NCERT Solutions for Class 11 Chemistry Chapter 2- Structure of Atom

NCERT Solutions for Class 11 Chemistry Chapter 2- Structure of Atom

Q.1. (i) Calculate the number of electrons which will together weigh one gram.
(ii) Calculate the mass and charge of one mole of electrons.

Ans:

(i)1 electron has mass = 9.108 x 10-28 g

So, 1 g will be the mass of = 1/(9.108 x 10-28) = 1.098 x 1027 electrons

(ii) 1 mole of electron has mass = 9.108 x 10-28 x 6.022 x 1023= 5.48 x 10-4 g

1 mole of electron has charge = 1.6 x 10-19 C x 6.022 x 1023= 9.63 x 104 C

Q.2. (i) Calculate the total number of electrons present in one mole of methane.
(ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of 14C.
(Assume that mass of a neutron = 1.675 × 10–27 kg).
(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH3
at STP.
Will the answer change if the temperature and pressure are changed?

Ans:

(i) We know that 1 molecule of methane contains 10 electrons.

Hence, 1 mole of methane will contain= 10 x 6.022 x 1023 = 6.022 x 1024 electrons.

(ii) (a) We know that mass of 1 mole of 14C = 14 g 

Now, no. of neutrons in 14000 mg of Carbon = 6.022 x 1023 x 8 neutrons

Therefore, 7 mg of 14C will have neutrons = {(6.022 x 1023 x 8)/(14000)} x7

 = 2.4088 x 1021 neutrons.

(b)Given mass of 1 neutron = 1.675 x 10-27 kg

Hence, mass of 2.4088 x 1021 neutrons = (2.4088 x 1021 x 1.67 x 10-27)= 4.0347 x 10-6 kg

(iii) (a)

We know that,

Gram molecular mass of ammonia (NH3) = 17 g = 17000 mg

17000 mg of NH3 have molecules = 6.022 x 1023 (1 mole of NH3)

Therefore,

34 mg of NH3 will have molecules = (6.022 x 1023x 34)/ 17000 = 12.044 x 1020 molecules.

b) We know that no. of protons present in one molecule of NH3 = 7 + 3 = 10

So, total protons present in 12.044 x 1020 molecules of NH3 will be = 12.044 x 1020 x 10

= 1.2044 x 1022 protons

Also, mass of one proton = 1.67 x 10-27 kg

Therefore, mass of 1.2044 x 1022 Protons = (1.67 x 10-27 kg) x 1.2044 x 1022

= 2.01 x 10-5 kg

Also, changing the temperature and pressure will not have any effect because only the protons and mass of protons are considered. Hence answer will not change.

Q.3. How many neutrons and protons are there in the following nuclei?

Ans:

Here, Carbon has mass number = 13

Also, Carbon has number of protons = Atomic number given = 6

Also, mass number (A) = number of protons (Z) + number of neutrons (n)

So, no. of neutrons (n) = mass number – no. of protons = 13-6 = 7.

Here, Oxygen has mass number = 16

Also, Oxygen has number of protons = Atomic number given = 8

Also, mass number (A) = number of protons (Z) + number of neutrons (n)

So, no. of neutrons (n) = mass number – no. of protons = 16-8 = 8.

Here, Magnesium has mass number = 24

Also, Magnesium has number of protons = Atomic number given = 12

Also, mass number (A) = number of protons (Z) + number of neutrons (n)

So, no. of neutrons (n) = mass number – no. of protons = 24-12 = 12.

Here, Iron has mass number = 56

Also, Iron has number of protons = Atomic number given = 26

Also, mass number (A) = number of protons (Z) + number of neutrons (n)

So, no. of neutrons (n) = mass number – no. of protons = 56-26 = 30.

Here, Strontium has mass number = 88

Also, Strontium has number of protons = Atomic number given = 38

Also, mass number (A) = number of protons (Z ) + number of neutrons (n)

So, no. of neutrons (n) = mass number – no. of protons = 88-38 = 50.

Q.4. Write the complete symbol for the atom with the given atomic number (Z) and atomic mass (A)

(I)Z = 17, A = 35

(II)Z = 92, A = 233

(III)Z = 4, A = 9

Ans:

(I) Given that atomic number is 17 and mass number is 35, we can say that this atom is Chlorine. Hence, its complete

(II) Given that atomic number is 92 and mass number is 233, we can say that this atom is Uranium. Hence, its complete

(III) Given that atomic number is 4 and mass number is 9, we can say that this atom is Beryllium. Hence, its complete

Q.5. Yellow light emitted from a sodium lamp has a wavelength (λ) of 580 nm. Calculate the frequency (ν) and wave number (ν ) of the yellow light.

Ans: We know that frequency, wavelength and speed of light are related to each other by the equation:

­λ = c/ν where ­λ is the wavelength and c is the speed of electromagnetic radiation in vacuum and ν is the frequency.

Given ­λ = 580nm = 580× 10-9 m/s  and  c = 3 × 108 m/s

Q.6. Find the energy of each of the photons which
(i) correspond to light of frequency 3×1015 Hz.
(ii) have a wavelength of 0.50 Å.

Ans:

(i)The energy (E) of a photon is given by:

E=ℎν

Where, ‘h’ is Planck’s constant equal to 6.626×10–34 J s.

Given that frequency of the light, ν=3×1015Hz

Thus, we get the E of the photons as:

E = (6.626×10–34) × (3×1015) = 1.988×10-18 J

(ii)The energy (E) of a photon is given by:

E= hν= hc/λ

Where, ‘h’ is Planck’s constant equal to 6.626×10–34 J s, c is speed of light equal to

3 ×108 m/s.

Given wavelength = 0.5 Å = 0.5 × 10-10 m

Hence, E = {(6.626×10–34) × (3 ×108)} / (0.5 × 10-10) = 3.98 × 10-15 J 

Q.7. Calculate the wavelength, frequency and wavenumber of a light wave whose period is 2.0 × 10–10 s.

Ans: Frequency of the light wave, ν, is given by the equation,

ν = 1/Period

Therefore, ν = 1/ (2.0 × 10–10) = 5 × 109 s-1

Wavelength, λ = c/ν = (3 ×108) / (5 × 109) = 6 × 10-2 m

Also, wavenumber = 1/λ = 1 / (6 × 10-2) = 1.66 × 101 m-1 = 16.66 m-1

Q.8. What is the number of photons of light with a wavelength of 4000 pm that provides 1J of energy?

Ans: Energy of a photon is given by the equation:

E = hc/λ

Given that,

λ = 4000 pm = 4 × 10-9 m

c = 3 ×108 m/s.

h = 6.626×10–34 J s

thus, by substituting the respective values in eq., we have,

E = {(6.626×10–34)×(3 ×108)}/(4 × 10-9) = 4.969 x 10-17 J

Now, 4.969 x 10-17 J is the energy of 1 photon

Therefore,

1 J will be the energy of = 1/ (4.969 x 10-17) = 2.01 × 1016 photons

Q.9. A photon of wavelength 4 × 10–7 m strikes on metal surface, the work function of the metal is 2.13 eV. Calculate (i) the energy of the photon (eV), (ii) the kinetic energy of the emission, and (iii) the velocity of the photoelectron (1 eV= 1.6020 × 10–19 J).

Ans:

(i)We know that,

E = hc/λ

Given that,

λ = 4 × 10-7 m

c = 3 ×108 m/s.

h = 6.626×10–34 J s

thus, by substituting the respective values in eq., we have,

E = {(6.626×10–34)×(3 ×108)}/(4 × 10-7) = 4.97 x 10-19 J

(ii)

The kinetic energy is given by,

Ek= h(ν – ν0)= hν – hν0

    = (E – W) eV = [ {(4.97 x 10-19) / (1.602x 10-19)} – 2.13] eV

    = (3.102 – 2.13)eV

    = 0.972 eV.

(iii)The velocity of the photoelectron (v) is given as,

1/2mv2= hν – hν0

Thus, v = {(2 x0.972×1.602x 10-19) / 9.1 x 10-31}1/2 = 5.84 x 105 m/s.

Q.10. Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the
sodium atom. Calculate the ionisation energy of sodium in kJ mol–1.

Ans: 

Ionisation energy of one Na Atom is given as = hc/λ

Ionisation energy of one mole of Na Atom is given as = NAhc/λ (where NA is one mole of Na atoms)

= NAhc/λ= {(6.023×1023)×(6.626×10–34)×(3 ×108)}/(242× 10-9) 

= 494.7 × 103 J/mol

= 494 KJ/mol

Q.11. A 25-watt bulb emits monochromatic yellow light of the wavelength of 0.57µm. Calculate the rate of emission of quanta per second.

Ans: The power of the bulb is given as = 25 watt = 25 J/s

Also, the energy of a photon is given as

= hc/λ = {(6.626×10–34)×(3 ×108)}/(0.57× 10-6) = 34.873×10–20 J

Now, 34.873×10–20 J is energy of 1 photon per second

So, 25 J will be the energy = 25 / (34.873×10–20) = 7.16 ×1019 s-1

Q.12. Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency (ν0) and work function (W0) of the metal.

Ans: Wavelength of the radiation is given as = 6800 Å = 6.8× 10-7 m

Now, threshold frequency(ν0)of metal is given as = c/λ0= (3 ×108) / (6.8× 10-7)

                                                                                              = 4.4 × 1014 s-1

Now, the work function w0is given as = hν0 = (6.626×10–34)×(4.4 × 1014)

                                                                                  = 2.92× 10-19 J

Q.13. What is the wavelength of light emitted when the electron in a hydrogen atom undergoes the transition from an energy level with n = 4 to an energy level with n = 2?

Ans: 

The energy involved in transition when electron goes from initial orbit ni to final orbit nfcan be calculated as below:

E = 2.18 × 10-18[1/ni2– 1/nf2]

E = 2.18 × 10-18[1/42– 1/22]= 2.18 × 10-18×(-3/16) = -(4.0875 × 10-19) J

Q.14. How much energy is required to ionise a H atom if the electron occupies n = 5 orbit? Compare your answer with the ionization enthalpy of H atom (energy required to remove the electron from n =1 orbit).

Ans: 

Ionization energy is given by,

Where Z denotes the atomic number and n is the principal quantum number

For the ionization from

So, the required energy for the ionization of hydrogen from n = 5 to n = ∞  is

Therefore, a lower amount of energy will be required to ionize electrons in the 5th orbital of a hydrogen atom when compared to that in the ground state.

Q.15. What is the maximum number of emission lines when the excited electron of a H atom in n = 6 drops to the ground state?

Ans.

We know that the total number of spectral lines that will be emitted when an electron in the nth level drops to the ground state is given by {n(n-1)}/2

Since n = 6, total no. spectral lines =6(6−1)/2=15

Hence, the maximum number of emission lines when the excited electron of a H atom in n = 6 drops to the ground state is 15

Q.16. (i) The energy associated with the first orbit in the hydrogen atom is –2.18 × 10–18 J atom–1. What is the energy associated with the fifth orbit? (ii) Calculate the radius of Bohr’s fifth orbit for the hydrogen atom.

Ans. We know that the energy associated with nth orbit in hydrogen atom is

En = (-2.18 x 10-18)/n2 J/atom

(i) Hence, energy associated with Hydrogen’s fifth orbit is:

E5=−(2.18×10−18)/52=−8.72×10−20 J/atom

(ii) also, the radius of Bohr’s hydrogen atom5th orbitis given as,

r5=(0.0529nm)×52=1.3225nm

Q.17. Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen.

Ans. For the Balmer series of the hydrogen emission spectrum, for having longest wavelength transition, we need to consider the minimum energy emitted for transition from n=3 to n=2.

Hence, wave number is given by,

ν = 1.09677 ×107(1/22 – 1/32)

= 1.09677 ×107(5/36)

= 1.523 ×106 m-1

Q.18. What is the energy in joules, required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns to the ground state? The ground state electron energy is –2.18 × 10–11 ergs.

Ans. 

The energy of electron is given as En= (–2.18 × 10–11)/n2 ergs = (–2.18 × 10–18)/n2 J.

Now, energy in 1st orbit = (–2.18 × 10–18)/12 J = (–2.18 × 10–18)J

energy in 5th orbit = (–2.18 × 10–18)/52 J = (–2.18 × 10–18)/25 J

∆E = E5 – E1

Now, wavelength is calculated as:

λ = hc/∆E = {(6.626x 10-34)x(3 x 108)} / (2.09 × 10–18)= 9.50 x10-8= 950 Å

Q.19. The electron energy in hydrogen atom is given by En= (–2.18 × 10–18)/n2 J. Calculate the energy required to remove an electron completely from the n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?

Ans.

The energy is given as En= (–2.18 × 10–18)/n2 J.

So, the energy required for the ionization from n = 2 orbit is given as:

∆E = E∞ – E2

      = {(–2.18 × 10–18)/∞2 – (–2.18 × 10–18)/22}

      = {(2.18 × 10–18)/4 – 0}

      = 5.45 × 10–19 J

Now,λ is given by,

λ= hc/∆E = {(6.626x 10-34)x(3 x 108)} / (5.45 × 10–19)= 3647 x 1010 = 3647 Å

Q.20. Calculate the wavelength of an electron moving with a velocity of 2.05 × 107 m s–1

Ans. 

From the Broglie’s equation, we have,

λ = h/mv

Substituting the values, we get,

λ = (6.626x 10-34) / {(9.1 x 10-31)x(2.05 x 107)}=3.551 x 10-11 m

Q.21. The mass of an electron is 9.1 × 10–31 kg. If its K.E. is 3.0 × 10–25 J, calculate its wavelength.

Ans.

From the Broglie’s equation, we have,

λ = h/mv

Now, the K.E. of electron 3.0 x 10-25 J

 So, ½(mv2) = 3.0 x 10-25 J

v = {2(3.0 x 10-25 J) / (9.1 x 10-31)kg}1/2 = 811.579 m/s

Hence λ = (6.626x 10-34) / {(9.1 x 10-31)x(811.579)}= 8.9625 x 10-7 m.

Q. 22. Which of the following are isoelectronic species i.e., those having the same number
of electrons?
Na+, K+, Mg2+, Ca2+, S2–, Ar.

Ans:

First, we will calculate different number of electrons present in different species as:

11Na+ = 11 – 1 = 10;           

 19K+ = 19 – 1 = 18;

12Mg2+ = 12 – 2 = 10;

20Ca2+ = 20 – 2 = 18;         

16S2- = 16 + 2 = 18

Ar = 18

Hence,

(i) Na+ and Mg2+are isoelectronic species. (having number of electrons = 10)

(ii) K+, Ca2+, S2- and Arare isoelectronic species. (having number of electrons = 18)

Q.23. (I)Write the electronic configurations of the following ions:

(a)H–

(b)Na+

(c)O2–

(d)F –

(II) What are the atomic numbers of elements whose outermost electrons are represented by

(a) 3s1

(b) 2p3 and

(c) 3p5?

(III) Which atoms are indicated by the following configurations?

(a)[He] 2s1

(b)[Ne] 3s2 3p3

(c)[Ar] 4s2 3d1 .

Ans:

(I) (a) H– ion

The ground state electronic configuration of the Hydrogen 1s1. The single negative charge on this atom indicates that there is a gain of an electron. Hence, the electronic configuration of H– will be = 1s2

(b) Na+ ion

We have ground state electronic configuration of Na = 1s2 2s2 2p6 3s1. Here, the +ve charge indicates there is loss of an electron.

Hence, the electronic configuration of Na+will be = 1s2 2s2 2p6

(c) O2– ion

We have ground state electronic configuration of oxygen = 1s 2 2s 2 2p 4.

Now, having charge -2 indicates that it has acquired 2 electrons.

Hence,electronic configuration of O2– ion will be = 1s 2 2s 2 2p 6

(d) F – ion

We have ground state electronic configuration of Fluorine as = 1s22s22p5. The single negative charge means that it has gained one electron.

Hence, theelectronic configuration of F– ion will be = 1s 2 2s 2 2p 6

(II) (a) 3s1

The complete electronic configuration is given as : 1s 2 2s 2 2p 6 3s 1.

So, we get the total number of electrons in the atom as = 2 + 2 + 6 + 1 = 11

∴ The element’s atomic number is 11

(b) 2p 3

The complete electronic configuration is given as: 1s 2 2s 2 2p 3

So, we get the total number of electrons in the atom as=2 + 2 + 3 = 7

∴The element’s atomic number is 7

(c) 3p 5

The complete electronic configuration is given as: 1s 2 2s 2 2p 6 3s2 3p5 

So, we get the total number of electrons in the atom as= 2 + 2 + 6 + 2 + 5 = 17

∴ The element’s atomic number is 17

(III)(a)[He] 2s 1

The complete electronic configuration after replacing [He] is given as: 1s 2 2s 1

So, we get the total number of electrons in the atom as(2+1) = 3

So, number of electrons = number of protons = atomic number = 3

Hence, the element is lithium (Li).

(b)[Ne] 3s 2 3p3

The complete electronic configuration after replacing [Ne] is given as1s 2 2s 2 2p 6 3s 2 3p 3 .

So, we get the total number of electrons in the atom as(2+2+6+2+3) = 15.

So, number of electrons = number of protons = atomic number = 15

Hence, the element is phosphorus (P).

(c)[Ar] 4s 2 3d1

The complete electronic configuration after replacing [Ar] is given as: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 1

So, we get the total number of electrons in the atom as(2+2+6+2+6+2+1) = 21.

So, number of electrons = number of protons = atomic number = 21

Hence, the element is scandium (Sc).

Q.24. What is the lowest value of n that allows g orbitals to exist?

Ans. We know that for f-orbitals, we have, l = 3.

So, for g-orbitals, we have, l=4.

Since for any given value of ‘n’, we know that the possible values of ‘l’ range from 0 to (n-1).

Thus, for g orbitals to exist, the least value of n required will be = 5.

Q.25. An electron is in one of the 3d orbitals. Give the possible values of n, l and ml for this electron.

Ans: For the 3d orbital:

The possible values of the Principle quantum number (n) = 3

The possible values of the Azimuthal quantum number (l), is given by, (n-1) = 2

The possible values of the Magnetic quantum number (ml), is given by(2l +1) = 5

 = – 2, – 1, 0, 1, 2 (these are the 5 values)

Q.26. An atom of an element contains 29 electrons and 35 neutrons. Deduce (i) the number of protons and (ii) the electronic configuration of the element.

Ans:

(i)We know that number of protons = number of electrons in a neutral atom.

Hence,number of protons present in the atom = no. of electrons = 29

(ii)The electronic configuration of this element will be 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 3d10.

Q.27.Give the number of electrons in the species, H2+, H2 and O2+

Ans: We know that a H atom contains 1 electron.

So, No. electrons present in H2 = 1 + 1 = 2.

∴ Number of electrons in H2+ = 2 – 1 = 1

We know that number of electrons present in O atom is 8.

Hence, number of electrons in O2 = 8 + 8 = 16.

∴ Number of electrons in O2+= 16 – 1 = 15

Q.28. (I)An atomic orbital has n = 3. What are the possible values of l and ml ?

(II)List the quantum numbers (ml and l) of electrons for 3d orbital.

(III) Which of the following orbitals are possible? 1p, 2s, 2p and 3f

Ans.

(I)We know that the possible values of Azimuthal quantum number ‘l’ range from 0 to (n – 1). Thus, for n = 3, the possible values of l will be 0, 1, and 2.

Now, the total number of possible values for magnetic quantum number, mlis given as = (2l + 1). Its values range from -l to l.

For n = 3 and l = 0, 1, 2:

m0 = 0

m1 = – 1, 0, 1

m2 = – 2, – 1, 0, 1, 2

(II)

For 3d orbitals, n = 3 and l = 2. For l = 2, possible values of m2 = –2, –1, 0, 1, 2

(III)

It is possible for the 2s and 2p orbitals to exist. The 1p and 3f cannot exist.

For the 1p orbital, n=1 and l=1, which is not possible since the value of l must always be lower than that of n.

Similarly, for the 3f orbital, n =3 and l = 3, which is not possible.

Q.29. Using s, p and d notations, describe the orbital with the following quantum numbers.

(a)n = 1, l = 0;

(b)n = 3; l =1

(c) n = 4; l = 2;

(d) n = 4; l =3.

Ans:

(a)Since n = 1and l = 0 , it is 1s orbital.

(b)Since n = 3 and l = 1,it is 3p orbital.

(c)Since n = 4 and l = 2,it is 4d orbital.

(d)Since n = 4 and l = 3,it is 4f orbital.

Q.30. Explain, giving reasons, which of the following sets of quantum numbers are not possible.

a) n = 0, l = 0, ml= 0, ms =+1/2

b) n = 1, l = 0, ml= 0, ms =-1/2

c) n = 1, l = 1, ml= 0, ms =+1/2

d) n = 2, l = 1, ml= 0, ms =-1/2

e) n = 3, l = 3, ml= -3, ms =+1/2

f) n = 3, l = 1, ml= 0, ms =+1/2

Ans. (a) Not possible since value of principal quantum number n cannot be 0.

(b) Possible.

(c) Not possible since for a given value of n, azimuthal quantum number,l can have values ranging from 0 to n – 1.

(d) Possible.

(e) Not possible since for a given value of n, azimuthal quantum number,l can have values ranging from 0 to n – 1.

(f) Possible.