Magnesium has three naturally occurring isotopes - VCE - SSCE Chemistry - Question 2 - 2005 - Paper 1

To determine the relative atomic mass of magnesium, use the formula:

ext{Relative Atomic Mass} = rac{ ext{(abundance of }^{24} ext{Mg)} imes ext{(mass of }^{24} ext{Mg)} + ext{(abundance of }^{25} ext{Mg)} imes ext{(mass of }^{25} ext{Mg)} + ext{(abundance of }^{26} ext{Mg)} imes ext{(mass of }^{26} ext{Mg)}}{100}

By substituting in the values:

= rac{(78.99 imes 23.985) + (10.00 imes 24.986) + (11.01 imes 25.983)}{100}

Calculating this gives:

• For 24Mg: 78.99 × 23.985 = 1897.53
• For 25Mg: 10.00 × 24.986 = 249.86
• For 26Mg: 11.01 × 25.983 = 285.95

Sum: 1897.53 + 249.86 + 285.95 = 2433.34 Thus, ext{Relative Atomic Mass} = rac{2433.34}{100} = 24.33

Therefore, the relative atomic mass of magnesium is approximately 24.31, rounded to two decimal places.

Mendeleev placed calcium and magnesium in the same vertical group because they exhibit similar chemical properties. Both elements belong to Group 2 of the periodic table, known as the alkaline earth metals. They have similar valence electron configurations, which leads to comparable reactivity and bonding characteristics.

Magnesium has a higher electronegativity than calcium because of its smaller atomic radius and higher effective nuclear charge. As the size of the atom decreases, the ability of the nucleus to attract bonding electrons increases, resulting in higher electronegativity.

The electron configuration for the calcium atom (Ca), which has an atomic number of 20, is:

$1s^2 2s^2 2p^6 3s^2$ This indicates that calcium has 2 electrons in its outermost shell, making it part of Group 2.

The electron configuration for the Ca²⁺ ion, which has lost two electrons, is:

$1s^2 2s^2 2p^6 3s^2 \rightarrow 1s^2 2s^2 2p^6$ This shows that the Ca²⁺ ion has 18 electrons, resulting in a stable, noble gas configuration.

The calcium atom is significantly larger than the Ca²⁺ ion because it has more electrons, which leads to increased electron-electron repulsion in the outer shell. When calcium loses two electrons to form Ca²⁺, the effective nuclear charge experienced by the remaining electrons increases, pulling them closer to the nucleus and resulting in a smaller ionic radius.