Loose Leaf for Chemistry: The Molecular Nature of Matter and Change
Loose Leaf for Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781260151749
Author: Silberberg Dr., Martin; Amateis Professor, Patricia
Publisher: McGraw-Hill Education
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Chapter 11, Problem 11.1P

(a)

Interpretation Introduction

Interpretation:

The hybridization of the central atom that corresponds to a trigonal planar arrangement is to be determined.

Concept introduction:

The atomic orbital is the wave function that is used to find the probability to find an electron around the nucleus of an atom. It is the space around the nucleus of an atom where the electrons are supposed to be found.

Hybridization is the process of intermixing of atomic orbital of slightly different energies to form hybrid orbitals that have similar energy. These orbital have lower energy and more stability than the atomic orbital.

(a)

Expert Solution
Check Mark

Answer to Problem 11.1P

The hybridization of the central atom in trigonal planar is sp2.

Explanation of Solution

The molecule that has trigonal planar geometry is BF3. The atomic number of boron is 5 so its electronic configuration is 1s22s22p1.

The partial orbital diagram of an isolated B atom is as follows:

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change, Chapter 11, Problem 11.1P , additional homework tip  1

The partial orbital diagram of a hybridized B atom is as follows:

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change, Chapter 11, Problem 11.1P , additional homework tip  2

The half-filled sp2 orbitals are filled with three electrons of three fluorine atoms. So the hybridization of B in BF3 is sp2.

Conclusion

The hybridization of the central atom in trigonal planar is sp2.

(b)

Interpretation Introduction

Interpretation:

The hybridization of the central atom that corresponds to the octahedral arrangement is to be determined.

Concept introduction:

The atomic orbital is the wave function that is used to find the probability to find an electron around the nucleus of an atom. It is the space around the nucleus of an atom where the electrons are supposed to be found.

Hybridization is the process of intermixing of atomic orbital of slightly different energies to form hybrid orbitals that have similar energy. These orbital have lower energy and more stability than the atomic orbital.

(b)

Expert Solution
Check Mark

Answer to Problem 11.1P

The hybridization of the central atom in octahedral is sp3d2.

Explanation of Solution

The molecule that has octahedral geometry is SF6. The atomic number of sulfur is 16 so its electronic configuration is [Ne]3s23p4.

The partial orbital diagram of an isolated B atom is as follows:

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change, Chapter 11, Problem 11.1P , additional homework tip  3

The partial orbital diagram of a hybridized B atom is as follows:

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change, Chapter 11, Problem 11.1P , additional homework tip  4

The six half-filled sp3d2 orbitals are filled with six electrons of six fluorine atoms. So the hybridization of S in SF6 is sp3d2.

Conclusion

The hybridization of the central atom in octahedral is sp3d2.

(c)

Interpretation Introduction

Interpretation:

The hybridization of the central atom that corresponds to a linear arrangement is to be determined.

Concept introduction:

The atomic orbital is the wave function that is used to find the probability to find an electron around the nucleus of an atom. It is the space around the nucleus of an atom where the electrons are supposed to be found.

Hybridization is the process of intermixing of atomic orbital of slightly different energies to form hybrid orbitals that have similar energy. These orbital have lower energy and more stability than the atomic orbital.

(c)

Expert Solution
Check Mark

Answer to Problem 11.1P

The hybridization of the central atom in a linear arrangement is sp.

Explanation of Solution

The molecule that has a linear arrangement is BeCl2. The atomic number of beryllium is 4 so its electronic configuration is 1s22s2.

The partial orbital diagram of an isolated Be atom is as follows:

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change, Chapter 11, Problem 11.1P , additional homework tip  5

The partial orbital diagram of a hybridized Be atom is as follows:

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change, Chapter 11, Problem 11.1P , additional homework tip  6

The two half-filled sp orbitals are filled with two electrons of two chlorine atoms. So the hybridization of Be in BeCl2 is sp.

Conclusion

The hybridization of the central atom in a linear arrangement is sp.

(d)

Interpretation Introduction

Interpretation:

The hybridization of the central atom that corresponds to a tetrahedral arrangement is to be determined.

Concept introduction:

The atomic orbital is the wave function that is used to find the probability to find an electron around the nucleus of an atom. It is the space around the nucleus of an atom where the electrons are supposed to be found.

Hybridization is the process of intermixing of atomic orbital of slightly different energies to form hybrid orbitals that have similar energy. These orbital have lower energy and more stability than the atomic orbital.

(d)

Expert Solution
Check Mark

Answer to Problem 11.1P

The hybridization of the central atom in a tetrahedral arrangement is sp3.

Explanation of Solution

The molecule that has a tetrahedral arrangement is CH4. The atomic number of carbon is 6 so its electronic configuration is 1s22s22p2.

The partial orbital diagram of an isolated C atom is as follows:

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change, Chapter 11, Problem 11.1P , additional homework tip  7

The partial orbital diagram of a hybridized C atom is as follows:

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change, Chapter 11, Problem 11.1P , additional homework tip  8

The four half-filled sp3 orbitals are filled with four electrons of four hydrogen atoms. So the hybridization of C in CH4 is sp3.

Conclusion

The hybridization of the central atom in a tetrahedral arrangement is sp3.

(e)

Interpretation Introduction

Interpretation:

The hybridization of the central atom that corresponds to a trigonal bipyramidal arrangement is to be determined.

Concept introduction:

The atomic orbital is the wave function that is used to find the probability to find an electron around the nucleus of an atom. It is the space around the nucleus of an atom where the electrons are supposed to be found.

Hybridization is the process of intermixing of atomic orbital of slightly different energies to form hybrid orbitals that have similar energy. These orbital have lower energy and more stability than the atomic orbital.

(e)

Expert Solution
Check Mark

Answer to Problem 11.1P

The hybridization of the central atom in a trigonal bipyramidal arrangement is sp3d.

Explanation of Solution

The molecule that has a trigonal bipyramidal arrangement is PCl5. The atomic number of phosphorus is 15 so its electronic configuration is [Ne]3s23p3.

The partial orbital diagram of an isolated P atom is as follows:

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change, Chapter 11, Problem 11.1P , additional homework tip  9

The partial orbital diagram of a hybridized P atom is as follows:

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change, Chapter 11, Problem 11.1P , additional homework tip  10

The five half-filled sp3d orbitals are filled with five electrons of five chlorine atoms. So the hybridization of P in PCl5 is sp3d.

Conclusion

The hybridization of the central atom in a trigonal bipyramidal arrangement is sp3d.

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Chapter 11 Solutions

Loose Leaf for Chemistry: The Molecular Nature of Matter and Change

Ch. 11 - Prob. 11.3PCh. 11 - Prob. 11.4PCh. 11 - Prob. 11.5PCh. 11 - Give the number and type of hybrid orbital that...Ch. 11 - What is the hybridization of nitrogen in each of...Ch. 11 - What is the hybridization of carbon in each of the...Ch. 11 - Prob. 11.9PCh. 11 - Prob. 11.10PCh. 11 - Prob. 11.11PCh. 11 - Prob. 11.12PCh. 11 - Phosphine (PH3) reacts with borane (BH3) as...Ch. 11 - The illustrations below depict differences in...Ch. 11 - Use partial orbital diagrams to show how the...Ch. 11 - Use partial orbital diagrams to show how the...Ch. 11 - Prob. 11.17PCh. 11 - Prob. 11.18PCh. 11 - Methyl isocyanate, , is an intermediate in the...Ch. 11 - Are these statements true or false? Correct any...Ch. 11 - Prob. 11.21PCh. 11 - Identify the hybrid orbitals used by the central...Ch. 11 - Prob. 11.23PCh. 11 - Identify the hybrid orbitals used by the central...Ch. 11 - Prob. 11.25PCh. 11 - Prob. 11.26PCh. 11 - Certain atomic orbitals on two atoms were combined...Ch. 11 - Prob. 11.28PCh. 11 - Antibonding MOs always have at least one node. Can...Ch. 11 - Prob. 11.30PCh. 11 - Prob. 11.31PCh. 11 - The molecular orbitals depicted are derived from...Ch. 11 - The molecular orbitals depicted below are derived...Ch. 11 - Prob. 11.34PCh. 11 - Use an MO diagram and the bond order you obtain...Ch. 11 - Prob. 11.36PCh. 11 - Prob. 11.37PCh. 11 - Prob. 11.38PCh. 11 - Prob. 11.39PCh. 11 - Epinephrine (or adrenaline; below) is a naturally...Ch. 11 - Prob. 11.41PCh. 11 - Isoniazid (below) is an antibacterial agent that...Ch. 11 - Prob. 11.43PCh. 11 - Prob. 11.44PCh. 11 - Prob. 11.45PCh. 11 - Prob. 11.46PCh. 11 - Tryptophan is one of the amino acids found in...Ch. 11 - Prob. 11.48PCh. 11 - Prob. 11.49PCh. 11 - Prob. 11.50PCh. 11 - Prob. 11.51PCh. 11 - Prob. 11.52PCh. 11 - Sulfur forms oxides, oxoanions, and halides. What...Ch. 11 - Prob. 11.54PCh. 11 - Use an MO diagram to find the bond order and...Ch. 11 - Acetylsalicylic acid (aspirin), the most widely...Ch. 11 - Prob. 11.57P
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