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Provide a brief explanation for each of the following: (a) \(\mathrm{O}^{2-}\) is larger than O. (b) \(\mathrm{S}^{2-}\) is larger than \(\mathrm{O}^{2-}\). (c) \(\mathrm{S}^{2-}\) is larger than \(\mathrm{K}^{+}\). (d) \(\mathrm{K}^{+}\)is larger than \(\mathrm{Ca}^{2+}\).

Short Answer

Expert verified
(a) O虏鈦 is larger than O because as O gains 2 electrons to form O虏鈦, the greater repulsive forces between electrons expand the electron cloud. (b) S虏鈦 is larger than O虏鈦 because S has an additional energy level of electrons compared to O. (c) S虏鈦 is larger than K鈦 because S虏鈦 has an extra energy level of electrons and is an anion while K鈦 is a cation. (d) K鈦 is larger than Ca虏鈦 because Ca虏鈦 has a higher effective nuclear charge resulting in a smaller ionic size compared to K鈦.

Step by step solution

01

(a) Why O虏鈦 is larger than O

Adding electrons to an atom forms a negatively charged ion or anion. In this case, when O gains 2 electrons to form O虏鈦, there are more electrons than protons. This causes greater repulsive forces between the electrons, pushing them away from each other, causing the electron cloud to expand, and making O虏鈦 larger than O.
02

(b) Why S虏鈦 is larger than O虏鈦

O and S are both in the same group in the periodic table (Group 6), but S is one period lower than O. This means that S has one additional energy level of electrons compared to O. When both elements form a 2- ion, this extra energy level makes S虏鈦 larger than O虏鈦.
03

(c) Why S虏鈦 is larger than K鈦

When an atom loses an electron, it becomes a positively charged ion or cation (in this case, K鈦). This causes an increase in the effective nuclear charge experienced by the remaining electrons, causing them to be drawn closer to the nucleus, making the ion smaller than the neutral atom. Comparing S虏鈦 and K鈦, S虏鈦 has an extra energy level of electrons and is an anion, so it is larger than K鈦, which is a cation.
04

(d) Why K鈦 is larger than Ca虏鈦

K鈦 and Ca虏鈦 are both in the same period of the periodic table (Period 4), but K is one group to the left of Ca. When forming cations, K loses one electron while Ca loses two, resulting in Ca虏鈦 having a higher effective nuclear charge than K鈦. The remaining electrons in Ca虏鈦 are drawn more tightly to the nucleus due to this higher charge, resulting in a smaller ionic size compared to K鈦.

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Key Concepts

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

Anions and Cations
Anions and cations are types of ions, which are atoms or molecules that have gained or lost one or more electrons, resulting in a net charge. Anions are negatively charged ions, formed when an atom gains electrons. For example, the oxygen atom (O) gains two electrons to become the oxide ion (\(\mathrm{O}^{2-}\)), turning it into an anion. This increase in electrons leads to more electron-electron repulsion, which causes the ionic radius to expand.
Anions can be remembered as seeking 'additional electrons', and therefore, they are usually larger than their neutral parent atoms because of the additional repulsion between the extra electrons.
Cations, on the other hand, are positively charged ions formed when an atom loses electrons. For instance, potassium (K) loses an electron to become the potassium ion (\(\mathrm{K}^{+}\)). The loss of an electron means that there are fewer electrons compared to protons, resulting in a stronger pull from the nucleus on the remaining electrons. Consequently, cations tend to be smaller than their neutral counterparts.
  • Anions are larger due to addition of electrons.
  • Cations are smaller due to electron loss and stronger nuclear pull.
Periodic Table Trends
Periodic table trends help us understand how elements behave and change their properties across different periods and groups. Trends in ionic size, for instance, evolve predictably. As we move down a group in the periodic table, atoms have more energy levels, and their atomic and ionic radii increase. For example, sulfur (\(\mathrm{S}\)) is below oxygen (\(\mathrm{O}\)) in the same group, which means that \(\mathrm{S}^{2-}\) has more electron shells than\(\mathrm{O}^{2-}\), resulting in a larger ionic radius.
Across a period, moving from left to right, atoms generally become smaller. This is because the effective nuclear charge increases, pulling the electrons closer to the nucleus. For identical charges, such as \(\mathrm{K}^{+}\) and \(\mathrm{Ca}^{2+}\) in the same period, the cation with a higher effective nuclear charge will be smaller. Thus,\(\mathrm{Ca}^{2+}\) is smaller than \(\mathrm{K}^{+}\).

### Key points about periodic trends:
  • Atomic and ionic sizes increase down a group.
  • Sizes decrease across a period from left to right.
  • More electron shells contribute to larger ionic radii.
Effective Nuclear Charge
Effective nuclear charge is a crucial concept that influences the size of ions. The effective nuclear charge is the net positive charge experienced by electrons in an atom. It plays a critical role in determining how strongly electrons are held in orbit by the nucleus.
For cations, like \(\mathrm{Ca}^{2+}\), losing electrons reduces electron shielding, increasing the effective nuclear charge. The remaining electrons are pulled in more tightly, making the ion smaller. On the other hand, anions, such as \(\mathrm{O}^{2-}\), experience more electron-electron repulsion due to added electrons. These repulsive forces lead to an expanded electron cloud.
  • Anions: Increased repulsion = expanded size.
  • Cations: Higher effective nuclear charge = smaller size.
The effective nuclear charge can be roughly calculated by taking the total charge of an atom's nucleus and subtracting the shielding or screening effect of the inner electrons. This helps predict the size and chemical properties of ions and how they compare when ranked together, such as more tightly packed cations versus expanded anions.

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Most popular questions from this chapter

Arrange the following oxides in order of increasing acidity: $$ \mathrm{CO}_{2}, \mathrm{CaO}, \mathrm{Al}_{2} \mathrm{O}_{3}, \mathrm{SO}_{3}, \mathrm{SiO}_{2}, \mathrm{P}_{2} \mathrm{O}_{5} \text {. } $$

(a) Why is calcium generally more reactive than magnesium? (b) Why is calcium generally less reactive than potassium?

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.

Consider the isoelectronic ions \(\mathrm{F}^{-}\)and \(\mathrm{Na}^{+}\). (a) Which ion is smaller? (b) Using Equation \(7.1\) and assuming that core electrons contribute \(1.00\) and valence electrons contribute \(0.00\) to the screening constant, \(S\), calculate \(Z_{\text {eff }}\) for the \(2 p\) electrons in both ions. (c) Repeat this calculation using Slater's rules to estimate the screening constant, \(S\). (d) For isoelectronic ions, how are effective nuclear charge and ionic radius related?

Write equations that show the processes that describe the first, second, and third ionization energies of an aluminum atom. Which process would require the least amount of energy?

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