Molecular Geometry and Polarity

The structure of benzene, including its two resonance forms, is shown below. How many unhybridized p orbitals are present on each of the C atoms in the resonance structures? Use a whole number in your answer.

1 Each of the carbon atoms is bonded to three other atoms with no lone pairs, so the correct hybridization is sp2. That leaves one unhybridized p orbital on each of the carbon atoms, which form the π-bonds that contribute to the resonance.

Which of the following is the most likely bond angle in a molecule where the central atom is sp hybridized?

180∘ Molecules with central atoms that are sp hybridized tend to form linear geometry, which corresponds to a bond angle of 180∘

How many σ-bonds are there in the molecule ethene, C2H4, whose orbital structure is shown below? Use a whole number in your answer.

5 In the ethene molecule, C2H4, there are five σ-bonds. One C−C σ-bond results from overlap of an sp2 hybrid orbital on one carbon atom with an sp2 hybrid orbital on the other carbon atom. Four bonds result from the overlap between the atoms' sp2 orbitals with s orbitals of each of the individual hydrogen atoms.

pi

A _____ bond is a covalent bond where the regions of orbital overlap are above and below the internuclear axis.

What happens to bond energy at distances shorter than bond distance?

At distances shorter than the bond distance, the positive charges on the two nuclei repel each other, and the overall energy increases. The specific average bond distance is the distance with the lowest energy. At distances less than the bond distance, the repulsion between the two positively charged nuclei begins to overcome the attraction between each electron and the opposite nucleus, and the overall energy increases.

Which molecule has a central atom that is sp3d hybridized?

BrF3 With two lone pairs and three single bonds, bromine has five electron domains which will make it sp3d hybridized.

Consider the following Lewis structure for*Urea*, a compound produced by the metabolism of amino acids. H−N⋅⋅|H−C−||:O:N⋅⋅|H−H Which atom(s) in the molecule will be at the center of a local trigonal planar molecular structure?

C The carbon atom has three electron domains, all of which are bonds to atoms, and so it will display trigonal planar electron-pair geometry and molecular geometry. The nitrogen atoms each have four electron domains and one lone pair, and so they will exhibit tetrahedral electron pair geometry and trigonal pyramidal molecular geometry

Which molecule will have no net dipole? CH3Cl CH2Cl2 CHCl3 CCl4

CCl4 The four polar bonds will cancel out, leaving the molecule nonpolar. CCl4 and CH4 have tetrahedral shape and are symmetric so they have dipole moment zero...

Which of the following has the least lattice energy? NaF CsI CaO

CsI The charges are CsI (+1,−1), NaF (+1,−1), CaO (+2,−2). Na and F ions are smaller than Cs and I, and since lattice energy is inversely proportional to the size of the ions, CsI would have the lowest value.

Which bond is the shortest? F-F Cl-Cl Br-Br I-I

F-F The smaller the atoms involved, the shorter the covalent bond.

Which molecule has bond angles that are not reflective of hybridization? H2Te OF2 NH3 CH4

H2Te Tellurium is a large atom, relatively speaking, so hybridization will not come into play for its valence shell, and it will exhibit 90∘ bond angles that are sufficiently described by pure valence bond theory.

Which of the following statements about hybridization are true? Hybrid orbitals exist in isolated atoms. Hybrid orbitals within the same atom have the same energy and shape. Hybrid orbitals are described mathematically as a linear combination of atomic orbitals. An atom can have both hybridized and unhybridized orbitals at the same time.

Hybrid orbitals within the same atom have the same energy and shape. Hybrid orbitals are described mathematically as a linear combination of atomic orbitals. An atom can have both hybridized and unhybridized orbitals at the same time. Hybridization is a model used to describe the observed bonding in molecules. These hybrid orbitals are described mathematically by a process called LCAO, or the linear combination of atomic orbitals. The hybrid orbitals are equal in size, shape, and energy. sp and sp2 hybridization mixes 1 or 2 p orbitals with the s orbitals, while the remaining p orbital(s) do not form hybrid orbitals.

Which molecule has a central atom that is sp3d2 hybridized? Select the correct answer below: ClF3 SF4 PCl5 IF5

IF5 The iodine atom in IF5 has six electron domains, in the form of five single bonds and one lone pair, so it will be sp3d2 hybridized. ClF3 ,SF4, and PCl5 each have five electron domains, making the central atoms sp3d hybridized. CLF3 sp3d

What is the correct neutral ionic compound when magnesium (Mg) ions and chloride (Cl) ions are combined?

MgCl2 One Mg2+ ion and two Cl− ions will result in MgCl2, a neutral compound.

Which of the following has the largest dipole moment? NCl3 BF3 CO2 CF4

NCl3 The molecule NCl3 is the only molecule listed above that has a dipole moment. BF3, CO2 and CF4 are all symmetrical molecules in which all the polar covalent bonds are identical. Therefore, BF3, CO2 and CF4 are all non-polar molecules, and NCl3 has the largest dipole moment.

Which of the following molecules has a nitrogen atom with an sp2 hybridization? NO3− HCN N2 NH3

NO3− The NO3− molecule contains a nitrogen with 3 electron regions so its hybridization is sp2.

Which of the following molecules will contain a node, or nodal plane, along the internuclear axis? O2 H2 N2 HCl Br2

O2 N2 In a π bond, the regions of orbital overlap lie on opposite sides of the internuclear axis. Along the axis itself, there is a node: a plane in which there is no probability of finding an electron. Of the molecules listed above, only O2 and N2 will contain π bonds, and so only O2 and N2 will contain a node along the internuclear axis.

What statement is NOT correct about σ bonding?

Rotation around single σ-bonds does not occur easily. Rotation around single (σ) bonds occurs easily because the end-to-end orbital overlap does not depend on the relative orientation of the orbitals on each atom in the bond. Rotation around the internuclear axis does not change the extent to which the bonding orbitals overlap because the bonding electron density is symmetric about the axis. Hybrid orbitals on carbon are used to form the σ-bonds between carbon atoms and between carbon and hydrogen atoms.

pi bonds in carbon?

The atom has one unhybridized p orbital, with a blue lobe and a red lobe, that is in the plane of the page and oriented vertically. Since π-bonds are formed from unhybridized p orbitals, this atom may form only one π-bond; it might represent, for example, the π-bond of the carbon-carbon double bond in ethene.

The orbital structure of the molecule acetylene, C2H2, is shown below. What do the dashed lines signify?

The four dashed lines indicate the formation of *two* π-bonds that are formed from the four unhybridized p orbitals. In the acetylene molecule, C2H2, there are two π-bonds, which is indicated by the four dashed lines in the orbital structure shown. There are two dashed lines for each of the π-bonds because each of the two unhybridized 2p orbitals on a carbon atom has two areas of electron density, one on either side of the atom, that overlap with the similar areas of electron density on the two unhybridized 2p orbitals on the other carbon atom. This makes a total of four 2p orbitals and two π-bonds.

Which of the following must be true if a nitrogen in a compound displays a tetrahedral molecular geometry? : The nitrogen atom must form three bonds and have one lone pair. The nitrogen atom must form four bonds. The nitrogen atom must form two bonds and have two lone pairs None of the above; a nitrogen atom can not display a tetrahedral molecular geometry.

The nitrogen atom must form four bonds. In order for a nitrogen atom to display a tetrahedral molecular geometry, it must four bonds and no lone pairs. Consider the ammonium cation NH+4. This ion is a simple example of a nitrogen atom with four bonds and no lone pairs, and so it will display a tetrahedral molecular geometry. Include some information on CADMS studies on N4 supposed disproval of the Td tetrahedral model of N atoms having equivalency in the molecular structure.

CO2

The structure of carbon dioxide is O=C=O. This molecule has two carbon-oxygen double bonds. Each of these double bonds consists of one C−O σ bond and one C−O π bond. Therefore, there are 2 σ bonds and 2 π bonds.

The orbital structure of the molecule ethene, C2H4, is shown below. What do the dashed lines signify?

The two dashed lines indicate the formation of one π-bond that is formed from the two unhybridized p orbitals. In the ethene molecule, C2H4, there is one π-bond, which is indicated by the two dashed lines in the orbital structure shown. There are two dashed lines because each of the unhybridized 2p orbitals has two areas of electron density, one above and one below the plane of the molecule, but only one π-bond.

No element in the second row can form more than a triple bond.

What is the maximum bond order that is possible for a bond between elements in the second row of the periodic table? 3!

Which molecule has a central atom that is sp3d2 hybridized?

XeF4 will have six electron domains, in the form of four bonds and two lone pairs, so it will be sp3d2 hybridized.

Which of the following produces sigma bonds? the overlap of an s and p orbital the overlap of two s orbitals the end-to-end overlap of two p orbitals all of the above

all of the above A sigma bond is a covalent bond that can be formed through the overlap of two p orbitals (end-to-end), two s orbitals, or an s and a p orbital. The electron density for all of these is concentrated along the internuclear axis, allowing for a sigma bond to form.

C-O-H

bent The oxygen atom has four electron domains so the electron pair geometry is tetrahedral. Two of the four regions are due to nonbonding pairs so the molecular structure is bent.

In the structure illustrating the bonding in ethene, C2H4, the two arrows point to what type of bond or bonds?

one π-bond with one lobe above and one lobe below the plane of the molecule The π-bond is formed by the side-by-side overlap of the two unhybridized p orbitals in the two carbon atoms of ethene. The two lobes of the π-bond are above and below the plane of the molecule.

If the portion of one orbital and the portion of another occupy the same region of space, this is known as:

overlap Overlap occurs between the orbitals of two different atoms when a portion of each orbital occupies the same space, resulting in a covalent bond.

Which type of orbital overlap results in the strongest covalent bond?

direct overlap Direct overlap maximizes the volume shared by the orbitals and thus maximizes bond strength.

In an energy diagram, sp3 hybridized orbitals are: below 2s orbitals above 2p orbitals in between 2s and 2p orbitals

in between 2s and 2p orbitals sp3 hybridized orbitals have an energy in between 2s and 2p orbitals.

A σ bond

is formed by the direct overlap of hybrid orbitals

What must be true about an atom that is $_sp^{3}d$_ hybridized?

it can be found in the third period In order to become $_sp^{3}d$_ hybridized, an atom must have access to $_d$_ orbitals, which begin at $_n = 3$_. Therefore, of the available choices, only the third period is possible.

Given the equation ΔHlattice= (C(Z+)(Z−))/Ro, which of the following will result in a stronger lattice?

larger charges and closer ions As the interionic distance (Ro) decreases, the denominator will decrease so the lattice energy will be higher. Similarly, if the charges (Z+ and Z−) increase, the numerator will increase so the lattice energy will be higher.

In order for a pi bond to form, there must be

overlap of unhybridized orbitals Pi bonds are created by the lateral overlap of unhybridized orbitals that contain at least one node along the internuclear plane, while sigma bonds occur through overlap of hybridized orbitals. Since s orbitals do not contain nodes, they cannot be involved in making pi bonds.

Unhybridized orbitals overlap to form:

pi bonds Unhybridized orbitals overlap to form pi bonds. Pi bonds form parallel to the bonded atoms and do not form a new hybrid orbital (unlike sigma bonds that form new hybrid orbitals linearly). The orbitals that overlap to form the sigma bonds must overlap head to head or end to end. The hybrid orbitals about a central atom always are directed at the bonded atoms. Hybrid orbitals will always overlap head to head to form sigma bonds. The unhybridized p atomic orbitals are used to form π bonds.

Hybridization predicts geometry best for:

small atoms Hybridization begins to break down in its ability to predict geometry as atoms get larger.

What is the hybridization of the carbon atom shown in the structure below?

sp The atom has two unhybridized p orbitals, each with a "blue" lobe and a "red" lobe. One p orbital is in the plane of the page and oriented vertically, and one p orbital is perpendicular to the plane of the page. There are two identical orbitals in the plane of the page shown in orange, one pointing to the left and one pointing to the right; these are the two identical sp hybrid orbitals.

What is the hybridization of S in H2S?

sp3 S in H2S has an sp3 hybridization because of its 4 regions of electron density (2 bonding, 2 nonbonding).

An sp3 hybridized atom exhibits which geometry?

tetrahedral

Which of the following determines the strength of a covalent bond?

the amount of overlap for the orbitals involved When overlap occurs between two atomic orbitals, the positively charged nuclei and the negatively charged electrons for the atoms involved create a covalent bond. The more overlap that occurs between two orbitals, the stronger the attraction between the positively charged nuclei and the negatively charged electrons. Therefore, the more overlap that occurs, the stronger the covalent bond.

An ionic compound is stable due to:

the electrostatic attraction between its positive and negative ions The strong electrostatic attraction between the cations (+) and anions (−) in an ionic compound due to their opposite charges allows for an extremely stable bond to form.

pi bonds in acetyline?

two π-bonds In acetylene, the carbon atoms have sp hybridization. The two unhybridized p orbitals per carbon are positioned such that they overlap side by side and, hence, form two π-bonds. The four arrows in the figure point to the four lobes of the two π-bonds; there is one π-bond with a red and blue lobe in the plane of the page above and below the C−C bond, and one π-bond with a red and blue lobe in the plane perpendicular to the plane of the page.

ACETYLNE

two π-bonds formed from p orbitals and one σ-bond formed from sp hybrid orbitals The sp hybrid orbitals of the two carbon atoms overlap end to end to form a σ-bond between the carbon atoms. The remaining hybridized sp orbitals form σ bonds with the two hydrogen atoms at the far ends of the two carbons. The two unhybridized p orbitals on each carbon are positioned such that they overlap side by side and, hence, form two π-bonds. The two carbon atoms of acetylene are thus bound together by one σ-bond and two π-bonds, giving a triple bond.

A hybridized atomic orbital can contain or participate in

σ-bonds lone pairs of electrons single unpaired electrons (radicals) A hybrid orbital can make σ-bonds, and both lone pairs and single unpaired electrons can occupy hybrid orbitals. π-bonds are made by overlap of unhybridized p orbitals.