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The correct value of ionisation energies (in \(\mathrm{kJ} \mathrm{mol}^{-1}\) ) of Si, P, CI and S respectively are (a) \(786,1012,999,1256\) (b) \(1012,786,999,1256\) (c) \(786,1012,1256,999\) (d) \(786,999,1012,1256\)

Short Answer

Expert verified
(d) 786, 999, 1012, 1256 is the correct order of ionization energies for Si, P, Cl, and S.

Step by step solution

01

Understanding Ionization Energy Trend

Ionization energy generally increases across a period from left to right due to increasing nuclear charge and decreasing atomic radius. This means that for elements in the same period, the element with higher atomic number will typically have a higher ionization energy.
02

Arranging Given Elements

Arrange Si, P, Cl, and S in order of atomic number: Si (Z=14), P (Z=15), S (Z=16), Cl (Z=17). Cl is expected to have the highest ionization energy and Si the lowest, following the atomic number trend in the same period.
03

Identify Anomalies in Ionization Energies

In most cases, ionization energy increases as we move across the period. However, there can be anomalies due to electronic configurations, such as when an electron is removed from a half-filled or fully filled subshell, which is more stable. For this set, closer elements like P and S show such slight anomaly where S has lower ionization energy compared to P due to subshell repulsion in S.
04

Order and Compare with Choices

Based on the trend and the anomaly, the order should be: Si < S (due to anomaly, S lower than P) < P < Cl. Therefore, the increasing order should be 786 (Si), 999 (S), 1012 (P), 1256 (Cl).
05

Selecting the Correct Answer

Compare each multiple choice option against the trend found in Step 4. (d) matches our computed trend as it states: 786 (Si), 999 (S), 1012 (P), 1256 (Cl).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Periodic Table Trends
When we look at the periodic table, various trends can help us predict how elements will behave. Ionization energy is one such property. It generally increases across a period from left to right. Why? Because as you move across a period, atomic numbers increase. This leads to more protons in the nucleus, which creates a greater positive charge. Consequently, the electrons are held more tightly and more energy is required to remove one. Furthermore, the atomic radius decreases across a period, meaning the outer electrons are closer to the nucleus. This results in stronger attraction, thus higher ionization energy. However, there are exceptions to these trends, such as when electron configuration creates a more stable state for certain atoms, leading to anomalies.
Electronic Configuration
Electronic configuration plays a vital role in determining ionization energy. Electrons are arranged in shells and subshells around the nucleus, with each shell having a specific energy level. When a shell or subshell is fully or half filled, it is particularly stable. As a result, it requires more energy to remove an electron from such stable configurations compared to others. In our specific case, consider phosphorus (P) and sulfur (S) where sulfur, though having a higher atomic number, has a lower ionization energy than phosphorus. This is due to an electron being removed from the semi-filled p subshell in phosphorus, which is a stable configuration, compared to sulfur where electron-electron repulsion makes it less stable.
Atomic Number
The atomic number signifies the number of protons in an atom's nucleus. This number, represented by 'Z', not only determines the identity of the element but also indicates the element's position in the periodic table. The atomic number increases across a period. The order based on atomic number for silicon (Si), phosphorus (P), sulfur (S), and chlorine (Cl) is: Si (Z=14), P (Z=15), S (Z=16), Cl (Z=17). Generally, as the atomic number increases across a period, so does the ionization energy due to increased nuclear charge. However, as mentioned before, this is not an absolute rule due to electronic configurations causing exceptions.
Ionization Energy Anomalies
While periodic trends provide a general guideline, anomalies can occur. Such anomalies often arise due to unique electronic configurations that disrupt the typical trend. For instance, in the case of sulfur and phosphorus, despite being adjacent in the periodic table and belonging to the same period, sulfur exhibits a lower ionization energy. This anomaly is due to electron-electron repulsion within the filled subshells, making it easier to remove an electron from sulfur compared to the more stable half-filled subshell in phosphorus. These kinds of exceptions highlight the importance of assessing individual electronic configurations alongside periodic trends when predicting ionization energies.

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Most popular questions from this chapter

In the following questions two statements (Assertion) (A) and Reason (R) are given. Mark (a) If both \(\mathrm{A}\) and \(\mathrm{R}\) are correct and \(\mathrm{R}\) is the correct explanation of \(\mathrm{A}\). (b) If both \(\mathrm{A}\) and \(\mathrm{R}\) are correct but \(\mathrm{R}\) is not the correct expalnation of \(\mathrm{A}\). (c) \(\mathrm{A}\) is true but \(\mathrm{R}\) is false. (d) A is false but \(R\) is true. (e) \(\mathrm{A}\) and \(\mathrm{R}\) both are false. Assertion: The electron gain enthalpies have large negative values toward the upper right of the periodic table preceding the noble gases. Reason: The effective nuclear charge increases from left to right across a period and consequently it will be easier to add an electron to a smaller atom since the added electron on an average would be closer to the positively charged nucleus.

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