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Which of the following elements will have highest ionisation energy? (a) \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{1}\) (b) \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{3}\) (c) \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{4}\) (d) \(1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{1}\)

Short Answer

Expert verified
Element (a), with an outer electron in the 3s orbital and a higher effective nuclear charge due to less electron shielding, will have the highest ionisation energy.

Step by step solution

01

Identify the Elements

Each electron configuration corresponds to a different element in the periodic table. Determine which elements these configurations represent.
02

Understand Ionisation Energy

Ionisation energy is the energy required to remove an electron from an atom in the gas phase. The atom's position in the periodic table influences its ionisation energy.
03

Compare Electron Configurations

The element with the highest effective nuclear charge and the closest outermost electron to the nucleus will typically have the highest ionisation energy.
04

Apply Periodic Trends

Ionisation energy increases across a period (from left to right) and decreases down a group (from top to bottom) in the periodic table. Apply these trends to the given elements.
05

Determine the Element with the Highest Ionisation Energy

Based on their electron configurations and their positions in the periodic table, identify which element has the highest ionisation energy.

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Key Concepts

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

Electron Configuration
Understanding electron configuration is essential for predicting how atoms will interact and how reactive they may be. At its core, electron configuration is the distribution of electrons in an atom's orbitals, which are layers of space that surround an atom's nucleus. Electrons fill these orbitals in a specific order, following the 'aufbau principle,' which tells us they occupy the lowest energy levels first. For instance, the first two electrons will fill the 1s orbital, designated as the 1s2 configuration.

When all the orbitals in a shell (or a layer) are filled, such as the 2p6 configuration, it indicates a stable, less reactive atom, because it takes more energy to remove an electron from a completely filled shell. On the other hand, an element with one electron in a new shell, such as 3s1, may lose that electron more easily, indicating it might have lower ionisation energy compared to more filled configurations.
Periodic Trends
The periodic table is not just a chart of elements; it's a map of periodic trends that dictate elemental properties, one of which is ionisation energy. Ionisation energy tends to increase across a period, or row, from left to right. This happens because as protons are added to the nucleus, the overall positive charge increases, pulling electrons closer.

Conversely, as you move down a group, or column, in the periodic table, ionisation energy decreases. This is because with each additional layer of electron orbitals, the outer electrons are further from the nucleus and feel a lesser attraction, thus requiring less energy to be removed. These trends help us predict which element in a given set might have the highest or lowest ionisation energy, greatly narrowing down the analysis needed when comparing similar elements.
Effective Nuclear Charge
The effective nuclear charge (Zeff) is the net positive charge experienced by an electron in a multi-electron atom. The more closely an electron is held by the nucleus, the higher the effective nuclear charge. Zeff increases across a period because each successive element has an additional proton in the nucleus, increasing the nuclear charge without a corresponding increase in electron shielding.

The concept of electron shielding is important here; inner electrons can shield outer electrons from the full charge of the nucleus. Thus, as the inner electrons 'shield' the outer electrons from the nucleus, the effective nuclear charge that these outer electrons experience is less than the actual charge of the nucleus. This concept is key in determining why some atoms have higher ionisation energies than others.
Periodic Table
The periodic table is not merely a list of elements but a profound tool that encapsulates four key areas of chemical properties: atomic size, ionisation energy, electronegativity, and electron affinity. Relative positions of elements within this table give insight into how their electrons are configured and how they react with other substances.

For example, elements on the right side of the table generally have higher ionisation energies due to their near-complete electron shells, making them less eager to give an electron away. This is in stark contrast to elements on the left side, which tend to lose electrons more readily given their sparser electron shells. Understanding these nuances helps us to not only predict chemical reactivities but also to delve into the reasons behind these predictions, enabling better grasp of complex chemical phenomena.

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Most popular questions from this chapter

Match the atomic numbers of the elements given in column I with the periods given in column II and mark the appropriate choice. \begin{tabular}{|c|c|c|c|} \hline \multicolumn{2}{|c|} { Column I (Atomic number) } & \multicolumn{2}{c|} { Column II (Period) } \\ \hline (A) & 31 & (i) & 5 \\ \hline (B) & 50 & (ii) & 3 \\ \hline (C) & 56 & (iii) & 4 \\ \hline (D) & 14 & (iv) & 6 \\ \hline \end{tabular} (a) (A) \(\rightarrow\) (i), (B) \(\rightarrow\) (ii), (C) \(\rightarrow\) (iii), (D) \(\rightarrow\) (iv) (b) (A) \(\rightarrow\) (ii), (B) \(\rightarrow\) (i), (C) \(\rightarrow(\mathrm{iv}),(\mathrm{D}) \rightarrow\) (iii) (c) (A) \(\rightarrow\) (iii), (B) \(\rightarrow(\mathrm{iv}),(\mathrm{C}) \rightarrow\) (i), (D) \(\rightarrow\) (ii) (d) (A) \(\rightarrow\) (iii), (B) \(\rightarrow\) (i), (C) \(\rightarrow\) (iv), (D) \(\rightarrow\) (ii)

First and second ionisation enthalpies (in \(\mathrm{kJ} /\) mol) of few elements are given below: \begin{tabular}{|c|c|c|} \hline Element & \(\boldsymbol{I E}_{\mathbf{1}}\) & \(\boldsymbol{I E}_{2}\) \\ \hline (i) & 520 & 7300 \\ \hline (ii) & 900 & 1760 \\ \hline (iii) & 1680 & 3380 \\ \hline (iv) & 2080 & 3963 \\ \hline \end{tabular} Which of the above elements will form halides with formula \(M X_{2} ?\) (a) (i) and (ii) (b) (i) and (iii) (c) (ii) and (iii) (d) (i) and (iv)

Which is correct increasing order of their tendency of the given elements to form \(M^{3}\) ion? (a) \(\mathrm{Bi}>\mathrm{Sb}>\mathrm{As}>\mathrm{P}>\mathrm{N}\) \(2 e^{\circ}\) (b) \(\mathrm{Bi}<\mathrm{Sb}<\mathrm{As}<\mathrm{P}<\mathrm{N}\) (c) \(\quad \mathrm{N}<\mathrm{P}<\mathrm{Sb}<\mathrm{Bi}<\mathrm{As}\) (d) \(\mathrm{Bi}>\mathrm{Sb} \sim \mathrm{N} \sim \mathrm{P}>\mathrm{As}\)

lonization enthalpies of transition metals are (a) intermediate between those of \(s\) - and \(p\)-block elements (b) more than \(p\)-block elements (c) highest in all the elements (d) lower than \(s\)-block elements.

An element with atomic number 117 is known as (a) nihonium (b) flerovium (c) tennessine (d) roentgenium

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