ATOMIC STRUCTURE-ANSWER

SESSION 2020/2021 TOPIC 2: ATOMIC STRUCTURE
CHEMISTRY SK015

TUTORIAL 2

2.1: Bohr’s Atomic Model

1. a) Bohr’s Atomic Model Description
• Electrons in an atom move in certain circular orbits.
• An electron moving in an orbit does not radiate or absorb energy.
• The moving electron has a specific amount of energy and its energy is quantized.
• Energy is emitted or absorbed only when the electron moves to another energy
level.
• Electron at its excited states is unstable. It will fall back to lower energy level and
emit a specific amount of energy in the form of light.

b) Differences 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) Differences between line spectrum and continuous spectrum

Line spectrum Continuous spectrum

A spectrum consists of discontinuous and A spectrum consists of radiation

discrete lines with specific wavelength. distributed over all wavelengths

without any blank spot.

Emission spectrum an element. Rainbow

2. a) Reason:
• Electron in an atom exists at its lower energy level (ground state).
• Each electron will absorb different amount of energy and excited to higher energy
level and become unstable.
• The electrons than will fall back to lower energy level.
• During this transition, they will release specific amount of energy in the form of
photon.
• Lots of electrons emit lots of photon in form a series of line.
• Each lines represent specific wavelength and frequency.

b) Differences between Lyman and Balmer series

Lyman Balmer

Formed when electron transit from higher Formed when electron transit from

energy level to n=1. higher energy level to n=2.

Emits a specific amount of energy with Emits a specific amount of energy with

specific wavelength and frequency which specific wavelength and frequency which

formed line that falls in the ultraviolet region. formed line that falls in the visible region.

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SESSION 2020/2021 TOPIC 2: ATOMIC STRUCTURE
CHEMISTRY SK015
3. a) Transition
Blue; n=5 to n=2.
Red; n=3 to n=2.

b) Shortest wavelength: Violet.

c) Lowest frequency: Red.

d) energy released
transition from ni=6 to nf=2
∆ = ( 1 2 − 1 2 )
= 2.18 × 10−18 (612 − 212)
= − . × −

4. Given:
Balmer series, nf=2
Wavelength = 656.3 nm = 656.3×10−9 m

a) Frequency
ѵ =



= 3.00 x 108 ms-1
656.3 x 10-9 m

= 4.571 x 1014 s-1

b) Energy emitted
∆E = hc
λ
= (6.63 x 10-34 Js-1)(3.00 x 108 ms-1)
656.3 x 10-9 m
= 3.03 x 10-19 J

c) transition,

1 = ( 1 12 − 1 22) , 1 < 2


656.3 1 m = 1.097 × 107 −1 (212 − 1 22)
x 10−9

n2 = 3

∴ n=3 to n=2

5. a) Given:
ni = 5, nf = 1

wavelength

1 = ( 1 12 − 1 22) , 1 < 2


= 1.097 × 107 −1 (112 − 512)

λ = 94.96 nm

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SESSION 2020/2021 TOPIC 2: ATOMIC STRUCTURE
CHEMISTRY SK015

frequency
ѵ =



= 3.00 x 108 ms-1

94.96 x 10-9 m

= 3.2 x 1015 s-1

b) Given:
ni = 5, nf = 3

wavelength

1 = ( 1 12 − 1 22) , 1 < 2


= 1.097 × 107 −1 (312 − 512)

λ = 1282 nm

frequency
ѵ =



= 3.00 x 108 ms-1

1282 x 10-9 m

= 2.3 x 1014 s-1

6. 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. a) Given:
Transition from ni=4 to nf=3

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

= 2.18 × 10−18 (412 − 312)
= -1.06 x 10-19 J

b) Frequency
∆E = hѵ
1.06 x 10-19 J = (6.63 x 10-34 Js-1) ѵ
ѵ = 1.60 x 1014 s-1

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SESSION 2020/2021 TOPIC 2: ATOMIC STRUCTURE
CHEMISTRY SK015
wavelength
∆E = hc

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

λ
λ = 1876 nm

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 =

4π

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.

4π

(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.

m

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

mv

2.2: Quantum Mechanical Model & 2.3: Electronic Configuration

1. a) Definition
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) quantum numbers
i. n (principal quantum number : energy level)
ii. l (angular momentum quantum number : shape)
iii. m (magnetic quantum number : spatial orientation)
iv. s (electron spin quantum number : spinning of electrons)

c) one set of possible quantum numbers

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

4p: n=4, l=1, m=-1, s=+1/2 or n=4, l=1, m=-1, s=-1/2

n=4, l=1, m=0, s=+1/2 or n=4, l=1, m=0, s=-1/2

n=4, l=1, m=+1, s=+1/2 or n=4, l=1, m=+1, s=-1/2

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SESSION 2020/2021 TOPIC 2: ATOMIC STRUCTURE
CHEMISTRY SK015

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

2. Orbitals and number of electrons at each sub-energy level

Set n l Orbitals No. of
electrons
I 2 0 2s
II 3 2 3d 2
III 5 1 5p
10

6

3. a) 2d (not allowed) z z
b) 3f (not allowed)
c) 4d (allowed)
d) 7s (allowed)

4. Shape of orbitals
z

x yx yx y
1s 2s
z z 3py
z

x y x yx y

3dxz 3dx2-y2 3dz2

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

b) Order of increasing energy
1s < 2s < 2py < 3s < 3pz, 3py < 4s < 3dxy, 3dyz < 4pz < 4dxy

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SESSION 2020/2021 TOPIC 2: ATOMIC STRUCTURE
CHEMISTRY SK015

6. Given:
Atom X = 15 electrons

a) Definition
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) Orbital diagram
X : 1s22s22p63s23p3

1s 2s 2p 3s 3p
c) sets of quantum numbers for highest energy electrons

3p 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
n=3, l=1, m=+1, s=+1/2 or
Choose either 3 sets.

7. Orbital diagram
a) Cl−: 1s22s22p63s23p6

1s 2s 2p 3s 3p
b) Al3+: 1s22s22p6

1s 2s 2p

c) Ni2+: 1s22s22p63s23p63d8

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

8. Electronic configuration
Fe2+: 1s22s22p63s23p63d6
Fe3+: 1s22s22p63s23p63d5

Stability
Fe3+ is more stable because it contains half-filled 3d orbital. Half-filled 3d orbital in Fe3+ is more
stable compare to partially-filled 3d orbital in Fe2+.

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SESSION 2020/2021 TOPIC 2: ATOMIC STRUCTURE
CHEMISTRY SK015
Values of n, l, m, s for outermost shell of Fe2+ ion,
3d6
n=3, l=0, m=0, s=+1/2 n=3, l=0, m=0, s=-1/2
n=3, l=1, m=-1, s=+1/2 n=3, l=1, m=-1, s=-1/2
n=3, l=1, m=0, s=+1/2 n=3, l=1, m=0, s=-1/2
n=3, l=1, m=+1, s=+1/2 n=3, l=1, m=+1, s=-1/2
n=3, l=2, m=+2, s=+1/2 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=0, s=+1/2 n=3, l=2, m=-2, s=+1/2

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SESSION 2020/2021 TOPIC 2: ATOMIC STRUCTURE
CHEMISTRY SK015

MEKA 2

2.1: Bohr’s Atomic Model

1. Formation of line in Lyman series,
• When energy is absorbed, the electron at the ground state will absorb the energy and
excited to higher energy level.
• This excited electron is unstable. So it will fall back to lower energy level (n=1) and
emit certain amount of energy with certain wavelength and frequency.

2. (a) Transition to produced line z,
n=5 to n=1

(b) Region of line z,
Ultraviolet region

(c) Energy corresponding to line y,
ni= 4, nf = 1

∆ = ( 1 2 − 1 2 )
= 2.18 × 10−18 (412 − 112)
= -2.04 x 10-18 J

(d) Wavelength of line y

E = hc


( )( )2.04 10−18 J = 6.6310−34 J s−1 3.00 108 m s−1

λ = 9.73 x 10-8 m

Frequency of line y,
∆E = hѵ
2.04×10-18 J = (6.63×10-34 J s-1) ѵ
ѵ = 3.08×1015 s-1

3. Wavelength of fourth line in Balmer series,
ni = 6, nf = 2

1 = RH  1 − 1  , n1  n2
  n12 
 n 2 
2

= 1.097 × 107 −1 (212 − 612)
λ = 4.102 x 10-7 m

2.2: Quantum Mechanical Model & 2.3: Electronic Configuration

1 (a) Set of four quantum number for each valence electrons (4 valence electrons)
n=3, l=0, m=0, s=+1/2
n=3, l=0, m=0, s=-1/2
n=3, l=1, m=0, s=+1/2
n=3, l=1, m=-1, s=+1/2

(b) Electronic configuration of X
1s22s22p63s23p2

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SESSION 2020/2021 TOPIC 2: ATOMIC STRUCTURE
CHEMISTRY SK015
(c) Shape of orbitals of valence electrons
zz z

x yx yx y
3s
3px 3py

2. (a) Orbital diagram for,
V2+:

1s 2s 2p 3s 3p 3d
V3+:

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

(b) Set of four quantum for one electrons with the highest energy (3d orbital) for V2+

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

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

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

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

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

choose 1 only.

3. (a) Expected and actual electronic configuration,

Chromium
Expected: 1s22s22p63s23p64s23d4
Actual: 1s22s22p63s23p64s13d5

Copper
Expected: 1s22s22p63s23p64s23d9
Actual: 1s22s22p63s23p64s13d10

(b) Reason for the anomaly,

Chromium: half-filled 3d orbital is more stable than partially-filled 3d orbital.
Copper: completely-filled 3d orbital is more stable than partially-filled 3d
orbital.

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SESSION 2020/2021 TOPIC 2: ATOMIC STRUCTURE
CHEMISTRY SK015
KUMBE 2
2.1 Bohr’s Atomic Model n=5
1. Energy level diagram to show the electronic transitions, n=4
n=3
energy n=2

(d)

(c)
(b)

(a) n=1

2. (a) Formation of line spectrum

• When energy is absorbed, the electron at the ground state will absorb the
energy and excited to higher energy level.

• This excited electron is unstable. So it will fall back to lower energy level
(n=3) and released certain amount of energy with certain wavelength and
frequency.

(b) wavelength,

ni = 6, nf = 3

1 = ( 1 12 − 1 22) , 1 < 2


= 1.097 × 107 −1 (312 − 612)
λ = 1094 nm

(c) Direction of energy increase
energy increase from left to right

(d) Reason lines increasingly closer from left to right.
The difference in the energy levels of orbital becomes smaller at higher energy
levels.

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SESSION 2020/2021 TOPIC 2: ATOMIC STRUCTURE
CHEMISTRY SK015
3. (a) series for the lines form (colour)
Balmer series. n=6
n=5
(b) electronic transition on energy level diagram n=4
energy n=3
n=2

n=1

red cyan blue violet

(c) energy of blue line
ni = 5, nf = 2

∆ = ( 1 2 − 1 2 )
= 2.18 × 10−18 (512 − 212)
= -4.58 x 10-19 J

4. (a) energy electron fall (Paschen series)
n=3

(b) energy of electron at excited state

E= −
2
= - (2.18 x 10-18 J)

72

= -4.45 x 10-20 J

5. (a) Name of the series (visible region)
Balmer series.

(b) electronic transition to produced the line

1 = ( 1 12 − 1 22) , 1 < 2


486.3 1 m = 1.097 × 107 −1 (212 − 1 22)
x 10−9

n2 = 4

∴ n=4 to n=2

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SESSION 2020/2021 TOPIC 2: ATOMIC STRUCTURE
CHEMISTRY SK015
6. frequency
∆ = ( 1 2 − 1 2 )
= 2.18 × 10−18 (412 − 212)
= -4.088 x 10-19 J

∆E = hѵ
4.088 x 10-19 J = (6.63 x 10-34 Js-1) ѵ
ѵ = 6.17 x 1014 s-1

2.2: Quantum Mechanical Model & 2.3: Electronic Configuration

1. permissible quantum numbers
(a) n=2, l=1, m=+1, s= +½
permissible

(b) n=1, l=0, m= -1, s= -½
Not permissible
If l=0, m should be 0 only.

(c) n=3, l=3, m= -1, s= +½
Not permissible
If n=3, l should be 0, 1 and 2 only.

(d) not permissible
If n=1 l=0 m only 0

2. (a) electronic configuration
A: 1s22s22p63s23p3
B: 1s22s22p63s23p64s23d3

(b) number of unpaired electron
A: 3
B: 3

(c) Shape of orbital of valence electrons in A z z

zz

x yx yx yx y
3s
3px 3py 3pz

Set of quantum number of valence electrons in A
n=3, l=0, m= 0, s= +½
n=3, l=0, m= 0, s= -½
n=3, l=1, m= -1, s= +½
n=3, l=1, m= 0, s= +½
n=3, l=1, m= +1, s= +½

3. (a) orbital diagram

1s 2s 2p 3s 3p

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SESSION 2020/2021 TOPIC 2: ATOMIC STRUCTURE
CHEMISTRY SK015
(b) charge of ion
2- z
Electronic configuration
X2-: 1s22s22p63s23p6

4. (a) Quantum number that represents shape
Angular Momentum Quantum Number, l.

(b) i. shape of orbital
3d orbital and 4s orbital

z zz

x yx yx yx y
4s
3dx2-y2 3dxy 3dxz

ii. electronic configuration
1s22s22p63s23p64s23d3

iii. electronic configuration of the ion
J5+: 1s22s22p63s23p6

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