Chapter 10: Problem 10
Comment on the following radii of the \(\mathrm{H}^{-}\)ion calculated from structures of ionic compounds: $$ \begin{array}{lccccccc} \hline & \text { LiH } & \mathrm{NaH} & \mathrm{KH} & \mathrm{CsH} & \mathrm{MgH}_{2} & \mathrm{CaH}_{2} & \mathrm{BaH}_{2} \\ \text { Radius/pm } & 114 & 129 & 134 & 139 & 109 & 106 & 111 \\ \hline \end{array} $$
Short Answer
Step by step solution
Analyze the given data
Group the compounds by metal type
Observe trends for Group 1 metals
Observe trends for Group 2 metals
Compare the findings
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Ionic Compounds
- Metal ions in these compounds donate electrons.
- Hydride ions possess an extra electron, giving them a negative charge.
- The ionic bond occurs due to the attraction between the positive metal ion and the negative hydride ion.
Alkali Metals
- They form strong ionic bonds with the hydride ions.
- The size of the alkali metal ions increases down the group from \( \mathrm{Li} \) to \( \mathrm{Cs} \).
- This increase in size causes the hydride ion to also appear larger as the metal ion becomes less able to compress the \( \mathrm{H}^{-} \) ions due to reduced electron cloud overlap.
Alkaline Earth Metals
- They tend to form ionic compounds with stronger ionic interactions compared to alkali metals.
- The ionic radii of the hydride ions in these compounds are observed to be smaller compared to alkali hydrides.
- As alkaline earth metals form divalent cations, they often create more compact crystalline structures.
Hydride Ions
- The \( \mathrm{H}^{-} \) ion's size can be influenced by the nature of the metal it is bonded with.
- The more positive charge a metal ion has, the more it can influence the electron cloud distribution of the \( \mathrm{H}^{-} \) ion.
- Larger metal ions tend to result in larger hydride ions due to weaker electrostatic forces.