To arrange the given alkaline-earth-metal iodates in order of decreasing solubility in water, it's helpful to consider the trends in solubility of ionic compounds. The solubility of a compound typically depends on the sizes of the involved ions and their lattice energies.
### Step-by-Step Solution:
1. Understand the Ionic Sizes:
- Alkaline-earth metals (Group 2 elements) consist of beryllium (Be), calcium (Ca), strontium (Sr), and radium (Ra).
- As you go down the group in the periodic table, the ionic size of the cations increases.
2. General Solubility Trends:
- For the same anion, the solubility of a metal compound decreases as the size of the cation increases. This is because larger cations form stronger ionic bonds with large anions like IO₃⁻, making the compound less soluble in water.
3. Evaluate the Iodates Given:
- Be(IO₃)₂: Beryllium iodate
- Ca(IO₃)₂: Calcium iodate
- Sr(IO₃)₂: Strontium iodate
- Ra(IO₃)₂: Radium iodate
4. Arrange Based on the Cation Size:
- Beryllium (Be) has the smallest ionic radius among the given cations.
- Calcium (Ca) is next larger.
- Strontium (Sr) follows.
- Radium (Ra) has the largest ionic radius.
Following the solubility trend (smaller cations form more soluble compounds with large anions):
- Most soluble: Be(IO₃)₂
- Less soluble: Ca(IO₃)₂
- Even less soluble: Sr(IO₃)₂
- Least soluble: Ra(IO₃)₂
### Conclusion:
The iodates arranged in order of decreasing solubility in water are:
1. Be(IO₃)₂
2. Ca(IO₃)₂
3. Sr(IO₃)₂
4. Ra(IO₃)₂
So, the order is:
Be(IO₃)₂, Ca(IO₃)₂, Sr(IO₃)₂, Ra(IO₃)₂.
Hence, the correct arrangement from highest to lowest solubility in water is:
Be(IO₃)₂, Ca(IO₃)₂, Sr(IO₃)₂, Ra(IO₃)₂.
