Chem 105 Chapter 5 Homework

Calculate the formal charge of the N . Enter the formal charge, including the magnitude and sign.

+1

Calculate the formal charge of the O . Enter the formal charge, including the magnitude and sign.

-1

How well does your estimated bond length agree with the bond length given in the textbook (bond length of HF= 92 pm , bond length of HBr = 141 pm )? Express your answers using two significant figures separated by a comma.

8.2,6.3%

In which cases do you expect deviations from the idealized bond angle? OF2 NF3 CF4 H2S

OF2 NF3 H2S NF3 has three bonding groups and one lone pair whereas OF2 and H2S have two bonding groups and two lone pairs. Lone pairs and double bonds take up more room than single bonds and they repel the bonding groups to a greater extent, resulting in smaller bond angles. For example, NH3 has bond angles of 107∘. Thus, only CF4 will have the idealized bond angles of 109.5∘.

Determine whether each molecule is polar or nonpolar.

The geometry of a molecule is a key element in determining the polarity of a molecule. A polar molecule must have polar bonds oriented in an asymmetric fashion. However, even molecules with extremely electronegative atoms, such as fluorine, can be nonpolar if the geometry negates the individual dipole moments.

Which statement best captures the fundamental idea behind VSEPR theory?

a. The angle between two or more bonds is determined primarily by the repulsions between the electrons within those bonds and other (lone pair) electrons on the central atom of a molecule. Each of these electron groups (bonding electrons or lone pair electrons) lowers its potential energy by maximizing its separation from other electron groups, thus determining the geometry of the molecule.

Suppose that a molecule has four bonding groups and one lone pair on the central atom. Suppose further that the molecule is confined to two dimensions. (This is a purely hypothetical assumption for the sake of understanding the principles behind VSEPR theory.) Estimate the bond angles between electron groups. Express your answer using two significant figures.

α =72∘ According to VSEPR theory, the geometry of a molecule is determined by the configuration where areas of electron density are as far apart as possible. For a molecule confined to two dimensions, whose central atom has five electron groups (four bonding groups and one lone pair) around, the preferred electron geometry would be a pentagonal planar, with the central atom in the center and five electron groups at the corners of a pentagon. In this case, the bond angle is estimated as ideal bond angle=360∘/5=72∘ The molecular geometry will be different than that of the electron geometry.

What is the magnitude of the dipole moment formed by separating a proton and an electron by 100 pm ? Express your answer to two significant figures and include the appropriate units.

μ = 4.8 D

What is the magnitude of the dipole moment formed by separating a proton and an electron by 200 pm ? Express your answer to two significant figures and include the appropriate units.

μ = 9.6 D

In the resonance structure with the electron geometry "C tetrahedral; 1st N trigonal planar; 2nd N linear" would you expect the two nitrogennitrogen bond angles and bond lengths to be the same or different? the C-N-N bond angle will be closer to 109 ∘ ; the N-N-N bond angle will be 180 ∘ ; the two N-N bond lengths will be the same the C-N-N bond angle will be closer to 120 ∘ ; the N-N-N bond angle will be 180 ∘ ; the two N-N bond lengths will be the same the C-N-N bond angle will be closer to 120 ∘ ; the N-N-N bond angle will be 180 ∘ ; the N1−N2 bond length will be longer than the N2−N3 bond length the C-N-N bond angle will be closer to 109 ∘ ; the N-N-N bond angle will be 180 ∘ ; the N1−N2 bond length will be longer than the N2−N3 bond length

the C-N-N bond angle will be closer to 120 ∘ ; the N-N-N bond angle will be 180 ∘ ; the two N-N bond lengths will be the same

Determine the molecular geometry for each molecule. CF4 NF3 OF2 H2S

Although each of these molecules has four electron groups and a tetrahedral electron geometry, the molecular geometries are not identical because of the combinations of bonding groups and lone-pair electrons. CF4 has only four bonding groups and no lone pairs, making its electron geometry and molecular geometry the same, whereas NF3 has three bonding groups and one lone pair, resulting in a trigonal pyramidal molecular geometry. The three bonding groups form a triangular base with the central atom above this plane and the lone pair vertically up, which results in the observed three-dimensional shape. Both OF2 and H2S have two bonding groups and two lone pairs. Two of the bonding groups occupy two of the tetrahedron's corners, and the two lone pairs occupy the other two of the tetrahedron's corners. Thus, the two bonding groups around the central atom form an obtuse bonding angle that determines the molecular shape, and the observed three-dimensional shape is bent.

Identify whether each molecule given below is polar or nonpolar. BrF3 OF2 CS2 SiH3Br

BrF3 and OF2 have an unsymmetric distribution of bonding pairs and lone pairs around the central atom and both are polar. SiH3Br has an unsymmetrical distribution of atoms (electronegativities) around the central atom and is polar. CS2 has a symmetric distribution of bonding pairs around the central atom and is nonpolar.

Determine the electron geometry about each interior atom for all resonance structures. Check all that apply. C trigonal planar; 1st N tetrahedral; 2nd N linear C tetrahedral; 1st N tetrahedral; 2nd N linear C tetrahedral; 1st N trigonal planar; 2nd N linear C tetrahedral; 1st N bent; 2nd N linear C tetrahedral; 1st N linear; 2nd N bent C tetrahedral; 1st N linear; 2nd N linear

C tetrahedral; 1st N tetrahedral; 2nd N linear C tetrahedral; 1st N trigonal planar; 2nd N linear C tetrahedral; 1st N linear; 2nd N linear

How do you determine whether a molecule is polar? Choose all that apply. Determine whether the molecule has resonance structures. Draw the Lewis structure for the molecule and determine the molecular geometry. Determine whether the molecule contains polar bonds. Determine whether the polar bonds add together to form a net dipole moment. Determine whether the molecule contains nonpolar bonds.

Draw the Lewis structure for the molecule and determine the molecular geometry. Determine whether the molecule contains polar bonds. Determine whether the polar bonds add together to form a net dipole moment.

Draw the Lewis structure for H3PO4. If necessary, expand the octet on appropriate atoms to lower formal charge.

Elements such as phosphorus and sulfur can expand their octet to accommodate additional bonds and lone pairs of electrons. This structure also has a resonance form in which the octet rule is obeyed on phosphorus, but that resonance form would have two atoms with formal charges, so this structure is the more stable form.

Which species has the smaller bond angle, CH4 or H3C− ?

Lone pairs of electrons are more repulsive than bonding pairs. Since H3C− has one lone pair of electrons on the central carbon atom and CH4 has zero lone pairs of electrons, H3C− will have the smaller bond angle because lone pair-bonding pair repulsions are greater than bonding pair-bonding pair repulsions.

In which cases do you expect deviations from the idealized bond angle? Check all that apply. PF3 CH4 COCl2 SBr2

PF3 COCl2 SBr2 Since PF3 and SBr2 both have lone pairs of electrons on the central atom, their bond angles should deviate from the idealized bond angle as lone pairs are more repulsive than bonding pairs of electrons. Large atoms, such as Cl or Br , will cause bond angles to deviate from the ideal, and the presence of a multiple bond will cause a deviation as well owing to the increased electron density, so COCl2 should deviate from the idealized bond angle. CH4 has identical atoms symmetrically surrounding a central atom with no lone pairs on it, so CH4 should not deviate from the idealized bond angle.

Determine the idealized bond angles for each molecule.

PF3 has three bonding pairs and one lone pair (four total) surrounding the central phosphorus, so the electron geometry is tetrahedral with idealized bond angles of 109.5 ∘ . SBr2 has two bonding pairs and two lone pairs (four total) surrounding the central sulfur, so the electron geometry is tetrahedral with idealized bond angles of 109.5 ∘ . CH4 has a central carbon surrounded by four bonding pairs and no lone pairs (four total), so the electron geometry is tetrahedral with idealized bond angles of 109.5 ∘ . COCl2 has three bonding pairs and no lone pairs surrounding the central carbon, so the electron geometry is trigonal planar with idealized bond angles of 120 ∘ .

Determine the electron geometry for each molecule.

PF3 has three bonding pairs and one lone pair (four total) surrounding the central phosphorus, so the electron geometry is tetrahedral. SBr2 has two bonding pairs and two lone pairs (four total) surrounding the central sulfur, so the electron geometry is tetrahedral as well. CH4 has a central carbon surrounded by four bonding pairs and no lone pairs (four total), so the electron geometry is tetrahedral. COCl2 has three bonding pairs and no lone pairs (three total) surrounding the central carbon, so the electron geometry is trigonal planar.

Determine the molecular geometry for each molecule.

PF3 has three bonding pairs and one lone pair surrounding the central phosphorus, so the molecular geometry is trigonal pyramidal. SBr2 has two bonding pairs and two lone pairs surrounding the central sulfur, so the molecular geometry is bent. CH4 has a central carbon surrounded by four bonding pairs and no lone pairs, so the molecular geometry is the same as the electron geometry of tetrahedral. COCl2 has three bonding pairs and no lone pairs surrounding the central carbon, so the molecular geometry is trigonal planar.

Determine the electron geometry for each molecule. CF4 NF3 OF2 H2S

Tetrahedral: CF4, NF3, OF2, H2S Each of these molecules has four electron groups. CF4 has four bonding groups and zero lone pairs, NF3 has three bonding groups and one lone pair, and both OF2 and H2S have two bonding groups and two lone pairs. Therefore, the electron geometry (tetrahedral) is the same for each, as summarized in the table:

Draw the Lewis structure (including resonance structures) for diazomethane (CH2N2) . For each resonance structure, assign formal charges to all atoms that have formal charge. Draw the molecules by placing atoms on the canvas and connecting them with bonds. Include all lone pairs of electrons. Show the formal charges of all atoms in the correct structures.

The formal charge is calculated as the number of valence electrons minus the number of nonbonding electrons minus half of the bonding electrons. The central nitrogen atom carries a formal charge of +1 in both structures because it only contains bonding electrons ( 5−0−0.5(8)=1 ). The resonance forms alternate between carbon carrying the formal charge of − 1 and nitrogen carrying the formal charge of − 1, where the more stable form is when nitrogen carries the negative formal charge.

Determine the idealized bond angle for each molecule. CF4 NF3 OF2 H2S

The idealized bond angles for a tetrahedral electron geometry are 109.5 ∘ ; however, the actual bond angles for molecules may differ depending on how strongly each electron group repels other groups.

Draw the Lewis structure (including resonance structures) for the acetate ion (CH3COO−) . For each resonance structure, assign formal charges to all atoms that have formal charge.

The skeletal structure for the acetate ion has the two carbon atoms bonded to each other with one carbon bound to three hydrogen atoms and the other carbon to two oxygen atoms. The total number of electrons for the Lewis structure is the sum of the valence electrons for each atom in the ion plus one extra electron for the 1 − charge. There are a total of 24 valence electrons for the structure. Distribute the electrons among the atoms, creating full valence shells for as many atoms as possible. The carbon attached to two oxygen atoms does not have a complete octet and forms a double bond with one oxygen. The double bond can form with either oxygen atom so there are two resonance structures for the ion.

Water alone does not easily remove grease from dishes or hands because grease is nonpolar and water is polar. The addition of soap to water, however, allows the grease to dissolve. Study the structure of sodium stearate (a soap) and describe how its structure allows it to interact with both nonpolar grease and polar water.

The soap molecule has a nonpolar hydrocarbon end and a polar anionic end when it is dissolved in water. When it dissolves in water, the sodium stearate congregates to form small spheres (called micelles) with the hydrocarbon end on the inside and the anionic end on the surface. The anionic end interacts with the polar water molecules, while the hydrocarbon end can attract and interact with the nonpolar grease. This allows the soapy water to remove the grease by trapping the grease inside the micelle. Owing to the nature of the interaction with water molecules, the nonpolar hydrocarbon end is often called hydrophobic, while the polar ionic end is called hydrophilic. Chemical compounds possessing both hydrophilic and hydrophobic properties are known as amphiphilic. Hydrocarbon-based surfactants, phospholipids, cholesterol, and fatty acids are examples of amphiphilic compounds.

Draw four Lewis structures corresponding to HSCN with C as the central atom, HSCN with N as the central atom, N2O with N as the central atom, and N2O with O as the central atom. Use formal charges for each of these four Lewis structures to explain why nitrogen is the central atom of N2O but carbon is the central atom of HSCN . Match the items in the left column to the appropriate blanks in the sentences on the right.

When C is the central atom of HSCN , the formal charge on the central C atom is 0 When N is the central atom of HSCN , the formal charge on the central N atom is 1+ When N is the central atom of N2O , the formal charge on the central N atom is 1+ When O is the central atom of N2O , the formal charge on the central O atom is 2+ An N atom is more electronegative than C and less electronegative than O. Since structures that place positive formal charges on the more electronegative atoms are less stable, for both HSCN and N2O the structure observed has the less electronegative atom as the central atom.

Use the dipole moments of HF and HBr ( μHF= 1.82 D , μHBr = 0.79 D , 1D=3.34×10−30C⋅m ) together with the percent ionic character of each bond ( HF 45 % ionic, HBr 11 % ionic) to estimate the bond length in each molecule. Express your answers using two significant figures separated by a comma.

bond length of HF , HBr =84,150pm

In the resonance structure with the electron geometry "C tetrahedral; 1st N tetrahedral; 2nd N linear" would you expect the two nitrogennitrogen bond angles and bond lengths to be the same or different? the C-N-N bond angle will be closer to 109 ∘ ; the N-N-N bond angle will be 180 ∘ ; the N1−N2 bond length will be longer than the N2−N3 bond length the C-N-N bond angle will be closer to 109 ∘ ; the N-N-N bond angle will be 180 ∘ ; the two N-N bond lengths will be the same the C-N-N bond angle will be closer to 120 ∘ ; the N-N-N bond angle will be 180 ∘ ; the two N-N bond lengths will be the same the C-N-N bond angle will be closer to 120 ∘ ; the N-N-N bond angle will be 180 ∘ ; the N1−N2 bond length will be longer than the N2−N3 bond length

the C-N-N bond angle will be closer to 109 ∘ ; the N-N-N bond angle will be 180 ∘ ; the N1−N2 bond length will be longer than the N2−N3 bond length