum number .
Therefore, for ,
, magnetic quantum number can have values.
i.e.,
is NOT possible because for , the value of is zero.
is possible because, when . .
is possible because when . .
is NOT possible because for , the value of .
Here, is principal quantum number and is azimuthal quantum number.
Then the orbital with given quantum numbers is which can have a maximum of 2 electrons.
Here, is principal quantum number and is azimuthal quantum number.
Then the orbital with given quantum numbers is which can have a maximum of 6 electrons.
Here, is principal quantum number and is azimuthal quantum number.
Then the orbital with given quantum numbers is which can have a maximum of 10 electrons.
Here, is principal quantum number and is azimuthal quantum number.
Then the orbital with given quantum numbers is which can have a maximum of 14 electrons.
Given quantum numbers :
NOT possible, because n cannot be equal to zero.
Given quantum numbers :
It is possible and it is 1s orbital.
Given quantum numbers :
It is NOT possible because when .
Given quantum numbers :
It is possible and it is 2p orbital.
Given quantum numbers :
It is NOT possible because when , .
Given quantum numbers :
It is possible and it is 3p orbital.
The total number of electrons in an atom for a value of n is given by:
Therefore, the total no. of electrons when ,
and half of them i.e. 16 will have .
When then it is orbital which can have 2 electrons.
According to the Bohr's postulate of angular momentum,
which can be written as: .....................................(1)
Then according to de Broglie's equation for wavelength,
......................................(2)
Now, substituting the values of equation (2) in equation (1) we get,
Thus, the circumference of the Bohr's orbit for the hydrogen atom is an integral multiple of de Broglie's wavelength associated with the electron revolving around the orbit.
For the transition of Hlike particles,
For transition spectrum,
, , and
Therefore,
Then for the hydrogen spectrum,
and
Therefore,
The values of can be found by the hit and trial method in the above equation.
So, we get and , i.e., the transition is from to .
The ionization energy for the H atom in the ground state is
For the hydrogenlike particles,
For Hatom, Ionization energy:
For the given process, the energy required will be:
Given the diameter of a carbon atom which is
Then the number of carbon atoms which can be placed side by side in a straight line across the length of the scale of length long will be:
The arrangement length is given which is .
and the number of atoms of carbon which are arranged in this length is given .
Let the radius of carbon atom be then,
or
Hence the radius of carbon atom is .
If the diameter of zinc atom is then, its radius would be:
The number of atoms present in a length of will be:
As the charge carried by one electron is
Therefore, the number of electrons present in particle carrying charge will be:
Given charge on the oil drop is
and the charge carried by one electron is
Therefore, the number of electrons present on the oil drop carrying charge is:
The thin foil of heavy atoms, like gold, platinum, etc. have a nucleus carrying a large amount of positive charge. Therefore, some particles will easily get deflected back.
These particles also deflect through small angles because of the large number of a positive charge.
Hence if we use light atoms, their nuclei will have a small positive charge, hence the number of particles getting deflected even through small angles will be negligible.
The general way to represent an element along with its atomic mass (A) and atomic number (Z) is . Here the atomic number of an element is fixed. However, its mass number is not fixed as it depends upon the isotope taken.
Hence, it is essential to indicate the mass number.
Let the number of protons of an atom be .
Then the number of neutrons will be,
and the mass number is (Given).
Mass number = number of neutrons + number of protons.
Therefore,
Thus, there are numbers of protons which is also its atomic number.
Hence, the symbol for the element is .
Given an ion has mass number 37 and possesses one unit of negative charge.
Let the number of electrons be then,
the number of neutrons will be:
The number of electrons in the neutral atom (ion possesses one unit of negative charge).
Therefore, the number of protons will be .
Mass number = number of protons + number of neutrons
therefore,
Therefore, the number of protons is equal to the atomic number.
Hence, the symbol for an ion will be:
Given an ion has mass number 56 and possesses three units of negative charge.
Let the number of electrons be then,
the number of neutrons will be:
The number of electrons in the neutral atom (ion possesses three units of positive charge).
Therefore, the number of protons will be .
Mass number = number of protons + number of neutrons
therefore,
Therefore, the number of protons is equal to the atomic number.
Hence, the symbol for an ion will be:
The increasing order of frequency of radiations will be:
Radiation from FM radio < amber light from traffic signal < radiation from microwave oven < Xrays < cosmic rays from outer space.
The energy emitted by the nitrogen laser is:
The wavelength of neon gas is or
Hence the frequency of this radiation will be:
The velocity of neon gas radiation is .
Therefore, the distance travelled in will be:
The energy of quantum will be:
If it produces 2J of energy then, the number of quanta present in it will be .
Therefore,
or ,
Where and from the previous part.
Let the number of photons received by the detector be .
Then, the total energy it receives from the radiation of will be:
Or
Where,
,
,
and
Substituting the values in the equation above, we get
number of photons is received by the detector.
Given the duration of a radiation source and the number of pulse source is , then its frequency will be:
and the energy of the source for the given frequency will be:
We have two wavelengths of and .
Calculating the frequency for each:
Therefore, the energy difference between two excited states will be:
Given the work function for the Caesium atom is .
i.e., Or
As
Therefore, the threshold wavelength is 6.53× 10^{7} m.
To find threshold frequency:
Where
h = Planck’s constant
= threshold frequency
Finding the kinetic energy of the ejected electrons:
K.E of the ejected photoelectron:
Finding the Velocity of the ejected electrons:
Where,
= mass of electron
= velocity of electron
Therefore, the velocity is given by,
Let us assume the threshold wavelength to be and the kinetic energy of the radiation is given as:
.................................(1)
Similarly, we can also write,
...................................(2)
...................................(3)
Now, dividing the equations (3) with (1),
Therefore, the wavelength is .
We have the threshold wavelength.
Then substituting this value in any of the equation ( look in the previous part), we get
Taking the mass of an electron to be .
approximately.
Given work function of the metal, and the Wavelength,
From the Law of conservation of energy, the energy of an incident photon E is equal to the sum of the work function W of radiation and its kinetic energy K.E i.e.,
The energy of incident radiation
Or
Since the potential applied gives the kinetic energy to the radiation, therefore K.E of the electron
Therefore, Work Function
Given the wavelength of a photon which strikes an atom is .
Then the energy associated with this photon will be:
Given the velocity of ejected inner bounded electron: .
Then, the energy associated with this electron will be, Kinetic energy.
Hence finding
Where
m = mass of electron, v = velocity of electron
Hence the energy with which the electrons are bounded to the nucleus is:
Given transition in the Paschen series end at orbit and starts from orbit n: .........................(1)
.........................(2)
Equating both (1) and (2) equations: we get
Therefore, the radiation corresponding to 1285 nm lies in the infrared region.
The radius of orbit of Hlike particles is given by:
Or
Here, starting radius,
Ending radius,
Therefore,
If and , then the transition is from orbit to orbit.
Therefore, it belongs to the Balmer Series.
Frequency is given by:
Wavelength :
.
Therefore, it lies in the visible range.
According to de Broglie's equation for the wavelength.
Given the velocity of electron
and mass of electron
So, the wavelength will be:
or
Given the wavelength of neutron:
and the mass of neutron
So, According to the deBroglie's equation,
Substituting the values in above equation:
Given the velocity of the electron in Bohr's first orbit is .
And we know the mass of electron which is
Hence the deBroglie's wavelength associated with the electron will be:
Given a proton is moving with velocity .
And if the hockey ball of mass is also moving with the same velocity, then
According to deBroglie's equation we have,
We have given the uncertainty in position, i.e., .
According to Heisenberg's Uncertainty Principle:
Where,
is uncertainty in the position of the electron.
is uncertainty in the momentum of the electron.
Then,
Or
The actual momentum of the electron:
Therefore, it cannot be defined because the actual magnitude of the momentum is smaller than the uncertainty.
Quantum number provides the entire information about an electron of a particular atom.
Principal quantum number
Azimuthal quantum number
Magnetic quantum number
Spin quantum number .
The orbitals occupied by the electrons are:
1. 4dorbital
2. 3dorbital
3. 4porbital
4. 3dorbital
5. 3porbital
6. 4porbital
For the same orbitals, electrons will have the same energy and higher the value of value higher is the energy.
Therefore, the increasing order of energies:
As the porbital is farthest from the nucleus hence the electrons in (4p)subshell experiences the lowest effective nuclear charge.
Nuclear charge is defined as the net positive charge experienced by an electron in the orbital of an atom exerted by the nucleus of the atom.
Closer orbitals experience more nuclear charge than outer orbitals.
Therefore, (i) 2s and 3s
The 2s orbital is more closer to the nucleus than 3s orbital hence 2s will experience a larger effective nuclear charge compared to 3s.
Nuclear charge is defined as the net positive charge experienced by an electron in the orbital of an atom exerted by the nucleus of the atom.
Closer orbitals experience more nuclear charge than outer orbitals.
Therefore, (ii) 4d and 4f
The 4d orbital is more closer to the nucleus than 4f orbital hence 4d will experience a larger effective nuclear charge compared to 4f.
Nuclear charge is defined as the net positive charge experienced by an electron in the orbital of an atom exerted by the nucleus of the atom.
Closer orbitals experience more nuclear charge than outer orbitals.
Therefore, (ii) 3d and 3p
The 3p orbital is more closer to the nucleus than 3d orbital hence 3p will experience a larger effective nuclear charge compared to 3d.
Nuclear charge is defined as the net positive charge experienced by an electron in the orbital of an atom exerted by the nucleus of the atom.
Silicon has a greater nuclear charge than aluminium .
Hence, the effective nuclear charge exerted on the unpaired 3p electron of silicon would be greater as compared to that of aluminium.
The electronic configuration of P: .
Hence, the number of unpaired electrons are 3 in 3p orbital.
The electronic configuration of Si : .
Hence, the number of unpaired electrons are 2 in orbital.
The electronic configuration of Cr : .
Hence, the number of unpaired electrons are 6 (1 in 4s and 5 in 3d).
The electronic configuration of Fe : .
Hence, the number of unpaired electrons are 4 (in 3d ).
As Krypton (Kr) is a noble gas whose atomic number is 36 and have all orbitals filled.
Hence, there are no unpaired electrons in Kr element.
For a given value of , can have values from to .
Therefore, for ,
can have values from to .
i.e., .
Thus, four subshells are associated with , which are
The number of orbitals in the shell is equal to .
So, for , there are orbitals present.
As each orbital has one electron with spin .
Hence, there will be electrons with .
Chapter 1 
NCERT solutions for class 11 chemistry chapter 1 Some Basic Concepts of Chemistry 
Chapter2 
CBSE NCERT solutions for class 11 chemistry chapter 2 Structure of Atom 
Chapter3 

Chapter4 
NCERT solutions for class 11 chemistry chapter 4 Chemical Bonding and Molecular Structure 
Chapter5 
CBSE NCERT solutions for class 11 chemistry chapter 5 States of Matter 
Chapter6 
Solutions of NCERT class 11 chemistry chapter 6 Thermodynamics 
Chapter7 
NCERT solutions for class 11 chemistry chapter 7 Equilibrium 
Chapter8 
CBSE NCERT solutions for class 11 chemistry chapter 8 Redox Reaction 
Chapter9 

Chapter10 
NCERT solutions for class 11 chemistry chapter 10 The SBlock Elements 
Chapter11 
CBSE NCERT solutions for class 11 chemistry chapter 11 The PBlock Elements 
Chapter12 

Chapter13 
NCERT solutions for class 11 chemistry chapter 13 Hydrocarbons 
Chapter14 
CBSE NCERT solutions for class 11 chemistry chapter 14 Environmental Chemistry 
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