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The three species \(\mathrm{NH}_{2}^{-}, \mathrm{NH}_{3}\), and \(\mathrm{NH}_{4}{ }^{+}\)have \(\mathrm{H}-\mathrm{N}-\mathrm{H}\) bond angles of \(105^{\circ}, 107^{\circ}\), and \(109^{\circ}\), respectively. Explain this variation in bond angles.

Short Answer

Expert verified
The variation in bond angles among NH2-, NH3, and NH4+ can be explained by considering the molecular geometry and electron pair repulsions in each species. In NH2-, there is one lone pair of electrons, and the bond angle is 105° due to lone pair-bonding pair repulsion. In NH3, the trigonal pyramidal shape has one lone pair, leading to a bond angle of 107°. In NH4+, the tetrahedral shape has no lone pairs, resulting in an equal amount of repulsion between bonding pairs and a bond angle of 109°.

Step by step solution

01

1. Draw Lewis Structures

First, we will draw the Lewis structures of the three species to visualize the electron distribution and lone pairs. For NH2-: The nitrogen atom has 5 valence electrons and it is bonded to two hydrogen atoms, with one lone pair remaining in addition to the extra electron from the negative charge. Thus, in total, it has 3 electron pairs around it. For NH3: The nitrogen atom in ammonia has 5 valence electrons as well, but it is bonded to three hydrogen atoms, leaving one lone pair. There are a total of 4 electron pair regions around it. For NH4+: In ammonium, the nitrogen atom donates an electron to form a bond with an additional hydrogen atom, giving it a positive charge. Therefore, there are no lone pairs, and the 4 bonding pairs form a tetrahedral structure.
02

2. Determine Molecular Geometry

Next, using VSEPR theory, identify the molecular geometry of each species. For NH2-: It has 3 electron pair regions (2 bonding pairs and 1 lone pair) which will repel each other in order to minimize electron-electron repulsion. This results in a bent molecular shape. For NH3: It has 4 electron pair regions (3 bonding pairs and 1 lone pair), leading to a trigonal pyramidal molecular shape. For NH4+: There are 4 bonding pairs and no lone pairs in NH4+, giving it a tetrahedral molecular shape.
03

3. Evaluate Electron Pair Repulsions

Electron pair repulsions between bonding pairs and lone pairs are responsible for the difference in bond angles. In general, the repulsion strength follows this order: lone pair-lone pair > lone pair-bonding pair > bonding pair-bonding pair. For NH2-, the bent molecular shape results from one lone pair of electrons and two bonding pairs. The lone pair-bonding pair repulsion leads to a bond angle less than 109.5° (the angle for a tetrahedral arrangement of electron pair regions), leading to a bond angle of 105°. For NH3, the trigonal pyramidal shape results from one lone pair and three bonding pairs. The lone pair-bonding pair repulsion slightly compresses the angle of the bonding pairs, resulting in a bond angle of 107°. For NH4+, the tetrahedral molecular shape consists of only bonding pairs, with an equal amount of repulsion between each pair. Therefore, the bond angle is close to the ideal 109.5°, giving a bond angle of 109°. #Conclusion# The variation in bond angles among NH2-, NH3, and NH4+ is explained by the molecular geometry of each species and the electron pair repulsions present in each case. Lone pair-bonding pair repulsions increase the distortion in bond angles, while the absence of lone pairs in NH4+ allows for a bond angle closer to the ideal tetrahedral angle of 109.5°.

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Key Concepts

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

Lewis Structures
Lewis structures play a foundational role in predicting the molecular geometry of compounds. By drawing Lewis structures, students can visualize the arrangement of valence electrons around a molecule's atoms. For each element, tally up the total number of valence electrons, including any charges as additional electrons or fewer electrons. Connect atoms with single lines that represent shared electron pairs or bonds, and mark unshared electrons or lone pairs with dots. For instance, \(\text{NH}_2^{-}\) has a nitrogen atom with five valence electrons, plus one more for the negative charge. It bonds to two hydrogens, completing its octet with one lone pair.
On the other hand, \(\text{NH}_4^{+}\) does not have any lone pairs because nitrogen shares all of its valence electrons with four hydrogen atoms to form a stable positively charged ammonium ion. Recognizing these patterns helps to determine the electronic framework that governs molecular shape and ultimately, the molecules' physical and chemical properties.
Molecular Geometry
The molecular geometry of a substance is the three-dimensional arrangement of its atoms in space, profoundly influencing its reactivity and physical properties. The VSEPR (Valence Shell Electron Pair Repulsion) theory asserts that electron pairs will arrange themselves to minimize the repulsion between them, leading to specific molecular shapes. For example, with two bond pairs and one lone pair,
\(\text{NH}_2^{-}\) adopts a bent shape. In contrast,
\(\text{NH}_3\), with three bond pairs and one lone pair, has a trigonal pyramidal geometry. Lastly,
\(\text{NH}_4^{+}\) forms a tetrahedral shape with four bond pairs. By identifying these shapes, students can begin to predict molecular behavior and interactions.
Electron Pair Repulsion
The principle of electron pair repulsion is that electron pairs around a central atom will organize themselves to have the maximum separation possible. This is due to the negative nature of electrons that causes them to repel each other. In the case of
\(\text{NH}_2^{-}\), the lone pair exerts a greater repulsion than the bonding pairs, resulting in a compressed bond angle of 105°. For
\(\text{NH}_3\), the lone pair still plays a role in pushing the bonding pairs closer together, leading to the slightly wider bond angle of 107°. The absence of lone pairs in
\(\text{NH}_4^{+}\) allows the bonds to arrange symmetrically, resulting in a bond angle of 109°, which is the ideal angle for a tetrahedral arrangement.
Bond Angles
Bond angles are the angles formed between adjacent bonds on the same atom. They are crucial for determining a molecule's shape and are directly influenced by the electron pair repulsion discussed earlier. The bond angles in
\(\text{NH}_2^{-}\),
\(\text{NH}_3\), and
\(\text{NH}_4^{+}\) differ due to the varying number of lone pairs and bonding pairs. As lone pairs exert greater repulsion than bonding pairs, molecules with lone pairs have smaller bond angles than those without. For instance, the 109° bond angle in tetrahedral \(NH_4^{+}\) decreases to 107° in the trigonal pyramidal \(NH_3\) due to the presence of a lone pair, which compresses the angles between the bonding pairs.

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Most popular questions from this chapter

Shown here are three pairs of hybrid orbitals, with each set at a characteristic angle. For each pair, determine the type of hybridization, if any, that could lead to hybrid orbitals at the specified angle.

The vertices of a tetrahedron correspond to four alternating corners of a cube. By using analytical geometry, demonstrate that the angle made by connecting two of the vertices to a point at the center of the cube is \(109.5^{\circ}\), the characteristic angle for tetrahedral molecules.

Consider the Lewis structure for glycine, the simplest amino acid: (a) What are the approximate bond angles about each of the two carbon atoms, and what are the hybridizations of the orbitals on each of them? (b) What are the hybridizations of the orbitals on the two oxygens and the nitrogen atom, and what are the approximate bond angles at the nitrogen? (c) What is the total number of \(\sigma\) bonds in the entire molecule, and what is the total number of \(\pi\) bonds?

(a) Explain why \(\mathrm{BrF}_{4}{ }^{-}\)is square planar, whereas \(\mathrm{BF}_{4}{ }^{-}\)is tetrahedral. (b) How would you expect the \(\mathrm{H}-\mathrm{X}-\mathrm{H}\) bond angle to vary in the series \(\mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2} \mathrm{~S}, \mathrm{H}_{2} \mathrm{Se}\) ? Explain. (Hint: The size of an electron pair domain depends in part on the electronegativity of the central atom.)

Sulfur tetrafluoride \(\left(\mathrm{SF}_{4}\right)\) reacts slowly with \(\mathrm{O}_{2}\) to form sulfur tetrafluoride monoxide \(\left(\mathrm{OSF}_{4}\right)\) according to the following unbalanced reaction: $$ \mathrm{SF}_{4}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{OSF}_{4}(g) $$ The \(\mathrm{O}\) atom and the four \(\mathrm{F}\) atoms in \(\mathrm{OSF}_{4}\) are bonded to a central \(\mathrm{S}\) atom. (a) Balance the equation. (b) Write a Lewis structure of \(\mathrm{OSF}_{4}\) in which the formal charges of all atoms are zero. (c) Use average bond enthalpies (Table 8.4) to estimate the enthalpy of the reaction. Is it endothermic or exothermic? (d) Determine the electron-domain geometry of \(\mathrm{OSF}_{4}\), and write two possible molecular geometries for the molecule based on this electron-domain geometry. (e) Which of the molecular geometries in part (d) is more likely to be observed for the molecule? Explain.

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