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The charge/size ratio of a cation determines its polarizing power. Which one of the following sequences represents the increasing order of the polarizing power of the cationic species, \(\mathrm{K}^{+}, \mathrm{Ca}^{2+}, \mathrm{Mg}^{2+}, \mathrm{Be}^{2+}\) ? (a) \(\mathrm{Be}^{2+}<\mathrm{K}^{+}<\mathrm{Ca}^{2+}<\mathrm{Mg}^{2+}\) (b) \(\mathrm{K}^{+}<\mathrm{Ca}^{2+}<\mathrm{Mg}^{2+}<\mathrm{Be}^{2+}\) (c) \(\mathrm{Ca}^{2+}<\mathrm{Mg}^{2+}<\mathrm{Be}^{2+}<\mathrm{K}^{+}\) (d) \(\mathrm{Mg}^{2+}<\mathrm{Be}^{2+}<\mathrm{K}^{+}<\mathrm{Ca}^{2+}\)

Short Answer

Expert verified
(b) \( \mathrm{K}^{+}<\mathrm{Ca}^{2+}<\mathrm{Mg}^{2+}<\mathrm{Be}^{2+} \).

Step by step solution

01

Understanding Polarizing Power

Polarizing power of a cation is determined by its charge-to-size ratio. A higher charge and a smaller ionic radius result in a higher polarizing power.
02

Analyzing Given Cations

Consider the cation charge: - \( \mathrm{K}^{+} \) has a charge of +1.- \( \mathrm{Ca}^{2+} \), \( \mathrm{Mg}^{2+} \), and \( \mathrm{Be}^{2+} \) each have a charge of +2. Now, consider the size: Be is smallest, followed by Mg, Ca, and then K, which is the largest.
03

Arranging Cations by Charge to Size Ratio

The polarizing power increases with increase in charge and decrease in size:1. \( \mathrm{K}^{+} \) (largest size, lowest charge-to-size ratio)2. \( \mathrm{Ca}^{2+} \)3. \( \mathrm{Mg}^{2+} \)4. \( \mathrm{Be}^{2+} \) (smallest size, highest charge-to-size ratio)
04

Selecting the Correct Sequence

Based on the order from the previous step, the sequence of increasing polarizing power is: \( \mathrm{K}^{+}<\mathrm{Ca}^{2+}<\mathrm{Mg}^{2+}<\mathrm{Be}^{2+} \). This corresponds to option (b).

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Key Concepts

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

Charge-to-Size Ratio
The charge-to-size ratio is a critical factor in understanding the polarizing power of cations. In chemical terminology, this ratio refers to the magnitude of a cation's charge in relation to its ionic radius. Here's why it's important and what you need to know:

• **Charge**: Cations can carry different charges, often depending on the number of electrons lost. For example, calcium (\(\mathrm{Ca}^{2+}\)) has lost two electrons and carries a +2 charge. The larger the charge, the more pronounced the polarizing power.

• **Size**: Refers to the ionic radius of the cation. Smaller ions can get closer to anions, enhancing their ability to distort the electron cloud of the anions they approach. To sum up, when trying to determine which cation has more polarizing power, look for a higher charge and a smaller size. High charge-to-size ratio means the cation will have more polarizing power.
Ionic Radius
The ionic radius is essentially the radius of an ion in its gaseous state. It plays a pivotal role in determining the characteristics of the ion, including its polarizing power.

Let's break down what affects ionic radius and why it matters:

• **Electron Configuration**: When a neutral atom forms a cation, it loses an electron which can cause the electron cloud to contract, leading to a smaller radius compared to the neutral atom.

• **Nuclear Charge**: As the number of protons in the nucleus increases without a corresponding increase in electron shielding, the effective nuclear charge felt by the electrons increases. This can pull the electron cloud closer to the nucleus, reducing the ionic radius.

In our exercise, accounting for ionic radii helps to understand why \(\mathrm{K}^{+}\) is the least polarizing compared to smaller cations like \(\mathrm{Be}^{2+}\). Since \(\mathrm{Be}^{2+}\) has the smallest ionic radius, its ability to polarize adjacent anions will be stronger.
Cationic Species
Cations are positively charged ions formed when an atom loses one or more electrons. Their characteristics, such as charge and size, are key to understanding their chemical behavior, particularly polarizing power.

In a cation, the following features are influenced:
  • **Charge**: Cations like \(\mathrm{Ca}^{2+}\), \(\mathrm{Mg}^{2+}\), and \(\mathrm{Be}^{2+}\) have lost two electrons, gaining a +2 charge. The higher the charge, the stronger the potential polarizing power.
  • **Size Reduction**: When an atom becomes a cation, the loss of electrons causes a reduction in ionic radius compared to its neutral state. This reduction makes smaller cations more effective in polarizing surrounding anions because the positive charge is concentrated in a smaller volume.
To predict and evaluate how different cations will interact with other particles, understanding these features is essential. The exercise emphasizes \(\mathrm{Be}^{2+}\) having the greatest polarizing power due to its high charge and small size, a common trait in highly effective cations.

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Most popular questions from this chapter

Consider the following statements: 1\. the bond order of \(\mathrm{NO}\) is \(2.5\) 2\. the bond order of \(\mathrm{NO}^{+}\)is 3 3\. the bond order of \(\mathrm{O}_{2}\) is \(1.5\) 4\. the bond order of \(\mathrm{CO}\) is 3 Which of these statements are correct? (a) 1,2 and 3 (b) 2,3 and 4 (c) 1,3 and 4 (d) 1,2 and 4

Of the following, the number of species having two and more than two electrons in the antibonding molecular orbital is__ \(\mathrm{He}_{2}, \mathrm{He}_{2}^{+}, \mathrm{B}_{2}, \mathrm{O}_{2}^{-}, \mathrm{N}_{2}^{-}\)

Species having the same bond order are (a) \(\mathrm{N}_{2}\) (b) \(\mathrm{N}_{2}^{+}\) (c) \(\mathrm{N}_{2}^{-}\) (d) \(\mathrm{N}_{2}^{2-}\)

The electronegativity difference between \(\mathrm{N}\) and \(\mathrm{F}\) is greater than that between \(\mathrm{N}\) and \(\mathrm{H}\), yet the dipole moment of \(\mathrm{NH}_{3}(1.5 \mathrm{D})\) is larger than that of \(\mathrm{NF}_{3}\) \((0.2 \mathrm{D})\) This is because (a) in \(\mathrm{NH}_{3}\) as well as \(\mathrm{NF}_{3}\) the atomic dipole and bond dipole are in opposite directions (b) in \(\mathrm{NH}_{3}\) the atomic dipole and bond dipole are in the opposite directions whereas in \(\mathrm{NF}_{3}\) these are in the same direction (c) in \(\mathrm{NH}_{3}\) as well as in \(\mathrm{NF}_{3}\) the atomic dipole and bond dipole are in same direction (d) in \(\mathrm{NH}_{3}\) the atomic dipole and bond dipole and in the same direction whereas in \(\mathrm{NF}_{3}\) these are in opposite directions

According to molecular orbital theory which of the following statement about the magnetic character and bond order is correct regarding \(\mathrm{O}_{2}^{+} ?\) (a) paramagnetic and bond order \(<\mathrm{O}_{2}\) (b) paramagnetic and bond order \(>\mathrm{O}_{2}\) (c) diamagnetic and bond order \(<\mathrm{O}_{2}\) (d) diamagnetic and bond order \(>\mathrm{O}_{2}\)

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