Question 1

Valence bond theory predicts that tin will use _____ hybrid orbitals in SnCl₃^-.

Accepted Answer

To determine the hybridization of a central atom, we need to count the number of regions of electron density around it.

For our molecule, we would count: 
- 1 region from the lone pair on tin
- 4 regions from the surrounding atoms (2 each from Cl and S)

This gives a total of 5 regions of electron density.

To accommodate 5 regions, tin would need to hybridize its orbitals. The hybridization would involve one s and three p orbitals, giving four sp3 hybrid orbitals.

Therefore, the correct hybridization for tin in this molecule is spS1P13S1S1P0. The correct answer is C.

Question 2

According to valence bond theory, which of the following molecules involves sp² hybridization of orbitals on the central atom? (central atom is bold)

Accepted Answer

In order to determine the hybrid orbitals on the central atom, we need to count the number of electron domains (bonded atoms and lone pairs) around it. The central atom in option B, carbon, has two bonded atoms (S and H), two lone pairs, and one double bond, giving it a total of five electron domains. To accommodate these domains, the carbon atom undergoes spS1U1P12S1S1P0 hybridization, which means it forms one sigma bond with sulfur using an sp orbital, three sigma bonds with hydrogen and sulfur using three spS1U1P12 hybrid orbitals, and one pi bond with oxygen using a p orbital. The other options have either a different number of electron domains or a different type of hybridization.

Question 3

Carbon uses ______ hybrid orbitals in ClCN.

Accepted Answer

In ClCN, carbon is bonded to both chlorine and nitrogen. Chlorine is a larger atom than nitrogen, so it will have more influence on the hybridization of carbon. Chlorine has three lone pairs and one bond, so its electronic geometry is tetrahedral. To accommodate this, carbon must have sp3 hybridization. However, since there are only two atoms bonded to carbon, only two of the hybrid orbitals are used for bonding, leaving the other two as lone pairs. Therefore, carbon uses two sp hybrid orbitals for bonding in ClCN.

Question 4

Valence bond theory predicts that sulfur will use _____ hybrid orbitals in sulfur dioxide, SO₂.

Accepted Answer

The answer of Valence bond theory predicts that sulfur will...

Question 5

Valence bond theory predicts that tin will use _____ hybrid orbitals in SnF₅^-.

Accepted Answer

According to valence bond theory, the central atom (tin, Sn) in a molecule forms hybrid orbitals by combining its valence atomic orbitals. The number and type of hybrid orbitals formed depends on the number of bonding and lone pairs of electrons around the central atom. In SnF₅^-, there are one sulfur atom and two fluorine atoms bonded to tin, and one lone pair on sulfur. This means that tin will form four hybrid orbitals to bond with the surrounding atoms and lone pair. The hybridization for the four hybrid orbitals can be determined by the following formula: hybridization = number of atoms bonded to Sn + number of lone pairs on Sn = 3 + 1 = 4 Since there are four hybrid orbitals, the hybridization is sp3. Therefore, the correct choice is C, which shows the correct hybridization of sp³. The other choices do not correctly show the sp3 hybridization needed for four hybrid orbitals.

Question 6

Valence bond theory predicts that xenon will use _____ hybrid orbitals in XeOF₄.

Accepted Answer

Xenon in XeOF4 forms five bonds (four with oxygen and fluorine atoms and one with a lone pair), requiring sp^3d^2 hybridization to accommodate the five regions of electron density around the xenon atom.

Question 7

Valence bond theory predicts that carbon will use _____ hybrid orbitals in the carbonate anion, CO₃^2-.

Accepted Answer

In the carbonate anion, CO3^2-, carbon is bonded to three oxygen atoms and has no lone pairs, which requires sp^2 hybridization to form three equivalent sigma bonds and accommodate the trigonal planar geometry.

Question 8

For which one of the following molecules is the indicated type of hybridization not appropriate for the central atom?

Accepted Answer

The answer of For which one of the following molecules...

Question 9

A molecule with the formula AX₄E₂ uses _________ to form its bonds.

Accepted Answer

The formula AX₄E₂ indicates a molecule with six regions of electron density, which corresponds to sp³d² hybridization.

Question 10

According to valence bond theory, which orbitals on N and H overlap in the NH₃molecule?

Accepted Answer

In NH₃, nitrogen undergoes sp³ hybridization, and these sp³ orbitals overlap with the 1s orbitals of hydrogen atoms to form N-H bonds.

Question 11

When PCl₅solidifies it forms PCl₄⁺ cations and PCl₆^-anions. According to valence bond theory, what hybrid orbitals are used by phosphorus in the PCl₄⁺ cations?

Accepted Answer

Phosphorus in the PClS1U1B14S1U1B0S1U1P1+S1S1P0 cation has a total of four hybrid orbitals due to the bonding with the four surrounding atoms (three chlorine atoms and one phosphorus atom). The central phosphorus atom is sp3 hybridized, meaning it has three hybrid orbitals in the plane of the three chlorine atoms (S1U1P13S1S1P0), and one hybrid orbital perpendicular to that plane (S1U1P0).

Question 12

Valence bond theory predicts that bromine will use _____ hybrid orbitals in BrF₅.

Accepted Answer

Valence bond theory suggests that bromine in BrF5 uses sp3d2 hybrid orbitals to accommodate the five bonds with fluorine and one lone pair, resulting in an octahedral electron geometry.

Question 13

Valence bond theory predicts that bromine will use _____ hybrid orbitals in BrF₃.

Accepted Answer

The Br atom is bonded to one F, one S, and two B atoms in BrF₃. To accommodate these four bonds, Br needs four hybrid orbitals. The hybridization of Br in this molecule can be determined using the formula: hybridization = number of sigma bonds + number of lone pairs on central atom. In this case, Br has four sigma bonds and no lone pairs, so its hybridization is sp3. The C option provides four hybrid orbitals: one s orbital, three p orbitals, and one d orbital. All other options provide either too few or too many hybrid orbitals for sp3 hybridization.

Question 14

Which one of the following statements about orbital hybridization is incorrect?

Accepted Answer

The answer of Which one of the following statements about...

Question 15

A molecule with the formula AX₂ uses _________ to form its bonds.

Accepted Answer

A molecule with the formula AX₂ typically forms linear geometry, which is associated with sp hybridization.

Question 16

A molecule with the formula AX₃E uses _________ to form its bonds.

Accepted Answer

The answer of A molecule with the formula AX₃E uses...

Question 17

A molecule with the formula AX₄uses _________ to form its bonds.

Accepted Answer

The formula indicates that there are two central atoms, A and B, with 1 and 14 bonded atoms respectively. The S1U1 and S1 in the formula represent lone pairs of electrons on A and B, respectively. The P13 and P12 indicate the number of valence electrons on A and B, respectively. To form the bonds in this molecule, A and B would hybridize their valence orbitals, which in this case would be s, p, and d orbitals. However, based on the formula, there are not enough electrons to form d orbitals on A or B, so we can eliminate answer D and E. The remaining choices involve sp hybrid orbitals, which would not allow for the necessary number of bonds to be formed, so we can eliminate answer A. The correct answer is therefore C, which involves sp ³ hybrid orbitals.

Question 18

A molecule with the formula AX₃ uses __________ to form its bonds.

Accepted Answer

The answer of A molecule with the formula AX₃ uses...

Question 19

A molecule with the formula AX₄E uses _________ to form its bonds.

Accepted Answer

The answer of A molecule with the formula AX₄E uses...

Question 20

Valence bond theory predicts that iodine will use _____ hybrid orbitals in ICl₂^-.

Accepted Answer

In ICl₂^-, iodine has three regions of electron density due to the bonds and lone pair. Valence bond theory predicts that iodine will use hybrid orbitals to form these bonds and the lone pair. Since there are three regions of electron density, it requires a hybridization of sp2 (one s orbital and two p orbitals). Therefore, the hybridization of iodine in this molecule would be sp³. Choice C has the correct hybridization but also includes an unnecessary d orbital, which does not play a role in the hybridization of iodine in this molecule.

Quiz 11: Theories of Covalent Bonding

Valence bond theory predicts that tin will use _____ hybrid orbitals in SnCl3-.

According to valence bond theory, which of the following molecules involves sp2 hybridization of orbitals on the central atom? (central atom is bold)

Carbon uses ______ hybrid orbitals in ClCN.

Valence bond theory predicts that sulfur will use _____ hybrid orbitals in sulfur dioxide, SO2.

Valence bond theory predicts that tin will use _____ hybrid orbitals in SnF5-.

Valence bond theory predicts that xenon will use _____ hybrid orbitals in XeOF4.

Valence bond theory predicts that carbon will use _____ hybrid orbitals in the carbonate anion, CO32-.

For which one of the following molecules is the indicated type of hybridization not appropriate for the central atom?

A molecule with the formula AX4E2 uses _________ to form its bonds.

According to valence bond theory, which orbitals on N and H overlap in the NH3 molecule?

When PCl5 solidifies it forms PCl4+ cations and PCl6- anions. According to valence bond theory, what hybrid orbitals are used by phosphorus in the PCl4+ cations?

Valence bond theory predicts that bromine will use _____ hybrid orbitals in BrF5.

Valence bond theory predicts that bromine will use _____ hybrid orbitals in BrF3.