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Chapter 8: Problem 114

Write the formulas of the binary hydrides for the second-period elements from \(\mathrm{Li}\) to \(\mathrm{F}\). Identify the bonding in each as covalent, polar covalent, or ionic.

Short Answer

Expert verified
The formulas are \(\mathrm{LiH, BeH_2, BH_3, CH_4, NH_3, H_2O, HF}\). Types: \(\mathrm{LiH}\) is ionic; others are polar covalent, except \(\mathrm{CH_4}\), which is covalent.

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01

Identify Second-Period Elements

The second-period elements in the periodic table from lithium (\(\mathrm{Li}\)) to fluorine (\(\mathrm{F}\)) are: lithium (\(\mathrm{Li}\)), beryllium (\(\mathrm{Be}\)), boron (\(\mathrm{B}\)), carbon (\(\mathrm{C}\)), nitrogen (\(\mathrm{N}\)), oxygen (\(\mathrm{O}\)), and fluorine (\(\mathrm{F}\)).
02

Write Binary Hydride Formulas

For each of these second-period elements, combine them with hydrogen (\(\mathrm{H}\)) to form binary hydrides: \(\mathrm{LiH}\), \(\mathrm{BeH_2}\), \(\mathrm{BH_3}\), \(\mathrm{CH_4}\), \(\mathrm{NH_3}\), \(\mathrm{H_2O}\), \(\mathrm{HF}\).
03

Determine Bonding Type

Analyze the type of bonding based on electronegativity differences between the element and hydrogen:1. \(\mathrm{LiH}\) - Ionic, due to a large electronegativity difference.2. \(\mathrm{BeH_2}\) - Polar Covalent, electronegativity difference is moderate.3. \(\mathrm{BH_3}\) - Polar Covalent, due to moderate electronegativity difference.4. \(\mathrm{CH_4}\) - Covalent, with negligible electronegativity difference making it mostly nonpolar.5. \(\mathrm{NH_3}\) - Polar Covalent, due to a moderate difference in electronegativity.6. \(\mathrm{H_2O}\) - Polar Covalent, significantly polar due to higher electronegativity of oxygen.7. \(\mathrm{HF}\) - Polar Covalent, highly polar due to very high electronegativity of fluorine.

Key Concepts

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

second-period elements
The second-period elements of the periodic table play a pivotal role in chemistry, particularly when it comes to forming compounds like binary hydrides. These elements range from lithium (\(\text{Li}\)) to fluorine (\(\text{F}\)), each showcasing unique properties and behaviors. In the formation of binary hydrides, each of these elements combines with hydrogen (\(\text{H}\)), the simplest element.
  • Lithium (\(\text{Li}\))
  • Beryllium (\(\text{Be}\))
  • Boron (\(\text{B}\))
  • Carbon (\(\text{C}\))
  • Nitrogen (\(\text{N}\))
  • Oxygen (\(\text{O}\))
  • Fluorine (\(\text{F}\))
Understanding these elements is crucial as we delve into the types of bonding behaviors they exhibit, particularly in comparison to hydrogen's relatively low electronegativity.
covalent bonding
Covalent bonding occurs when electrons are shared between atoms relatively equally. For second-period elements bonded with hydrogen, this mostly happens when the difference in electronegativity between the bonded atoms is small. A famous example here is methane (\(\text{CH}_4\)).
  • Characteristic of Covalent Bonds: Involves electron sharing.
  • Non-Polar Nature: When the electronegativity difference between atoms such as carbon and hydrogen is negligible, the molecule remains nonpolar.
Covalent bonding typically results in stable molecules with specific geometric shapes, which influence the physical properties of the compound.
polar covalent bonding
Polar covalent bonds represent a middle ground where there is an unequal sharing of electrons due to differences in electronegativity between atoms. This is a common feature among the binary hydrides of some second-period elements, such as ammonia (\(\text{NH}_3\)) and water (\(\text{H}_2\text{O}\)).
  • Polarity in Molecules: The more electronegative atom pulls the shared electrons closer, creating a partial negative charge.
  • Examples in Hydrides: Ammonia and water both demonstrate polar covalent bonding. The atoms involved have differing electronegativities, causing a dipole moment.
  • Impact on Physical Properties: The polar nature affects solubility and boiling/melting points due to intermolecular forces.
ionic bonding
Ionic bonding is a type of chemical bond formed when there is a complete transfer of electrons from one atom to another, resulting in a significant charge difference. This typically occurs when there is a large electronegativity difference between an element and hydrogen, such as seen in lithium hydride (\(\text{LiH}\)).
  • Nature of Ionic Bonds: Features electrostatic attraction between positively and negatively charged ions.
  • Formation in Hydrides: In \(\text{LiH}\), lithium loses an electron to become \(\text{Li}^+\), while hydrogen gains an electron, becoming \(\text{H}^-\).
  • Properties: Ionic compounds often have high melting and boiling points and can conduct electricity when dissolved in water or melted.
Understanding ionic bonding is essential for recognizing how different types of chemical interactions can result in vastly differing properties and behaviors of compounds.

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Most popular questions from this chapter

(a) From the following data calculate the bond enthalpy of the \(\mathrm{F}_{2}^{-}\) ion. $$ \begin{array}{ll} \mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{~F}(g) & \Delta H_{\mathrm{rxn}}^{\circ}=156.9 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{F}^{-}(g) \longrightarrow \mathrm{F}(g)+e^{-} & \Delta H_{\mathrm{rxn}}^{\circ}=333 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{F}_{2}^{-}(g) \longrightarrow \mathrm{F}_{2}(g)+e^{-} & \Delta H_{\mathrm{rxn}}^{\circ}=290 \mathrm{~kJ} / \mathrm{mol} \end{array} $$ (b) Explain the difference between the bond enthalpies of \(\mathrm{F}_{2}\) and \(\mathrm{F}_{2}^{-}\).

Draw reasonable resonance structures for the following ions: (a) \(\mathrm{HSO}_{4}^{-},\) (b) \(\mathrm{PO}_{4}^{3-}\), (c) \(\mathrm{HPO}_{3}^{2-},(\mathrm{d}) \mathrm{IO}_{3}^{-}\)

Draw Lewis structures for the following molecules: (a) \(\mathrm{ICl},\) (b) \(\mathrm{PH}_{3},\) (c) \(\mathrm{P}_{4}\) (each \(\mathrm{P}\) is bonded to three other \(\mathrm{P}\) atoms), (d) \(\mathrm{H}_{2} \mathrm{~S},\) (e) \(\mathrm{N}_{2} \mathrm{H}_{4}\) (f) \(\mathrm{HClO}_{3}\)

The species \(\mathrm{H}_{3}^{+}\) is the simplest polyatomic ion. The geometry of the ion is that of an equilateral triangle. (a) Draw three resonance structures to represent the ion. (b) Given the following information $$ \begin{aligned} 2 \mathrm{H}+\mathrm{H}^{+} \longrightarrow & \mathrm{H}_{3}^{+} & & \Delta H^{\circ}=-849 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{H}_{2} \longrightarrow & 2 \mathrm{H} & & \Delta H^{\circ}=436.4 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ calculate \(\Delta H^{\circ}\) for the reaction $$ \mathrm{H}^{+}+\mathrm{H}_{2} \longrightarrow \mathrm{H}_{3}^{+} $$

Classify the following bonds as covalent, polar covalent, or ionic, and explain: (a) the SiSi bond in \(\mathrm{Cl}_{3} \mathrm{SiSiCl}_{3}\), (b) the \(\mathrm{SiCl}\) bond in \(\mathrm{Cl}_{3} \mathrm{SiSiCl}_{3}\), (c) the CaF bond in \(\mathrm{CaF}_{2},(\mathrm{~d})\) the \(\mathrm{NH}\) bond in \(\mathrm{NH}_{3} .\)
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