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Chapter 9: Problem 47

Indicate the hybridization of the central atom in (b) \(\mathrm{AlCl}_{4}^{-}\), (c) \(\mathrm{CS}_{2}\) (d) \(\mathrm{KrF}_{2}\), (e) \(\mathrm{PF}_{6}^{-}\). (a) \(\mathrm{BCl}_{3}\),

Short Answer

Expert verified
The hybridizations of the central atoms for the given molecules are as follows: (a) BCl\(_3\): sp^2 (b) AlCl\(_4^-\): sp^3 (c) CS\(_2\): sp (d) KrF\(_2\): sp^3d (e) PF\(_6^-\): sp^3d^2

Step by step solution

01

(a) Determine the hybridization of BCl\(_3\)

Step 1: Calculate the total number of valence electrons - B has 3 valence electrons (since it belongs to Group 13) - Cl has 7 valence electrons (since it belongs to Group 17) - BCl\(_3\) has one B atom and three Cl atoms, so total valence electrons = 3 + (3 脳 7) = 24 Step 2: Determine the electron geometry - The boron atom is the central atom in BCl\(_3\) - There are three electron domains (one for each Cl atom) around the boron atom - With three electron domains, the electron geometry is trigonal planar Step 3: Determine the hybridization - In a trigonal planar electron geometry, the central atom is sp^2 hybridized - Thus, the hybridization of the central atom (boron) in BCl\(_3\) is sp^2
02

(b) Determine the hybridization of AlCl\(_4^-\)

Step 1: Calculate the total number of valence electrons - Al has 3 valence electrons (since it belongs to Group 13) - Cl has 7 valence electrons (since it belongs to Group 17) - AlCl\(_4^-\) has one Al atom, four Cl atoms, and one extra electron (due to the -1 charge) - Total valence electrons = 3 + (4 脳 7) + 1 = 32 Step 2: Determine the electron geometry - The aluminum atom is the central atom in AlCl\(_4^-\) - There are four electron domains (one for each Cl atom) around the aluminum atom - With four electron domains, the electron geometry is tetrahedral Step 3: Determine the hybridization - In a tetrahedral electron geometry, the central atom is sp^3 hybridized - Thus, the hybridization of the central atom (aluminum) in AlCl\(_4^-\) is sp^3
03

(c) Determine the hybridization of CS\(_2\)

Step 1: Calculate the total number of valence electrons - C has 4 valence electrons (since it belongs to Group 14) - S has 6 valence electrons (since it belongs to Group 16) - CS\(_2\) has one C atom and two S atoms, so total valence electrons = 4 + (2 脳 6) = 16 Step 2: Determine the electron geometry - The carbon atom is the central atom in CS\(_2\) - There are two electron domains (one for each S atom) around the carbon atom - With two electron domains, the electron geometry is linear Step 3: Determine the hybridization - In a linear electron geometry, the central atom is sp hybridized - Thus, the hybridization of the central atom (carbon) in CS\(_2\) is sp
04

(d) Determine the hybridization of KrF\(_2\)

Step 1: Calculate the total number of valence electrons - Kr has 8 valence electrons (since it belongs to Group 18) - F has 7 valence electrons (since it belongs to Group 17) - KrF\(_2\) has one Kr atom and two F atoms, so total valence electrons = 8 + (2 脳 7) = 22 Step 2: Determine the electron geometry - The krypton atom is the central atom in KrF\(_2\) - There are two electron domains (one for each F atom) around the krypton atom, and there are three lone pairs of electrons - With five electron domains, the electron geometry is trigonal bipyramidal Step 3: Determine the hybridization - In a trigonal bipyramidal electron geometry, the central atom is sp^3d hybridized - Thus, the hybridization of the central atom (krypton) in KrF\(_2\) is sp^3d
05

(e) Determine the hybridization of PF\(_6^-\)

Step 1: Calculate the total number of valence electrons - P has 5 valence electrons (since it belongs to Group 15) - F has 7 valence electrons (since it belongs to Group 17) - PF\(_6^-\) has one P atom, six F atoms, and one extra electron (due to the -1 charge) - Total valence electrons = 5 + (6 脳 7) + 1 = 48 Step 2: Determine the electron geometry - The phosphorus atom is the central atom in PF\(_6^-\) - There are six electron domains (one for each F atom) around the phosphorus atom - With six electron domains, the electron geometry is octahedral Step 3: Determine the hybridization - In an octahedral electron geometry, the central atom is sp^3d^2 hybridized - Thus, the hybridization of the central atom (phosphorus) in PF\(_6^-\) is sp^3d^2

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Key Concepts

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

Electron Geometry
Electron geometry is the three-dimensional arrangement of electron pairs around a central atom. It considers both bonding pairs of electrons and lone pairs. Understanding electron geometry is key to predicting molecular shapes because it determines how atoms and lone pairs repel each other. Common geometries include linear, trigonal planar, tetrahedral, trigonal bipyramidal, and octahedral. By knowing the number of electron domains around a central atom, you can determine the electron geometry. For instance, if an atom has four electron domains, the electron geometry is typically tetrahedral.
Valence Electrons
Valence electrons are the outermost electrons of an atom and play a vital role in chemical bonding. They are primarily involved in forming bonds with other atoms. To determine a molecule's total valence electrons, sum the valence electrons of each atom in the molecule. For example, in \( ext{BCl}_3\), boron has 3 valence electrons and each chlorine atom has 7, resulting in a total of 24 valence electrons for the molecule. Recognizing the number and arrangement of valence electrons helps predict how atoms will interact and bond.
Trigonal Planar
Trigonal planar geometry occurs when three electron domains surround a central atom, forming a flat, triangular shape with 120-degree bond angles. A common example is \( ext{BCl}_3\), where the central boron atom has three bonding pairs with chlorine atoms and no lone pairs. This arrangement leads to \( ext{sp}^2\) hybridization, wherein one s orbital and two p orbitals mix to form three equivalent hybrid orbitals. These hybrid orbitals result in a trigonal planar shape.
Tetrahedral
A tetrahedral geometry consists of four electron domains around a central atom, creating a shape with 109.5-degree bond angles. This configuration can be seen in molecules like \( ext{AlCl}_4^-\). The aluminum atom at the center has four bonding pairs with chlorine atoms. No lone pairs ensure the ideal tetrahedral structure, resulting in \( ext{sp}^3\) hybridization, where one s and three p orbitals mix to form four equivalent hybrid orbitals, accommodating the four domains.
Linear Geometry
Linear geometry features two bonds arranged at 180 degrees from each other, creating a straight-line shape. An example molecule is \( ext{CS}_2\), where the central carbon forms two double bonds with sulfur atoms. Since there are only two domains and no lone pairs, linear geometry is achieved. The central carbon exhibits \( ext{sp}\) hybridization, involving one s and one p orbital, resulting in two hybrid orbitals that align in a linear formation.
Trigonal Bipyramidal
Trigonal bipyramidal geometry involves five electron domains around a central atom. This structure includes three equatorial and two axial positions, creating 90 and 120-degree angles. A classic example is \( ext{KrF}_2\), with two bonds and three lone pairs around krypton. This arrangement leads to \( ext{sp}^3d\) hybridization, where one s, three p, and one d orbital merge to form five hybrid orbitals. The molecular shape adjusts to accommodate both bonding and non-bonding domains.
Octahedral
Octahedral geometry is seen in molecules with six electron domains around a central atom, leading to 90-degree bond angles. \( ext{PF}_6^-\) is a prime example, with the phosphorus atom bonded to six fluorine atoms. The absence of lone pairs on phosphorus ensures a perfect octahedral shape. This leads to \( ext{sp}^3d^2\) hybridization, involving one s, three p, and two d orbitals forming six hybrid orbitals to suit the six domains, balancing the spatial arrangement.

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Most popular questions from this chapter

The lactic acid molecule, \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}\), gives sour milk its unpleasant, sour taste. (a) Draw the Lewis structure for the molecule, assuming that carbon always forms four bonds in its stable compounds. (b) How many \(\pi\) and how many \(\sigma\) bonds are in the molecule? (c) Which \(\mathrm{CO}\) bond is shortest in the molecule? (d) What is the hybridization of atomic orbitals around each carbon atom associated with that short bond? (e) What are the approximate bond angles around each carbon atom in the molecule?

An \(\mathrm{AB}_{2}\) molecule is described as linear, and the \(\mathrm{A}-\mathrm{B}\) bond length is known. (a) Does this information completely describe the geometry of the molecule? (b) Can you tell how many nonbonding pairs of electrons are around the A atom from this information?

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) \(\mathrm{CH}_{3} \mathrm{~F}\) ?

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 \(S\) 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 electrondomain 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.

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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