TUTORIAL #4 – Quantum Theory & Electronic Structure of Atoms – Part 2
1.
Draw the orbital diagram for the electron configuration of oxygen. How many
unpaired electrons does an oxygen atom possess? Is oxygen paramagnetic or
diamagnetic element?
Answer:
O (8e) 1s2
2s2
2p4
↑
↑
↓
↓
Oxygen has two unpaired electrons and it is paramagnetic.
↑↓
2.
↑↓
What is the characteristic valence electron configuration of the Group 17 element, the
Halogens? Use two elements as examples in your explanation.
Answer:
3.
↑↓
Two elements form Group 17 are Fluorine and Chlorine, their electron
configurations are as follows:
F (9e) 1s2 2s2 2p5
Cl (17e) 1s2 2s2 2p6 3s2 3p5
The general similarity between these configurations is the valence
electron: ns2 np5 – there are 7 electrons in the outer shell.
(a) What are ‘valence electrons’?
(b) What are ‘core electrons’?
(c) What does each box in an orbital diagram represents?
(d) What quantity is represented by the direction (up or down) of the arrows in an
orbital diagram?
Answer:
(a) Valence electrons are the electrons in the outermost shell that take part in
chemical reaction.
(b) Core electrons are inner shell electrons that have the electron configuration of the
nearest noble gas element.
(c) Each box represents an orbital.
(d) The direction of the arrow indicates the spin of the electron.
4.
*Boron (atomic number = 5) occurs naturally as two isotopes: 10B and 11B, with
natural abundance of 19.9% and 80.1% respectively.
(a) Does the electronic configuration of 10B differ from that of 11B?
(b) Draw the orbital diagram for an atom of 11B. Which electrons are the valence
electrons?
(c) Indicate three (3) major ways in which the 1s electrons in boron differ from its 2s
electrons.
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(d) Elemental boron reacts with fluorine to form BF3, a gas. Write a balanced
chemical equation for the reaction of solid boron with fluorine gas.
(e) ∆𝐻𝑓0 for BF3 (g) is –1135.6 kJ/mol. Calculate the standard enthalpy change in the
reaction of boron with fluorine.
Answer:
(a) The two isotopes have identical electron configuration, 1s2 2s2 2p1 because each
has five electrons.
(b)
11
B (5e)
1s2
2s2
↑↓
↑↓
2p1
↑
The valance electrons are the ones in the outermost occupied shell: the 2s2 and 2p1
electrons.
(c) The 1s and 2s orbitals are both spherical, but they differences are:
 1s orbital is lower in energy than the 2s orbital.
 The average distance of the 2s electrons from the nucleus is greater than
the 1s electrons
 The 2s orbital has one node but the 1s orbital has no node
(d) 2 B (s) + 3 F2 (g)  2 BF3 (g)
(e) ∆Ho
=  ∆H (product) –  ∆H (reactant)
= 2(–1135.6 kJ/mol) – [(2 x 0) + (3 x 0)]
= –2271.2 kJ/mol
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