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Chapter 9: Problem 48

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 sp鲁 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 sp虏 hybridized. (d) ICl2-: The I atom has five electron domains (T-shape geometry). Therefore, it is sp鲁d hybridized. (e) BrF4: The Br atom has six electron domains (square-planar geometry). Therefore, it is sp鲁d虏 hybridized. The hybridizations of the central atoms in the given molecules are as follows: (a) SiCl4: sp鲁 (b) HCN: sp (c) SO3: sp虏 (d) ICl2-: sp鲁d (e) BrF4: sp鲁d虏

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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

The phosphorus trihalides \(\left(\mathrm{PX}_{3}\right)\) show the following variation in the bond angle \(\mathrm{X}-\mathrm{P}-\mathrm{X}: \mathrm{PF}_{3}, 96.3^{\mathrm{a}} ; \mathrm{PCl}_{3}\) \(100.3^{\circ} ; \mathrm{PBr}_{3}, 101.0^{\circ} ; \mathrm{Pl}_{3}, 102.0^{\circ} .\) The trend is generally at- tributed to the change in the electronegativity of the halogen. (a) Assuming that all electron domains are the same size, what value of the \(\mathrm{X}-\mathrm{P}-\mathrm{X}\) angle is predicted by the VSEPR model? (b) What is the general trend in the \(\mathrm{X}-\mathrm{P}-\mathrm{X}\) angle as the electronegativity increases? (c) Using the VSEPR model, explain the observed trend in \(X-P-X\) angle as the electronegativity of \(X\) changes. (d) Based on your answer to part (c), predict the structure of \(\mathrm{PBrCI}_{4}\)

The \(\mathrm{PF}_{3}\) molecule has a dipole moment of \(1.03 \mathrm{D}\), but \(\mathrm{BF}_{3}\) has a dipole moment of zero. How can you explain the difference?

Write the electron configuration for the first excited state for \(\mathrm{N}_{2}\) - that is, the state with the highest-energy electron moved to the next available energy level. (a) Is the nitrogen in its first excited state diamagnetic or paramagnetic? (b) Is the \(\mathrm{N}-\mathrm{N}\) bond strength in the first excited state stronger or weaker compared to that in the ground state? Explain.

(a) Draw a picture showing how two \(p\) orbitals on two different atoms can be combined to make a sigma bond. (b) Sketch a \(\pi\) bond that is constructed from \(p\) orbitals. (c) Which is generally stronger, a \(\sigma\) bond or a \(\pi\) bond? Explain. (d) Can two s orbitals make a \(\pi\) bond? Explain.

The reaction of three molecules of fluorine gas with a Xe atom produces the substance xenon hexafluoride, \(\mathrm{XeF}_{6}\) : $$ \mathrm{Xe}(g)+3 \mathrm{~F}_{2}(g) \longrightarrow \mathrm{XeF}_{6}(s) $$ (a) Draw a Lewis structure for \(\mathrm{XeF}_{6}\). (b) If you try to use the VSEPR model to predict the molecular geometry of \(\mathrm{XeF}_{6 r}\) you run into a problem. What is it? (c) What could you do to resolve the difficulty in part (b)? (d) Suggest a hybridization scheme for the Xe atom in \(\mathrm{XeF}_{6}\). (e) The molecule \(\mathrm{IF}_{7}\) has a pentagonal- bipyramidal structure (five equatorial fluorine atoms at the vertices of a regular pentagon and two axial fluorine atoms). Based on the structure of \(\mathrm{IF}_{7}\), suggest a structure for \(\mathrm{XeF}_{6}\)
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