Problem 42
Question
For the following groups of elements, select the one that has the property noted: (a) the largest atom: \(\mathrm{Mg}, \mathrm{Mn}, \mathrm{Mo}, \mathrm{Ba}, \mathrm{Bi}, \mathrm{Br}.\) (b) the lowest first ionization energy: \(\mathrm{B}, \mathrm{Sr}, \mathrm{Al}, \mathrm{Br}\) \(\mathrm{Mg}_{\ell} \mathrm{Pb}.\) (c) the most negative electron affinity: \(\mathrm{As}, \mathrm{B}, \mathrm{Cl}\) \(\mathrm{K}, \mathrm{Mg}, \mathrm{S}.\) (d) the largest number of unpaired electrons: \(\mathrm{F}, \mathrm{N}, \mathrm{S}^{2-}, \mathrm{Mg}^{2+}, \mathrm{Sc}^{3+}, \mathrm{Ti}^{3+}.\)
Step-by-Step Solution
Verified Answer
The largest atom is Ba. The element with the lowest first ionization energy is Sr. The element with the most negative electron affinity is Cl. The element with the largest number of unpaired electrons is Ti3+.
1Step 1: Identifying the Largest Atom
Larger atoms are located towards the bottom of the periodic table. Out of the options Mg, Mn, Mo, Ba, Bi, and Br, Ba (Barium) is the furthest down the periodic table, making it the largest atom.
2Step 2: Finding the Element with the Lowest First Ionization Energy
The first ionization energy increases across a period and up a group in the periodic table. Thus, the element with the lowest first ionization energy is typically found towards the bottom left of the periodic table. Out of the options B, Sr, Al, Br, Mg, and Pb, Sr (Strontium) is furthest to the left and down, indicating it has the lowest first ionization energy.
3Step 3: Determining the Element with the Most Negative Electron Affinity
Electron affinity becomes more negative (indicating a stronger attraction for an added electron) as you move from left to right across a period. Thus, out of options As, B, Cl, K, Mg, and S, Cl (Chlorine) will have the most negative electron affinity.
4Step 4: Recognizing the Element with the Largest Number of Unpaired Electrons
The number of unpaired electrons typically corresponds to an element's group number in the periodic table. However, some ions also need to be considered here. For F, N, S, Mg, Sc, and Ti, it's best to refer to the electron configuration. The Ti3+ ion (Titanium) will have an electron configuration of [Ar] 3d1, thus having the largest number of unpaired electrons.
Key Concepts
Atomic SizeIonization EnergyElectron AffinityUnpaired Electrons
Atomic Size
Atomic size refers to the volume of an atom, which is crucial in determining how atoms bond and interact. As you move down a group in the periodic table, the atomic size increases. This is due to the addition of electron shells, which increases the distance between the outermost electrons and the nucleus.
Conversely, as you move across a period from left to right, atomic size tends to decrease. This shrinkage is because additional protons in the nucleus pull the electrons closer due to a stronger effective nuclear charge.
Conversely, as you move across a period from left to right, atomic size tends to decrease. This shrinkage is because additional protons in the nucleus pull the electrons closer due to a stronger effective nuclear charge.
- For example, Barium (Ba) is much larger than Magnesium (Mg) because it is located further down the periodic table.
Ionization Energy
Ionization energy is the energy required to remove an electron from an atom. It's a key factor in understanding an element's reactivity. Typically, ionization energy decreases as you move down a group because the electrons are further from the nucleus and experience less nuclear pull, making them easier to remove.
Conversely, moving across a period from left to right, ionization energy increases as additional protons enhance the effective nuclear charge, holding electrons more tightly.
Conversely, moving across a period from left to right, ionization energy increases as additional protons enhance the effective nuclear charge, holding electrons more tightly.
- For instance, Strontium (Sr) has a lower ionization energy than Aluminum (Al) because it's further down the table, meaning it's easier to remove an electron from Sr.
Electron Affinity
Electron affinity measures how much energy is released when an atom gains an electron. A more negative electron affinity indicates a stronger attraction between the atom and the added electron. Generally, electron affinity becomes more negative across a period from left to right, reflecting increasing nuclear charge which attracts additional electrons.
The general trend for electron affinity is less clear down a group, but can show some decrease as additional shielding from inner electrons occurs.
The general trend for electron affinity is less clear down a group, but can show some decrease as additional shielding from inner electrons occurs.
- An example is Chlorine (Cl) having a more negative electron affinity than Boron (B) or Sulfur (S) due to its position in the periodic table as a halogen.
Unpaired Electrons
Unpaired electrons in an atom or ion play a significant role in determining its magnetic properties and reactivity. These are electrons which occupy an orbital alone, rather than being paired with another electron.
The electron configuration is a helpful guide to identify elements with a high number of unpaired electrons. Elements or ions with partially filled d or f orbitals typically have more unpaired electrons.
The electron configuration is a helpful guide to identify elements with a high number of unpaired electrons. Elements or ions with partially filled d or f orbitals typically have more unpaired electrons.
- In the case of Titanium (Ti3+), its electron configuration of [Ar] 3d1 demonstrates it has unpaired electrons, contributing to its potential for magnetism and chemical reactivity.
Other exercises in this chapter
Problem 37
Must all atoms with an odd atomic number be paramagnetic? Must all atoms with an even atomic number be diamagnetic? Explain.
View solution Problem 38
Neither \(\mathrm{Co}^{2+}\) nor \(\mathrm{Co}^{3+}\) has \(4 \mathrm{s}\) electrons in its electron configuration. How many unpaired electrons would you expect
View solution Problem 47
Which of the following ions are unlikely to be found in chemical compounds: \(\mathrm{K}^{+}, \mathrm{Ga}^{4+}, \mathrm{Fe}^{6+} \mathrm{S}^{2-}, \mathrm{Ge}^{5
View solution Problem 48
Which of the following ions are likely to be found in chemical compounds: \(\mathrm{Na}^{2+}, \mathrm{Li}^{+}, \mathrm{Al}^{4+}, \mathrm{F}^{2-},\) or \(\mathrm
View solution