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What is the hybridization of the central atom in (a) \(\mathrm{SiCl}_{4}\) (b) \(\mathrm{HCN},(\mathrm{c}) \mathrm{SO}_{3}\), (d) \(\mathrm{ICl}_{2}^{-}\), (e) \(\mathrm{BrF}_{4}\) ?

Short Answer

Expert verified
The hybridizations of the central atoms in the given molecules are: (a) SiCl4: \(sp^3\) (b) HCN: \(sp\) (c) SO3: \(sp^2\) (d) ICl2-: \(sp^3d\) (e) BrF4: \(sp^3d^2\)

Step by step solution

01

Identify central atoms and their bonding species

For each molecule, we first identify the central atom and the bonded atoms as follows: (a) SiCl4: Si is the central atom bonded to four Cl atoms. (b) HCN: C is the central atom bonded to one H atom and one N atom. (c) SO3: S is the central atom bonded to three O atoms. (d) ICl2-: I is the central atom bonded to two Cl atoms (with one extra electron added to the I atom due to the negative charge). (e) BrF4: Br is the central atom bonded to four F atoms.
02

Draw Lewis structures and identify electron domains

For each molecule, we draw the Lewis structure and identify the electron domains around the central atom. (a) SiCl4: Si is bonded to four Cl atoms and has no unshared electron pairs (tetrahedral geometry). (b) HCN: C forms a single bond with H and a triple bond with N. There is one sigma bond with H, and one sigma bond and two pi bonds with N (linear geometry). (c) SO3: S forms double bonds with each of the three O atoms and has no unshared electron pairs (trigonal planar geometry). (d) ICl2-: I forms single bonds with each of the two Cl atoms and has three unshared electron pairs, giving a total of five electron domains (T-shape geometry). (e) BrF4: Br forms single bonds with four F atoms and has two unshared electron pairs, giving a total of six electron domains (square-planar geometry).
03

Determine hybridization of the central atom

Based on the electron domain count around each central atom, we can now determine their hybridization. (a) SiCl4: The Si atom has four electron domains (tetrahedral geometry). Therefore, it is sp³ hybridized. (b) HCN: The C atom has two electron domains (linear geometry). Therefore, it is sp hybridized. (c) SO3: The S atom has three electron domains (trigonal planar geometry). Therefore, it is sp² hybridized. (d) ICl2-: The I atom has five electron domains (T-shape geometry). Therefore, it is sp³d hybridized. (e) BrF4: The Br atom has six electron domains (square-planar geometry). Therefore, it is sp³d² hybridized. The hybridizations of the central atoms in the given molecules are as follows: (a) SiCl4: sp³ (b) HCN: sp (c) SO3: sp² (d) ICl2-: sp³d (e) BrF4: sp³d²

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Key Concepts

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

Central Atom
In a molecule, the central atom is the main atom to which other atoms are bonded. It is usually the least electronegative element, as it can form the most bonds. For example:
  • In \(\text{SiCl}_4\), silicon (Si) is the central atom.
  • In \(\text{HCN}\), carbon (C) serves as the central atom.
  • Sulfur (S) is the central atom in \(\text{SO}_3\).
Choosing the central atom correctly is important because it influences the molecule's geometry and properties.
Electron Geometry
Electron geometry considers all electron domains around the central atom, including both bonding pairs and lone pairs. This geometry helps in determining how electrons around the central atom are organized. Some basic types include:
  • Tetrahedral: Four electron domains, as in \(\text{SiCl}_4\).
  • Trigonal planar: Three electron domains, found in \(\text{SO}_3\).
  • Linear: Two electron domains, such as in \(\text{HCN}\).
Understanding electron geometry is crucial in predicting the molecule's reactivity and interactions.
Lewis Structure
The Lewis structure is a diagram that shows the arrangement of atoms and the distribution of electrons in a molecule. It highlights the bonding between atoms and any lone electron pairs. Creating a Lewis structure involves:
  • Identifying valence electrons.
  • Drawing single, double, or triple bonds as needed.
  • Distributing remaining electrons to satisfy each atom's octet rule (where applicable).
For instance, in \(\text{SO}_3\), sulfur forms double bonds with oxygen atoms, resulting in a complete octet for each atom.
Molecular Geometry
Molecular geometry refers to the 3D arrangement of atoms in a molecule. Unlike electron geometry, it focuses only on bonded atoms. Key geometries include:
  • Square planar: Seen in \(\text{BrF}_4\), with two lone pairs and four bonded pairs.
  • T-shaped: Present in \(\text{ICl}_2^-\), with three lone pairs and two bonded pairs.
  • Linear: As in \(\text{HCN}\), where the atoms form a straight line.
Molecular geometry affects the molecule's polarity and potential reactions.
Electron Domains
Electron domains around the central atom include all areas where electrons are located, like:
  • Bonding pairs - shared between atoms.
  • Non-bonding pairs (lone pairs) - not shared.
The number of electron domains helps determine the hybridization of the central atom:
  • Four domains (such as in \(\text{SiCl}_4\)) indicate \(sp^3\) hybridization.
  • Five domains (like \(\text{ICl}_2^-\)) suggest \(sp^3d\) hybridization.
  • Six domains (for \(\text{BrF}_4\)) lead to \(sp^3d^2\) hybridization.
Identifying electron domains is essential for understanding the molecule's structure and bonding.

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Most popular questions from this chapter

Antibonding molecular orbitals can be used to make bonds to other atoms in a molecule. For example, metal atoms can use appropriate \(d\) orbitals to overlap with the \(\pi_{2 p}^{*}\) orbitals of the carbon monoxide molecule. This is called \(d-\pi\) backbonding. (a) Draw a coordinate axis system in which the \(y\) -axis is vertical in the plane of the paper and the \(x\) -axis horizontal. Write \(^{\prime} \mathrm{M}^{\prime \prime}\) at the origin to denote a metal atom. (b) Now, on the \(x\) -axis to the right of \(M\), draw the Lewis structure of a CO molecule, with the carbon nearest the \(\mathrm{M}\). The \(\mathrm{CO}\) bond axis should be on the \(x\) -axis. (c) Draw the CO \(\pi_{2 p}^{*}\) orbital, with phases (see the Closer Look box on phases) in the plane of the paper. Two lobes should be pointing toward \(\mathrm{M}\). (d) Now draw the \(\mathrm{d}_{x y}\) orbital of \(\mathrm{M}\), with phases. Can you see how they will overlap with the \(\pi_{2 p}^{*}\) orbital of CO? (e) What kind of bond is being made with the orbitals between \(\mathrm{M}\) and \(\mathrm{C}, \sigma\) or \(\pi ?\) (f) Predict what will happen to the strength of the CO bond in a metal-CO complex compared to CO alone.

(a) The \(\mathrm{PH}_{3}\) molecule is polar. How does this offer experimental proof that the molecule cannot be planar? (b) It tums out that ozone, \(\mathrm{O}_{3}\), has a small dipole moment. How is this possible, given that all the atoms are the same?

Propylene, \(\mathrm{C}_{3} \mathrm{H}_{4}\) is a gas that is used to form the important polymer called polypropylene. Its Lewis structure is (a) What is the total number of valence electrons in the propylene molecule? (b) How many valence electrons are used to make \(\sigma\) bonds in the molecule? (c) How many valence electrons are used to make \(\pi\) bonds in the molecule? (d) How many valence electrons remain in nonbonding pairs in the molecule? (e) What is the hybridization at each carbon atom in the molecule?

How many nonbonding electron pairs are there in each of the following molecules: (a) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{~S} ;\) (b) \(\mathrm{HCN}\); (c) \(\mathrm{H}_{2} \mathrm{C}_{2}\) (d) \(\mathrm{CH}_{3} \mathrm{~F}\) ?

The molecule shown here is difluaromethane \(\left(\mathrm{CH}_{2} \mathrm{~F}_{2}\right)\) which is used as a refrigerant called R-32. (a) Based on the structure, how many electron domains surround the \(C\) atom in this molecule? (b) Would the molecule have a nonzero dipole moment? (c) If the molecule is polar, in what direction will the overall dipole moment vector point in the molecule? [Sections \(9.2\) and 9.3]

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