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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

Using only the periodic table, arrange each set of atoms in order from largest to smallest: (a) K, Li, Cs; (b) Pb, Sn, Si; (c) F, O, N.

When magnesium metal is burned in air (Figure 3.6), two products are produced. One is magnesium oxide, \(\mathrm{MgO}\). The other is the product of the reaction of \(\mathrm{Mg}\) with molecular nitrogen, magnesium nitride. When water is added to magnesium nitride, it reacts to form magnesium oxide and ammonia gas. (a) Based on the charge of the nitride ion (Table 2.5), predict the formula of magnesium nitride. (b) Write a balanced equation for the reaction of magnesium nitride with water. What is the driving force for this reaction? (c) In an experiment, a piece of magnesium ribbon is burned in air in a crucible. The mass of the mixture of \(\mathrm{MgO}\) and magnesium nitride after burning is \(0.470 \mathrm{~g}\). Water is added to the crucible, further reaction occurs, and the crucible is heated to dryness until the final product is \(0.486 \mathrm{~g}\) of \(\mathrm{MgO}\). What was the mass percentage of magnesium nitride in the mixture obtained after the initial burning? (d) Magnesium nitride can also be formed by reaction of the metal with ammonia at high temperature. Write a balanced equation for this reaction. If a 6.3-g Mg ribbon reacts with \(2.57 \mathrm{~g} \mathrm{NH}_{3}(g)\) and the reaction goes to completion, which component is the limiting reactant? What mass of \(\mathrm{H}_{2}(g)\) is formed in the reaction? (e) The standard enthalpy of formation of solid magnesium nitride is \(-461.08 \mathrm{~kJ} / \mathrm{mol}\). Calculate the standard enthalpy change for the reaction between magnesium metal and ammonia gas.

(a) What is meant by the term effective nuclear charge? (b) How does the effective nuclear charge experienced by the valence electrons of an atom vary going from left to right across a period of the periodic table?

(a) Does metallic character increase, decrease, or remain unchanged as one goes from left to right across a row of the periodic table? (b) Does metallic character increase, decrease, or remain unchanged as one goes down a column of the periodic table? (c) Are the periodic trends in (a) and (b) the same as or different from those for first ionization energy?

Mercury in the environment can exist in oxidation states \(0,+1\), and \(+2\). One major question in environmental chemistry research is how to best measure the oxidation state of mercury in natural systems; this is made more complicated by the fact that mercury can be reduced or oxidized on surfaces differently than it would be if it were free in solution. XPS, X-ray photoelectron spectroscopy, is a technique related to PES (see Exercise 7.111), but instead of using ultraviolet light to eject valence electrons, \(\mathrm{X}\) rays are used to eject core electrons. The energies of the core electrons are different for different oxidation states of the element. In one set of experiments, researchers examined mercury contamination of minerals in water. They measured the XPS signals that corresponded to electrons ejected from mercury's \(4 f\) orbitals at \(105 \mathrm{eV}\), from an X-ray source that provided \(1253.6 \mathrm{eV}\) of energy. The oxygen on the mineral surface gave emitted electron energies at \(531 \mathrm{eV}\), corresponding to the \(1 s\) orbital of oxygen. Overall the researchers concluded that oxidation states were \(+2\) for \(\mathrm{Hg}\) and \(-2\) for \(\mathrm{O}\). (a) Calculate the wavelength of the \(\mathrm{X}\) rays used in this experiment. (b) Compare the energies of the \(4 f\) electrons in mercury and the \(1 s\) electrons in oxygen from these data to the first ionization energies of mercury and oxygen from the data in this chapter. (c) Write out the ground-state electron configurations for \(\mathrm{Hg}^{2+}\) and \(\mathrm{O}^{2-}\); which electrons are the valence electrons in each case? (d) Use Slater's rules to estimate \(Z_{\text {eff }}\) for the \(4 f\) and valence electrons of \(\mathrm{Hg}^{2+}\) and \(\mathrm{O}^{2-}\); assume for this purpose that all the inner electrons with \((n-3)\) or less screen a full \(+1\).

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