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

To find the relative atomic mass of magnesium, we use the formula:

ext{Relative Atomic Mass} = rac{ ext{(Abundance of Isotope 1) x (Isotopic mass of Isotope 1)} + ( ext{Abundance of Isotope 2) x (Isotopic mass of Isotope 2)} + ( ext{Abundance of Isotope 3) x (Isotopic mass of Isotope 3)}}{100}

Applying the values from the table:

1. Calculate for each isotope:

   • For $^{24} ext{Mg}$: $78.99 imes 23.985$
   • For $^{25} ext{Mg}$: $10.00 imes 24.986$
   • For $^{26} ext{Mg}$: $11.01 imes 25.983$

2. Add the results and divide by 100: ext{Relative Atomic Mass} = rac{(78.99 imes 23.985) + (10.00 imes 24.986) + (11.01 imes 25.983)}{100} This calculation results in a relative atomic mass of approximately 24.305, which can be reported to 5 significant figures: 24.305.

Mendeleev placed calcium and magnesium in the same vertical group because they exhibit similar chemical properties due to having the same number of valence electrons. Both elements belong to group 2 of the periodic table, which is characterized by the tendency to form +2 ions and similar reactivity patterns.

The difference in electronegativity between magnesium (1.31) and calcium (1.00) can be attributed to the atomic size. As you move down a group in the periodic table, the atomic radius increases, leading to a larger distance between the nucleus and the valence electrons in calcium. This increased distance reduces the effective nuclear charge experienced by the valence electrons, making it easier for calcium to lose its outer electrons and resulting in a lower electronegativity compared to magnesium.

The electron configuration of a neutral calcium atom, which has 20 electrons, is written as:

$ext{Ca: } 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2$

The electron configuration for the Ca$^{2+}$ ion, after the removal of 2 electrons from its outermost shell, is:

$ext{Ca}^{2+}: 1s^2 2s^2 2p^6 3s^2 3p^6$

The calcium atom is significantly larger than the Ca$^{2+}$ ion because the Ca$^{2+}$ ion has lost two valence electrons. This loss results in fewer electron-electron repulsions in the valence shell, allowing the remaining electrons to be held more closely to the nucleus due to the increased effective nuclear charge. Consequently, the ionic radius of Ca$^{2+}$ is smaller than the atomic radius of neutral calcium.