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Ionization energy is always positive for...

Ionization energy is always positive for an element and successive ionization energies always gradually increase.
Choose the incorrect ionization energy order

A

`Al gt Ga`

B

`TI gt In`

C

`Pb gt Sn`

D

`Y gt Sc`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the question regarding the incorrect ionization energy order, we can analyze the ionization energies of the elements mentioned: aluminum (Al), gallium (Ga), indium (In), thallium (Tl), lead (Pb), tin (Sn), scandium (Sc), and yttrium (Y). ### Step-by-Step Solution: 1. **Understanding Ionization Energy**: - Ionization energy is defined as the energy required to remove the outermost electron from an atom in its gaseous state. It is always a positive value because energy must be supplied to overcome the attraction between the nucleus and the electron. **Hint**: Remember that ionization energy is always positive because it requires energy to remove an electron. 2. **Analyzing the Group Trends**: - As we move down a group in the periodic table, the ionization energy generally decreases. This is due to the increase in atomic size and the shielding effect, which makes it easier to remove the outermost electron. **Hint**: Consider how atomic size and shielding affect ionization energy as you move down a group. 3. **Comparing Aluminum and Gallium**: - Aluminum (Al) has a lower atomic number than Gallium (Ga). Ga has d-electrons that provide less shielding, leading to a stronger nuclear charge felt by the outermost electron. Therefore, Ga has a higher ionization energy than Al. **Hint**: Think about the effect of d-electrons on ionization energy when comparing Al and Ga. 4. **Comparing Indium and Thallium**: - Indium (In) and Thallium (Tl) are in the same group. Tl has f-electrons that also provide less shielding, resulting in a higher ionization energy for Tl compared to In. **Hint**: Remember that the presence of f-electrons can affect the ionization energy. 5. **Comparing Lead and Tin**: - Lead (Pb) and Tin (Sn) are also in the same group. Pb has f-electrons, which leads to a higher nuclear charge and smaller atomic size compared to Sn, resulting in a higher ionization energy for Pb. **Hint**: Consider how the addition of f-electrons influences the ionization energy trend in the carbon family. 6. **Comparing Scandium and Yttrium**: - Scandium (Sc) and Yttrium (Y) are in the same group. As we move down the group, Yttrium has more electron shells than Scandium, leading to a larger atomic size and lower ionization energy for Y compared to Sc. **Hint**: Focus on the number of electron shells and how they influence ionization energy. 7. **Identifying the Incorrect Order**: - From the analysis, we find that: - Ga > Al (correct) - Tl > In (correct) - Pb > Sn (correct) - Sc > Y (incorrect, as Y's ionization energy should be less than Sc) Therefore, the incorrect ionization energy order is Scandium (Sc) having a higher ionization energy than Yttrium (Y). ### Final Conclusion: The incorrect ionization energy order is Scandium (Sc) > Yttrium (Y).
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Knowledge Check

  • Ionization energy is always positive for an element and successive ionization energies always gradually increase. The successive ionization energy for an element is given below in mJ/mole. 0.7865, 1.5771,3.2316, 4.3555, 16.091, 19.785, 23.786, 29252 Predict the number of valence shell electrons in the given element.

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  • Ionization energy is always positive for an element and successive ionization energies always gradually increase. The successive ionization energy for an element is given below in mJ/mole. 0.7865, 1.5771,3.2316, 4.3555, 16.091, 19.785, 23.786, 29252 Predict the number of valence shell electrons in the given element.

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