CHEM 123 Sapling Learning Chapter 11

A 0.08410 g sample of gas occupies 10.0 mL at 293.0 K and 1.10 atm. Upon further analysis, the compound is found to be 13.068% C and 86.932% Br. What is the molecular formula of the compound? molecular formula: Draw the Lewis structure of the compound. Identify the geometry around each carbon atom. Is this compound polar?

C_2X_2 linear The compound is nonpolar. Solution To find the molecular formula, first convert each percentage to grams, assuming 100 g of the compound. This gives 3.3 g H, 19.3 g C, and 77.4 g O. Now, convert the masses of H, C, and O to moles by dividing by the molar masses of these elements. 3.3 g H×1 mol/1.008 g H=3.3 mol H 19.3 g C×1 mol/12.01 g C=1.61 mol C 77.4 g O×1 mol/15.999 g O=4.84 mol O This gives a formula H3.27C1.61O4.84. A molecular formula should be expressed as a ratio of whole numbers. To obtain the number of moles of H, C, and O in whole numbers, divide each mole value by the smallest number of moles calculated. Then, round to the nearest whole number. H=3.3/1.61=2.05≈2 C=1.61/1.61=1 O=4.84/1.61=3.01≈3 The empirical formula for the compound is H2C1O3. To determine if this is also the molecular formula, determine the molar mass of this empirical formula. H=2 mol×1.008 g/mol=2.016 g C=1 mol×12.01 g/mol=12.01 g O=3 mol×15.999 g/mol=48.00 g 2.016+12.00+48.00=62.02 g Since the molar mass of the compound is approximately 60 g/mol, the empirical formula is also the molecular formula for this compound. In the Lewis structure, the carbon atom is the central atom because it is the least electronegative atom that is not hydrogen. All three of the oxygen atoms are bonded to the carbon atom. The two hydrogen atoms are each bonded to oxygen atoms. To determine the number of valence electrons for a compound, find the sum of the valence electrons for each neutral atom in the compound. For H2CO3, the hydrogen atoms each have 1 valence electron, the carbon atom has 4 valence electrons, and each oxygen atom has 6 valence electrons, for a total of 24 valence electrons. Start by drawing single bonds between the carbon and each oxygen atom, and between two of the oxygen atoms and a hydrogen atom. This accounts for 10 of the valence electrons. Then, add lone pairs of atoms around each atom until each atom has a full octet. The oxygens bonded to hydrogen atoms will each get two lone pairs of electrons. This brings the count up to 18 valence electrons, leaving 6 valence electrons left. If these 6 valence electrons are added to the remaining oxygen atom, all of the valence electrons will be used up, but the carbon atom will not have a full octet. In this case, remove one of the lone pairs from the third oxygen atom and use them to make a double bond to the carbon atom. Now, all og the atoms have a full octet and all the valence electrons are used. A central carbon atom is bonded to two oxygen atoms through single bonds and one oxygen atom through a double bond. There are two lone pairs on each oxygen atom. The single bonded oxygen atoms are bonded to a hydrogen aotm. This means the central carbon atom has three electron groups around it. All of these groups are atoms, so the geometry around the carbon atom is trigonal planar. Since the central carbon atom is attached to three atoms, and participates in one double bond, the hybridization around this atom is sp2. The leftover 𝑝 orbital is used to make the double bond to the oxygen atom. A σ bond results from the end‑on overlap of orbitals. A σ bond is present in all covalent bonds. A π bond results from the side‑on overlap of orbitals. They are present in double and triple bonds. Overall, this molecule has 5 σ bonds connecting the atoms and 1 π bond in the double bond.

A Lewis structure is a two-dimensional representation of a molecule that does not necessarily show what shape that molecule would take in three dimensions. From the given Lewis structure and what you know about VSEPR theory, identify the shape of the molecule. PICTURED: A molecule with atom Y single bonded with 2 X substituents. The Y atom has one lone pair.

bent Solution The way that the atoms are arranged in the Lewis structure may not match the true shape of the molecule. When determining the shape of the molecule, consider only that the molecule has two outer atoms and one lone pair, and ignore the shape suggested by the Lewis structure. For a molecule with two outer atoms and one lone pair, you would expect a bent geometry with approximate bond angles of 120 degrees. Note that because a lone pair is present on the central atom, the actual bond angle will be slightly less than 120 degrees.

The molecule 2-butene is able to undergo a process called cis-trans isomerization, where the molecule switches from being a cis-alkene to a trans-alkene. This transformation can be induced by light. The molecule on the left side of the reaction arrow is composed of a four carbon chain. The two end carbons each have three hydrogens attached. The central two carbons are connected with a double bond. The two C H 3 groups are on the same side of the double bond. The reaction arrow is reversible and has light written above it. The molecule on the right side of the reaction arrow is composed of a four carbon chain. The two end carbons each have three hydrogens attached. The central two carbons are connected with a double bond. The two C H 3 groups are on opposite sides of the double bond. What is the hybridization of the two central carbon atoms in 2-butene? The isomerization requires breaking the π bond. Use the table of bond energies to determine the approximate amount of energy (in joules) required to break the C−C π bond in 2-butene, both per mole and per molecule. What is the wavelength of light that would be required to perform the cis-trans isomerization of one molecule of 2-butene?

𝑠𝑝2 2.61×10^5 3.44*10^-19 458 VISIBLE Solution The two central carbon atoms in 2-butene are the carbon atoms in the double bond. The hybridization of the carbon atoms can be determined from the number of electron groups (bonded groups and lone pairs) around the atoms. Since there are three electron groups around each carbon atom, the hybridization is 𝑠𝑝2. In the table of bond energies, the energies listed are to fully break each bond, whether a single or double bond. When 2-butene isomerizes, only the π bond in the double bond breaks. Therefore, to calculate the energy needed to break the C−C π bond of 2-butene, you need to subtract the energy of breaking a C−C single bond, 350 kJ/mol, from the energy of breaking a C−C double bond, 611 kJ/mol. 611 kJ/mol−350 kJ/mol=261 kJ/mol Convert the energy to joules per mole using the conversion 1000 J=1 kJ. 261 kJ/1 mol×1000 J/1 kJ=2.61×105 J/mol To determine the energy in joules per molecule, use Avogadro's number to convert moles into molecules. 2.61×105 J/1 mol×1 mol/6.022×1023 molecules=4.33×10−19 J/molecule Next, calculate the wavelength of light that has enough energy to break the π bond in 1 molecule of 2-butene. The equation that relates energy (𝐸) to wavelength (𝜆) is 𝐸=ℎ𝑐𝜆 where ℎ is Plank's constant, 6.626×10−34 J·s, and 𝑐 is the speed of light, 2.998×108 m/s. Rearrange the equation to solve for wavelength. 𝜆=ℎ𝑐𝐸 𝜆=(6.626×10−34 J·s)(2.998×108m/s)/4.33×10−19J=4.58×10−7 m A wavelength of 4.58×10−7m is equal to a wavelength of 458 nm. Looking at the electromagnetic spectrum, a wavelength of 458 nm falls in the visible light region.

What is the ground state electron configuration for O−2? What is the bond order for O−2? bond order:

(𝜎1𝑠)2(𝜎1𝑠*)2(𝜎2𝑠)2(𝜎2𝑠*)2(𝜎2𝑝)2(𝜋2𝑝)4(𝜋2𝑝*)3 1.5

Select the correct value for the indicated bond angle in each of the compounds. O−S−O angle of SO2 O−S−O angle of SO3 Cl−S−Cl angle of SCl2 Cl−Be−Cl angle of BeCl2 F−P−F angle of PF3 Cl−Si−Cl angle of SiCl4

<120° 120° <109.5° 180° <109.5° 109.5°

Fill in the molecular orbitals in the molecular orbital diagram for CO. One 2 s and three 2 p orbitals from carbon and one 2 s and three 2 p orbitals combine to form eight molecular orbitals in C O. The molecular orbitals in order of increasing energy are one sigma 2 s, one sigma 2 s star, two pi 2 p, one sigma 2 p, two pi 2 p star, and one sigma 2 p star.

ALL BOTTOM 4 ROWS HAVE EQUILIBRIUM ARROWS Solution Start by determining the number of valence electrons that each atom has. A carbon atom has six electrons, four of which are considered valence, and an oxygen atom has eight electrons, six of which are considered valence. The total number of valence electrons is total valence electrons=4+6=10 e− Following the Aufbau and Pauli exclusion principles, begin filling each orbital from the lowest to the highest energy. Each MO can hold up to two electrons, and because there is a total of 10 electrons, the first five MOs will be filled with paired sets of electrons.

My Attempt Classify these diatomic molecules as diamagnetic or paramagnetic. Diamagnetic Paramagnetic

C2 AND N2 AND F2 B2 AND O2

Open the Molecule Shapes interactive and select Real Molecules. Check the box to Show Bond Angles. Classify each molecule by whether its real bond angles are the same as or different than its model (ideal) bond angles. In other words, do the bond angles change when you switch between Real and Model mode at the top of the page? a) Same (angles do not change) b) Different (angles change)

a) CO2 XeF2 BF3 CH4 XeF4 PCL5 SF6 b) H20 SO2 ClF3 NH3 SF4 BrF5

Construct the molecular orbital diagram for H+2 . A 1 s orbital from an H atom and a 1 s orbital from an H plus cation combine to form the molecular orbitals sigma 1 s and sigma 1 s star for the molecular H 2 plus. Sigma 1 s is lower in energy than the atomic orbitals and sigma 1 s star is higher in energy than the atomic orbitals. Identify the bond order.

0.5 Solution Start by determining the number of electrons that each atom or ion has. A hydrogen atom has one electron (1𝑠1) and a hydrogen cation has zero electrons. Thus, there is 1 electron total in the molecule H+2 . The electron occupies the lowest energy molecular orbital, 𝜎1𝑠 , which can hold two electrons. The bond order is determined using the equation bond order=12(bonding electrons−antibonding electrons) There is one electron in the bonding molecular orbital (𝜎1𝑠) and zero electrons in the antibonding molecular orbital (𝜎∗1𝑠) . bond order=12(1−0)=0.5

A Lewis structure is a two-dimensional representation of a molecule that does not necessarily show what shape that molecule would take in three dimensions. Based on the Lewis structure and your knowledge of VSEPR theory, approximate the smallest bond angle in this molecule. A molecule with atom Y single bonded with 2 X substituents. The Y atom has two lone pairs. smallest bond angle:

109.5∘ Solution The way that the atoms are arranged in the Lewis structure may not match the true shape of the molecule. Consider only that the molecule has two outer atoms and two lone pairs, and ignore the shape suggested by the Lewis structure. For a molecule with two outer atoms and two lone pairs, you would expect a bent geometry with approximate bond angles of 109.5∘. Note that since lone pairs are present on the central atom, the actual bond angle will be slightly less than 109.5∘.

Determine the bond order from the molecular electron configurations. (𝜎1𝑠)2(𝜎1𝑠*)2(𝜎2𝑠)2(𝜎2𝑠*)2(𝜋2𝑝)4 bond order: (𝜎1𝑠)2(𝜎1𝑠*)2(𝜎2𝑠)2(𝜎2𝑠*)2(𝜎2𝑝)2(𝜋2𝑝)4(𝜋2𝑝*)4 bond order:

2 1 Solution To calculate bond order, determine the number of bonding electrons and antibonding electrons from the molecular electron configuration. The Greek symbols sigma (𝜎) and pi (𝜋) describe molecular orbitals in the molecular electron configuration. If there is an asterisk (*) after the molecular orbital, it indicates that it is an antibonding orbital. If there is no asterisk after the molecular orbital, it indicates that it is a bonding orbital. The number of electrons in a molecular orbital is shown as a superscript after the parentheses around each orbital. Once the total number of bonding and antibonding electrons have been determined, calculate the bond order using the equation bond order=number of bonding electrons−number of antibonding electrons2 The molecular electron configuration of (𝜎1𝑠)2(𝜎1𝑠*)2(𝜎2𝑠)2(𝜎2𝑠*)2(𝜋2𝑝)4 has eight bonding electrons and four antibonding electrons. Therefore, it has a bond order of bond order= (8−4)/2=2 The molecular electron configuration of (𝜎1𝑠)2(𝜎1𝑠*)2(𝜎2𝑠)2(𝜎2𝑠*)2(𝜎2𝑝)2(𝜋2𝑝)4(𝜋2𝑝*)4 has 10 bonding electrons and 8 antibonding electrons. Therefore, it has a bond order of bond order=(10−8)/2=1

Determine the total number of valence electrons in bromine pentafloride, BrF5. total number of valence electrons: Identify the molecular geometry of BrF5. What are the approximate bond angles in BrF5? A BrF5 molecule is

42 electrons square pyramidal 90 degrees polar. Solution Begin by counting the number of valence electrons in BrF5. Fluorine and bromine atoms each have seven valence electrons. Thus, a BrF5 molecule has a total of 42 valence electrons, 7+7(5)=42, as shown in the Lewis structure of BrF5. PICTURED: Five fluorine atoms are single bonded to one central bromine atom. Each fluorine atom has three non-bonding lone pairs of electrons. The bromine atom has one non-bonding lone pair of electrons. Next, compare the electron groups surrounding the central atom to identify the molecular geometry of BrF5. Of the six electron groups, five are bonding and one is a nonbonding lone pair of electrons which produces square pyramidal molecular geometry. In square pyramidal geometry, four bonding electron groups form the square plane around the central atom, whereas the fifth bonding group lies above the plane to form the top of the pyramid. Connecting any two bonding groups through the cental atom forms a right triangle. Thus, square planar molecules have bond angles of approximately 90 degrees. Finally, the polarity of BrF5 depends on the molecular geometry and dipole moments of each Br−F bond. Each Br−F bond is polar because the electronegativity of fluorine of the Br−F bonds that form the square plane will cancel each other out because they are equivalent in magnitude, but opposite in direction. The fifth Br−F dipole moment is not canceled because it is opposite the nonbonding lone pair of electrons. Consequently, a BrF5 molecule is polar.

Escitalopram is a drug used to treat depression and anxiety. PICTURED: The molecule escitalopram contains 40 single bonds, six double bonds, and one triple bond. Determine the number of σ and π bonds in the molecule. σ bonds: π bonds:

47 8 Solution A σ bond results from the end‑on overlap of orbitals. A σ bond is present in all covalent bonds. A π bond results from the side‑on overlap of orbitals. They are present in double and triple bonds. A single bond consists of one σ bond. A double bond consists of one σ bond and one π bond. A triple bond consists of one σ bond and two π bonds. In escitalopram, there are 47 σ bonds holding the atoms together in the structure. There are 3 π bonds in the benzene ring at the top of the structure, 3 π bonds in the benzene ring on the bottom left of the structure, and 2 π bonds in the triple bond between carbon and nitrogen in the bottom left of the structure. This makes 8 π bonds total.

How many 𝜎 and 𝜋 bonds are in this molecule? PICTURED: The molecule H C C C H O. Note that there is a carbon oxygen double bond and a carbon carbon triple bond. number of 𝜎 bonds: number of 𝜋

5 3 Solution A 𝜎 bond results from the end‑on overlap of orbitals. A 𝜎 bond is present in all covalent bonds. A 𝜋 bond results from the side‑on overlap of orbitals. They are only present in double and triple bonds. Each single bond consists of one 𝜎 bond. Each double bond consists of one 𝜎 and one 𝜋 bond. Each triple bond consists of one 𝜎 and two 𝜋 bonds. There is a 𝜎 bond in each of the five bonds in this molecule. There is one 𝜋 bond in the double bond and two in the triple bond.

How many 𝜎 and 𝜋 bonds are in this molecule? PICTURED: A chain of five carbon atoms. There is a double bond between the first and second carbon atoms and a triple bond between the fourth the fifth carbon atoms. There are single bonds between the remaining carbon atoms. There are two hydrogen atoms bonded to the first carbon atom through single bonds, and a single hydrogen atom bonded to both the second and fifth carbon atoms through single bonds. There is an oxygen atom bonded to the third carbon atom through a double bond. number of 𝜎 bonds: number of 𝜋 bonds:

9 4 Solution Each single bond consists of one 𝜎 bond. Each double bond consists of one 𝜎 and one 𝜋 bond. Each triple bond consists of one 𝜎 and two 𝜋 bonds. There is a 𝜎 bond in each of the nine bonds in this molecule: six single, two double, and one triple. There is a 𝜋 bond in each of the two double bonds, and there are two 𝜋 bonds in the triple bond.

My Attempt Which Lewis structures represent a molecule that would assume a linear geometry?

A molecule with atom Y single bonded with 2 X substituents. The Y atom has three lone pairs. A molecule with atom Y single bonded with 2 X substituents. No lone pairs of electrons are shown. A molecule with atom Y single bonded with 2 X substituents. No lone pairs of electrons are shown. With two bonded groups and zero lone pairs on a central atom, the bonded groups can get as far apart as possible, that is, directly across from each other at 180∘. A molecule with atom Y single bonded with 2 X substituents. The Y atom has one lone pair. With two bonded groups and one lone pair on the central atom, there are three regions of electron density forming a trigonal planar shape. Thus, the bonded groups are at a 120∘ angle. A molecule with atom Y single bonded with 2 X substituents. The Y atom has two lone pairs. With two bonded groups and two lone pairs on the central atom, there are four regions of electron density forming a tetrahedral shape. Thus, the bonded groups are at a 109.5∘ angle. A molecule with atom Y single bonded with 2 X substituents. The Y atom has three lone pairs. With two bonded groups and three lone pairs on the central atom, there are five regions of electron density forming a trigonal bipyramidal shape. Since the lone pairs will occupy all of the equatorial positions, the bonds will be in the axial positions at 180∘.

Complete the molecular orbital diagram for CN−. Note that the 1𝑠 orbitals are not shown. Identify the bond order of CN−. The atomic orbitals on the left side of the molecular orbital diagram are those of The atomic orbitals on the right side of the molecular orbital diagram are those of

ALL BOTTOM 4 ROWS HAVE EQUILIBRIUM ARROWS 3 CARBON NITROGEN Solution A carbon atom has 4 valence electrons. A nitrogen atom has 5 valence electrons. The −1 charge indicates one extra electron for a total of 4+5+1=10 valence electrons in this ion. The electrons are added to the lowest‑energy molecular orbitals. Because nitrogen is more electronegative than carbon, the atomic orbitals of nitrogen are lower in energy than the atomic orbitals of carbon. The bond order is determined using the equation bond order=1/2(bonding electrons−antibonding electrons) Antibonding electrons are in molecular orbitals marked with an asterisk (*). Bonding electrons are in molecular orbitals that are not marked with an asterisk. There are 8 valence electrons in the bonding orbitals and 2 valence electrons in the antibonding orbitals. bond order=1/2(8−2)=3

Construct the molecular orbital diagram for N2. The molecular orbital diagram for N 2. The molecular orbitals from lowest energy to highest energy are one sigma 2 s orbital, one sigma 2 s star orbital, two pi 2 p orbitals, one sigma 2 p orbital, two pi 2 p star orbitals, one sigma 2 p star orbital.

ALL EQUILIBRIUM SIGNS FOR THE BOTTOM 4 ROWS 3 Solution Start by determining the number of valence electrons that each N atom has individually. A nitrogen atom has 7 total electrons (1𝑠22𝑠22𝑝3), but only 5 are valence electrons. Thus, there are 10 valence electrons in the molecule N2. The electrons fill the 5 lowest energy molecular orbitals: 𝜎1s, 𝜎∗1s, 𝜎2s, 𝜎∗2s, 𝜋2p, and 𝜎2p. The bond order is determined using the equation bond order=12(bonding electrons−antibonding electrons) Examining the valence electrons, there are 8 electrons in the bonding molecular orbitals (𝜎2s, 𝜋2p, and 𝜎2p) and 2 electrons in the antibonding molecular orbital (𝜎∗2s). bond order=1/2(8−2)=3

Select the molecules that are polar.

AsF3 IOF5 CH2Cl2 Solution For a molecule to be polar, it must have at least one polar bond, and polar bonds must be asymmetric. In other words, all of the polar bonds should not cancel. PICTURED: A central carbon atom is bonded to four fluorine atoms. Each carbon fluorine bond is a single band. Each fluorine atom has three lone pairs of electrons. AX4; tetrahedral; nonpolar PICTURED: A central arsenic atom is bonded to three fluroine atoms. each arsenic fluorine bond is a single bond. There is one lone pair of electrons on the arsenic atom and three lone pairs of electrons on each fluorine atom. AX3E; trigonal pyramidal; polar PICTURED: A central iodine atom is bonded to five fluorine atoms and one oxygen atom. The iodine fluorine bonds are single bonds and the iodine oxygen bond is a double bond. Each fluorine atom has three lone pairs of electrons and the oxygen atom has two lone pairs of electrons. AX6; octahedral; polar AX4; tetrahedral; polarA central carbon atom is bonded to two chlorine atoms and two hydrogen atoms. Each bond is a single bond. There are three lone pairs of electrons on each chlorine atom. In the 2D Lewis structure, it may look like the polar bonds cancel. However, if you look at a 3D representation, you will see the asymmetry.

Which of the molecules are linear?

BeCl2 XeF2 Solution A xenon atom has 8 valence electrons. A fluorine atom has 7 valence electrons. Thus, an XeF2 molecule has 8+2(7)=22 valence electrons, which is equal to 11 electron pairs (in this case, 2 bonds and 9 lone pairs). The Lewis structure of this compound is A central xenon atom is connected to two fluorine atoms through single bonds. There are three lone pairs of electrons around the xenon atom and each fluorine atom. With two bonded groups and three lone pairs on the central atom, there are five regions of electron density, forming a trigonal bipyramidal shape. Since the lone pairs will occupy all of the equatorial positions, the bonds will be in the axial positions at 180°. XeF2 is linear. A ball and stick model of X e F 2. A central xenon atom is connected to two fluorine atoms through single bonds. The molecule is linear. A sulfur atom has 6 valence electrons. A fluorine atom has 7 valence electrons. Thus, an SF2 molecule has 6+2(7)=20 valence electrons, which is equal to 10 electron pairs (in this case, 2 bonds and 8 lone pairs). The Lewis structure of this compound is A central sulfur atom is connected to two fluorine atoms through single bonds. There are two lone pairs of electrons on the sulfur atom and three lone pairs of electrons on each fluorine atom. With two bonded groups and two lone pairs on the central atom, there are four regions of electron density, forming a tetrahedral shape. Thus, the bonded groups are at a 109.5° angle. SF2 is not linear. A ball and stick model of S F 2. A central sulfur atom is connected to two fluorine atoms through single bonds. The molecule is bent. A nitrogen atom has 5 valence electrons. An oxygen atom has 6 valence electrons. A −1 charge indicates one extra electron beyond what the atoms contribute. Thus, an NO−2 molecule has 5+2(6)+1=18 valence electrons, which is equal to 9 electron pairs (in this case, 3 bonds and 6 lone pairs). The Lewis structure of this compound is A central nitrogen atom is connected to two oxygen atoms. One oxygen atom is connected through a double bond and the other is connected through a single bond. There is one lone pair of electrons on the nitrogen atom. The double bonded oxygen has two lone pairs of electrons. The single bonded oxygen has three lone pairs of electrons and a negative 1 charge. With two bonded groups and one lone pair on the central atom, there are three regions of electron density, forming a trigonal planar shape. Thus, the bonded groups are at a 120° angle. NO−2 is not linear. A ball and stick model of N O 2 minus. The two oxygen atoms are connected to the nitrogen atom through a single line. The molecule is bent. Only single-bond connectivity is shown. A beryllium atom has 2 valence electrons. A chlorine atom has 7 valence electrons. Thus, an BeCl2 molecule has 2+2(7)=16 valence electrons, which is equal to 8 electron pairs (in this case, 2 bonds and 6 lone pairs). The Lewis structure of this compound is A central beryllium atom is connected to two chlorine atoms through single bonds. There are three lone pairs of electrons on each chlorine atom. With two bonded groups and zero lone pairs on a central atom, the bonded groups can get as far apart as possible, that is, directly across from each other at 180°. BeCl2 is linear. A ball and stick model of B e C l 2. A central beryllium atom is connected to two chlorine atoms through single bonds. The molecule is linear.

The Lewis structures of four compounds are given. PICTURED: In the first molecule sufur is the central atom. There are two oxygen atoms bonded to the sulfur atom through double bonds. There is one lone pair of electrons on the sulfur atom and two lone pairs of electrons on each oxygen atom. In the second molecule carbon is the central atom. There are two oxygen atoms bonded to the sulfur atom through double bonds. There are two lone pairs of electrons on each oxygen atom. In the third molecule carbon is the central atom. There are two chlorine atoms and two hydrogen atoms bonded to the carbon atom through single bonds. Each chlorine atom contains three lone pairs of electrons. In the four molecule phosphorus is the central atom. There are three chlorine atoms bonded to the phosphorus atom through single bonds. There is one lone pair of electrons on the phosphorus atom and three lone pairs of electrons on each chlorine atom. Which of these molecules are polar?

CH2Cl2 SO2 PCl3 Solution SO2 and CO2 each have two polar bonds of equal magnitude. The only way for such a molecule to be nonpolar is if those two bonds are directly across from each other. This is the case for the linear CO2 molecule, but not for the bent SO2 molecule. Therefore, SO2 is polar. CH2Cl2 has four bonds in a tetrahedral shape, but the C−H bonds have a different polarity than the C−Cl bonds. This molecule would be nonpolar if the C−H bonds were directly across from each other and if the C−Cl bonds were directly across from one another, so that the dipole moments canceled. In a tetrahedral molecule, none of the bonds are directly across from each other, so this molecule is polar. PCl3 has three polar bonds of equal magnitude. Since those bonds are arranged asymmetrically in a trigonal pyramidal shape, this molecule is polar. The 3‑D structures of the four molecules are useful when determining whether it is polar or not. Note that only single‑bond connectivity is displayed in the 3‑D structures of the molecules. SO2 A large ball representing a sulfur atom is connected to two smaller balls, each representing an oxygen atom. Each oxygen atom is connected to the sulfur atom through a single line. The O S O bond angle is approximately 120 degrees. CO2 A ball representing a carbon atom is connected to two balls of similar size, each representing an oxygen atom. Each oxygen atom is connected to the carbon atom through a single line. The O C O bond angle is 180 degrees. CH2Cl2 A ball representing a carbon atom is connected to two smaller balls, each representing a hydrogen atom, and two larger balls, each representing a chlorine atom. Each hydrogen and chlorine atom is connected to the carbon atom through a single line. The H C H and Cl C Cl bond angles are approximately 109.5 degrees. PCl3 A ball representing a phosphorus atom is connected to three balls of similar size, each representing a chlorine atom. Each chlorine atom is connected to the phosphorus atom by a single line. Each Cl P Cl bond angle is approximately 109.5 degrees.

Use the molecular orbital theory to determine the ground state electron configuration of F2 and F+2. Molecule Ground state electron configuration F2 (𝜎1𝑠) (𝜎1𝑠*) (𝜎2𝑠) (𝜎2𝑠*) (𝜎2𝑝) (𝜋2𝑝) (𝜋2𝑝*) F+2 (𝜎1𝑠) (𝜎1𝑠*) (𝜎2𝑠) (𝜎2𝑠*) (𝜎2𝑝) (𝜋2𝑝) (𝜋2𝑝*) Then, use the molecular orbital theory to determine the bond order and magnetism of the molecules. F2 bond order: F+2 bond order: F2 is F+2 is

F2: 2222244 F+2: 2222243 1 1.5 DIAMAGNETIC PARAMAGNETIC Solution Start by determining the number of electrons in each molecule. F2 has 18 electrons. F+2 has 17 electrons. Then, use the image as a guide for filling the electrons into molecular orbitals from lowest to highest energy. An M O diagram has the following orbitals from lowest energy to highest: a sigma 1 s, an antibonding sigma 1 s, a sigma 2 s, an antibonding sigma 2 s, a sigma 2 p, 2 pi 2 p orbitals, 2 antibonding pi 2 p orbitals, and an antibonding sigma 2 p orbital. A sigma, 𝜎, orbital can hold a maximum of two electrons. A set of pi, 𝜋, orbitals can hold a maximum of four electrons. To find the bond order, you need to find the total number of bonding electrons and antibonding electrons. Antibonding orbitals are indicated with an asterisk (*). Once the total number of bonding and antibonding electrons has been determined, calculate the bond order using the equation Bond order=number of bonding electrons−number of antibonding electrons/2 For F2, there are 10 bonding electrons and 8 antibonding electrons. Therefore, the bond order is 10−8)/2=1. For F+2, there are 10 bonding electrons and 7 antibonding electrons. Therefore, the bond order is 10−7)/2=1.5. Finally, diamagnetic species have all paired electrons. Paramagnetic species have at least one unpaired electron. Species with an odd number of electrons are always paramagnetic. Even‑electron species may be either diamagnetic or paramagnetic, so you will need to fill in the diagram to know for sure. F2 has no unpaired electrons and is diamagnetic. F+2 has one unpaired electron and is paramagnetic.

Diazene, N2H2 (also known as diimine), is commonly used as a reagent for organic syntheses. Label each image with the type of bond in diazene that it shows. Three images of overlapping orbitals. In the first image, an orbital has two lobes. A second identical orbital is parallel to the first such that corresponding lobes overlap. In the second image, an orbital has three lobes. A second identical orbital has one lobe overlapping with one lobe from the first orbital such that the orbitals are mirror images of each other. In the third image, an orbital has three lobes. A spherical lobe overlaps with one of the lobes.

N-N pi N-N omega N-H omega

If the symbol X represents a central atom, Y represents outer atoms, and Z represents lone pairs on the central atom, the structure A central X atom has two lone pairs. Two Y atoms are attached to X with single bonds. could be abbreviated as XY2Z2. Classify these structures by the hybridization of the central atom. 𝑠𝑝 𝑠𝑝2 𝑠𝑝3

XY2 XY3 AND XY2Z XY4 AND XY2Z2 AND XY3Z

If the symbol X represents a central atom, Y represents outer atoms, and Z represents lone pairs on the central atom, the structure The central X atom is single bonded to two Y atoms, which are 180 degrees apart. The X atom has two lone pairs. could be abbreviated as XY2Z2. Classify each molecule according to its shape. Linear Bent (≈120∘) Bent (≈109∘) Trigonal pyramidal T‑shaped See‑saw Square planar Square pyramidal

XY2Z3 XY2Z XY2Z2 XY3Z XY3Z2 XY4Z XY4Z2 XY5Z Solution With 5 total electron groups around the central atom, XY2Z3 has an electron geometry of trigonal bipyramidal. The three lone pairs occupy the equatorial positions, leaving the outer atoms in the axial positions. The result is a linear molecular geometry. With three total electron groups around the central atom, XY2Z has an electron geometry of trigonal planar. Because one of the electron groups is a lone pair, the molecular geometry is bent. In a trigonal planar molecule where all three electron groups are atoms, the bond angles are 120∘. The lone pair occupies a larger amount of space than an atom, which pushes the other two atoms closer together resulting in a bond angle that is slightly less than 120∘. With four total electron groups around the central atom, XY2Z2 has an electron geometry of tetrahedral. Lone pairs occupy two of the positions around the central atom, so the molecular geometry is bent. In a tetrahedral molecule where all four electron groups are atoms, the bond angles are 109.5∘. The lone pairs occupy a larger amount of space than an atom, which pushes the other two atoms closer together resulting in a bond angle that is slightly less than 109.5∘. With four total electron groups around the central atom, XY3Z also has an electron geometry of tetrahedral. In this molecule though, only one of the groups is a lone pair, so the molecular geometry is trigonal pyramidal. With five total electron groups around the central atom, XY3Z2 has an electron geometry of trigonal bipyramidal. The two lone pairs occupy two of the equatorial positions resulting in a T‑shaped molecular geometry. With five total electron groups around the central atom, XY4Z also has an electron geometry of trigonal bipyramidal. The lone pair occupies one of the equatorial positions, resulting in a see‑saw molecule geometry. With six total electrons groups around the central atom, XY4Z2 has an electron geometry of octahedral. The two lone pairs occupy positions that are directly across from each other. The result is a square planar molecular geometry. With six total electron groups around the central atom, XY5Z also has an electron geometry of octahedral. Because one of the electron groups is a lone pair, the molecular geometry is square pyramidal.

Which of the species contains a π bond? Which of the species contains a delocalized π bond?

a) CO2−3 HCN O3 b) O3 CO2−3 Solution The Lewis structure for each species is shown. PICTURES: Lewis structure for H 2 O. A central O atom has two lone pairs of electrons and is connected to two hydrogen atoms through single bonds. The Lewis structure for O 3. A central O atom has one lone pair of electrons. It is connected to one O atom by a single bond and another O atom by a double bond. The single bonded O atom has three lone pairs of electrons and the double bonded oxygen atom has two lone pairs of electrons. The Lewis structure for H C N. A central C atom is bonded to a H atom by a single bond and a N atom by a triple bond. There is a lone pair of electrons on the N atom. The Lewis structure for C O 3 2 minus. A central C atom is bonded to three oxygen atoms, two through a single bond and one through a double bond. The double bonded O atom has two lone pairs of electrons. The single bonded O atoms each have three lone pairs of electrons and a negative charge. The overall charge is 2 minus. Each single bond is a σ bond. Each double bond is made up of one σ bond and one π bond. Each triple bond is made up of one σ and two π bonds. Thus, all of the molecules contain a π bond except for H2O. Delocalized π bonds occur when π orbitals extend over more than two atoms. In terms of Lewis structures, this occurs with resonance structures involving double and triple bonds. For the species in this question, O3 and CO32− have resonance structures. PICTURED: The resonance structures for O 3. In the first Lewis structure, a central O atom has one lone pair of electrons. It is connected to one O atom by a single bond and another O atom by a double bond. The single bonded O atom has three lone pairs of electrons and the double bonded oxygen atom has two lone pairs of electrons. In the second strucutre, the double bond is between the central O atom and the other outer O atom. The resonance structures for C O 3 2 minus. In the first Lewis structure, a central C atom is bonded to three oxygen atoms, two through a single bond and one through a double bond. The double bonded O atom has two lone pairs of electrons. The single bonded O atoms each have three lone pairs of electrons and a negative charge. The overall charge is 2 minus. In the other two structures, the double bond is between the C atom and one of the other O atoms. HCN, the other compound with π bonds, cannot have delocalized π bonds because a second valid resonance structure cannot be drawn. PICTURED: In the first Lewis structure for H C N a central C atom is bonded to a H atom by a single bond and a N atom by a triple bond. There is a lone pair of electrons on the N atom. In the second Lewis structure, a central C atom is bonded to a H atom and an N atom by double bonds. The H atom has a positive charge. The N atom has two lone pairs of electrons and a negative charge. This is not a valid Lewis structure. The two structures are not equivalent. Therefore, O3 and CO32− have delocalized π bonds and HCN and H2O do not.

a) PICTURED: A central beryllium atom is connected to two chlorine atoms through single bonds. Each chlorine atom has three lone pairs of electrons. bonds molecules b) PICTURED: A central oxygen atom is connected to two hydrogen atoms through single bonds. There are two lone pairs of electrons around the oxygen atom. bonds molecules

a) polar nonpolar b) polar polar Solution Each Be−Cl bond is polar because the two atoms have different electronegativities. The number of outer atoms (2) and lone pairs on the central atom (0) indicate that this molecule has a linear geometry. The bonds in a linear molecule are symmetric, so their dipoles cancel out, meaning the molecule is nonpolar. Each O−H bond is polar because the two atoms have different electronegativities. The number of outer atoms (2) and lone pairs on the central atom (2) indicate that this molecule has a bent geometry. The bonds in a bent molecule are asymmetric, so their dipoles do not cancel out, meaning the molecule is polar. In addition, the asymmetric arrangement of the lone pairs on O further contributes to the dipole of this molecule.

Identify the molecular shape of each of the molecules. Molecule Geometry Molecule Geometry a) SiH4 b) BrF5 c) SF4 d) PCl5 e) PICTURED: Four atoms bonded to a central atom. Two of the outer atoms are 180 degrees from each other and 90 degrees from the other two outer atoms, which are 120 degrees from each other. f) PICTURED: Five atoms are bonded to a central atom. Two of the outer atoms are 180 degrees from each other and 90 degrees from the other three outer atoms, which are 120 degrees from each other. g) PICTURED: A central carbon atom is connected to four fluorine atoms through single bonds. h) PICTURED: A central xenon atom is connected to four fluorine atoms through single bonds. There are two lone pairs of electrons on the xenon atom.

a) tetrahedral b) square pyramidal c) seesaw d) trigonal bipyramidal e) seesaw f) trigonal bipyramidal g) tetrahedral h) square planar

Indicate the electron pair geometry and the molecular geometry for each of the six compounds. Compound Electron pair geometry Molecular geometry a) A sulfur atom is double bonded to an oxygen atom on the left and the right, and has a lone pair. Each oxygen atom has two lone pairs. b) A sulfur atom is bonded to a chlorine atom on the left and the right, and has two lone pairs. Each chlorine atom has three lone pairs. tetrahedral c) A beryllium atom is bonded to two chlorine atoms 180 degrees apart. Each chlorine atom has three lone pairs. d)A phosphorous atom is bonded to a fluorine atom on the left, the right, and the bottom, and has one lone pair. Each fluorine atom has three lone pairs. e)A boron atom is bonded to a fluorine atom on the left, the right, and the bottom. Each fluorine atom has three lone pairs. f)A carbon atom is bonded to a hydrogen atom on the left, the right, the top, and the bottom.

a) trigonal planar bent b) tetrahedral bent c) linear linear d) tetrahedral trigonal pyramidal e) trigonal planar trigonal planar f) tetrahedral tetrahedral Solution Analyze the Lewis structure of each compound to determine the number of electron groups around the central atom. An electron group consists of a lone pair of electrons, a single bonded atom, a double bonded atom, or a triple bonded atom. The electron pair geometry is determined from the arrangement of all the groups around the central atom. To minimize repulsion, the groups are arranged as far away from each other as possible. If there are no lone pairs of electrons, the molecular geometry is the same as the electron pair geometry. With one or more lone pairs of electrons, the molecular geometry is determined by the positions of the bonded atoms only. Formula Lewis structure Comment SO2 A sulfur atom is double bonded to an oxygen atom on the left and the right, and has a lone pair. Each oxygen atom has two lone pairs. Sulfur dioxide has three electron groups around the central atom, so the electron pair geometry is trigonal planar. There is one lone pair, so the molecular geometry is bent. SCl2 A sulfur atom is bonded to a chlorine atom on the left and the right, and has two lone pairs. Each chlorine atom has three lone pairs. Sulfur dichloride has four electron groups around the central atom, so the electron pair geometry is tetrahedral. There are two lone pairs, so the molecular geometry is bent. BeCl2 A beryllium atom is bonded to two chlorine atoms 180 degrees apart. Each chlorine atom has three lone pairs. Beryllium dichloride has two electron groups around the central atom, so the electron pair geometry is linear. There are no lone pairs, so the molecular geometry is also linear. PF3 A phosphorous atom is bonded to a fluorine atom on the left, the right, and the bottom, and has one lone pair. Each fluorine atom has three lone pairs. Phosphorus trifluoride has four electron groups around the central atom, so the electron pair geometry is tetrahedral. There is one lone pair, so the molecular geometry is trigonal pyramidal. BF3 A boron atom is bonded to a fluorine atom on the left, the right, and the bottom. Each fluorine atom has three lone pairs. Boron trifluoride has three electron groups around the central atom, so the electron pair geometry is trigonal planar. There are no lone pairs, so the molecular geometry is also trigonal planar. CH4 A carbon atom is bonded to a hydrogen atom on the left, the right, the top, and the bottom. Methane has four electron groups around the central atom, so the electron pair geometry is tetrahedral. There are no lone pairs, so the molecular geometry is also tetrahedral.

Predict the molecular shape of these compounds. a) ammonia, NH3 PICTURED: A central N atom bonded to three H atoms and a lone pair. b) ammonium, NH4+ PICTURED: central N atom with a positive charge bonded to four H atoms. c) beryllium fluoride, BeF2 PICTURED: A central B e atom bonded to two F atoms that each have six valence electrons. d) hydrogen sulfide, H2S PICTURED: A central S atom with two lone pairs bonded to two H atoms.

a) trigonal pyramidal b) tetrahedral c) linear d) bent Solution The most stable geometry is an arrangement that keeps the atoms or electrons bonded to the central atom as far apart as possible. Ammonia, NH3, has a central nitrogen atom surrounded by three hydrogen atoms and a lone pair of electrons. It has a tetrahedral electron geometry and a trigonal pyramidal molecular shape. PICTURED: 3 D model of N H 3. It has a trigonal pyramidal shape. Ammonium, NH4+, has a central nitrogen atom surrounded by four hydrogen atoms. It has a tetrahedral electron geometry and a tetrahedral molecular shape. PICTURED: 3 D model of N H 4 plus. It has a tetrahedral shape. Beryllium fluoride, BeF2, has a central beryllium atom surrounded by two fluoride atoms. It has a linear electron geometry and a linear molecular shape. PICTURED: 3 D model of B e F 2. It has a linear shape. Hydrogen sulfide, H2S, has a central sulfur atom surrounded by two hydrogen atoms and two lone pairs of electrons. It has a tetrahedral electron geometry and a bent molecular shape. PICTURED: 3 D model of H 2 S. It has a bent shape.

Where, approximately, is the negative pole on each of these molecules? Lewis structure for C O F 2. A central C atom is single bonded to two F atoms and double bonded to one O atom. There are two lone pairs of electrons on the O atom and three lone pairs of electrons on each F atom. The Lewis structure for C O F H. A central C atom is single bonded to an H and F atom and double bonded to an O atom. There are two lone pairs of electrons on the oxygen atom and three lone pairs of electrons on the F atom. Which molecule should have the higher dipole moment, and why?

between the F atoms between the O and F atoms COFH because the polar bonds in COF2 nearly cancel each other out. Solution Think of each polar bond as a vector pointing toward the more electronegative atom. Larger electronegativity differences warrant longer vectors. In COF2, both O and F are more electronegative than C, so the vectors point towards the O and F atoms in each bond. The electronegativity difference between C and F is slightly greater than the electronegativity difference between O and C, so the net vector is between the two F atoms. Similarly in COFH, both O and F are more electronegative than C, so the vectors point towards the O and F atoms in each bond. C is slightly more electronegative than H so the vector point towards the C atom. Because the electronegativity difference between C and F and C and O is similar and because the C−H vector points towards C, the net vector is between the O and F atoms. PICTURED: Lewis structure for C O F 2. A central C atom is single bonded to two F atoms and double bonded to one O atom. There are two lone pairs of electrons on the O atom and three lone pairs of electrons on each F atom. Both O and F are more electronegative than C, so the vectors point towards the O and F atoms in each bond. The electronegativity difference between C and F is slightly greater than the electronegativity difference between O and C, so the net vector is between the two F atoms. The Lewis structure for C O F H. A central C atom is single bonded to an H and F atom and double bonded to an O atom. There are two lone pairs of electrons on the oxygen atom and three lone pairs of electrons on the F atom. both O and F are more electronegative than C, so the vectors point towards the O and F atoms in each bond. C is slightly more electronegative than H so the vector point towards the C atom. Because the electronegativity difference between C and F and C and O is similar and because the C - H vector points towards C, the net vector is between the O and F atoms. The polar bonds in COF2 nearly cancel each other out, so COFH has the greater dipole moment.

Open the Molecule Shapes interactive and select Model mode. Build a molecule with at least three single bonds, and then answer these questions. Which of these actions will change the molecule geometry? Which of these actions will change the electron geometry?

changing a single bond to a double bond neither Solution The type of bond (single, double, triple) has no effect on either the molecule geometry or the electron geometry. The type of electron domain (bond versus lone pair) affects the molecule geometry but not the electron geometry.

Diagram B 𝜋2𝑝>𝜎2𝑝 M O diagram B has the following orbitals from lowest energy to highest: sigma 2 s, antibonding sigma 2 s, sigma 2 p, pi 2 p, antibonding pi 2 p, and antibonding sigma 2 p. Diagram A 𝜎2𝑝>𝜋2𝑝 M O diagram A has the following orbitals from lowest energy to highest: sigma 2 s, antibonding sigma 2 s, pi 2 p, sigma 2 p, antibonding pi 2 p, and antibonding sigma 2 p. For each of these molecules, identify the proper MO diagram and the number of valence electrons. The 1𝑠 orbital is not shown. Identify the MO diagram for B2. B2 valence e−: Identify the MO diagram for C2. C2 valence e−: Identify the MO diagram for N2 N2 valence e−: Identify the MO diagram for O2. O2 valence e−: Identify the MO diagram for F2. F2 valence e−:

diagram A 6 diagram A 8 diagram A 10 diagram B 12 diagram B 14 Solution In B2, C2, and N2, the 𝜎2𝑝 orbital is higher in energy than the 𝜋2𝑝 orbitals as shown in diagram A. In O2 and F2, the 𝜋2𝑝 orbitals are higher in energy than the 𝜎2𝑝 orbital as shown in diagram B. The number of valence electrons per atom of an element is related to the group number in the periodic table. Boron, in group 3A, has 3 valence electrons per atom. Diatomic boron has 2(3)=6 valence electrons. Carbon, in group 4A, has 4 valence electrons per atom. Diatomic carbon has 2(4)=8 valence electrons. Nitrogen, in group 5A, has 5 valence electrons per atom. Diatomic nitrogen has 2(5)=10 valence electrons. Oxygen, in group 6A, has 6 valence electrons per atom. Diatomic oxygen has 2(6)=12 valence electrons. Fluorine, in group 7A, has 7 valence electrons per atom. Diatomic fluorine has 2(7)=14 valence electrons.

Draw the Lewis structure of SF4 showing all lone pairs. Identify the molecular geometry of SF4. What is the hybridization of the central atom? What are the ideal bond angles of this geometry? An SF4 molecule is

see‑saw 𝑠𝑝3𝑑 180° AND 120° AND 90° polar.

Select the nitrogen and hydrogen orbitals that overlap to form each N−H σ bond in NH3 . It may be useful to consult the periodic table. nitrogen orbital: hydrogen orbital:

sp3 s Solution Begin by identifying the valence electron configurations of each nitrogen and hydrogen atom. The valence electron configuration of hydrogen is 1𝑠1, and the valence electron configuration of nitrogen is 2𝑠22𝑝3. Based on the valence electron configuration, hydrogen is only able to bond with one other atom by contributing a half‑filled 𝑠 orbital to the bond. Similarly, nitrogen should be able to contribute three half‑filled 𝑝 orbitals to three bonds. However, valence bond theory states that the atomic orbitals (𝑠, 𝑝, etc.) of the central atom in a molecule with three or more atoms will mix to form hybrid orbitals (𝑠𝑝, 𝑠𝑝2, 𝑠𝑝3). Because nitrogen is the central atom in NH3 , the atomic orbitals of nitrogen will mix to produce hybrid orbitals. To determine the type of hybrid orbitals produced, consider the Lewis structure of NH3 . PICTURED: A central N atom is bonded to three H atoms. There is one lone pair of electrons on the N atom. NH3 has a tetrahedral electron geometry and four electron groups around the central N atom: one nonbonding lone pair and three single bonds. Recall that when a central atom exhibits tetrahedral electron geometry, the 𝑠 and 𝑝 orbitals mix to form four equivalent 𝑠𝑝3 hybrid orbitals. Hybrid orbitals, like atomic orbitals, can only hold two electrons, so one 𝑠𝑝3 hybrid orbital on nitrogen holds the lone pair of electrons and the other three are half‑filled. Each half‑filled 𝑠𝑝3 orbital is then able to overlap with the 𝑠 orbitals of the three hydrogen atoms to produce the three N−H σ bonds in NH3 .

PICTURED: What is the hybridization of phosphorus in each of the molecules or ions? Lewis structure of P C l 3. The central P atom has one lone pair and is bonded to three C l atoms. PICTURED: Lewis structure of P 4 O 6. Each P atom has a lone pair and is bonded to three O atoms. Each O atom has two lone pairs and is bonded to two P atoms. PICTURED: Lewis structure of P 4. Each P atom has a lone pair and is bonded to three other P atoms. PICTURED: Lewis Structure of P O 4 3 minus. A central P atom with no lone pairs is bonded to 4 oxygen atoms, 3 through single bonds and one through a double bond. The double bonded O atom has two lone pairs of electrons. The single bonded O atoms have three lone pairs of electros and a negative charge.

sp3 sp3 sp3 sp3

Anthracene is a yellow, crystalline solid found in coal tar. Complete the structure for anthracene, C14H10,C14H10, by adding bonds and hydrogen atoms as necessary. What type of hybrid orbitals are utilized by carbon in anthracene? How many σ bonds and π bonds are there in an anthracene molecule? σ bonds: π bonds: How many valence electrons occupy σ‑bond orbitals, and Zhow many occupy π‑bond orbitals? valence e−e− in σ‑bond orbitals: valence e−e− in π‑bond orbitals:

sp^2 26 7 52 14 Solution The alternating single and double bonds in anthracene give each carbon atom an octet. Each carbon atom is bonded to three other atoms in anthracene. When bonded to three other atoms, carbon is 𝑠𝑝2 hybridized. Single bonds are σ bonds. Double bonds are made up of one σ bond and one π bond. There are 19 single bonds and 7 double bonds in this molecule, so there are 19+7=26 σ bonds. There are 7 double bonds, each containing a π bond, so there are 7 π bonds. The numbers of σ and π electrons are double the numbers of σ and π bonds, respectively because each bond is made of two electrons. Therefore, there are 52 σ electrons and 14 π electrons for a total of 52 σ electrons+14 π electrons=66 electrons which matches the expected number of valence electrons for C14H10.

Draw the Lewis structure of XeCl4. Include all lone pairs. Identify the molecular geometry of XeCl4. What are the approximate bond angles in XeCl4? An XeCl4 molecule is

square planar 90° AND 180° nonpolar. Solution An Xe atom has eight valence electrons. A Cl atom has seven valence electrons. Therefore, an XeCl4 molecule has 8+4(7)=36 valence electrons. The structure has four Cl atoms bonded to a central Xe atom. Three lone pairs are added to each Cl to complete their octets. The remaining four electrons are placed around the central Xe atom, which can have an expanded valence shell. There are six total electron regions around Xe. Therefore, the electron geometry is octahedral. Four of the groups are bonded atoms and two of the groups are lone pairs, so the molecular geometry is square planar. In this geometry the bond angles are 90° between adjacent atoms and 180° between atoms that are directly across from each other. The Xe−Cl bonds are polar, but because the polar bonds are arranged symmetrically around the central atom, the bond dipoles cancel and the molecule is nonpolar. A three-dimensional structure of XeCl4 is shown. Click on the structure to rotate it, or use the controls provided.

What is the molecular geometry of the left carbon atom (circled) in acetic acid? PICTURED: The left carbon atom is attached to three hydrogen atoms and one carbon atom through single bonds. The right carbon atom is attached to a carbon atom and an oxygen atom through single bonds and a second oxygen atom through a double bond. What is the molecular geometry of the right carbon atom (circled) in acetic acid? PICTURED: The left carbon atom is attached to three hydrogen atoms and one carbon atom through single bonds. The right carbon atom is attached to a carbon atom and an oxygen atom through single bonds and a second oxygen atom through a double bond.

tetrahedral trigonal planar Solution The 3 D structure of acetic acid. The left carbon atom is attached to three hydrogen atoms and the right carbon atom through single bonds and forms a tetrahedral geometry. The right carbon atom is attached to the left carbon atom and one oxygen atom through single bonds and a second oxygen atom through a double bond forming a trigonal planar geometry. The single bonded oxygen atom is also bonded to a hydrogen atom.In acetic acid, the left carbon atom has four bonded groups and no lone electron pairs, which form a tetrahedral geometry. The right carbon atom has three bonded groups and no lone electron pairs, which form a trigonal planar geometry. Trigonal planar is sometimes called triangular. In the 3D image of the structure of acetic acid, only single bond connectivity is shown.

Draw the Lewis structure for PF3, including lone pairs. What is the molecular shape of PF3? What is the F−P−F bond angle? The P−F bond in PF3 is The molecule PF3 is

trigonal pyramidal <109.5∘ polar. polar. Solution PICTURED: A central P atom is bonded to three F atoms through single bonds. There are three lone pairs on each F atom and one lone pair on the P atom. There are 26 valence electrons in the structure, and P is the central atom. There are four electron groups around the central atom, so the electron pair geometry is tetrahedral. One of the pairs is a lone pair, making the molecular geometry trigonal pyramidal. The lone pair has a slightly stronger repulsion than the bonding pairs, so the bond angles are less than the tetrahedral bond angle of 109.5∘. The electronegativity of phosphorus is 2.1 and the electronegativity of fluorine is 4.0, so the P−F bond is polar and the three fluorine atoms have a partial negative charge, giving some charge separation between the base of the pyramid and the phosphorus atom at its apex. As a result, the molecule is polar.

Caffeine is the active ingredient in coffee, tea, and some carbonated beverages. Add lone pairs, as needed, to the structure of caffeine. Indicate the hybridization of the specified atoms. Be sure to consider any lone pairs you added in the first part. The structure of caffeine. Atom A is a carbon atom bonded to three hydrogen atoms and one nitrogen atom through single bonds. Atom B is a carbon atom bonded to one carbon atom through a double bond and one carbon and nitrogen atom through single bonds. Atom C is a nitrogen atom that is bonded to one carbon atom by a single bond and one carbon atom through a double bond. What is the hybridization of atom A? What is the hybridization of atom B? What is the hybridization of atom C?

𝑠𝑝3 𝑠𝑝2 𝑠𝑝2