Draw The Lewis Structure For Of2. What Is The Hybridization On The O Atom?

Chapter 8. Advanced Theories of Covalent Bonding

8.2 Hybrid Atomic Orbitals

Learning Objectives

By the cease of this section, you will be able to:

• Explicate the concept of diminutive orbital hybridization
• Decide the hybrid orbitals associated with various molecular geometries

Thinking in terms of overlapping diminutive orbitals is one mode for us to explain how chemical bonds form in diatomic molecules. However, to understand how molecules with more than 2 atoms form stable bonds, we crave a more detailed model. Equally an example, let us consider the water molecule, in which we accept 1 oxygen atom bonding to two hydrogen atoms. Oxygen has the electron configuration is ²2s ²iip ⁴, with 2 unpaired electrons (one in each of the ii twop orbitals). Valence bond theory would predict that the two O–H bonds grade from the overlap of these two 2p orbitals with the 1s orbitals of the hydrogen atoms. If this were the case, the bond angle would be 90°, equally shown in Figure i, considering p orbitals are perpendicular to each other. Experimental evidence shows that the bond bending is 104.5°, not 90°. The prediction of the valence bail theory model does not match the existent-world observations of a h2o molecule; a unlike model is needed.

Effigy 1.The hypothetical overlap of two of the 2p orbitals on an oxygen cantlet (cherry-red) with the 1southward orbitals of 2 hydrogen atoms (blue) would produce a bond angle of 90°. This is not consistent with experimental prove.^[1]

Quantum-mechanical calculations suggest why the observed bond angles in H₂O differ from those predicted by the overlap of the 1due south orbital of the hydrogen atoms with the twop orbitals of the oxygen atom. The mathematical expression known equally the wave function, ψ, contains data well-nigh each orbital and the wavelike backdrop of electrons in an isolated cantlet. When atoms are bound together in a molecule, the wave functions combine to produce new mathematical descriptions that accept different shapes. This process of combining the wave functions for atomic orbitals is called hybridization and is mathematically accomplished by the linear combination of diminutive orbitals, LCAO, (a technique that we volition run across again later). The new orbitals that consequence are called hybrid orbitals. The valence orbitals in an isolated oxygen cantlet are a 2s orbital and iii 2p orbitals. The valence orbitals in an oxygen atom in a water molecule differ; they consist of four equivalent hybrid orbitals that signal approximately toward the corners of a tetrahedron (Figure 2). Consequently, the overlap of the O and H orbitals should outcome in a tetrahedral bond angle (109.5°). The observed bending of 104.v° is experimental evidence for which quantum-mechanical calculations give a useful explanation: Valence bail theory must include a hybridization component to give accurate predictions.

Figure 2. (a) A water molecule has four regions of electron density, so VSEPR theory predicts a tetrahedral organization of hybrid orbitals. (b) Two of the hybrid orbitals on oxygen contain lone pairs, and the other two overlap with the 1s orbitals of hydrogen atoms to form the O–H bonds in H_twoO. This description is more consistent with the experimental structure.

The post-obit ideas are of import in agreement hybridization:

1. Hybrid orbitals do not be in isolated atoms. They are formed simply in covalently bonded atoms.
2. Hybrid orbitals take shapes and orientations that are very different from those of the atomic orbitals in isolated atoms.
3. A ready of hybrid orbitals is generated by combining atomic orbitals. The number of hybrid orbitals in a ready is equal to the number of atomic orbitals that were combined to produce the prepare.
4. All orbitals in a set of hybrid orbitals are equivalent in shape and free energy.
5. The type of hybrid orbitals formed in a bonded atom depends on its electron-pair geometry equally predicted past the VSEPR theory.
6. Hybrid orbitals overlap to form σ bonds. Unhybridized orbitals overlap to form π bonds.

In the following sections, we shall discuss the mutual types of hybrid orbitals.

sp Hybridization

The glucinium atom in a gaseous BeCl_two molecule is an example of a central atom with no lone pairs of electrons in a linear arrangement of three atoms. There are 2 regions of valence electron density in the BeCl₂ molecule that correspond to the ii covalent Be–Cl bonds. To adjust these two electron domains, two of the Be cantlet'due south iv valence orbitals will mix to yield two hybrid orbitals. This hybridization process involves mixing of the valence s orbital with ane of the valence p orbitals to yield two equivalent sp hybrid orbitals that are oriented in a linear geometry (Figure 3). In this effigy, the prepare of sp orbitals appears similar in shape to the original p orbital, but in that location is an important difference. The number of atomic orbitals combined always equals the number of hybrid orbitals formed. The p orbital is i orbital that can hold up to two electrons. The sp set is 2 equivalent orbitals that point 180° from each other. The two electrons that were originally in the south orbital are now distributed to the two sp orbitals, which are half filled. In gaseous BeCl₂, these half-filled hybrid orbitals volition overlap with orbitals from the chlorine atoms to form 2 identical σ bonds.

Figure 3. Hybridization of an s orbital (blue) and a p orbital (red) of the aforementioned cantlet produces two sp hybrid orbitals (majestic). Each hybrid orbital is oriented primarily in just one management. Note that each sp orbital contains one lobe that is significantly larger than the other. The gear up of two sp orbitals are oriented at 180°, which is consistent with the geometry for two domains.

We illustrate the electronic differences in an isolated Exist atom and in the bonded Be cantlet in the orbital free energy-level diagram in Figure 4. These diagrams correspond each orbital by a horizontal line (indicating its energy) and each electron by an arrow. Energy increases toward the top of the diagram. Nosotros utilize one upward arrow to bespeak ane electron in an orbital and two arrows (up and downwardly) to indicate 2 electrons of opposite spin.

Figure 4. This orbital energy-level diagram shows the sp hybridized orbitals on Be in the linear BeCl_ii molecule. Each of the ii sp hybrid orbitals holds one electron and is thus half filled and bachelor for bonding via overlap with a Cl 3p orbital.

When atomic orbitals hybridize, the valence electrons occupy the newly created orbitals. The Be atom had 2 valence electrons, so each of the sp orbitals gets ane of these electrons. Each of these electrons pairs up with the unpaired electron on a chlorine atom when a hybrid orbital and a chlorine orbital overlap during the formation of the Be–Cl bonds.

Any fundamental atom surrounded by simply 2 regions of valence electron density in a molecule will showroom sp hybridization. Other examples include the mercury atom in the linear HgCl_ii molecule, the zinc atom in Zn(CH₃)₂, which contains a linear C–Zn–C arrangement, and the carbon atoms in HCCH and CO_two.

Check out the Academy of Wisconsin-Oshkosh website to learn about visualizing hybrid orbitals in iii dimensions.

sp ² Hybridization

The valence orbitals of a central atom surrounded by iii regions of electron density consist of a gear up of three sp ² hybrid orbitals and 1 unhybridized p orbital. This organisation results from sp ² hybridization, the mixing of one s orbital and two p orbitals to produce three identical hybrid orbitals oriented in a trigonal planar geometry (Figure 5).

Effigy v. The hybridization of an s orbital (bluish) and ii p orbitals (cerise) produces iii equivalent sp ² hybridized orbitals (regal) oriented at 120° with respect to each other. The remaining unhybridized p orbital is not shown hither, simply is located along the z centrality.

Although quantum mechanics yields the "plump" orbital lobes every bit depicted in Effigy 5, sometimes for clarity these orbitals are drawn thinner and without the minor lobes, equally in Effigy 6, to avoid obscuring other features of a given illustration. We will utilise these "thinner" representations whenever the truthful view is also crowded to easily visualize.

Figure six. This alternate fashion of drawing the trigonal planar sp ² hybrid orbitals is sometimes used in more crowded figures.

The observed structure of the borane molecule, BH_3, suggests sp ² hybridization for boron in this compound. The molecule is trigonal planar, and the boron atom is involved in three bonds to hydrogen atoms (Figure 7). We tin can illustrate the comparing of orbitals and electron distribution in an isolated boron atom and in the bonded atom in BH₃ every bit shown in the orbital energy level diagram in Figure eight. We redistribute the three valence electrons of the boron atom in the 3 sp ⁱⁱ hybrid orbitals, and each boron electron pairs with a hydrogen electron when B–H bonds class.

Figure seven. BH₃ is an electron-deficient molecule with a trigonal planar structure.

Effigy 8. In an isolated B cantlet, there are one iis and three 2p valence orbitals. When boron is in a molecule with three regions of electron density, three of the orbitals hybridize and create a set of iii sp ⁱⁱ orbitals and one unhybridized 2p orbital. The three half-filled hybrid orbitals each overlap with an orbital from a hydrogen atom to form iii σ bonds in BH₃.

Any fundamental cantlet surrounded by three regions of electron density will exhibit sp ² hybridization. This includes molecules with a lone pair on the primal atom, such as ClNO (Figure 9), or molecules with two unmarried bonds and a double bond connected to the central cantlet, every bit in formaldehyde, CH₂O, and ethene, H₂CCH_ii.

Figure 9. The central cantlet(s) in each of the structures shown contain three regions of electron density and are sp ² hybridized. As we know from the discussion of VSEPR theory, a region of electron density contains all of the electrons that point in one management. A lone pair, an unpaired electron, a single bond, or a multiple bond would each count equally one region of electron density.

sp ³ Hybridization

The valence orbitals of an atom surrounded by a tetrahedral organisation of bonding pairs and lone pairs consist of a set of iv sp ³ hybrid orbitals. The hybrids outcome from the mixing of one south orbital and all 3 p orbitals that produces iv identical sp ³ hybrid orbitals (Figure 10). Each of these hybrid orbitals points toward a different corner of a tetrahedron.

Effigy x. The hybridization of an s orbital (bluish) and three p orbitals (red) produces four equivalent sp ³ hybridized orbitals (purple) oriented at 109.5° with respect to each other.

A molecule of methyl hydride, CH₄, consists of a carbon atom surrounded by four hydrogen atoms at the corners of a tetrahedron. The carbon cantlet in methane exhibits sp ³ hybridization. We illustrate the orbitals and electron distribution in an isolated carbon cantlet and in the bonded cantlet in CH₄ in Figure eleven. The four valence electrons of the carbon atom are distributed equally in the hybrid orbitals, and each carbon electron pairs with a hydrogen electron when the C–H bonds grade.

Figure xi. The iv valence atomic orbitals from an isolated carbon atom all hybridize when the carbon bonds in a molecule like CH₄ with four regions of electron density. This creates four equivalent sp ⁱⁱⁱ hybridized orbitals. Overlap of each of the hybrid orbitals with a hydrogen orbital creates a C–H σ bond.

In a methane molecule, the idue south orbital of each of the four hydrogen atoms overlaps with one of the four sp ⁱⁱⁱ orbitals of the carbon atom to class a sigma (σ) bond. This results in the germination of four potent, equivalent covalent bonds betwixt the carbon atom and each of the hydrogen atoms to produce the methane molecule, CH₄.

The structure of ethane, C_twoH_6, is similar to that of marsh gas in that each carbon in ethane has 4 neighboring atoms bundled at the corners of a tetrahedron—iii hydrogen atoms and ane carbon atom (Figure 12). However, in ethane an sp ^three orbital of i carbon atom overlaps terminate to end with an sp ³ orbital of a 2d carbon atom to form a σ bond between the two carbon atoms. Each of the remaining sp ^three hybrid orbitals overlaps with an due south orbital of a hydrogen atom to form carbon–hydrogen σ bonds. The construction and overall outline of the bonding orbitals of ethane are shown in Figure 12. The orientation of the two CH₃ groups is not stock-still relative to each other. Experimental testify shows that rotation effectually σ bonds occurs hands.

Figure 12. (a) In the ethane molecule, C₂H₆, each carbon has four sp ⁱⁱⁱ orbitals. (b) These four orbitals overlap to form seven σ bonds.

An sp ³ hybrid orbital tin likewise hold a alone pair of electrons. For example, the nitrogen atom in ammonia is surrounded by three bonding pairs and a lone pair of electrons directed to the four corners of a tetrahedron. The nitrogen atom is sp ³ hybridized with one hybrid orbital occupied by the alone pair.

The molecular structure of water is consequent with a tetrahedral arrangement of two lonely pairs and two bonding pairs of electrons. Thus we say that the oxygen atom is sp ³ hybridized, with 2 of the hybrid orbitals occupied by lone pairs and ii by bonding pairs. Since lone pairs occupy more space than bonding pairs, structures that contain lone pairs have bond angles slightly distorted from the ideal. Perfect tetrahedra have angles of 109.5°, but the observed angles in ammonia (107.3°) and water (104.five°) are slightly smaller. Other examples of sp ⁱⁱⁱ hybridization include CCl₄, PCl₃, and NCl₃.

sp ^three d and sp ³ d ² Hybridization

To draw the five bonding orbitals in a trigonal bipyramidal organisation, we must use five of the valence vanquish diminutive orbitals (the southward orbital, the three p orbitals, and ane of the d orbitals), which gives five sp ³ d hybrid orbitals. With an octahedral arrangement of half-dozen hybrid orbitals, nosotros must use six valence vanquish atomic orbitals (the s orbital, the 3 p orbitals, and two of the d orbitals in its valence shell), which gives six sp ^three d ^two hybrid orbitals. These hybridizations are but possible for atoms that have d orbitals in their valence subshells (that is, not those in the first or second period).

In a molecule of phosphorus pentachloride, PCl_v, there are v P–Cl bonds (thus five pairs of valence electrons around the phosphorus atom) directed toward the corners of a trigonal bipyramid. We use the 3south orbital, the iii 3p orbitals, and ane of the threed orbitals to form the set of five sp ⁱⁱⁱ d hybrid orbitals (Figure fourteen) that are involved in the P–Cl bonds. Other atoms that exhibit sp ³ d hybridization include the sulfur atom in SF₄ and the chlorine atoms in ClF₃ and in ClF_iv ⁺. (The electrons on fluorine atoms are omitted for clarity.)

Effigy 13. The three compounds pictured exhibit sp ³ d hybridization in the central atom and a trigonal bipyramid course. SF4 and ClF_iv ⁺ accept i solitary pair of elctrons on the fundamental atom, and ClF₃ has two lone pairs giving information technology the T-shape shown.

Figure 14. (a) The five regions of electron density around phosphorus in PCl₅ require five hybrid sp ³ d orbitals. (b) These orbitals combine to grade a trigonal bipyramidal structure with each big lobe of the hybrid orbital pointing at a vertex. Equally before, there are also small lobes pointing in the opposite management for each orbital (non shown for clarity).

The sulfur atom in sulfur hexafluoride, SF_half-dozen, exhibits sp ³ d ² hybridization. A molecule of sulfur hexafluoride has six bonding pairs of electrons connecting six fluorine atoms to a unmarried sulfur atom. In that location are no lone pairs of electrons on the central atom. To bond half-dozen fluorine atoms, the threes orbital, the three 3p orbitals, and two of the 3d orbitals form half dozen equivalent sp ³ d ² hybrid orbitals, each directed toward a dissimilar corner of an octahedron. Other atoms that showroom sp ³ d ² hybridization include the phosphorus atom in PCl₆ ⁻, the iodine atom in the interhalogens IF₆ ⁺, IF_five, ICl₄ ⁻, IF₄ ⁻ and the xenon atom in XeF₄.

Figure 15. (a) Sulfur hexafluoride, SF_vi, has an octahedral structure that requires sp ^three d ² hybridization. (b) The 6 sp ³ d ² orbitals class an octahedral structure around sulfur. Again, the minor lobe of each orbital is non shown for clarity.

Assignment of Hybrid Orbitals to Central Atoms

The hybridization of an atom is determined based on the number of regions of electron density that surround it. The geometrical arrangements characteristic of the various sets of hybrid orbitals are shown in Figure 16. These arrangements are identical to those of the electron-pair geometries predicted by VSEPR theory. VSEPR theory predicts the shapes of molecules, and hybrid orbital theory provides an explanation for how those shapes are formed. To discover the hybridization of a fundamental atom, we can use the post-obit guidelines:

1. Determine the Lewis structure of the molecule.
2. Determine the number of regions of electron density around an atom using VSEPR theory, in which single bonds, multiple bonds, radicals, and lone pairs each count every bit 1 region.
3. Assign the prepare of hybridized orbitals from Figure xvi that corresponds to this geometry.

Figure xvi. The shapes of hybridized orbital sets are consistent with the electron-pair geometries. For example, an atom surrounded by iii regions of electron density is sp ² hybridized, and the three sp ^two orbitals are arranged in a trigonal planar manner.

It is important to remember that hybridization was devised to rationalize experimentally observed molecular geometries. The model works well for molecules containing pocket-size central atoms, in which the valence electron pairs are close together in space. However, for larger central atoms, the valence-shell electron pairs are farther from the nucleus, and there are fewer repulsions. Their compounds exhibit structures that are ofttimes not consistent with VSEPR theory, and hybridized orbitals are not necessary to explicate the observed data. For case, we accept discussed the H–O–H bond angle in H₂O, 104.5°, which is more consequent with sp ³ hybrid orbitals (109.5°) on the central atom than with 2p orbitals (90°). Sulfur is in the same group as oxygen, and H₂S has a like Lewis structure. However, it has a much smaller bail angle (92.ane°), which indicates much less hybridization on sulfur than oxygen. Standing down the group, tellurium is even larger than sulfur, and for H_twoTe, the observed bond angle (xc°) is consequent with overlap of the 5p orbitals, without invoking hybridization. We invoke hybridization where information technology is necessary to explain the observed structures.

Example one

Assigning Hybridization
Ammonium sulfate is important as a fertilizer. What is the hybridization of the sulfur atom in the sulfate ion, SO₄ ²⁻?

Solution
The Lewis structure of sulfate shows there are four regions of electron density. The hybridization is sp ³.

Check Your Learning
What is the hybridization of the selenium atom in SeF₄?

Answer:

The selenium atom is sp ^three d hybridized.

Example 2

Assigning Hybridization
Urea, NH₂C(O)NH₂, is sometimes used equally a source of nitrogen in fertilizers. What is the hybridization of each nitrogen and carbon atom in urea?

Solution
The Lewis structure of urea is

The nitrogen atoms are surrounded past four regions of electron density, which arrange themselves in a tetrahedral electron-pair geometry. The hybridization in a tetrahedral organization is sp ^three (Figure sixteen). This is the hybridization of the nitrogen atoms in urea.

The carbon atom is surrounded past 3 regions of electron density, positioned in a trigonal planar arrangement. The hybridization in a trigonal planar electron pair geometry is sp ² (Figure sixteen), which is the hybridization of the carbon atom in urea.

Check Your Learning
Acetic acid, H₃CC(O)OH, is the molecule that gives vinegar its aroma and sour taste. What is the hybridization of the ii carbon atoms in acetic acid?

Answer:

H₃ C, sp ³; C(O)OH, sp ⁱⁱ

Cardinal Concepts and Summary

We tin can employ hybrid orbitals, which are mathematical combinations of some or all of the valence atomic orbitals, to describe the electron density effectually covalently bonded atoms. These hybrid orbitals either form sigma (σ) bonds directed toward other atoms of the molecule or contain lone pairs of electrons. We can determine the type of hybridization around a central atom from the geometry of the regions of electron density nigh it. Two such regions imply sp hybridization; iii, sp ² hybridization; four, sp ³ hybridization; five, sp ³ d hybridization; and half-dozen, sp ³ d ^two hybridization. Pi (π) bonds are formed from unhybridized atomic orbitals (p or d orbitals).

Chemistry Terminate of Chapter Exercises

1. Why is the concept of hybridization required in valence bond theory?
2. Requite the shape that describes each hybrid orbital gear up:

   (a) sp ^two

   (b) sp ⁱⁱⁱ d

   (c) sp

   (d) sp ⁱⁱⁱ d ²

3. Explain why a carbon cantlet cannot course five bonds using sp ⁱⁱⁱ d hybrid orbitals.
4. What is the hybridization of the fundamental cantlet in each of the post-obit?

   (a) BeH₂

   (b) SF₆

   (c) PO₄ ³⁻

   (d) PCl₅

5. A molecule with the formula AB₃ could have one of four different shapes. Give the shape and the hybridization of the central A atom for each.
6. Methionine, CH₃SCH₂CH_twoCH(NH₂)CO_twoH, is an amino acid plant in proteins. Depict a Lewis construction of this compound. What is the hybridization type of each carbon, oxygen, the nitrogen, and the sulfur?
   
7. Sulfuric acrid is manufactured by a series of reactions represented past the following equations:[latex]\text{South}_8(due south) + 8 \text{O}_2(g) \longrightarrow 8\text{SO}_2(g)[/latex]
   [latex]two\text{Then}_2(g) + \text{O}_2(m) \longrightarrow ii\text{SO}_3(g)[/latex]
   [latex]\text{SO}_3(grand) + \text{H}_2 \text{O}(50) \longrightarrow \text{H}_2 \text{SO}_4(l)[/latex]

   Depict a Lewis structure, predict the molecular geometry past VSEPR, and decide the hybridization of sulfur for the following:

   (a) round S₈ molecule

   (b) And so_ii molecule

   (c) SO_iii molecule

   (d) H₂SO₄ molecule (the hydrogen atoms are bonded to oxygen atoms)

8. 2 important industrial chemicals, ethene, C_iiH_iv, and propene, C_iiiH₆, are produced past the steam (or thermal) cracking process:

   [latex]2\text{C}_3 \text{H}_8(chiliad) \longrightarrow \text{C}_2\text{H}_4(g) + \text{C}_3\text{H}_6(g) + \text{CH}_4(g) + \text{H}_2(g)[/latex]

   For each of the four carbon compounds, do the post-obit:

   (a) Depict a Lewis construction.

   (b) Predict the geometry about the carbon atom.

   (c) Determine the hybridization of each type of carbon atom.

9. For many years after they were discovered, it was believed that the noble gases could not form compounds. Now nosotros know that conventionalities to exist wrong. A mixture of xenon and fluorine gases, confined in a quartz bulb and placed on a windowsill, is found to slowly produce a white solid. Analysis of the compound indicates that it contains 77.55% Xe and 22.45% F by mass.

   (a) What is the formula of the compound?

   (b) Write a Lewis structure for the compound.

   (c) Predict the shape of the molecules of the compound.

   (d) What hybridization is consistent with the shape you predicted?

10. Consider nitrous acid, HNO₂ (HONO).

    (a) Write a Lewis structure.

    (b) What are the electron pair and molecular geometries of the internal oxygen and nitrogen atoms in the HNO₂ molecule?

    (c) What is the hybridization on the internal oxygen and nitrogen atoms in HNO₂?

11. Strike-anywhere matches contain a layer of KClO₃ and a layer of P₄S₃. The heat produced by the friction of striking the match causes these two compounds to react vigorously, which sets burn down to the wooden stem of the match. KClO₃ contains the ClO_three ⁻ ion. P₄S₃ is an unusual molecule with the skeletal structure.
    

    (a) Write Lewis structures for P_ivDue south₃ and the ClO₃ ^– ion.

    (b) Depict the geometry nigh the P atoms, the Due south atom, and the Cl atom in these species.

    (c) Assign a hybridization to the P atoms, the S atom, and the Cl cantlet in these species.

    (d) Determine the oxidation states and formal charge of the atoms in P₄S_three and the ClO_three ^– ion.

12. Identify the hybridization of each carbon atom in the following molecule. (The arrangement of atoms is given; you need to decide how many bonds connect each pair of atoms.)
    
13. Write Lewis structures for NF₃ and PF_five. On the basis of hybrid orbitals, explicate the fact that NF₃, PF₃, and PF₅ are stable molecules, but NF₅ does not exist.
14. In improver to NF_three, two other fluoro derivatives of nitrogen are known: Northward_iiF_four and N₂F₂. What shapes practice you predict for these two molecules? What is the hybridization for the nitrogen in each molecule?

Glossary

hybrid orbital

orbital created past combining atomic orbitals on a central atom

hybridization

model that describes the changes in the atomic orbitals of an atom when information technology forms a covalent chemical compound

sp hybrid orbital

ane of a set of two orbitals with a linear arrangement that results from combining one s and one p orbital

sp ⁱⁱ hybrid orbital

one of a set of three orbitals with a trigonal planar arrangement that results from combining one s and two p orbitals

sp ³ hybrid orbital

ane of a set of four orbitals with a tetrahedral arrangement that results from combining one southward and three p orbitals

sp ⁱⁱⁱ d hybrid orbital

1 of a set of five orbitals with a trigonal bipyramidal system that results from combining 1 south, three p, and ane d orbital

sp ⁱⁱⁱ d ⁱⁱ hybrid orbital

i of a fix of six orbitals with an octahedral arrangement that results from combining one s, 3 p, and two d orbitals

Solutions

Answers to Chemical science Stop of Chapter Exercises

1. Hybridization is introduced to explain the geometry of bonding orbitals in valance bond theory.

3. There are no d orbitals in the valence shell of carbon.

five. trigonal planar, sp ²; trigonal pyramidal (ane lone pair on A) sp ³; T-shaped (ii lone pairs on A sp ³ d, or (three alone pairs on A) sp ³ d ^two

7. (a) Each S has a bent (109°) geometry, sp ³

(b) Bent (120°), sp ²

(c) Trigonal planar, sp ²

(d) Tetrahedral, sp ³

ix. a) XeF₂

(b)

(c) linear (d) sp ³ d

11. (a)

(b) P atoms, trigonal pyramidal; S atoms, bent, with 2 lone pairs; Cl atoms, trigonal pyramidal; (c) Hybridization virtually P, S, and Cl is, in all cases, sp ³; (d) Oxidation states P +ane, [latex]\text{S} - 1\frac{one}{3}[/latex], Cl +5, O –2. Formal charges: P 0; S 0; Cl +ii: O –ane

13.

Phosphorus and nitrogen tin form sp ³ hybrids to grade three bonds and hold ane lonely pair in PF₃ and NF_iii, respectively. However, nitrogen has no valence d orbitals, so it cannot form a fix of sp ^three d hybrid orbitals to bind 5 fluorine atoms in NF₅. Phosphorus has d orbitals and can bind five fluorine atoms with sp ³ d hybrid orbitals in PF₅.

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