Electronic Structure of Atoms ( STPM )

Electronic Structure of atoms ( STPM Revision )

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Electronic

Structure

Of

Atoms

Contents

1. MPM Specimen Paper

2. STPM 2011

3. STPM 2012

4. STPM 2013

5. STPM 2014

Alex Tan

STPM

Che

mistry

mailto:[email protected]



Objective questions

Question 1 ( MPM Specimen Paper )

What is the maximum number of emission lines possible for a hydrogen atom with

electronic energy levels n = 1, n = 2 and n = 3?

A 2

B 3

C 4

D 6

Answer : B

Explanation :

For Lyman series , there are 2 lines . ( n=2 → n=1 & n=3 → n=1 )

For Balmer series , there is 1 line . ( n=3 → n=2 )

Total = 3 lines

Alex Tan

STPM

Che

mistry




Question 2 ( STPM 2011 )

Electrons may fill s , p , d & f orbitals . Which statement is true of p orbitals ?

A. p orbitals can form ơ and π bonds .

B. 2p orbitals have the same energy as 2s orbitals .

C. The principal quantum numbers of p orbitals start with n=3 .

D. p orbitals are filled with electrons according to Aufbau principle .

Answer : A

Explanation :

Head-on overlap of two p orbitals leads to the formation of ơ bond .

Sideways overlap between two p orbitals results in the formation of a π-bond .

2p orbitals have higher energy level than 2s orbital

The principal quantum numbers of p orbitals start with n=2 .

p orbitals are filled with electrons according to Aufbau principle , Hund’s rule &

Pauli exclusion principle .

Alex Tan

STPM

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mistry





What is the total number of orbitals that has the principal quantum number n=3 ?

A 3

B 4

C 6

D 9

Answer : D

Explanation :

number of 3s orbital = 1

number of 3p orbital = 3

number of 3d orbital = 5

Total number of orbitals = 9

Alex Tan

STPM

Che

mistry





An atom M has seven valence electrons and forms a stable M2+

ion in an aqueous

solution . What is the electronic configuration of atom M ?

A 1s2 2s

2 2p

6 3s

2 3p

5

B 1s2 2s

2 2p

6 3s

2 3p

6 3d

6

C 1s2 2s

2 2p

6 3s

2 3p

6 3d

5 4s

2

D 1s2 2s

2 2p

6 3s

2 3p

6 3d

7 4s

2

Answer : C

Explanation

M has 7 valence electrons .

It cannot be a Group 17 elements because Group 17 elements do not form +2

ions .

Thus , M must be a transition element with the valence shell configuration of

d5s

2 .

Alex Tan

STPM

Che

mistry





Which orbital diagram shows the filling of electron(s) based on Hund’s rule ?

Answer : D

Explanation

Hund’s rule : when the electrons are added to the degenerate orbitals ,

electrons will be added as single electron in parallel spin before they are

paired in opposite spin .

Option A is wrong because it has only one orbital .

Option B is wrong because the electrons must fill the available orbitals singly

before pairing occurs .

Option C is wrong because one of the unpaired electron has an opposite spin

to the other .

Alex Tan

STPM

Che

mistry




Structured & Essays Question

Question 1 ( MPM Specimen Paper ) [ Edited from STPM 2003 E ]

Water is a hydride of oxygen. The bonding in water molecules is a result of the

overlapping of the orbitals of oxygen and hydrogen atoms.

(i) What is meant by orbitals? [1 mark]

Answer :

An orbital is a region in space which there is high probability of finding an

electron .

(ii) Draw a labelled diagram illustrating the shapes of all the orbitals of an oxygen

atom with quantum number n = 2. [3 marks]

Answer :

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STPM

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mistry




Question 2 ( STPM 2011 E )

The number of electrons occupying the different orbitals of atom L are shown in the

following table .

Write the electronic configuration of L , and explain how each of these orbitals is

filled with electrons . [8m]

Answer :

The electronic configuration of L is 1s2 2s

2 2p

6 3s

2 3p

6 3d

5 4s

1 .

According to Aufbau’s principle , the electrons will occupy the orbitals with

the lowest energy level first .

Thus , the electrons will occupy the orbitals increase in the order :

1s , 2s , 2p , 3s , 3p , 4s , 3d

According to Hund’s rule , when the electrons are added to the degenerate

orbitals , electrons will be added as single electron in parallel spin before they

are paired in opposite spin .

According to Pauli exclusion principle , each orbital can only be filled with 2

electrons with opposite spins .

The electronic configuration of L is not 1s2 2s

2 2p

6 3s

2 3p

6 3d

4 4s

2 .

This is because d5 configuration is more stable than the d

4 configuration and

all the five d orbitals are singly occupied by the electrons according to

Hund’s rule .

Alex Tan

STPM

Che

mistry




Question 3 ( STPM 2012 S )

(a) The frequencies of the first six lines in the Lyman series of a hydrogen atom is

shown in the table below .

(i) Complete the above table . [1m]

(ii) Plot a graph of ∆v against frequency to determine the convergence limit for the

Lyman series . [3m]

Alex Tan

STPM

Che

mistry




(iii) Calculate the ionization energy , kJ mol-1

, of the hydrogen atom . [3m]

Answer :

*EXTRA !!! : How to determine the frequency for convergence limit ???

Answer :

~ When ∆ν = 0 , we can get the frequency for convergence limit .

Alex Tan

STPM

Che

mistry





When excited electrons fall from a higher to a lower energy level , the excess energy

is emitted as radiation .

(a) State the energy level transition of an electron that can produce and give three

characteristics of the series . [4m]

Answer :

1. The transition of electrons occurs from a higher energy level to energy level

n=1 .

2. The Lyman series consists of discrete lines with frequencies in the ultraviolet

region .

3. The lines become closer as the frequency increases and finally converge .

4. The lines will converge when they reach convergence limit and form a

continuous spectrum with convergence frequency .

(b) The ionization energy of hydrogen can be determined by using the frequency of

the convergence limit , v∞ of the Lyman series . The convergence limit occurs when

the difference in frequency of successive lines , ∆v , is zero . Five frequencies with

their corresponding ∆v values are shown in the table below .

v / 1014

Hz ∆v / 1014

Hz

29.23 1.60

30.83 0.74

31.57 0.40

31.97 0.24

32.21 0.16

By plotting a graph of ∆v against v , determine the v∞ for hydrogen , and calculate

its ionization energy in kJ mol-1

. [6m]

Alex Tan

STPM

Che

mistry




Answer :

From the graph , v∞ = 32.45 × 1014

Hz

By using the equation ,

Alex Tan

STPM

Che

mistry




(c) (i) State Hund’s rule . [1m]

Answer :

It states that when the electrons are added to the degenerate orbitals ,

electrons will be added as single electron in parallel spin before they are

paired in opposite spin .

(ii) Write the electronic configurations of copper and chromium in their ground

states , and comment on any irregularities present in both their electronic

configurations . [4m]

Answer :

Electronic configuration of chromium is 1s2 2s

2 2p

6 3s

2 3p

6 3d

5 4s

1 .

The electronic configuration of chromium is not 1s2 2s

2 2p

6 3s

2 3p

6 3d

4 4s

2

because d5 configuration is more stable than the d

4 configuration .

Electronic configuration of copper is 1s2 2s

2 2p

6 3s

2 3p

6 3d

10 4s

1

The electronic configuration of copper is not 1s2 2s

2 2p

6 3s

2 3p

6 3d

9 4s

2

because d10

configuration is more stable than the d9 configuration .

Alex Tan

STPM

Che

mistry





Niels Bohr postulated that energy ∆E is released in the form of light when there is a

transition of electron between a higher energy level to a lower energy level .

(i) The Lyman and Balmer series in the atomic emission spectrum of hydrogen are

formed when there is a transition of electrons between energy levels . Draw an

energy level diagram that shows the formation of these two series . [4m]

Answer :

(ii) One of the lines in the Lyman series has a wavelength of 121.6 nm , Calculate the

energy of this transition in joules . [5m]

[The Planck’s constant , h , is 6.63 × 10-34

J s and the light of speed , c , is 3.00 × 108

ms-1

]

Answer :

∆E = hν

= h (

)

= (6.63 × 10-34 ) (3.00 × 10

8 / 121.6 × 10

-9 )

= 1.64 × 10-18

J

Alex Tan

STPM

Che

mistry




(b) An atom of element X has 7 protons in its nucleus .

(i) Explain how the electron configuration of X obeys Hund’s rule . [3m]

Answer :

The electron configuration of X is 1s2 2s

2 2p

3 .

According to Hund’s rule , when the electrons are added to the degenerate

orbitals , electrons will be added as single electron in parallel spin before they

are paired in opposite spin .

The three electrons in 2p subshell will occupy an orbital singly and these

electrons have the same spin .

(ii) Draw the orbitals of the valence shell of X . [3m]

Answer :

*Electronic configuration for X is 1s2 2s

2 2p

3 .

Alex Tan

STPM

Che

mistry


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Electronic Structure of Atoms ( STPM )