Chemistry: The Molecular Nature of Matter and Change
Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781259631757
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
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Chapter 10, Problem 10.95P

(a)

Interpretation Introduction

Interpretation:

The distance between H atoms in C2H2 is to be calculated.

Concept introduction:

The bond length or bond distance is the average distance between the nuclei of two bonded atoms in a molecule. When two similar atoms are bonded together, half of the bond length is known as the covalent radius. The bond length depends on the number of bonded of two atoms. The unit of bond length is picometer.

In single, double and triple bonds, the order of bond length is as follows:

  Triple bond <Double bond <Single bond

(a)

Expert Solution
Check Mark

Answer to Problem 10.95P

339 pm is the distance between the two hydrogen atoms in acetylene.

Explanation of Solution

The total number of electrons in C2H2 can be calculated as follows:

  Total valence electrons(V.E)=2(V.E of C)+2(V.E of H)=2(4)+2(1)=10

So C2H2 has a total of 10 valence electrons. Out of these electrons, 6 electrons are utilized in one CC single bond and two CH single bonds and therefore 4 electrons are left behind. If these electrons are given to one carbon atom to complete its octet, the octet of another carbon atom cannot be fulfilled. So the 4 electrons act as bonding pairs for the triple bond that exists between the two carbon atoms. Therefore, the shape of C2H2 is linear.

The Lewis structure of C2H2 is as follows:

Chemistry: The Molecular Nature of Matter and Change, Chapter 10, Problem 10.95P , additional homework tip  1

The two hydrogen atoms are separated by two CH single bonds and one CC triple bond. The distance between the carbon and the hydrogen atom in CH single bond is 109pm and that between the two carbon atoms in CC triple bond is 121pm.

The total distance between H atoms in C2H2 is calculated as follows:

  Total distance=2(109pm)+121pm=218 pm+121pm=339 pm.

Conclusion

The bond length between the two atoms in a molecule depends not only on the atoms but also on other factors like orbital hybridization and the steric nature of the substituents.

(b)

Interpretation Introduction

Interpretation:

The distance between F atoms in SF6 is to be determined.

Concept introduction:

The bond length or bond distance is the average distance between the nuclei of two bonded atoms in a molecule. When two similar atoms are bonded together, half of the bond length is known as the covalent radius. The bond length depends on the number of bonded of two atoms. The unit of bond length is picometer.

In single, double and triple bonds, the order of bond length is as follows:

  Triple bond <Double bond <Single bond

(b)

Expert Solution
Check Mark

Answer to Problem 10.95P

316 pm is the distance between the opposite fluorine atoms and 223 pm is the distance between the adjacent fluorine atoms.

Explanation of Solution

The total number of electrons in SF6 can be calculated as follows:

  Total valence electrons(V.E)=V.E of S+6(V.E of F)=6+6(7)=48

So SF6 has total of 48 valence electrons. Out of these electrons, 12 electrons are utilized in six SF single bonds and 36 electrons are left behind. These electrons are given to the six fluorine atoms to complete their octets. So, the shape of SF6 is octahedral.

The Lewis structure of SF6 is as follows:

Chemistry: The Molecular Nature of Matter and Change, Chapter 10, Problem 10.95P , additional homework tip  2

The two fluorine atoms that are opposite to the sulfur atom are separated by two SF single bonds. The distance between the sulfur and the fluorine atom in SF single bond is 158pm.

The total distance between F atoms opposite to the S atom in SF6 is calculated as follows:

  Total distance=2(158pm)=316 pm

The adjacent fluorine atoms are present at the two corners of a right-angled triangle. The two sides of the triangle are the bond lengths of SF single bonds and the distance between the two adjacent fluorine atoms is the hypotenuse of the right-angled triangle.

The distance between the adjacent F atoms in SF6 is calculated by the Pythagorean Theorem as follows:

  (FFbond distance)2=(SFbond distance)2+(SFbond distance)2        (1)

Or,

  FFbond distance=(SFbond distance)2+(SFbond distance)2        (2)

Substitute 158pm for the SFbond distance in equation (2).

  FFbond distance=(158pm)2+(158pm)2=49928pm2=223.4457pm223 pm.

Conclusion

The bond length between the two atoms in a molecule depends not only on the atoms but also on other factors like orbital hybridization and the steric nature of the substituents.

(c)

Interpretation Introduction

Interpretation:

The distance between equatorial F atoms in PF5 is to be determined.

Concept introduction:

The bond length or bond distance is the average distance between the nuclei of two bonded atoms in a molecule. When two similar atoms are bonded together, half of the bond length is known as the covalent radius. The bond length depends on the number of bonded of two atoms. The unit of bond length is picometer.

In single, double and triple bonds, the order of bond length is as follows:

  Triple bond <Double bond <Single bond

(c)

Expert Solution
Check Mark

Answer to Problem 10.95P

270pm is the distance between equatorial F atoms in PF5.

Explanation of Solution

The total number of electrons in PF5 can be calculated as follows:

  Total valence electrons(V.E)=V.E of P+5(V.E of F)=5+5(7)=40

So PF5 has total of 40 valence electrons. Out of these electrons, 10 electrons are utilized in five PF single bonds and 30 electrons are left behind. These electrons are given to the five fluorine atoms to complete their octets. So, the shape of PF5 is trigonal bipyramidal.

The Lewis structure of PF5 is as follows:

Chemistry: The Molecular Nature of Matter and Change, Chapter 10, Problem 10.95P , additional homework tip  3

The adjacent fluorine atoms are present at the corners of a triangle with FPF bond angle of 120°. The two sides of the triangle are the bond lengths of PF single bonds (156pm).

The distance between the adjacent F atoms in PF5 is calculated by the Law of cosines as follows:

  (FFbond distance)2=(PFbond distance)2+(PFbond distance)22(PFbond distance)(PFbond distance)cosθ        (1)

Or,

  FFbond distance=(PFbond distance)2+(PFbond distance)22(PFbond distance)(PFbond distance)cosθ        (2)

Substitute 156pm for the PFbond distance and 120° for θ in equation (2).

  FFbond distance=(156pm)2+(156pm)22(156pm)(156pm)cos120°=24336pm2+24336pm248672pm2(0.5)=270.1999pm270 pm.

Conclusion

The bond length between the two atoms in a molecule depends not only on the atoms but also on other factors like orbital hybridization and the steric nature of the substituents.

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Chapter 10 Solutions

Chemistry: The Molecular Nature of Matter and Change

Ch. 10.2 - Prob. 10.6AFPCh. 10.2 - Prob. 10.6BFPCh. 10.2 - Prob. 10.7AFPCh. 10.2 - Prob. 10.7BFPCh. 10.2 - Prob. 10.8AFPCh. 10.2 - Prob. 10.8BFPCh. 10.3 - Prob. 10.9AFPCh. 10.3 - Prob. 10.9BFPCh. 10.3 - Prob. B10.1PCh. 10.3 - Prob. B10.2PCh. 10 - Prob. 10.1PCh. 10 - When is a resonance hybrid needed to adequately...Ch. 10 - Prob. 10.3PCh. 10 - Prob. 10.4PCh. 10 - Draw a Lewis structure for (a) SiF4; (b) SeCl2;...Ch. 10 - Draw a Lewis structure for (a) ; (b) C2F4; (c)...Ch. 10 - Prob. 10.7PCh. 10 - Prob. 10.8PCh. 10 - Prob. 10.9PCh. 10 - Draw Lewis structures of all the important...Ch. 10 - Prob. 10.11PCh. 10 - Draw Lewis structures of all the important...Ch. 10 - Prob. 10.13PCh. 10 - Prob. 10.14PCh. 10 - Draw the Lewis structure with lowest formal...Ch. 10 - Draw the Lewis structure with lowest formal...Ch. 10 - Prob. 10.17PCh. 10 - Prob. 10.18PCh. 10 - Prob. 10.19PCh. 10 - Prob. 10.20PCh. 10 - These species do not obey the octet rule. Draw a...Ch. 10 - These species do not obey the octet rule. Draw a...Ch. 10 - Molten beryllium chloride reacts with chloride ion...Ch. 10 - Prob. 10.24PCh. 10 - Prob. 10.25PCh. 10 - Phosgene is a colorless, highly toxic gas that was...Ch. 10 - If you know the formula of a molecule or ion, what...Ch. 10 - In what situation is the name of the molecular...Ch. 10 - Prob. 10.29PCh. 10 - Prob. 10.30PCh. 10 - Consider the following molecular shapes. (a) Which...Ch. 10 - Use wedge-bond perspective drawings (if necessary)...Ch. 10 - Prob. 10.33PCh. 10 - Determine the electron-group arrangement,...Ch. 10 - Determine the electron-group arrangement,...Ch. 10 - Prob. 10.36PCh. 10 - Prob. 10.37PCh. 10 - Prob. 10.38PCh. 10 - Prob. 10.39PCh. 10 - Determine the shape, ideal bond angle(s), and the...Ch. 10 - Prob. 10.41PCh. 10 - Determine the shape around each central atom in...Ch. 10 - Prob. 10.43PCh. 10 - Prob. 10.44PCh. 10 - Prob. 10.45PCh. 10 - Prob. 10.46PCh. 10 - Arrange the following ACln species in order of...Ch. 10 - State an ideal value for each of the bond angles...Ch. 10 - Prob. 10.49PCh. 10 - Prob. 10.50PCh. 10 - Prob. 10.51PCh. 10 - Prob. 10.52PCh. 10 - How can a molecule with polar covalent bonds not...Ch. 10 - Prob. 10.54PCh. 10 - Consider the molecules SCl2, F2, CS2, CF4, and...Ch. 10 - Consider the molecules BF3, PF3, BrF3, SF4, and...Ch. 10 - Prob. 10.57PCh. 10 - Prob. 10.58PCh. 10 - Prob. 10.59PCh. 10 - Prob. 10.60PCh. 10 - Prob. 10.61PCh. 10 - Prob. 10.62PCh. 10 - Prob. 10.63PCh. 10 - Prob. 10.64PCh. 10 - Prob. 10.65PCh. 10 - Prob. 10.66PCh. 10 - When SO3 gains two electrons, forms. (a) Which...Ch. 10 - The actual bond angle in NO2 is 134.3°, and in it...Ch. 10 - Prob. 10.69PCh. 10 - Propylene oxide is used to make many products,...Ch. 10 - Prob. 10.71PCh. 10 - Prob. 10.72PCh. 10 - Prob. 10.73PCh. 10 - Prob. 10.74PCh. 10 - Prob. 10.75PCh. 10 - Prob. 10.76PCh. 10 - Prob. 10.77PCh. 10 - A gaseous compound has a composition by mass of...Ch. 10 - Prob. 10.79PCh. 10 - Prob. 10.80PCh. 10 - Prob. 10.81PCh. 10 - Prob. 10.82PCh. 10 - Pure HN3 (atom sequence HNNN) is explosive. In...Ch. 10 - Prob. 10.84PCh. 10 - Prob. 10.85PCh. 10 - Oxalic acid (H2C2O4) is found in toxic...Ch. 10 - Prob. 10.87PCh. 10 - Hydrazine (N2H4) is used as a rocket fuel because...Ch. 10 - Prob. 10.89PCh. 10 - Prob. 10.90PCh. 10 - Prob. 10.91PCh. 10 - Consider the following molecular shapes: Match...Ch. 10 - Prob. 10.93PCh. 10 - Prob. 10.94PCh. 10 - Prob. 10.95PCh. 10 - Phosphorus pentachloride, a key industrial...
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