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Identify each statement as true or false: (a) Cations are larger than their corresponding neutral atoms. (b) \(\mathrm{Li}^{+}\)is smaller than Li. (c) \(\mathrm{Cl}^{-}\)is bigger than \(\mathrm{I}^{-}\).

Short Answer

Expert verified
(a) False, (b) True, (c) False. Cations are smaller than their corresponding neutral atoms due to increased effective nuclear charge. \(\mathrm{Li}^{+}\) is smaller than Li for the same reason. \(\mathrm{Cl}^{-}\) is smaller than \(\mathrm{I}^{-}\) as iodine is located at a lower period in the periodic table and has a larger atomic radius.

Step by step solution

01

Statement (a): Cations are larger than their corresponding neutral atoms.

Cations are formed when an atom loses one or more electrons, reducing the electron cloud around the nucleus. This results in an increased effective nuclear charge, which causes the remaining electrons to be attracted more strongly towards the nucleus. Consequently, cations are usually smaller in size than their respective neutral atoms. Therefore, the statement (a) is false.
02

Statement (b): \(\mathrm{Li}^{+}\) is smaller than Li

Lithium atom (\(\mathrm{Li}\)) has three electrons in its electron configuration: 1s²2s¹. When it loses one electron to form the cation \(\mathrm{Li}^{+}\), the electron configuration becomes 1s². As a result, the effective nuclear charge experienced by the remaining electrons increase, making \(\mathrm{Li}^{+}\) smaller than the neutral lithium atom. Therefore, the statement (b) is true.
03

Statement (c): \(\mathrm{Cl}^{-}\) is bigger than \(\mathrm{I}^{-}\)

When forming anions, atoms gain electrons, which increases the overall electron-electron repulsion in the electron cloud. This repulsion leads to a slight increase in the size of anions compared to their respective neutral atoms. However, statement (c) compares the size of \(\mathrm{Cl}^{-}\) and \(\mathrm{I}^{-}\). Since iodine is in the same group as chlorine in the periodic table but is located at a lower period, it has more electron shells and a larger atomic radius. Therefore, \(\mathrm{I}^{-}\) is bigger than \(\mathrm{Cl}^{-}\), making statement (c) false.

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Key Concepts

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

Cations and Anions
Understanding the size difference between cations and anions compared to their neutral atoms is foundational in the field of chemistry. A cation forms when an atom loses one or more electrons, whereas an anion is created when an atom gains electrons.

The key to grasping this concept is to know that when a neutral atom loses an electron to become a cation, it reduces its electron cloud, resulting in a size decrease. The loss of negatively charged electrons means that the positive charge of the nucleus can pull the remaining electrons closer, leading to a smaller ion.

On the flip side, when an atom gains an extra electron(s) and becomes an anion, the increase in electron-electron repulsions within the electron cloud forces the electrons apart, resulting in a larger ion size compared to the neutral atom.

This trend holds true across the periodic table, leading to the general rule that cations are smaller and anions are larger than their respective neutral atoms.
Effective Nuclear Charge
The effective nuclear charge (ENC) refers to the net positive charge experienced by an electron in a polyelectronic atom. This concept is vital for explaining the differences in ion sizes.

ENC is influenced by the balance between the attraction from protons in the nucleus and the repulsion from other electrons. When an atom loses an electron to form a cation, the reduction in electron shielding allows the electrons to feel a stronger attraction to the nucleus, thereby increasing the ENC. This stronger pull by the nucleus results in a smaller atomic or ionic radius.

Conversely, the formation of an anion by the addition of one or more electrons to an atom's electron cloud causes an increase in electron-electron repulsion, which effectively reduces the ENC felt by each electron. As a result, the added electrons are not as strongly pulled towards the nucleus, and the atomic or ionic radius increases.
Electron-Electron Repulsion
The idea of electron-electron repulsion is simple yet profound in its implications for atomic and ionic sizes. Electrons, being negatively charged, repel each other. This repulsion becomes more significant as the number of electrons increases, particularly when an atom becomes an anion.

In anions, the additional electrons contribute to a greater repelling force within the electron cloud, pushing the electrons farther apart and resulting in a larger ionic radius. This concept also impacts the structure and shape of the molecule, affecting properties such as bond angles and bond lengths.

It's important to consider electron-electron repulsion along with ENC when predicting and comparing the sizes of ions in various elements and compounds.
Periodic Trends in Atomic Size
The periodic table is a map of periodic trends, including the trend in atomic size. As we move across a period (row) from left to right, the atomic size generally decreases. There's a net increase in ENC due to an increased charge in the nucleus without a corresponding increase in electron shielding.

However, as we move down a group (column), the atomic size increases. This is due to the addition of electron shells, which outweighs the increase in ENC. It helps to visualize the periodic table as a guide to the relative sizes of atoms: across a period, size shrinks, and down a group, size grows.

These trends explain why, for instance, a chloride ion is smaller than an iodide ion; despite both being anions, iodide has more electron shells due to its position further down the group in the periodic table, leading to a larger size.

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Most popular questions from this chapter

Explain the following variations in atomic or ionic radii: (a) \(\mathrm{I}^{-}>\mathrm{I}>\mathrm{I}^{+}\) (b) \(\mathrm{Ca}^{2+}>\mathrm{Mg}^{2+}>\mathrm{Be}^{2+}\) (c) \(\mathrm{Fe}>\mathrm{Fe}^{2+}>\mathrm{Fe}^{3+}\)

The following observations are made about two hypothetical elements \(A\) and \(B\) : The \(A-A\) and \(B-B\) bond lengths in elemental A and B are \(2.36\) and \(1.94 \AA\), respectively. A and B react to form the binary compound \(\mathrm{AB}_{2}\), which has a linear structure (that is \(\angle \mathrm{B}-\mathrm{A}-\mathrm{B}=180^{\circ}\) ). Based on these statements, predict the separation between the two \(\mathrm{B}\) nuclei in a molecule of \(\mathrm{AB}_{2}\).

An element \(\mathrm{X}\) reacts with oxygen to form \(\mathrm{XO}_{2}\) and with chlorine to form \(\mathrm{XCl}_{4} \cdot \mathrm{XO}_{2}\) is a white solid that melts at high temperatures (above \(1000^{\circ} \mathrm{C}\) ). Under usual conditions, \(\mathrm{XCl}_{4}\) is a colorless liquid with a boiling point of \(58^{\circ} \mathrm{C}\). (a) \(\mathrm{XCl}_{4}\) reacts with water to form \(\mathrm{XO}_{2}\) and another product. What is the likely identity of the other product? (b) Do you think that element \(\mathrm{X}\) is a metal, nonmetal, or metalloid? (c) By using a sourcebook such as the CRC Handbook of Chemistry and Physics, try to determine the identity of element X.

Consider the stable elements through lead \((Z=82)\). In how many instances are the atomic weights of the elements out of order relative to the atomic numbers of the elements?

Note from the following table that there is a significant increase in atomic radius upon moving from \(\mathrm{Y}\) to La whereas the radii of \(\mathrm{Zr}\) to \(\mathrm{Hf}\) are the same. Suggest an explanation for this effect. $$ \begin{array}{llll} \hline & {\text { Atomic Radii (Å) }} \\ \hline \mathrm{Sc} & 1.70 & \mathrm{Ti} & 1.60 \\ \mathrm{Y} & 1.90 & \mathrm{Zr} & 1.75 \\ \mathrm{La} & 2.07 & \text { Hf } & 1.75 \\ \hline \end{array} $$

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