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What hybridization do you expect for the atom that is underlined in each of the following species? (a) \(\underline{\mathrm{I}} \mathrm{O}_{2}^{-} ;(\mathbf{b}) \underline{\mathrm{N}} \mathrm{H}_{4}^{+} ;(\mathbf{c}) \mathrm{SC} \mathrm{N}^{-}\) (d) \(\underline{\mathrm{Br}} \mathrm{Cl}_{3}\)

Short Answer

Expert verified
Iodine: \( sp^2 \), Nitrogen: \( sp^3 \), Carbon: \( sp \), Bromine: \( sp^3d \)

Step by step solution

01

Understanding Hybridization

Hybridization is a concept used to explain the shape and bonding of molecules. It involves mixing atomic orbitals to form new hybrid orbitals. The type of hybridization depends on the number of regions of electron density surrounding the atom (sum of bonded atoms and lone pairs). Common types include: \( sp \), \( sp^2 \), and \( sp^3 \).
02

Hybridization of \( \underline{\mathrm{I}} \mathrm{O}_{2}^{-} \)

Iodine in \( \mathrm{I} \mathrm{O}_{2}^{-} \) has three regions of electron density: two oxygens (each as a bonding partner) and one lone pair. This corresponds to \( sp^2 \) hybridization, with the shape being trigonal planar.
03

Hybridization of \( \underline{\mathrm{N}} \mathrm{H}_{4}^{+} \)

Nitrogen in \( \mathrm{NH}_4^+ \) has four hydrogen atoms bonded to it and no lone pairs. This is typical for \( sp^3 \) hybridization, resulting in a tetrahedral shape.
04

Hybridization of \( \underline{\mathrm{C}} \) in \( \mathrm{SCN}^{-} \)

Carbon in \( \mathrm{SCN}^{-} \) is bonded to sulfur and nitrogen and has no lone pairs. Here, carbon uses \( sp \) hybridization due to two regions of electron density, and the geometry is linear.
05

Hybridization of \( \underline{\mathrm{Br}} \mathrm{Cl}_{3} \)

Bromine in \( \mathrm{BrCl}_3 \) has three bonding pairs with chlorine and two lone pairs of electrons. This adds up to five electron density regions, indicating \( sp^3d \) hybridization, which forms a trigonal bipyramidal shape.

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Key Concepts

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

Molecular Geometry
Molecular geometry is the three-dimensional shape of a molecule. It is defined by the relative positions of the atomic nuclei in a molecule. The geometry is crucial because it affects the properties and reactivity of the molecule.

Molecular geometry is determined by the arrangement of electron pairs around a central atom. This includes both bonding pairs of electrons, which form bonds between atoms, and lone pairs, which are not involved in bonding but still influence the shape.
  • Predicting geometry involves considering the number of electron pairs and using the VSEPR (Valence Shell Electron Pair Repulsion) theory.
  • The theory states that electron pairs will position themselves as far apart as possible to minimize repulsion.
  • For instance, if a central atom has four electron pairs, they arrange themselves in a shape called tetrahedral.
  • Different numbers of electron pairs result in different basic geometries like linear, trigonal planar, and trigonal bipyramidal.
Each geometry has specific angles between the bonds, which impact the molecule's properties. Understanding these shapes helps to predict how molecules might interact with each other or with other substances.
Electron Density
Electron density refers to the distribution of electrons around an atom or molecule. It is an essential factor in determining a molecule’s chemical properties, especially its reactivity and interactions with other molecules.

The regions of electron density include bonded atoms and lone pairs around the central atom. The total number of these regions often dictates the molecule's hybridization and geometry.
  • A single bond, double bond, and even a lone pair count as one region of electron density each.
  • For example, in the molecule \(\mathrm{I} \mathrm{O}_{2}^{-}\), iodine is surrounded by two single bonds to oxygens and one lone pair, totaling three regions.
  • This determines the type of hybridization and the geometric shape (such as trigonal planar for iodine).
The concept is crucial for understanding the shape and bond angles in molecules. It helps predict how the electrons and molecules will interact during chemical reactions.
Hybrid Orbitals
Hybrid orbitals are the result of combining standard atomic orbitals, such as \(s\) and \(p\) orbitals, to form new orbitals that can form covalent bonds in molecules.

This concept helps explain the observed molecular shapes and bond angles that are not easily described using simple atomic orbitals.
  • For example, \(sp^3\) hybridization, seen in the molecule \(\mathrm{NH}_4^+\), involves one \(s\) and three \(p\) orbitals combining to form four equivalent orbitals that create a tetrahedral shape.
  • The energy and shape of these hybrid orbitals make it possible for atoms to form specific angles, which lead to the most stable, energy-efficient structure.
Hybridization can vary depending on the number of electron density regions:
  • With two regions, \(sp\) hybridization results, usually creating a linear shape, as seen in the carbon atom of \(\mathrm{SCN}^{-}\).
  • Three regions contribute to \(sp^2\) hybridization, leading to planar structures like \(\mathrm{I} \mathrm{O}_{2}^{-}\).
  • Four regions involve \(sp^3\) hybrid orbitals, producing a tetrahedral geometry.
  • For five regions, \(sp^3d\) hybridization occurs, forming geometries like trigonal bipyramidal, shown in \(\mathrm{BrCl}_3\).
Understanding hybrid orbitals allows for deeper insight into molecular structures and how atoms bond within a molecule.

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Most popular questions from this chapter

Give the electron-domain and molecular geometries of a molecule that has the following electron domains on its central atom: (a) four bonding domains and no nonbonding domains, (b) three bonding domains and two nonbonding domains, (c) five bonding domains and one nonbonding domain, (d) four bonding domains and two nonbonding domains.

For each statement, indicate whether it is true or false. (a) In order to make a covalent bond, the orbitals on each atom in the bond must overlap. (b) A \(p\) orbital on one atom cannot make a bond to an \(s\) orbital on another atom. (c) Lone pairs of electrons on an atom in a molecule influence the shape of a molecule. \((\mathbf{d})\) The \(1 s\) orbital has a nodal plane. \((\mathbf{e})\) The \(2 p\) orbital has a nodal plane.

In the sulphate ion, \(\mathrm{SO}_{4}{ }^{2-}\), the sulphur atom is the central atom with the other 4 oxygen atoms attached to it. (a) Draw a Lewis structure for the sulphate ion. (b) What hybridization is exhibited by the \(S\) atom? (c) Are there multiple equivalent resonance structures for the ion? (d) How many electrons are in the \(\pi\) system of the ion?

An \(\mathrm{AB}_{2}\) molecule is described as having a tetrahedral geometry. (a) How many nonbonding domains are on atom A? (b) Based on the information given, which of the following is the molecular geometry of the molecule: (i) linear, (ii) bent, (iii) trigonal planar, or (iv) tetrahedral?

Consider a molecule with formula \(\mathrm{AX}_{2}\). Supposing the \(\mathrm{A}-\mathrm{X}\) bond is polar, how would you expect the dipole moment of the \(\mathrm{AX}_{2}\) molecule to change as the \(\mathrm{X}-\mathrm{A}-\mathrm{X}\) bond angle decreases from \(180^{\circ}\) to \(100^{\circ}\) ?

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