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Published by NOR IRWAN SHAH MOHAMED NOOR, 2022-09-26 21:13:07

CHEMISTRY_SK015_Tutorial_2.1_2.2_2.3

You can check the answers here. If you have any questions, please feel free to consult your Chemistry lecturer.

CHEMISTRY 1 SK015 | TUTORIAL 2.1 | 2022/2023

2.1: Bohr’s Atomic Model

1. Based on Bohr’s atomic model of the hydrogen atom,

a) explain the existence of energy levels in an atom.

Bohr’s postulate imply that the permitted orbit has a specific amount of energy
(quantized energy). The nearest orbit to the nucleus has the smallest value of energy
and as the orbit getting far from the nucleus, the energy of the orbit getting higher as
well. This explain the existence of energy levels in an atom.

b) state the difference between ground state and excited state.

Ground state: : The lowest energy state of an atom, which corresponds to the most
stable energy state.

Excited state: : Level that is higher in energy than the ground state. The higher the
excited state, the farther away the electron is from the nucleus and the
less tightly it is held by the nucleus.

c) distinguish a continuous spectrum and a line spectrum

Line spectrum Continuous spectrum

A spectrum consists of discontinuous and A spectrum consists of radiation distributed

discrete lines with specific wavelength, e,g over all wavelengths without any blank

emission spectrum an element. space/area, e.g rainbow

2. a) Describe the formation of a line in the hydrogen atom emission spectrum.
When energy is supplied to an electron at a ground state in a hydrogen atom, the
electron is promoted from a lower to a higher-energy level. The electron is at excited
state and unstable thus it will fall to lower energy level. When the electrons fall to lower
energy levels, photons energies are emitted in the form of light with specific wavelength
can be detected as a line spectrum

b) Compare and contrast the Lyman and the Balmer series.

Lyman series Balmer series

Formed when electrons transit from higher Formed when electrons transit from higher
energy levels to n=1 and emit photons with energy level to n=2 and emit photons with
specific wavelengths and frequencies that fall specific wavelengths and frequencies that fall
in the ultraviolet region. in the visible region.

3. The emission spectrum of hydrogen atom in the visible region is shown below:

λ (nm)

line spectrum X YZ
W

a) State the transition of electrons that produce line W and Y respectively.
Line W ,from n = 3 to n = 2 ; Line Y, from n=5 to n = 2

b) Which of the above lines has the lowest frequency?
Line W

c) Calculate the wavelength that corresponds to line X.

1 = ( 1 12 − 1 22) , 1 < 2


= 1.097 × 107 −1 (212 − 512)

λ = 434 nm

4. One of the lines in the Balmer has a frequency of 4.57 x 1014 Hz
a) Calculate its wavelength.
ѵ =



4.57 x 1014 s-1 = 3.00 x 108 ms-1
λ

= 6.56 x 10-7 m
c) Calculate the energy emitted.

∆E = hυ
= (6.63 x 10-34 Js-1) x 4.57 x 1014 s-1
= 3.03 x 10-19 J

c) Determine the transition of electron involved.

∆ = ( 1 2 − 1 2 )
3.03 x 10-19 J = 2.18 × 10−18 ( 1 2 − 212)

ni = 3

5. Calculate the wavelength (in nm) and frequency of the:

a) third line in the Paschen series.

Transition from ni=6 to nf=3 ∆ = ( 1 2 − 1 2 )
Energy of photon emitted : = 2.18 × 10−18 (612 − 312)

= -1.82 x 10-19 J

Wavelength

∆E = hc
λ

1.82 x 10-19 J = (6.63 x 10-34 Js-1)(3.00 x 108 ms-1)
λ

λ = 1092.9 nm

Frequency

∆E = hυ
1.82 x 10-19 J = (6.63 x 10-34 Js-1) υ

υ = 2.75 x 1014 s-1

b) second line in the Brackett series.

Transition from ni=6 to nf=4

Energy of photon emitted : ∆ = ( 1 2 − 1 2 )

= 2.18 × 10−18 (612 − 412)

= -7.57 x 10-20 J

Wavelength

∆E = hc
λ

7.57 x 10-20 J = (6.63 x 10-34 Js-1)(3.00 x 108 ms-1)
λ

λ = 2627.5 nm

Frequency

∆E = hυ
7.57 x 10-20 J = (6.63 x 10-34 Js-1) υ

υ = 1.14 x 1014 s-1

6. Calculate the ionization energy of hydrogen in kJ mol−1.
Ionization energy ni = 1, nf = ∞

∆ = ( 1 2 − 1 2 )

= 2.18 × 10−18 (112 − ∞12) = . × −

1 electron ≡ 2.18 × 10−18
1 mol electron ≡ (6.02 x 1023) (2.18 × 10−18 )

= 1312 kJ

7. If an emission spectrum of hydrogen atom emit photons with the average kinetic energy of
10370 J mol−1 is in the infrared region, calculate the frequency of IR wave formed.

6.023 x 1023 electron - 10370 J
1 electron
- 1 electron x 10370 J
6.023 x 1023 electron

= 2.276 x 10-24 J

∆E = hυ
2.276 x 10-24 J = (6.63 x 10-34 Js-1) υ

υ = 3.43 x 109 s-1

8. Two important concepts that relate to the behavior of electrons in an atom are the
Heisenberg uncertainty principle and the wave-particle duality of matter.

(a) State the Heisenberg uncertainty principle as it related to the determining the position
and momentum of an object.

For any moving mass, Heisenberg uncertainty principle states that it is impossible to
know at the same time, both momentum p (defined as mass times velocity) and position
of a particle with certain. Since electron is a moving mass, so it is impossible to know
both momentum and position of an electron with certain as Bohr’s claims in his
postulate.

(b) What aspect of the Bohr theory of the atom is considered unsatisfactory as a result of
the Heisenberg uncertainty principle?
Bohr’s postulate states that electrons are circling in orbits a nucleus of an atom. This
shows Bohr was certain about the position and momentum the electrons in their orbits
around the nucleus of an atom.

On the other hand, Heisenberg uncertainty principle is uncertain of the position and
momentum of an electron (a moving object).

(c) Explain why the uncertainty principle or the wave nature of particles is not significant
when describing the behavior of macroscopic objects, but it is very significant when
describing the behavior of electrons.

For macroscopic objects, the uncertainty principle is not significant since it is possible
and easy to determine the position and momentum of the objects. However, due to
electrons as the microscopic objects and the light mass of electrons, make the
uncertainty principle become significant.

ATAU

8. (a) Heisenberg Uncertainty Principle:

• It is impossible to determine (or measure) both the position and the momentum if any
particle (or object or body) simultaneously.

• The more exactly the position of a particle is known, the less exactly the momentum or
velocity of the particle can be known.

• ∆x ∆p ≥ h where h = Plank’s constant, x = uncertainty in position, p =



uncertainty in momentum.

(b) Bohr postulated that the electron in an H atom travels about the nucleus in a circular orbit and

has a fixed angular momentum. With a fixed radius of orbit and a fixed momentum (or

energy), ∆x ∆p < h , hence the Heisenberg principle is violated.


(c) The wavelength of a particle is given by the De Broglie relation λ = h . For masses
mv

of macroscopic objects, h is so small for any v that  is too small to be detectable. For an
m

electron, m is so small that h yields a detectable .

mv

CHEMISTRY 1 SK015 | TUTORIAL 2.2 & 2.3 | 2022/2023
2.2: Quantum Mechanical Model & 2.3: Electronic Configuration
1. a) Define orbit and orbital.

Orbit: The pathway where the electron moves around the nucleus.
Orbital: An orbital is a three-dimensional region in space around the nucleus whereby
the probability to find an electron is highest.

b) What do these quantum numbers represent?

i. n - principal quantum number that describe energy level and size of the orbitasl
where
n = 1, 2, 3.....

ii. l - angular momentum quantum number describe the shape of the orbitasl where
l = 0, 1, 2.....(n – 1)

iii. m -- magnetic quantum number describe spatial orientation the shape of the orbitasl
where
m = ( - l..... 0......+l )

iv. s - electron spin quantum number describe the spin of electrons as clockwise or
anti- clockwise where s= -½ or s= +½

c) Give one set of possible quantum numbers for an electron that occupy the 3p, 4d and
5s orbitals respectively.

3p : n=3, l=1, m=0, s=+1/2 or any appropriate set

4d : n=4, l=2, m=0, s=+1/2 or any appropriate set

5s : n=5, l=0, 1, m=0, s=+1/2 or any appropriate set

2. For the following quantum numbers of n and l below,

name of number of maximun

n l subshell orbitals no of

electron

I 3 1 3p 3 6

II 4 2 4d 5 10

III 5 0 5s 1 2

a) name the subshell and give the number of orbitals involved for each subshell.
b) determine the maximum number of electrons that occupy each subshell.

3. Explain the following subshell as allowed or not allowed.

a) 6s b) 4p c) 2d d) 3f

a) 6s ( allowed – value of l= 0 which is smaller than n )
b) 4p ( allowed – value of l= 1 which is smaller than n)
c) 2d (not allowed – value of l= 2 which is bigger than n)
d) 3f (not allowed – value of l= 3 which is identical as n)

4. Draw the shape of the following orbitals: z
1s, 2s, 3py, 3dxz, 3 d x2 −y2 and 3 d z2

zz

x yx yx y
1s 2s 3py

z
zz

x y x yx y

3dxz 22 2

3dx -y 3dz

5. a) State Aufbau principle.
Electrons must occupy available orbitals of lower energy first before they filled orbitals
of higher energy.

b) Arrange the following orbitals in the order of increasing energy.
4dxy, 3dxy, 3dyz, 4pz, 3pz, 3py, 2py, 3s, 2s, 1s, 4s

1s < 2s < 2py < 3s < 3pz, 3py < 4s < 3dxy, 3dyz < 4pz < 4dxy
6. Atom X has 15 electrons.

a) State Hund’s rule and Pauli Exclusion Principle

Hund’s rule: When electrons are filled into the orbital the orbital of equivalent energy
(degenerate orbitals), each orbital is filled singly with electron of the same spin before
it is paired.

Pauli Exclusion Principle: No two electrons in an atom can have the same set of four
quantum numbers.

b) Draw the orbital diagram for X.

15 X : 1s22s22p63s23p3

1s 2s 2p 3s 3p

c) Give a set of quantum numbers for the highest energy electrons.

Highest energy electron are electrons that occupy the 3p orbitals

n=3, l=1, m=-1, s=+1/2 or n=3, l=1, m=-1, s=-1/2 or
n=3, l=1, m=0, s=+1/2 or n=3, l=1, m=0, s=-1/2 or
n=3, l=1, m=+1, s=+1/2 or n=3, l=1, m=+1, s=-1/2
Any 3 sets.

7. Give the electronic configuration of the following species as orbital diagram and spdf
notation.

a) O2−

O2−: 1s22s22p6

1s 2s 2p

b) Al3+

Al3+: 1s22s22p6

1s 2s 2p

c) Ni2+

Ni2+: 1s22s22p63s23p63d8

1s 2s 2p 3s 3p 3d

8. a) Write the electronic configuration of iron(II) and iron(III) ions.
Fe2+: 1s22s22p63s23p63d6

Fe3+: 1s22s22p63s23p63d5

b) Which of these two species is more stable? Explain.
Fe3+ is more stable than Fe2+ because of its half-filled 3d orbital is more stable than the
partially-filled of 3d orbital in Fe2+.

c) Write the sets of quantum numbers for the electrons in the outermost orbital of iron(III)
ions.

n=3, l=2, m=+2, s=+1/2 n=3, l=2, m=-1 s=+1/2
n=3, l=2, m=+1, s=+1/2 n=3, l=2, m=-1, s=+1/2
n=3, l=2, m=0, s=+1/2


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