Problem 55
Question
The electronic configurations of some elements are given below. The element with highest electron affinity is (a) \(\mathrm{ls}^{2}, 2 \mathrm{~s}^{2}, 2 \mathrm{p}^{3}\) (b) \(1 \mathrm{~s}^{2}, 2 \mathrm{~s}^{2}, 2 \mathrm{p}^{4}\) (c) \(1 \mathrm{~s}^{2}, 2 \mathrm{~s}^{2}, 2 \mathrm{p}^{5}\) (d) \(\mathrm{ls}^{2}, 2 \mathrm{~s}^{2}, 2 \mathrm{p}^{2}\)
Step-by-Step Solution
Verified Answer
Element (c) with configuration \(1s^{2}, 2s^{2}, 2p^{5}\), Fluorine, has the highest electron affinity.
1Step 1: Understand Electron Affinity
Electron affinity is the energy change when an atom in the gas phase accepts an electron, forming a negative ion. This property typically becomes more negative across a period due to higher nuclear charge making it easier to add electrons.
2Step 2: Determine Electronic Configurations
Review the given electronic configurations: - (a) \( 1s^{2}, 2s^{2}, 2p^{3} \) corresponds to Nitrogen (N).- (b) \( 1s^{2}, 2s^{2}, 2p^{4} \) corresponds to Oxygen (O).- (c) \( 1s^{2}, 2s^{2}, 2p^{5} \) corresponds to Fluorine (F).- (d) \( 1s^{2}, 2s^{2}, 2p^{2} \) corresponds to Carbon (C).
3Step 3: Compare Electron Affinities
- Nitrogen (a) has a half-filled p-shell, leading to a lower electron affinity because it is relatively stable.
- Oxygen (b) will have a higher electron affinity than nitrogen but less than fluorine.
- Fluorine (c), being the most electronegative element, will have the highest electron affinity.
- Carbon (d) has an even fewer number of p-electrons, which results in even lower affinity than oxygen.
4Step 4: Identify the Element with Highest Electron Affinity
Fluorine, from option (c), has the highest electron affinity because it is more effective at stabilizing added electrons due to its high electronegativity and small atomic radius.
Key Concepts
Electronic ConfigurationPeriodic TrendsElectronegativity
Electronic Configuration
When we talk about the electronic configuration of an atom, we're discussing the arrangement of electrons in various shells and subshells around the nucleus of an atom. Each element has a unique electronic configuration that determines its chemical properties. Understanding this arrangement helps predict how an atom will interact with others.
- The electronic configurations are usually written using numbers to denote the energy level and letters to show the subshell (like s, p, d, f) followed by a superscript number indicating the number of electrons in that subshell.
Periodic Trends
Periodic trends are patterns seen within elements on the periodic table. One such trend is electron affinity, which generally increases across a period and decreases down a group.
This occurs because, as you move across a period from left to right, atoms have more protons in their nuclei. This increased "nuclear charge" pulls electrons closer, making it easier for the atom to attract additional electrons.
This occurs because, as you move across a period from left to right, atoms have more protons in their nuclei. This increased "nuclear charge" pulls electrons closer, making it easier for the atom to attract additional electrons.
- Across a period: Electronegativity and electron affinity increase.
- Down a group: Both properties generally decrease due to more electron shells and increased distance from the nucleus.
Electronegativity
Electronegativity is a measure of an atom's ability to attract and hold on to electrons within a chemical bond. Unlike electron affinity, which deals with a single atom in the gas phase, electronegativity concerns atoms in molecules.
Fluorine is the most electronegative element, explaining its high electron affinity. The small atomic radius of fluorine allows for strong attractions between the nucleus and added electrons.
Fluorine is the most electronegative element, explaining its high electron affinity. The small atomic radius of fluorine allows for strong attractions between the nucleus and added electrons.
- High electronegativity leads to higher electron affinity because atoms like fluorine can stabilize additional electrons effectively.
- Electronegativity is a key determinant of how atoms interact in compounds.
Other exercises in this chapter
Problem 53
The correct order of ionic radii is (a) \(\mathrm{I}^{+}>1>\mathrm{I}\) (b) \(\mathrm{I}>\mathrm{I}^{+}>\mathrm{I}\) (c) \(\mathrm{I}^{+}>\mathrm{I}>\mathrm{I}^
View solution Problem 54
The first ionization energy in electron volts of nitrogen and oxygen atoms are respectively given by (a) \(14.6,13.6\) (b) \(13.6,14.6\) (c) \(13.6,13.6\) (d) \
View solution Problem 56
First second and third ionization potential values are \(100 \mathrm{eV}, 150 \mathrm{eV}\) and \(1500 \mathrm{eV}\). Element can be (a) \(\mathrm{Be}\) (b) B (
View solution Problem 57
Arrange \(\mathrm{P}^{3-}, \mathrm{S}^{2-}, \mathrm{H}^{-}, \mathrm{I}^{-}\)in the order of increasing ionic radii (a) \(\mathrm{P}^{3-}, \mathrm{S}^{2-}, \math
View solution