Question

The atomic radius of \(\mathrm{K}\) is \(227 \mathrm{pm}\) and that of \(\mathrm{K}^{+}\) is \(138 \mathrm{pm} .\) Calculate the percent decrease in volume that occurs when \(\mathrm{K}(g)\) is converted to \(\mathrm{K}^{+}(g) .\) (The volume of a sphere is \(\frac{4}{3} \pi r^{3}\), where \(r\) is the radius of the sphere.)

Short answer

Calculate the volumes for K and K⁺ using \( V = \frac{4}{3} \pi r^3 \), find volume difference, and use \((\frac{\Delta V}{V_{\mathrm{K}}}) \times 100\)% for percent decrease.

Step-by-step solution

Calculate Initial Volume of K

The formula for the volume of a sphere is \( V = \frac{4}{3} \pi r^3 \). For potassium, \( r = 227 \text{ pm} \). Substitute to get the initial volume:\[V_{\mathrm{K}} = \frac{4}{3} \pi (227)^3\]

Calculate Volume of K⁺

Using the same volume formula, substitute \( r = 138 \text{ pm} \) for \( \mathrm{K}^+ \):\[V_{\mathrm{K}^+} = \frac{4}{3} \pi (138)^3\]

Calculate Volume Difference

Find the difference in volume between \( \mathrm{K} \) and \( \mathrm{K}^+ \):\[\Delta V = V_{\mathrm{K}} - V_{\mathrm{K}^+}\]

Calculate Percent Decrease in Volume

The percent decrease in volume is given by:\[\frac{\Delta V}{V_{\mathrm{K}}} \times 100\%\]Substitute \( \Delta V \) from Step 3 and \( V_{\mathrm{K}} \) from Step 1, and perform the calculation.

Key concepts

Volume of a Sphere

The concept of the volume of a sphere is fundamental in understanding atomic and ionic radii calculations. A sphere's volume is all the space it contains, calculated using the formula \( V = \frac{4}{3} \pi r^3 \), where \(V\) is the volume and \(r\) is the radius. In this formula, \(\pi\) is a special mathematical constant, approximately equal to 3.14159, which helps to relate a circle's circumference to its diameter. Thus, the spherical volume depends heavily on the radius, the most significant contributor to the volume.As the radius changes slightly, the volume can change significantly due to the cube factor in the radius. It shows why small differences in atomic or ionic sizes can lead to notable volume differences. Therefore, when calculating the volume of an atomic or ionic sphere, every value change for the radius, even in picometers, has a tangible impact on the volume outcome.

Percent Decrease in Volume

Understanding the percent decrease in volume involves calculating how much a volume has reduced in percentage terms. It is a way to compare the initial and final volumes, helping to quantify how significant the change is. To determine this percent decrease, we use the formula:

• Calculate the original volume of the first sphere, like our potassium atom with its radius.
• Calculate the volume of the second sphere, such as the potassium ion.
• Find the difference between these two volumes \(\Delta V = V_{original} - V_{new}\).
• Then, divide this difference by the original volume and multiply by 100 to convert it to a percentage: \(\frac{\Delta V}{V_{original}} \times 100\%\).

Through this sequence, we can measure how much the potassium ion's volume has decreased compared to the original potassium atom. This measurement allows for specific insights into how atomic transformations affect spatial occupancy.

Ionic Radius

The ionic radius describes the size of an ion, an atom that has gained or lost one or more electrons. When an atom gains electrons, it becomes a negatively charged ion or anion and tends to become larger because the added electron(s) increase electron-electron repulsion.Conversely, when an atom loses electrons, becoming a positively charged ion or cation, its radius generally decreases. This is because there is a lesser electron cloud, leading to reduced repulsion and a more compact size due to a stronger nuclear pull. This process is visible with our potassium example. The atomic radius of neutral potassium (K) is greater than that of its cation form \(\mathrm{K}^+\). When calculating these radii, we typically refer to measurements in picometers (pm), which are ideal for tiny atomic or ionic scales.Hence, the ionic radius helps in comprehending how chemical properties and ion size are interconnected, significantly affecting how ions interact with other particles in reactions.