Problem 123 - A Complete Resource Book in Chemistry for JEE Main

Question

One mole of magnesium in the vapour state absorbed \(1200 \mathrm{~kJ} \mathrm{~mol}^{-1}\) of energy. If the first and second ionization energies of \(\mathrm{Mg}\) are 750 and \(1450 \mathrm{~kJ} \mathrm{~mol}^{-1}\) respectively, the final composition of the mixture is (a) \(86 \% \mathrm{Mg}^{+}+14 \% \mathrm{Mg}^{2+}\) (b) \(36 \% \mathrm{Mg}^{+}+64 \% \mathrm{Mg}^{2+}\) (c) \(69 \% \mathrm{Mg}^{+}+31 \% \mathrm{Mg}^{2+}\) (d) \(31 \% \mathrm{Mg}^{+}+69 \% \mathrm{Mg}^{2+}\)

Step-by-Step Solution

Answer

The final composition is 69% Mg⁺ and 31% Mg²⁺ (option c).

Step 1: Understanding Energy Distribution

We have 1200 kJ/mol of energy available, which is first used to ionize Mg to Mg⁺, requiring 750 kJ/mol. This leaves us with act 450 kJ/mol to potentially form Mg²⁺.

Step 2: Calculating Ionization Possibility for Mg²⁺

To convert Mg⁺ to Mg²⁺, an additional 1450 kJ/mol is required, but only 450 kJ/mol remains. Thus, not all Mg atoms can be converted to Mg²⁺.

Step 3: Determining Maximum Mg²⁺ Formation

Using the remaining energy, convert a portion of Mg⁺ to Mg²⁺. The portion of Mg²⁺ formed is given by the ratio of available energy for the second ionization (450 kJ/mol) to the energy required for it (1450 kJ/mol): \[\text{Fraction of } \mathrm{Mg}^{2+} = \frac{450}{1450} = 0.31\]

Step 4: Calculating Percentage Composition

So, the percentage composition of Mg²⁺ is 31%, and the remaining is Mg⁺. Therefore, \[\text{Percentage of } \mathrm{Mg}^{2+} = 31 \ \text{Percentage of } \mathrm{Mg}^{+} = 100 - 31 = 69\]

Step 5: Final Composition

Thus, the final composition of the mixture is 69% Mg⁺ and 31% Mg²⁺, which matches option (c).

Key Concepts

First Ionization Energy

When we talk about first ionization energy, we refer to the amount of energy needed to remove the outermost electron from a neutral atom in its gaseous state. In the case of magnesium (Mg), this energy is quantified as 750 kJ/mol. This means that to transform a mole of neutral magnesium atoms into singly charged ions (Mg⁺), 750 kJ of energy per mole must be provided. This is an important step in understanding how atoms lose electrons to form ions, which is a key process in chemistry. Ionization energy can vary significantly between different elements, influenced by factors such as atomic size and nuclear charge.

Second Ionization Energy

Second ionization energy refers to the energy required to remove an electron from a singly charged ion, converting it into a doubly charged ion (Mg²⁺ in this case). For magnesium, this energy is much higher, at 1450 kJ/mol. This increase is due to the fact that the electron is being removed from an already positively charged ion, which holds onto its electrons more tightly due to increased attractive forces from the nucleus. Hence, more energy is needed. Understanding the second ionization energy is crucial because it highlights how successive ionization energies tend to increase as electrons are removed, illustrating the concept of increasing ion stability with greater positive charge.

Energy Distribution

Energy distribution in this context is about how the 1200 kJ/mol of energy is allocated to ionize magnesium atoms. Initially, 750 kJ/mol is used to convert Mg into Mg⁺, which leaves us with 450 kJ/mol. The remaining energy is not enough to convert all Mg⁺ into Mg²⁺ because the conversion requires an additional 1450 kJ/mol. Thus, only a fraction of the Mg atoms can convert into Mg²⁺, based on the proportion of available energy to the energy required. This concept helps us understand practical limitations in chemical reactions based on available energy, making it a vital topic for predicting reaction outcomes.

Percentage Composition

The percentage composition in this situation refers to the relative amounts of Mg⁺ and Mg²⁺ ions present in the final mixture. After using the available energy, we determined that 31% of the magnesium atoms have been converted to Mg²⁺ with the remaining 69% as Mg⁺ ions. Calculating percentage composition is a common step in chemical calculations, important for determining the final makeup of reaction products. In this example, the percentage of Mg²⁺ is calculated based on the fraction of energy available for second ionization compared to what is required, i.e., \(\frac{450}{1450} = 0.31\), resulting in 31% Mg²⁺. The rest, 69%, remains as Mg⁺ ions.