Problem 162
Question
The Ionic radii (in \(\mathrm{A}\) ) of \(\mathrm{N}^{-3}, \mathrm{O}^{-2}, \mathrm{~F}^{-}\)are respectively? [2015] (a) \(1.36,1.4\) and \(1.71\) (b) \(1.36,1.71\) and \(1.4\) (c) \(1.71,1.4\) and \(1.36\) (d) \(1.71,1.36\) and \(1.4\)
Step-by-Step Solution
Verified Answer
The correct answer is (c) 1.71, 1.4, and 1.36.
1Step 1: Understanding Ionic Radii
The ionic radius is the measure of an atom's ion in a crystal lattice. The size of the ion is influenced by the number of electrons surrounding the nucleus.
2Step 2: Ionic Size Order in Negative Ions
For anions (negatively charged ions), as the negative charge increases in the same period, the ionic size increases. Hence, the ion \( ext{N}^{-3}\) is larger than \( ext{O}^{-2}\), which is larger than \( ext{F}^{-}\).
3Step 3: Review Given Options
The question provides four options to choose the correct sequence of ionic radii: \( ext{(a)} 1.36, 1.4, 1.71 \), \( ext{(b)} 1.36, 1.71, 1.4 \), \( ext{(c)} 1.71, 1.4, 1.36 \), \( ext{(d)} 1.71, 1.36, 1.4 \).
4Step 4: Match with Correct Sequence
Based on increasing ionic charge, the correct order for the ionic radii is from largest to smallest: \( ext{N}^{-3} \), \( ext{O}^{-2} \), and \( ext{F}^{-} \). The numbers from the options represent the sizes in angstroms corresponding to this order.
5Step 5: Selecting the Correct Option
The correct order identified \( ext{(N}^{-3} \, ext{O}^{-2} \, ext{F}^{-} \)) corresponds to choice \( ext{(c)} 1.71, 1.4, 1.36 \).
Key Concepts
AnionsIonic SizeNegative Ions
Anions
Anions are ions with a negative charge. They form when an atom gains electrons, resulting in more electrons than protons. This imbalance between the number of protons and electrons gives the ion its negative charge. For instance, the nitrogen atom in
- Nitrogen atom typically has 7 protons and 7 electrons. However, the anion \[ \mathrm{N}^{-3} \] has gained 3 additional electrons, totaling 10.
- Similarly, an oxygen atom with 8 protons can gain 2 electrons to form \[ \mathrm{O}^{-2} \], resulting in a total of 10 electrons.
- Fluorine, usually with 9 protons, gains one electron to become \[ \mathrm{F}^{-} \] with 10 electrons.
Ionic Size
The ionic size, or ionic radius, refers to the size of an ion, often influenced by its electron configuration and the resulting interactions within the atom itself. When comparing anions like \[ \mathrm{N}^{-3}, \mathrm{O}^{-2}, \text{and} \mathrm{F}^{-} \], these have the same electron count, but their sizes differ due to their charges and nuclear pull. Here's why this matters:
- Higher negative charge increases electron-electron repulsion, which in turn expands the electron cloud. As a result, \[ \mathrm{N}^{-3} \] with its higher negative charge is largest among the three.
- The number of protons in \[ \mathrm{F}^{-} \] exerts a greater pull on its electrons compared to the other two, leading to a smaller radius.
- As a general rule, adding electrons (increasing negative charge) tends to increase ionic size.
Negative Ions
Negative ions, otherwise known as anions, are essential to both chemistry and the real world. They play vital roles in various chemical reactions and processes, including electricity conduction and cellular processes.
- Negatively charged ions move towards the anode in a process called electrolysis, influencing chemical changes in solutions.
- In physiological environments, anions like chloride and phosphates are critical for nerve impulses and cellular activities.
- In materials science, controlled amounts of anions are used to modify materials' properties, such as superconductors and semiconductors.
Other exercises in this chapter
Problem 160
The first ionisation potential of \(\mathrm{Na}\) is \(5.1 \mathrm{ev}\). The value of electron gain enthalpy of \(\mathrm{Na}^{+}\)will be? (a) \(-2.55 \mathrm
View solution Problem 161
Which of the following arrangement represents the increasing order of Ionic radii of the given species \(0^{-2}\), \(\mathrm{S}^{-2}, \mathrm{~N}^{-3}, \mathrm{
View solution Problem 159
The increasing order of the ionic radii of the given isoelectronic species is: (a) \(\mathrm{Ca}^{2+}, \mathrm{K}^{+}, \mathrm{Cl}^{-}, \mathrm{S}^{2-}\) (b) \(
View solution