Question
The periodic table provides information about electron configuration, and physical and
chemical properties of elements.
(a) Bismuth has atomic number 83. Deduce two pieces of information about the electron configuration of bismuth from its position on the periodic table.
▶️Answer/Explanation
Ans:
(a) Two pieces of information about the electron configuration of bismuth (\(Bi\)) that can be deduced from its position on the periodic table:
1. Bismuth is located in period 6 of the periodic table. This indicates that bismuth’s electron configuration involves six electron shells (or energy levels). The electron configuration of bismuth can be represented as \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^6 5s^2 4d^{10} 5p^6 6s^2 4f^{14} 5d^{10} 6p^3\).
2. Bismuth is located in group 15 (also known as group V-A) of the periodic table. This indicates that bismuth has five valence electrons in its outermost shell. The presence of five valence electrons contributes to the chemical properties of bismuth, particularly in the formation of chemical bonds and the types of compounds it can form.
Question
(b) Outline why aluminium is malleable.
▶️Answer/Explanation
Ans:
Metallic Bonding: Aluminium is a metal, and like other metals, it has metallic bonding. Metallic bonding occurs between metal atoms, where the valence electrons are delocalized and free to move throughout the structure. This results in a strong electrostatic attraction between the positively charged metal ions and the delocalized electrons, creating a network of positively charged ions surrounded by a “sea” of electrons.
Crystal Structure: Aluminium has a face-centered cubic (FCC) crystal structure. In this structure, each aluminium atom is surrounded by twelve nearest neighbors, forming a close-packed arrangement of atoms. The arrangement of atoms in the FCC crystal structure allows for the layers of atoms to slide past each other with relative ease when a force is applied, leading to the malleability of aluminium.
Dislocation Movement: When a force is applied to aluminium, dislocations can occur within the crystal structure. Dislocations are defects in the crystal lattice where atoms are out of their normal positions. The movement of dislocations allows layers of atoms to slide past each other, enabling aluminium to be easily shaped and deformed without breaking. This property of dislocation movement contributes to the malleability of aluminium.
Question
(c) An 11.98g block of pure aluminium was heated. Calculate the heat energy absorbed, in J, to increase its temperature from 18.0°C to 40.0°C. The specific heat capacity of aluminium is 0.902J \(g^{−1} K^{−1}\).
▶️Answer/Explanation
Ans:
To calculate the heat energy absorbed (\(Q\)) to increase the temperature of the aluminium block, we can use the formula:
\[Q = mc\Delta T\]
Given:
Mass of aluminium block (\(m\)) = 11.98 g
Specific heat capacity of aluminium (\(c\)) = 0.902 J g\(^{-1}\) K\(^{-1}\)
Change in temperature (\(\Delta T\)) = \(40.0^\circ C – 18.0^\circ C = 22.0^\circ C\)
First, we need to convert the temperature change from Celsius to Kelvin:
\[\Delta T = 22.0^\circ C = 22.0 K\]
\[Q = mc\Delta T\]
\[Q = (11.98 \, \text{g})(0.902 \, \text{J g}^{-1} \text{K}^{-1})(22.0 \, \text{K})\]
\[Q = 241.0956 \, \text{J}\]
Therefore, the heat energy absorbed to increase the temperature of the aluminium block from \(18.0^\circ C\) to \(40.0^\circ C\) is approximately \(241.1 \, \text{J}\).
Question
(d) Argon has three naturally occurring isotopes, \(^{36}Ar\), \(^{38}Ar\) and \(^{40}Ar\).
(i) Identify the technique used to determine the relative proportions of the isotopes of argon.
▶️Answer/Explanation
Ans:
The technique used to determine the relative proportions of the isotopes of argon is called mass spectrometry.
Question
The isotopic composition of a sample of argon is 0.34% of \(^{36}Ar\), 0.06% of \(^{38}Ar\) and 99.6% of \(^{40}Ar\).
(ii) Calculate the relative atomic mass of this sample, giving your answer to two decimal places.
▶️Answer/Explanation
Ans:
To calculate the relative atomic mass (\(A_r\)) of the sample of argon, we can use the following formula:
\[ A_r = \frac{(m_1 \times \text{isotopic abundance}_1) + (m_2 \times \text{isotopic abundance}_2) + (m_3 \times \text{isotopic abundance}_3)}{\text{Total isotopic abundance}} \]
where:
\(m_1\), \(m_2\), \(m_3\) are the masses of the isotopes (\(^{36}Ar\), \(^{38}Ar\), \(^{40}Ar\)) respectively
Isotopic abundance is given as a percentage, so it needs to be converted to a decimal (e.g., 0.34% = 0.0034)
Total isotopic abundance is the sum of the isotopic abundances
Given:
\(m_1 = 36\)
\(m_2 = 38\)
\(m_3 = 40\)
Isotopic abundances:
Isotopic abundance of \(^{36}Ar = 0.34\% = 0.0034\)
Isotopic abundance of \(^{38}Ar = 0.06\% = 0.0006\)
Isotopic abundance of \(^{40}Ar = 99.6\% = 0.9960\)
Now, substitute these values into the formula and calculate the relative atomic mass (\(A_r\)):
\[ A_r = \frac{(36 \times 0.0034) + (38 \times 0.0006) + (40 \times 0.9960)}{0.0034 + 0.0006 + 0.9960} \]
\[ A_r = \frac{0.1224 + 0.0228 + 39.84}{0.0034 + 0.0006 + 0.9960} \]
\[ A_r = \frac{40.9852}{1.0000} \]
\[ A_r = 40.9852 \]
Therefore, the relative atomic mass of the sample of argon is approximately \(40.99\) (rounded to two decimal places).