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The bond angle in \(\mathrm{H}_{2} \mathrm{Se}\) is best described as (a) between \(109^{\circ}\) and \(120^{\circ} ;\) (b) less than in \(\mathrm{H}_{2} \mathrm{S} ;\) (c) less than in \(\mathrm{H}_{2} \mathrm{S},\) but not less than \(90^{\circ} ;(\mathrm{d})\) less than \(90^{\circ}\)

Short Answer

Expert verified
The bond angle in \(\mathrm{H}_{2} \mathrm{Se}\) is between \(109^{\circ}\) and \(120^{\circ}\), hence option (a) in the original problem is the most accurate

Step by step solution

01

Understand the Electron Pair Repulsion Theory

Valence Shell Electron Pair Repulsion (VSEPR) Theory is a simple, qualitative model that allows us to predict the geometry of polyatomic ions, molecules, and atoms based on the repulsion between electrons in the outer shell of atoms. It assumes that the molecule or ion will adopt a shape that minimizes the repulsion between electrons in the valence shell of that atom.
02

Identify the Molecular Structure of $\mathrm{H}_{2} \mathrm{Se}$ and $\mathrm{H}_{2}\mathrm{S}$

Both \(\mathrm{H}_{2} \mathrm{Se}\) and \(\mathrm{H}_{2}\mathrm{S}\) are nonpolar and fall under the AX2E (where A is the central atom, X is a terminal atom, and E stands for a lone pair of electrons on the central atom) category based on VSEPR theory. In these structures, the central atom (Se or S) is bonded to two hydrogen atoms and has two lone pairs of electrons.
03

Determine the Theoretical Bond Angle for AX2E Structures

In an AX2E structure, the ideal bond angle would be \(109.5^{\circ}\), but due to the presence of the lone pair electrons on the central atom, there is more repulsion between the lone pair and the bonding pairs. This causes the bonding pairs to squeeze together. As such, the bond angle is slightly less than \(109.5^{\circ}\). Therefore the bond angle in \(\mathrm{H}_{2} \mathrm{Se}\) and \(\mathrm{H}_{2}\mathrm{S}\) will be less than \(109.5^{\circ}\).
04

Compare the Bond Angles of $\mathrm{H}_{2} \mathrm{Se}$ and $\mathrm{H}_{2}\mathrm{S}$

H2Se has larger central atom (Se) than H2S (S), thus the size of S is smaller than that of Se which means Se has inner shells that shield the lone pairs more than the smaller S atom. Therefore the lone pair in Se is more diffuse. This makes the bond angle in H2Se larger than in H2S

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Key Concepts

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

Valence Shell Electron Pair Repulsion Theory
The Valence Shell Electron Pair Repulsion (VSEPR) Theory is fundamental in understanding the shapes of molecules. This theory posits that electron pairs surrounding a central atom arrange themselves as far apart as possible to minimize repulsion. Whether these are bonding pairs (involved in covalent bonds) or lone pairs (non-bonding pairs), their spatial arrangement dictates the molecule's geometry. Using VSEPR, we can predict the three-dimensional structure of a molecule, providing insight into its physical and chemical properties. For instance, H2Se, a molecule pertinent to this exercise, adopts an AX2E structure. This means it has two atoms bonded to the central selenium atom, and one lone pair of electrons. These arrangements lead to specific bond angles by pushing electron pairs apart. An understanding of VSEPR is crucial in predicting how molecular structures are shaped.
Bond Angles
Bond angles are defined as the angles made by lines representing bonds. In molecular geometry, bond angles tell us about the shape of the molecule. In VSEPR theory, an AX2E molecular structure, like H2Se, has an ideal bond angle of approximately 109.5°. Imagine a three-dimensional tetrahedral shape, where if we only consider bond angles between atoms, this is the separation you would expect. However, things are not so simple. Due to electron pair repulsion, especially from lone pairs, these angles are adjusted. In reality, lone pairs occupy more space than bonding pairs because they aren't shared by nuclei. This adjust in H2Se results in a bond angle that is less than 109.5°, as lone pairs push the bonded pairs closer together. Bond angles help us understand the molecular shape and are refined by considering repulsions and steric factors in any given molecule.
Electron Pair Repulsion
Electron Pair Repulsion is a key concept in VSEPR Theory. It highlights how electrons, negatively charged particles surrounding the nucleus, repel each other. This repulsion affects the spatial arrangement of atoms in a molecule. The theory dictates that pairs of electrons in the valence shell of an atom repel each other and will move into positions that minimize this repulsion. In H2Se, we see this play out as the lone pair and bonding pairs around the selenium atom repel each other. Due to lone pairs occupying more space, they exert a greater repulsive force compared to bonding pairs. This greater force warps the structure slightly from its ideal shape, influencing the bond angles and therefore, the molecule’s overall geometry. Electron pair repulsion not only determines the shape but also affects molecule reactivity and interactions.

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Most popular questions from this chapter

The paramagnetism of gaseous \(\mathrm{B}_{2}\) has been established. Explain how this observation confirms that the \(\pi_{2 p}\) orbitals are at a lower energy than the \(\sigma_{2 p}\) orbital for \(\mathrm{B}_{2}\)

Furan, \(\mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O},\) is a substance derivable from oat hulls, corn cobs, and other cellulosic waste. It is a starting material for the synthesis of other chemicals used as pharmaceuticals and herbicides. The furan molecule is planar and the \(\mathrm{C}\) and \(\mathrm{O}\) atoms are bonded into a fivemembered pentagonal ring. The H atoms are attached to the C atoms. The chemical behavior of the molecule suggests that it is a resonance hybrid of several contributing structures. These structures show that the double bond character is associated with the entire ring in the form of a \(\pi\) electron cloud. (a) Draw Lewis structures for the several contributing structures to the resonance hybrid mentioned above. (b) Draw orbital diagrams to show the orbitals that are involved in the \(\sigma\) and \(\pi\) bonding in furan. [Hint: You need use only one of the contributing structures, such as the one with no formal charges.] (c) How many \(\pi\) electrons are there in the furan molecule? Show that this number of \(\pi\) electrons is the same, regardless of the contributing structure you use for this assessment.

Write plausible molecular orbital diagrams for the following heteronuclear diatomic species: (a) \(\mathrm{NO} ;\) (b) \(\mathrm{NO}^{+}\) (c) \(\mathrm{CO} ;\) (d) \(\mathrm{CN} ;\) (e) \(\mathrm{CN}^{-} ;\) (f) \(\mathrm{CN}^{+} ;\) (g) BN.

Match each of the following species with one of these hybridization schemes: \(s p, s p^{2}, s p^{3}, s p^{3} d, s p^{3} d^{2} .\) (a) \(\mathrm{PF}_{6}^{-}\) (b) \(\operatorname{COS} ;\) (c) \(\operatorname{SiCl}_{4} ;\) (d) \(\mathrm{NO}_{3}^{-}\);(e) AsF \(_{5}\)

The hybridization scheme for the central atom includes a \(d\) orbital contribution in (a) \(\mathrm{I}_{3}^{-} ;\) (b) \(\mathrm{PCl}_{3}\) (c) \(\mathrm{NO}_{3}^{-} ;\) (d) \(\mathrm{H}_{2}\) Se.

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