# NCERT Solutions for Class 11 Chemistry

Find 100% accurate solutions for NCERT Class XI Chemistry. All questions have been explained and solved step-by-step, to help you understand thoroughly. Free Download option available!

1.    (i) Mass of one electron = 9.10939 × 10-31 kg

Number of electrons that weigh 9.10939 × 10-31 kg = 1

Number of electrons that will weigh 1 g = (1 × 10-3 kg)

(ii) Mass of one electron = 9.10939 × 10-31 kg

Mass of one mole of electron = (6.022 × 1023) × (9.10939 × 10-31 kg)

= 5.49 × 10-7 kg

Charge on one electron = 1.6022 × 10-19 coulomb

Charge on one mole of electron = 1.6022 × 10-19 C × 6.022 × 1023

= 9.65 × 104C

2.    (i) Number of electrons present in 1 molecule of methane (CH4)

{1(6) + 4(1)} = 10

Number of electrons present in 1 mole i.e., 6.022 × 1023 molecules of methane

= 6.022 × 1023 × 10 = 6.022 × 1024

(ii) (a) Number of atoms of in 1 mole = 6.022 × 1023

Since 1 atom of 24C contains (14 – 6) i.e., 8 neutrons, the number of neutrons in 14 g of 14C is 6.022 ×1023 × 8 Or, 14 g of 14C contains (6.022 ×1023 × 8) neutrons.

Number of neutrons in 7 mg

(b) Mass of one neutron = 1.67493 ×10-27 kg

Mass of total neutrons in 7 g of 14C

= (2.4092 × 1021) (1.67493 × 10-27 kg)

= 4.0352 ×10-6 kg

(iii) (a) 1 mole of NH3 = {1(14) + 3(1)} g of NH3

= 17 g of NH3

= 6.022 ×1023 molecules of NH3

Total number of protons present in 1 molecule of NH3

= {1(7) + 3(1)} = 10

Number of protons in 6.023 ×1023 molecules of NH3

= (6.023 ×1023) (10)

= 6.023 ×1024

⇒ 17 g of NH3 contains (6.022 ×1024) protons

Number of protons in 34 mg of NH3

(b) Mass of one proton = 1.67493 ×10-27 kg

Total mass of protons in 34 mg of NH3

= (1.67493 ×10-27 kg) (1.2046 ×1022)

= 2.0176 ×10-5 kg

The number of protons, electrons, and neutrons in an atom is independent of temperature and pressure conditions. Hence, the obtained values will remain unchanged if the temperature and pressure is changed.

3.

Atomic mass = 13

Atomic number = Number of protons = 6

Number of neutrons = (Atomic mass) – (Atomic number)

= 13 – 6 = 7

Atomic mass = 16

Atomic number = 8

Number of protons = 8

Number of neutrons = (Atomic mass)-(Atomic number)

= 16 -8 = 8

Atomic mass = 24

Atomic number = Number of protons = 12

Number of neutrons = (Atomic mass)-(Atomic number)

= 24 – 12 = 12

Atomic mass = 56

Atomic number = Number of protons = 26

Number of neutrons = (Atomic mass) -(Atomic number)

= 56 – 26 = 30

Atomic mass = 88

Atomic number = Number of protons = 38

Number of neutrons = (Atomic mass)-(Atomic number)

= 88 – 38 = 50

4.    From the expression,

We get,

ν = frequency of yellow light

c = velocity of light in vacuum = 3 ×10m/s

λ = wavelength of yellow light = 580 nm = 580 ×109 m

Substituting the values in expression (i):

Thus, frequency of yellow light emitted from the sodium lamp

= 5.7 × 1014 s-1

Wave number of yellow light,

5.    (i) Energy (E) of a photon is given by the expression,

E = hv

Where,

h = Planck’s constant = 6.626 × 10-34 Js

v = frequency of light = 3 × 1015 Hz

Substituting the values in the given expression of E:

E = (6.626 × 10-34) (3 × 1015)

E = 1.988 × 10-18 J

(ii) Energy (E) of a photon having wavelength λ is given by the expression,

h = Planck’s constant = 6.626 × 10-34 Js

c = velocity of light in vacuum = 3 × 108 m/s

Substituting the values in the given expression of E:

6.     Frequency (v) of light

Wavelength (λ) of light

Where,

c = velocity of light in vacuum = 3 × 108 m/s

Substituting the value in the given expression of λ:

7.    Energy (E) of a photon = hv

Energy (En) of ‘n‘ photons = nhv

Where,

λ = wavelength of light = 4000 pm = 4000 × 10-12 m

c = velocity of light in vacuum = 3 × 108 m/s

h = Planck’s constant = 6.626 × 10-34 Js

Substituting the values in the given expression of n:

Hence, the number of photons with a wavelength of 4000 pm and energy of 1 J are  2.012 × 1016

8.    (i) Energy (E) of a photon = h

Where,

h = Planck’s constant = 6.626 × 10-34 Js

c = velocity of light in vacuum = 3 × 108 m/s

λ = wavelength of photon = 4 × 10-7 m

Substituting the values in the given expression of E:

Hence, the energy of the photon is 4.97 × 10-19 J.

The kinetic energy of emission Eis given by

Hence, the kinetic energy of emission is 0.97 eV.

(iii) The velocity of a photo-electron (v) can be calculated by the expression,

Where, (hv – hv0)is the kinetic energy of emission in Joules and ‘m‘ is the mass of the photoelectron. Substituting the values in the given expression of v:

Hence, the velocity of the photo-electron is 5.84 × 105 m/s-1.

9.    Energy of sodium (E)

10.  Power of bulb, P = 25 Watt = 25 Js-1

Energy of one photon, E = hv

Substituting the values in the given expression of E:

Rate of emission of quanta per second

11.   Threshold wavelength of radian (λ0) =

Threshold  frequency (v0) of the metal

Thus, the threshold frequency (v0) of the metal is 4.41 × 1014 s-1.

Hence, work function (W0) of the metal = hv0

= (6.626 × 10-34 Js) (4.41 × 1014 s-1)

= 2.922 × 10-19 J.

12.   The ni = 4 to nf = 2 transition will give rise to a spectral line of the Balmer series. The energy involved in the transition is given by the relation,

Substituting the values in the given expression of E:

E = -(4.0875 × 10-19 J)

Wavelength of light emitted

Substituting the values in the given expression of λ:

13.    The expression of energy is given by

Where,

Z = atomic number of the atom

= principal quantum number

For ionization from to n1 = 5 to n2 = ∞,

Hence, the energy required for ionization from n = 5 to n = ∞ is 8.72 × 10-20 J.

Energy required for n1 = 1 to n = ∞,

On comparing

Hence, less energy is required to ionize an electron in the 5thorbital of hydrogen atom as compared to that in the ground state.

14.   When the excited electron of an H atom in n= 6 drops to the ground state, the following transitions are possible:

Hence, a total number of (5 + 4 + 3 + 2 + 1) 15 lines will be obtained in the emission spectrum.

The number of spectral lines produced when an electron in the  level drops down to the ground state is given by .

Given,

= 6

Number of spectral lines  = 15

15.   (i) Energy associated with the fifth orbit of hydrogen atom is calculated as:

(ii) Radius for Bohr’s nth orbit for hydrogen atom is given by,

rn = (0.0529 nm)n2

For,

n = 5

r5 = (0.0529 nm) (5)2

r5 = 1.3225 nm.

16.   For the Balmer series, ni = 2. Thus, the expression of wave number () is given by,

Wave  number () is inversely proportional to wavelength of transition. Hence, for the longest wavelength transition,  has to be the smallest.

For  to be minimum, nf should be minimum. For the Balmer series, a transition from ni = 2 to nf = 3 is allowed. Hence, taking nf = 3, we get:

17.   Energy (E) of the nth Bohr orbit of an atom is given by,

Where,

Z = atomic number of the atom

Ground state energy = 2.18 × 10-11 ergs

= 2.18 × 10-11 × 10-7 J

= 2.18 × 10-18 J

Energy required to shift the electron from n = 1 to n = 5 is given as:

Wavelength of emitted light

18.   Given,

Energy required for ionization from n = 2 is given by,

Here, λ is the longest wavelength causing the transition.

19.   According to de Broglie’s equation,

Where,

λ = wavelength of moving particle

m = mass of particle

v = velocity of particle

h = Planck’s constant

Substituting the values in the expression of λ:

Hence, the wavelength of the electron moving with a velocity of

2.05 × 10ms-1 is 3.548 × 10-11 m.

20.  From de Broglie’s equation

Given,

Kinetic energy (K.E) of the electron = 3.0 × 10-25 J.

Since, K.E. =

Substituting the value in the expression of λ:

Hence, the wavelength of the electron is 8.9625 × 10-7 m.

21.   Isoelectronic species have the same number of electrons.

Number of electrons in sodium (Na) = 11

Number of electrons in (Na+) = 10

A positive charge denotes the loss of an electron.

Similarly,

Number of electrons in K+ = 18

Number of electrons in Mg2+ = 10

Number of electrons in Ca2+ = 18

A negative charge denotes the gain of an electron by a species.

Number of electrons in sulphur (S) = 16

∴ Number of electrons in S2- = 18

Number of electrons in argon (Ar) = 18

Hence, the following are isoelectronic species:

1) Na+ and Mg2+ (10 electrons each)

2) K+, Ca2+, S2- and Ar (18 electrons each)

22.   (i) (a) H ion

The electronic configuration of H atom is 1s1.

A negative charge on the species indicates the gain of an electron by it.

∴ Electronic configuration of H = 1s2

(b) Naion

The electronic configuration of Na atom is 1s2 2s2 2p6 3s1

A positive charge on the species indicates the loss of an electron by it.

∴ Electronic configuration of Na+ = 1s2 2s2 2p6 3s0 or 1s2 2s2 2p6

(c) O2- ion

The electronic configuration of 0 atom is 1s2 2s2 2p4.

A dinegative charge on the species indicates that two electrons are gained by it.

∴ Electronic configuration of O2- ion = 1s2 2s2 2p6

(d) F ion

The electronic configuration of F atom is 1s2 2s2 2p5.

A negative charge on the species indicates the gain of an electron by it.

∴ Electron configuration of F- ion = 1s2 2s2 2p6

(ii) (a) 3s1

Completing the electron configuration of the element as 1s2 2s2 2p6 3s1

Number of electrons present in the atom of the element

= 2 + 2 + 6 + 1 = 11

∴ Atomic number of the element = 11

(b) 2p3

Completing the electron configuration of the element as 1s2 2s2 2p3

∴ Number of electrons present in the atom of the element = 2 + 2 + 3 = 7

∴ Atomic number of the element = 7

(c) 3p5

Completing the electron configuration of the element as 1s2 2s2 2p5

∴ Number of electrons present in the atom of the element = 2 + 2 + 5 = 9

∴ Atomic number of the element = 9

(iii) (a) [He] 2s1

The electronic configuration of the element is [He] 2s1 = 1s2 – 2s1

∴ Atomic number of the element = 3

Hence, the element with the electronic configuration [He] 2s1 is lithium (Li).

(b) [Ne] 3s2 3p3

The electronic configuration of the element is [Ne] 3s2 3p2 = 1s2 2s2 2p3p2 3p3

∴ Atomic number of the element = 15

Hence, the element with the electronic configuration [Ne] 3s2 3p3 is phosphorus (P).

(c) [Ar] 4s2 3d1

The electronic configuration of the element is [Ar] 4s2 3d1 = 1s2 2s2 2p6 3s2 3p6 4s2 3d1

∴ Atomic number of the element = 21

Hence, the element with the electronic configuration [Ar] 4s2 3d1 is scandium (Sc).

23.  For g-orbitals, l = 4.

As for any value ‘n‘ of principal quantum number, the Azimuthal quantum number (l) can have a value from zero to (n – 1).

∴ For l = 4, minimum value of n = 5

24.  For the 3d orbital:

Principal quantum number (n) = 3

Azimuthal quantum number (l) = 2

Magnetic  quantum  number (ml) = – 2, – 1, 0, 1, 2

25.   (i) For an atom to be neutral, the number of protons is equal to the number of electrons.

∴ Number of protons in the atom of the given element = 29

(ii) The electronic configuration of the atom is 1s2 2s2 2p6 3s2 3p6 4s2 3d10

26.  Number of electrons present in hydrogen molecule (H2) = 1 + 1 = 2

∴ Number of electrons in = 2 – 1 = 1

H2: Number of electrons in H2 = 1 + 1 = 2

: Number of electrons present in oxygen molecule  = 8 + 8 = 16

∴ Number of electrons in  = 16- 1 = 15

27.   (i) n = 3 (Given)

For a given value of n, l can have values from 0 to (– 1).

∴ For n = 3

l = 0, 1, 2

For a given value of l, ml can have (2l + 1) values.

For l = 0, m = 0

l = 1, m = – 1, 0, 1

l = 2, m = – 2, – 1, 0, 1, 2

∴ For n = 3

l = 0, 1, 2

m0 = 0

m= – 1, 0, 1

m2 = – 2, – 1, 0, 1, 2

(ii) For 3d orbital, l = 2.

For a given value of l,  can have (2l + 1) values i.e., 5 values.

For l = 2

m2 = – 2, – 1, 0, 1, 2

(iii) Among the given orbitals only 2s and 2p are possible. 1p and 3f cannot exist.

For p-orbital, l = 1

∴ For l is equal to 1, the minimum value of n is 2.

Similarly,

For f-orbital, l = 4.

For l = 4, the minimum value of n is 5.

Hence, 1p and 3f do not exist.

28.   (a) n = 1, l = 0 (Given) The orbital is 1s.

(b) For n = 3 and l = 1 The orbital is 3

(c) For n = 4 and l = 2 The orbital is 4

(d) For n = 4 and l = 3 The orbital is 4f.

29.   (a) The given set of quantum numbers is not possible because the value of the principal quantum number (n) cannot be zero.

(b) The given set of quantum numbers is

(c) The given set of quantum numbers is not

For a given value of n, ‘l‘ can have values from zero to (n – 1).

For n = 1, l = 0 and not 1.

(d) The given set of quantum numbers is

(e) The given set of quantum numbers is not

For n = 3,

l = 0 to (3 – 1)

l = 0 to 2 i.e., 0, 1, 2

(f) The given set of quantum numbers is possible.

30.  (a) Total number of electrons in an atom for a value of n = 2n2

∴ For n = 4,

Total number of electrons = 2(4)= 32

The given element has a fully filled orbital as

1s2 2s2 2p6 3s2 3p6 4s2 3d10 Hence, all the electrons are paired.

∴ Number of electrons (having n = 4 and ) = 16

(b) n = 3, l = 0 indicates that the electrons are present in the 3s orbital. Therefore, the number of electrons having n = 3 and l = 0 is 2.

31.  Since a hydrogen atom has only one electron, according to Bohr’s postulate, the angular momentum of that electron is given by:

Where,

n = 1, 2, 3,….

According to de Broglie’s equation:

Substituting the value of ‘mv’ from expression (2) in expression (1):

Since ‘2πr’ represents the circumference of the Bohr orbit (r), it is proved by equation (3) that the circumference of the Bohr orbit of the hydrogen atom is an integral multiple of de Broglie’s wavelength associated with the electron revolving around the orbit.

32.   For He+ ion, the wave number  associated with the Balmer transition, n = 4 to n = 2 is given by:

Where,

n1 = 2

n2 = 4

Z = atomic number of helium = 2

According to the question, the desired transition for hydrogen will have the same wavelength as that of He+.

By hit and trail method, the equality given by equation (1) is true only when n1 = 2 and n2 = 4

∴ The transition for to n = 1 in hydrogen spectrum would have the same wavelength as Balmer transition n = 4 to n = 2 of He+ spectrum.

33.   Energy associated with hydrogen-like species is given by,

For ground state of hydrogen atom,

For the given process,

He+(g) → He2+(g) + e

An electron is removed from n = 1 to n = ∞.

∴ The energy required for the process 8.72 × 10-18 J.

34.   1 m = 100 cm

1 cm = 10-2 m

Length of the scale = 20 cm

= 20 × 10-2 m

Diameter of a carbon atom = 0.15 nm

= 0.15 × 10-9 m

One carbon atom occupies = 0.15 × 10-9 m.

∴ Number of carbon atoms that can be placed in a straight line

35.   Length of the given arrangement = 2.4 cm

Number of carbon atoms present = 2 × 108

∴ Diameter of carbon atom

36.   (a) Radius of zinc atom

(b) Length of the arrangement = 1.6 cm

= 1.6 × 10-2 m

Diameter of zinc atom = 2 × 10-10 m

∴ Number of zinc atoms present in the arrangement

37.   Charge on one electron = 1.6022 × 10-10 C

⇒ 1.6022 × 10-19 C charge is carried by 1 electron.

∴ Number of electrons carrying a charge of = 2.5 × 10-16 C

38.   Charge on the oil drop = 1.282 × 10-18 C

Charge on one electron = 1.6022 × 10-16 C

∴ Number of electrons present on the oil drop

39.   A thin foil of lighter atoms will not give the same results as given with the foil of heavier atoms.

Lighter atoms would be able to carry very little positive charge. Hence, they will not cause enough deflection of α-particles (positively charged).

40.   The general convention of representing an element along with its atomic mass (A) and atomic number (Z) is

Hence,  is acceptable but  is not acceptable.

can be written but  cannot be written because the atomic number of an element is constant, but the atomic mass of an element depends upon the relative abundance of its isotopes. Hence, it is necessary to mention the atomic mass of an element.

41.   Let the number of protons in the element be x.

∴ Number of neutrons in the element

= x + 31.7% of x

= x + 0.317 x

= 1.317 x

According to the question,

Mass number of the element = 81

∴ (Number of protons + number of neutrons) = 81

Hence, the number of protons in the element i.e, x is 35.

Since the atomic number of an atom is defined as the number of protons present in its nucleus, the atomic number of the given element is 35.

∴ The atomic symbol of the element is .

∴ The atomic symbol of the element is .

42.   Let the number of electrons in the ion carrying a negative charge be x.

Then,

Number of neutrons present

= x + 11.1% of x

= x + 0.111x

= 1.111x

Number of electrons in the neutral atoms = (x – 1)

(When an ion carries a negative charge, it carries an extra electron)

Number of protons in the neutral atom = x – 1

Given,

Mass number of the ion = 37

(x – 1) + 1.111x = 37

2.111x = 38

x = 18

The symbol of the ion is

43.   Let the number of electrons present in ion A3+ be x.

∴ Number of neutrons in it = x + 30.4% of x = 1.304x

Since the ion is tripositive,

⇒ Number of electrons in neutral atom = x + 3

∴ Number of protons in neutral atom

Given,

Mass number of the ion = 56

∴ Number of protons = x + 3 = 23 + 3 = 26

∴ The symbol of the ion

44.   The increasing order of frequency is as follows:

Radiation from FM radio < amber light < radiation from microwave oven < X- rays < cosmic rays

The increasing order of wavelength is as follows:

Cosmic rays < X-rays < radiation from microwave ovens < amber light < radiation of FM radio

45.  Power of laser =

Energy with which it emits photons

Power  Where,

N = number of photons emitted

h = Planck’s constant

Substituting the values in the given expression of Energy (E):

= 0.3302 × 107 J

= 3.33 × 106 J

Hence, the power of the laser is 3.33 × 106 J.

46.  Wavelength of radiation emitted = 616 nm  (Given)

Frequency of emission

Where,

Substituting the values in the given expression of (ν):

= 4.87 × 10× 109 × 10-3 s-1

ν = 4.87 × 1014 s6

Frequency of emission (v) = 4.87 × 1014 s-1

= 3.0 × 10ms-1

Distance travelled by this radiation in 30 s

= (3.0 × 10ms-1) (30s)

= 9.0 × 10ms-1

(c) Energy of quantum (E) = hv

(6.626 × 10-34 Js) (4.87 × 1014 s-1)

Energy of quantum (E) = 32.27 × 10-20 J

(d) Energy of one photon (quantum) = 32.27 × 10-20 J

Therefore, 32.27 × 10-20 J of energy is present in 1 quantum.

Number of quanta in 2 J of energy

47.   From the expression of energy of one photon (E),

Where,

h = Planck’s constant

Substituting the values in the given expression of E:

E = 3.313 × 10-19 J

Energy of one photon = 3.313 × 10-19 J

Number of photons received with 3.15 × 10-18 J

48.  Given time duration(T) = 2ns = 2*10-9 s

number of photons(N) = 2.5 × 1015

Energy (E) of source = Nhv

Where,

N = number of photons emitted

h = Planck’s constant

Substituting the values in the given expression of (E):

E = (2.5 × 1015) (6.626 × 10-34 Js) (5.0 × 108s-1)

E = 8.282 × 10-10 J

Hence, the energy of the source (E) is 8.282 × 10-10 J.

49.  For λ1 = 589 nm

Frequency of transition

Frequency of transition (ν1) = 5.093 × 1014 s-1.

Similarly, for λ2 = 589.6 nm

Frequency of transition

Frequency of transition (ν2) = 5.088 × 1014 s-1.

Energy difference (ΔE) between excited states = E1 – E2

Where,

E2 = energy associated with λ2

E1= energy associated with λ1

ΔE = hv1 – hv2

= h (v1 – v2)

= (6.626 × 10-34 Js) (5.093 × 1014 – 5.088 × 1014) s-1

= (6.626 × 10-34 J) (5.0 × 10-3 × 1014)

ΔE = 3.31 × 10-22 J

50.  It is given that the work function (W0) for caesium atom is 1.9 eV.

(a) From the expression,

λ0 = threshold wavelength

h = Planck’s constant

Substituting the values in the given expression of (λ0):

Hence, the threshold wavelength λ0 is 653 nm.

(b) From the expression, W0 = hv0, we get:

ν0 = threshold frequency

h = Planck’s constant

Substituting the values in the given expression of ν0:

(1eV = 1.602 × 10-19 J)

ν= 4.593 × 1014 s-1

Hence, the threshold frequency of radiation (ν0) is 4.593 × 1014 s-1

(c) According to the question:

Wavelength used in irradiation (λ) = 500 nm

Kinetic energy = h (ν – ν0)

Kinetic energy of the ejected photoelectron = 9.3149 × 10-20 J

Hence, the velocity of the ejected photoelectron (ν) is 4.52 × 105 ms-1.

51.   (a) Assuming the threshold wavelength to be λ0 nm (= λ0 × 10-9 m), the kinetic energy of the radiation is given as:

Three different qualities cab be formed by the given value as:

Similarly

Dividing equation (3) by equation (1):

∴ Threshold (λ0) wavelength = 540 nm

(b) Substitute Value of Threshold wavelength in eq(3), We get

52.   From the principle of conservation of energy, the energy of an incident photon (E) is equal to the sum of the work function (W0) of radiation and its kinetic energy (K.E) i.e.,

E = W0 + K.E

⇒ W0 = E – K.E

Energy of incident photon (E)

Where,

h = Planck’s constant

Substituting the values in the given expression of E:

E = 4.83 eV

The potential applied to silver metal changes to kinetic energy (K.E) of the photoelectron. Hence,

K.E = 0.35 V

K.E = 0.35 eV

∴ Work function, W0 = E – K.E

= 4.83 eV – 035eV

= 4.48 eV

53.   Energy of incident photon (E) is given by,

Energy of the electron ejected (K.E)

Hence, the energy with which the electron is bound to the nucleus can be obtained as:

54.   Wavelength of transition = 1285 nm

Substituting the value of ν in the given expression,

Hence, for the transition to be observed at 1285 nm, n = 5.

The spectrum lies in the infra-red region.

55.   The radius of the nth orbit of hydrogen-like particles is given by,

For radius (r1) = 1.3225 nm

= 1.32225 × 10-9 m

= 1322.25 × 10-12 m

= 1322.25 pm

Similarly,

Thus, the transition is from 5th the orbit to the 2nd orbit. It belongs to the Balmer series. Wave  number for the transition is given by,

∴ Wavelength (λ) associated with the emission transition is given by,

This transition belongs to Balmer series and comes in the visible region of the spectrum.

56.  From de Broglie’s equation,

∴ de Broglie’s wavelength associated with the electron is 455 pm.

57.   From de Brgolie’s equation,

Where,

ν = velocity of particle (neutron)

h = Planck’s constant

m = mass of particle (neutron)

λ = wavelength

Substituting the values in the expression of velocity (ν),

∴ Velocity associated with the neutron = 494 ms-1

58.   According to de Broglie’s equation,

Where,

λ = wavelength associated with the electron

= Planck’s constant

= mass of electron

ν = velocity of electron

Substituting the values in the expression of λ:

λ = 332 pm

∴ Wavelength associated with the electron = 332 pm

59.  According to de Broglie’s expression,

Substituting the values in the expression,

60.   From Heisenberg’s uncertainty principle,

Where,

Δx = uncertainty in position of the electron

Δp = uncertainty in momentum of the electron

Substituting the values in the expression of Δp:

∴ Uncertainty in the momentum of the electron = 2.637 × 10-21 kgms-1

Since the magnitude of the actual momentum is smaller than the uncertainty, the value cannot be defined.

61.  For n = 4 and l = 2, the orbital occupied is 4d.

For n = 3 and l = 2, the orbital occupied is 3d.

For n = 4 and l = 1, the orbital occupied is 4p.

Hence, the six electrons i.e., 1, 2, 3, 4, 5, and 6 are present in the 4d, 3d, 4p, 3d, 3p, and 4p orbitals  respectively.

Therefore, the increasing order of energies is

5(3p) < 2(3d) = 4(3d) < 3(4p) = 6(4p) < 1 (4d).

62.   Nuclear charge experienced by an electron (present in a multi-electron atom) is dependant upon the distance between the nucleus and the orbital, in which the electron is present. As the distance increases, the effective nuclear charge also decreases.

Among p-orbitals, 4p orbitals are farthest from the nucleus of bromine atom with (+35) charge. Hence, the electrons in the 4p orbital will experience the lowest effective nuclear charge. These electrons are shielded by electrons present in the 2p and 3porbitals along with the s-orbitals. Therefore, they will experience the lowest nuclear charge.

63.  Nuclear charge is defined as the net positive charge experienced by an electron in the orbital of a multi-electron atom. The closer the orbital, the greater is the nuclear charge experienced by the electron (s) in it.

(i) The electron(s) present in the 2s orbital will experience greater nuclear charge (being closer to the nucleus) than the electron(s) in the 3s orbit

(ii) 4d will experience greater nuclear charge than 4f ,because 4d is closer to the

(iii) 3p will experience greater nuclear charge since it is closer to the nucleus than 3f.

64.  Nuclear charge is defined as the net positive charge experienced by an electron in a multi-electron  atom.

The higher the atomic number, the higher is the nuclear charge. Silicon has 14 protons while aluminium has 13 protons. Hence, silicon has a larger nuclear charge of (+14) than aluminium, which has a nuclear charge of (+13). Thus, the electrons in the 3p orbital of silicon will experience a more effective nuclear charge than aluminium.

65.  (a) Phosphorus (P):

Atomic number = 15

The electronic configuration of P is:

1s2 2s2 2p6 3s2 3p3

The orbital picture of P can be represented as:

From the orbital picture, phosphorus has three unpaired electrons.

(b) Silicon (Si): Atomic number = 14

The electronic configuration of Si is:

1s2 2s2 2p6 3s2 3pThe orbital picture of Si can be represented as:

From the orbital picture, silicon has two unpaired electrons.

(c) Chromium (Cr):

Atomic number = 24

The electronic configuration of Cr is:

1s2 2s2 2p6 3s2 3p6 4s1 3d5 The orbital picture of chromium is:

From the orbital picture, chromium has six unpaired electrons.

(d) Iron (Fe):

Atomic number = 26

The electronic configuration is:

1s2 2s2 2p6 3s2 3p6 4s2 3d6

The orbital picture of chromium is:

From the orbital picture, iron has four unpaired electrons.

(e) Krypton (Kr):

Atomic number = 36

The electronic configuration is:

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6

The orbital picture of krypton is:

Since all orbitals are fully occupied, there are no unpaired electrons in krypton.

66.  (a) n = 4 (Given)

For a given value of ‘n‘, ‘l‘ can have values from zero to (n – 1).

l = 0, 1, 2, 3

Thus, four sub-shells are associated with n = 4, which are s, p, d and f.

(b) Number of orbitals in the nth shell = n2

For n = 4

Number of orbitals = 16

If each orbital is taken fully, then it will have 1 electron with ms value of .

∴ Number of electrons with mvalue of  is 16.

67.   (i)

(ii)

(iii) .

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