Electronic Structure of Atoms ( STPM )

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Electronic Structure of atoms ( STPM Revision )

Electronic Structure Of

Atoms

Contents

1. MPM Specimen Paper 2. STPM 2011 3. STPM 2012 4. STPM 2013 5. STPM 2014

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Electronic Structure of atoms ( STPM Revision ) 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? A2

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B3 C4 D6

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

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Electronic Structure of atoms ( STPM Revision ) 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 .

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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 .

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Electronic Structure of atoms ( STPM Revision ) Question 3 ( STPM 2012 ) What is the total number of orbitals that has the principal quantum number n=3 ? A3 B4

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C6 D9

Answer : D

Explanation :



number of 3s orbital = 1



number of 3p orbital = 3



number of 3d orbital = 5



Total number of orbitals = 9

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Electronic Structure of atoms ( STPM Revision ) Question 4 ( STPM 2013 ) 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 2s2 2p6 3s2 3p5

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B 1s2 2s2 2p6 3s2 3p6 3d6 C 1s2 2s2 2p6 3s2 3p6 3d5 4s2 D 1s2 2s2 2p6 3s2 3p6 3d7 4s2 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 d5s2 .

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Electronic Structure of atoms ( STPM Revision ) Question 5 ( STPM 2014 )

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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 .

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Electronic Structure of atoms ( STPM Revision ) 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.

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(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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Electronic Structure of atoms ( STPM Revision ) Question 2 ( STPM 2011 E ) The number of electrons occupying the different orbitals of atom L are shown in the

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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 2s2 2p6 3s2 3p6 3d5 4s1 .



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 2s2 2p6 3s2 3p6 3d4 4s2 .



This is because d5 configuration is more stable than the d4 configuration and all the five d orbitals are singly occupied by the electrons according to Hund’s rule .

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Electronic Structure of atoms ( STPM Revision ) Question 3 ( STPM 2012 S )

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(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]

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Electronic Structure of atoms ( STPM Revision ) (iii) Calculate the ionization energy , kJ mol-1 , of the hydrogen atom . [3m]

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Answer :

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

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

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Electronic Structure of atoms ( STPM Revision ) Question 4 ( STPM 2013 E ) 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

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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]

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Electronic Structure of atoms ( STPM Revision )

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Answer :

From the graph , v∞ = 32.45 × 1014 Hz By using the equation ,

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Electronic Structure of atoms ( STPM Revision )

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(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 2s2 2p6 3s2 3p6 3d5 4s1 .



The electronic configuration of chromium is not 1s2 2s2 2p6 3s2 3p6 3d4 4s2 because d5 configuration is more stable than the d4 configuration .



Electronic configuration of copper is 1s2 2s2 2p6 3s2 3p6 3d10 4s1



The electronic configuration of copper is not 1s2 2s2 2p6 3s2 3p6 3d9 4s2 because d10 configuration is more stable than the d9 configuration .

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Electronic Structure of atoms ( STPM Revision ) Question 5 ( STPM 2014 E ) 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

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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 × 108 / 121.6 × 10-9 ) = 1.64 × 10-18 J

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Electronic Structure of atoms ( STPM Revision ) (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 2s2 2p3 .



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

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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 2s2 2p3 .

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