How To Draw Mo Diagrams For Hybridization

Affiliate viii. Avant-garde Theories of Covalent Bonding

8.2 Hybrid Diminutive Orbitals

Learning Objectives

By the end 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 atomic orbitals is one way for us to explain how chemical bonds form in diatomic molecules. However, to understand how molecules with more than two atoms form stable bonds, we require a more than detailed model. As an example, let u.s. consider the water molecule, in which we have one oxygen atom bonding to 2 hydrogen atoms. Oxygen has the electron configuration 1s ²2southward ²twop ^iv, with two unpaired electrons (one in each of the two 2p orbitals). Valence bond theory would predict that the two O–H bonds form from the overlap of these two 2p orbitals with the 1s orbitals of the hydrogen atoms. If this were the example, the bond bending would be xc°, equally shown in Figure 1, considering p orbitals are perpendicular to each other. Experimental testify shows that the bail angle is 104.5°, non 90°. The prediction of the valence bond theory model does non match the real-world observations of a water molecule; a different model is needed.

Figure 1.The hypothetical overlap of two of the 2p orbitals on an oxygen atom (red) with the is orbitals of two hydrogen atoms (blueish) would produce a bond angle of 90°. This is non consistent with experimental prove.^[1]

Breakthrough-mechanical calculations suggest why the observed bond angles in H_twoO differ from those predicted past the overlap of the 1s orbital of the hydrogen atoms with the 2p orbitals of the oxygen atom. The mathematical expression known every bit the wave function, ψ, contains information about each orbital and the wavelike properties of electrons in an isolated cantlet. When atoms are bound together in a molecule, the moving ridge functions combine to produce new mathematical descriptions that have dissimilar shapes. This procedure of combining the wave functions for atomic orbitals is called hybridization and is mathematically accomplished by the linear combination of atomic orbitals, LCAO, (a technique that we will encounter again later). The new orbitals that effect are called hybrid orbitals. The valence orbitals in an isolated oxygen atom are a 2s orbital and three 2p orbitals. The valence orbitals in an oxygen atom in a water molecule differ; they consist of iv equivalent hybrid orbitals that point approximately toward the corners of a tetrahedron (Figure 2). Consequently, the overlap of the O and H orbitals should result in a tetrahedral bail angle (109.5°). The observed angle of 104.5° is experimental evidence for which breakthrough-mechanical calculations requite a useful explanation: Valence bond theory must include a hybridization component to requite accurate predictions.

Figure 2. (a) A water molecule has four regions of electron density, then VSEPR theory predicts a tetrahedral arrangement 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 course the O–H bonds in H₂O. This description is more than consistent with the experimental structure.

The following ideas are of import in agreement hybridization:

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

In the post-obit sections, we shall talk over the mutual types of hybrid orbitals.

sp Hybridization

The beryllium atom in a gaseous BeCl₂ molecule is an example of a central atom with no alone 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 two covalent Be–Cl bonds. To accommodate these two electron domains, two of the Exist atom'south four valence orbitals will mix to yield two hybrid orbitals. This hybridization process involves mixing of the valence s orbital with one of the valence p orbitals to yield 2 equivalent sp hybrid orbitals that are oriented in a linear geometry (Figure 3). In this figure, the prepare of sp orbitals appears like in shape to the original p orbital, but there is an important difference. The number of atomic orbitals combined ever equals the number of hybrid orbitals formed. The p orbital is ane orbital that tin hold upwardly to two electrons. The sp set is two 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 grade two identical σ bonds.

Figure 3. Hybridization of an due south orbital (blueish) and a p orbital (red) of the same atom produces two sp hybrid orbitals (purple). Each hybrid orbital is oriented primarily in just one direction. Notation that each sp orbital contains one lobe that is significantly larger than the other. The set of two sp orbitals are oriented at 180°, which is consequent with the geometry for two domains.

We illustrate the electronic differences in an isolated Be atom and in the bonded Exist cantlet in the orbital energy-level diagram in Figure iv. These diagrams represent each orbital by a horizontal line (indicating its energy) and each electron by an arrow. Free energy increases toward the top of the diagram. We use one upward pointer to indicate one electron in an orbital and two arrows (upward and down) to indicate two electrons of contrary spin.

Effigy iv. This orbital energy-level diagram shows the sp hybridized orbitals on Exist in the linear BeCl_two molecule. Each of the two sp hybrid orbitals holds 1 electron and is thus half filled and bachelor for bonding via overlap with a Cl iiip orbital.

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

Any primal atom surrounded past just 2 regions of valence electron density in a molecule volition exhibit sp hybridization. Other examples include the mercury atom in the linear HgCl₂ molecule, the zinc cantlet in Zn(CH_iii)_two, which contains a linear C–Zn–C arrangement, and the carbon atoms in HCCH and CO_ii.

Check out the University of Wisconsin-Oshkosh website to larn about visualizing hybrid orbitals in three dimensions.

sp ² Hybridization

The valence orbitals of a primal atom surrounded past three regions of electron density consist of a set of iii sp ² hybrid orbitals and one unhybridized p orbital. This arrangement results from sp ⁱⁱ hybridization, the mixing of one south orbital and two p orbitals to produce three identical hybrid orbitals oriented in a trigonal planar geometry (Figure five).

Figure 5. The hybridization of an s orbital (bluish) and two p orbitals (red) produces three equivalent sp ² hybridized orbitals (purple) oriented at 120° with respect to each other. The remaining unhybridized p orbital is not shown hither, just is located forth the z centrality.

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

Effigy 6. This alternate way 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 can illustrate the comparison of orbitals and electron distribution in an isolated boron cantlet and in the bonded atom in BH_three as shown in the orbital energy level diagram in Figure viii. We redistribute the 3 valence electrons of the boron atom in the iii sp ² hybrid orbitals, and each boron electron pairs with a hydrogen electron when B–H bonds form.

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

Figure 8. In an isolated B cantlet, at that place are ane 2south and 3 2p valence orbitals. When boron is in a molecule with iii regions of electron density, three of the orbitals hybridize and create a fix of three sp ² 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 BH_three.

Any central atom surrounded by three regions of electron density will exhibit sp ² hybridization. This includes molecules with a lonely pair on the cardinal cantlet, such every bit ClNO (Figure 9), or molecules with two single bonds and a double bond continued to the central cantlet, as in formaldehyde, CH₂O, and ethene, H_twoCCH₂.

Figure nine. The primal atom(southward) in each of the structures shown incorporate 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 i direction. A alone pair, an unpaired electron, a single bail, or a multiple bond would each count equally one region of electron density.

sp ³ Hybridization

The valence orbitals of an atom surrounded past a tetrahedral system of bonding pairs and lone pairs consist of a set of 4 sp ^three hybrid orbitals. The hybrids result from the mixing of one s orbital and all three 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 10. The hybridization of an s orbital (blueish) 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 methane, CH₄, consists of a carbon cantlet surrounded by four hydrogen atoms at the corners of a tetrahedron. The carbon atom in methyl hydride exhibits sp ³ hybridization. Nosotros illustrate the orbitals and electron distribution in an isolated carbon atom and in the bonded atom in CH₄ in Figure xi. The four valence electrons of the carbon atom are distributed every bit in the hybrid orbitals, and each carbon electron pairs with a hydrogen electron when the C–H bonds course.

Figure 11. The 4 valence atomic orbitals from an isolated carbon cantlet all hybridize when the carbon bonds in a molecule like CH₄ with iv regions of electron density. This creates four equivalent sp ³ hybridized orbitals. Overlap of each of the hybrid orbitals with a hydrogen orbital creates a C–H σ bond.

In a methane molecule, the 1s orbital of each of the 4 hydrogen atoms overlaps with one of the four sp ³ orbitals of the carbon atom to form a sigma (σ) bond. This results in the formation of four strong, equivalent covalent bonds between the carbon atom and each of the hydrogen atoms to produce the methane molecule, CH₄.

The structure of ethane, C_iiH_6, is similar to that of marsh gas in that each carbon in ethane has four neighboring atoms arranged at the corners of a tetrahedron—iii hydrogen atoms and ane carbon atom (Figure 12). All the same, in ethane an sp ³ orbital of i carbon atom overlaps end to end with an sp ^three orbital of a 2d carbon cantlet to form a σ bond between the two carbon atoms. Each of the remaining sp ³ hybrid orbitals overlaps with an due south orbital of a hydrogen cantlet 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 non fixed relative to each other. Experimental testify shows that rotation around σ bonds occurs easily.

Figure 12. (a) In the ethane molecule, C_iiH₆, each carbon has 4 sp ³ orbitals. (b) These four orbitals overlap to class vii σ bonds.

An sp ^three hybrid orbital can also concord a lone pair of electrons. For case, the nitrogen cantlet in ammonia is surrounded past iii bonding pairs and a lone pair of electrons directed to the four corners of a tetrahedron. The nitrogen atom is sp ³ hybridized with i hybrid orbital occupied by the alone pair.

The molecular structure of water is consistent with a tetrahedral arrangement of two lone pairs and 2 bonding pairs of electrons. Thus we say that the oxygen cantlet is sp ³ hybridized, with two of the hybrid orbitals occupied by lone pairs and two by bonding pairs. Since lonely pairs occupy more infinite than bonding pairs, structures that contain lone pairs accept bond angles slightly distorted from the ideal. Perfect tetrahedra accept angles of 109.5°, but the observed angles in ammonia (107.3°) and water (104.5°) are slightly smaller. Other examples of sp ³ hybridization include CCl_four, PCl_three, and NCl_iii.

sp ³ d and sp ³ d ² Hybridization

To depict the five bonding orbitals in a trigonal bipyramidal arrangement, we must apply five of the valence beat atomic orbitals (the s orbital, the three p orbitals, and one of the d orbitals), which gives five sp ³ d hybrid orbitals. With an octahedral system of six hybrid orbitals, we must utilize six valence shell atomic orbitals (the s orbital, the three p orbitals, and two of the d orbitals in its valence shell), which gives 6 sp ⁱⁱⁱ d ² hybrid orbitals. These hybridizations are but possible for atoms that take d orbitals in their valence subshells (that is, not those in the first or 2d period).

In a molecule of phosphorus pentachloride, PCl_five, there are five P–Cl bonds (thus 5 pairs of valence electrons around the phosphorus atom) directed toward the corners of a trigonal bipyramid. We use the threes orbital, the three 3p orbitals, and 1 of the 3d orbitals to course the prepare of five sp ³ d hybrid orbitals (Figure 14) that are involved in the P–Cl bonds. Other atoms that showroom sp ³ d hybridization include the sulfur atom in SF₄ and the chlorine atoms in ClF₃ and in ClF₄ ⁺. (The electrons on fluorine atoms are omitted for clarity.)

Effigy 13. The three compounds pictured exhibit sp ³ d hybridization in the central cantlet and a trigonal bipyramid class. SF4 and ClF₄ ⁺ have ane lone pair of elctrons on the central atom, and ClF₃ has two lone pairs giving information technology the T-shape shown.

Figure fourteen. (a) The 5 regions of electron density around phosphorus in PCl_five crave 5 hybrid sp ³ d orbitals. (b) These orbitals combine to form a trigonal bipyramidal structure with each large lobe of the hybrid orbital pointing at a vertex. As before, there are also small lobes pointing in the opposite management for each orbital (not shown for clarity).

The sulfur cantlet in sulfur hexafluoride, SF₆, exhibits sp ^three d ⁱⁱ hybridization. A molecule of sulfur hexafluoride has half dozen bonding pairs of electrons connecting half dozen fluorine atoms to a single sulfur atom. In that location are no alone pairs of electrons on the cardinal atom. To bond six fluorine atoms, the 3s orbital, the three iiip orbitals, and two of the 3d orbitals form 6 equivalent sp ³ d ² hybrid orbitals, each directed toward a dissimilar corner of an octahedron. Other atoms that exhibit sp ³ d ² hybridization include the phosphorus atom in PCl_half-dozen ⁻, the iodine atom in the interhalogens IF₆ ⁺, IF₅, ICl_iv ⁻, IF_four ⁻ and the xenon atom in XeF₄.

Effigy xv. (a) Sulfur hexafluoride, SF₆, has an octahedral structure that requires sp ³ d ⁱⁱ hybridization. (b) The six sp ⁱⁱⁱ d ² orbitals grade an octahedral structure effectually sulfur. Once more, the minor lobe of each orbital is not shown for clarity.

Assignment of Hybrid Orbitals to Key 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 diverse 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 find the hybridization of a central cantlet, we can employ the following guidelines:

1. Determine the Lewis structure of the molecule.
2. Determine the number of regions of electron density effectually an atom using VSEPR theory, in which single bonds, multiple bonds, radicals, and lone pairs each count as ane region.
3. Assign the set of hybridized orbitals from Figure 16 that corresponds to this geometry.

Figure 16. The shapes of hybridized orbital sets are consistent with the electron-pair geometries. For case, an atom surrounded past three regions of electron density is sp ² hybridized, and the 3 sp ² orbitals are bundled in a trigonal planar fashion.

It is important to think that hybridization was devised to rationalize experimentally observed molecular geometries. The model works well for molecules containing minor cardinal atoms, in which the valence electron pairs are shut together in infinite. 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 often non consistent with VSEPR theory, and hybridized orbitals are non necessary to explain the observed data. For case, we have discussed the H–O–H bond angle in H₂O, 104.5°, which is more consequent with sp ³ hybrid orbitals (109.5°) on the key atom than with 2p orbitals (90°). Sulfur is in the same grouping as oxygen, and H₂Due south has a similar Lewis structure. However, it has a much smaller bond angle (92.1°), which indicates much less hybridization on sulfur than oxygen. Continuing down the group, tellurium is even larger than sulfur, and for H_twoTe, the observed bail angle (90°) is consistent with overlap of the 5p orbitals, without invoking hybridization. We invoke hybridization where it is necessary to explain the observed structures.

Example i

Assigning Hybridization
Ammonium sulfate is of import as a fertilizer. What is the hybridization of the sulfur atom in the sulfate ion, Then₄ ²⁻?

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

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

Reply:

The selenium atom is sp ⁱⁱⁱ d hybridized.

Example 2

Assigning Hybridization
Urea, NH_twoC(O)NH₂, is sometimes used every bit 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 by four regions of electron density, which arrange themselves in a tetrahedral electron-pair geometry. The hybridization in a tetrahedral system is sp ³ (Figure 16). This is the hybridization of the nitrogen atoms in urea.

The carbon atom is surrounded by three regions of electron density, positioned in a trigonal planar organization. The hybridization in a trigonal planar electron pair geometry is sp ² (Figure 16), which is the hybridization of the carbon cantlet in urea.

Cheque Your Learning
Acetic acrid, H₃CC(O)OH, is the molecule that gives vinegar its odor and sour gustation. What is the hybridization of the 2 carbon atoms in acetic acid?

Reply:

H₃ C, sp ⁱⁱⁱ; C(O)OH, sp ²

Fundamental Concepts and Summary

We tin use hybrid orbitals, which are mathematical combinations of some or all of the valence diminutive orbitals, to describe the electron density around covalently bonded atoms. These hybrid orbitals either form sigma (σ) bonds directed toward other atoms of the molecule or contain lone pairs of electrons. Nosotros can determine the type of hybridization around a key atom from the geometry of the regions of electron density near it. Two such regions imply sp hybridization; three, sp ² hybridization; 4, sp ³ hybridization; five, sp ^three d hybridization; and half dozen, sp ³ d ² hybridization. Pi (π) bonds are formed from unhybridized atomic orbitals (p or d orbitals).

Chemistry End of Chapter Exercises

1. Why is the concept of hybridization required in valence bail theory?
2. Give the shape that describes each hybrid orbital prepare:

   (a) sp ²

   (b) sp ^three d

   (c) sp

   (d) sp ^three 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 atom in each of the post-obit?

   (a) BeH_ii

   (b) SF_vi

   (c) PO₄ ⁱⁱⁱ⁻

   (d) PCl_v

5. A molecule with the formula AB₃ could have one of four unlike shapes. Give the shape and the hybridization of the central A atom for each.
6. Methionine, CH₃SCH_twoCH_twoCH(NH_two)CO₂H, is an amino acid found in proteins. Draw a Lewis construction of this chemical compound. What is the hybridization type of each carbon, oxygen, the nitrogen, and the sulfur?
   
7. Sulfuric acid is manufactured past a serial of reactions represented by the following equations:[latex]\text{S}_8(south) + eight \text{O}_2(g) \longrightarrow 8\text{SO}_2(g)[/latex]
   [latex]two\text{SO}_2(g) + \text{O}_2(grand) \longrightarrow 2\text{Then}_3(g)[/latex]
   [latex]\text{SO}_3(grand) + \text{H}_2 \text{O}(50) \longrightarrow \text{H}_2 \text{SO}_4(l)[/latex]

   Draw a Lewis structure, predict the molecular geometry past VSEPR, and determine the hybridization of sulfur for the post-obit:

   (a) circular S₈ molecule

   (b) SO₂ molecule

   (c) SO₃ molecule

   (d) H₂SO_iv molecule (the hydrogen atoms are bonded to oxygen atoms)

8. Ii important industrial chemicals, ethene, C₂H₄, and propene, C_threeH₆, are produced by the steam (or thermal) cracking process:

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

   For each of the 4 carbon compounds, practice the following:

   (a) Describe a Lewis structure.

   (b) Predict the geometry about the carbon atom.

   (c) Decide 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 class compounds. Now nosotros know that belief to be incorrect. 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 past 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 y'all predicted?

10. Consider nitrous acid, HNO_ii (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 comprise a layer of KClO₃ and a layer of P_ivDue south₃. The heat produced by the friction of striking the lucifer causes these 2 compounds to react vigorously, which sets fire to the wooden stem of the match. KClO_three contains the ClO_iii ⁻ ion. P_fourS₃ is an unusual molecule with the skeletal structure.
    

    (a) Write Lewis structures for P₄Southward_three and the ClO₃ ^– ion.

    (b) Describe the geometry near the P atoms, the S cantlet, and the Cl atom in these species.

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

    (d) Determine the oxidation states and formal charge of the atoms in P_ivS₃ and the ClO₃ ^– ion.

12. Identify the hybridization of each carbon atom in the following molecule. (The arrangement of atoms is given; y'all need to determine how many bonds connect each pair of atoms.)
    
13. Write Lewis structures for NF₃ and PF₅. On the footing of hybrid orbitals, explicate the fact that NF₃, PF₃, and PF₅ are stable molecules, only NF_five does not be.
14. In addition to NF_three, two other fluoro derivatives of nitrogen are known: Due north₂F₄ and N₂F₂. What shapes exercise you predict for these ii molecules? What is the hybridization for the nitrogen in each molecule?

Glossary

hybrid orbital

orbital created by combining atomic orbitals on a central atom

hybridization

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

sp hybrid orbital

one of a set of 2 orbitals with a linear arrangement that results from combining 1 southward and ane p orbital

sp ² hybrid orbital

one of a set of three orbitals with a trigonal planar organisation that results from combining i s and ii p orbitals

sp ³ hybrid orbital

one of a set of 4 orbitals with a tetrahedral arrangement that results from combining one s and iii p orbitals

sp ³ d hybrid orbital

1 of a set of 5 orbitals with a trigonal bipyramidal organisation that results from combining one due south, three p, and one d orbital

sp ³ d ² hybrid orbital

one of a set of six orbitals with an octahedral system that results from combining one southward, three p, and two d orbitals

Solutions

Answers to Chemistry End of Chapter Exercises

1. Hybridization is introduced to explicate the geometry of bonding orbitals in valance bail theory.

3. At that place are no d orbitals in the valence shell of carbon.

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

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

(b) Bent (120°), sp ^two

(c) Trigonal planar, sp ²

(d) Tetrahedral, sp ^three

9. a) XeF_ii

(b)

(c) linear (d) sp ³ d

11. (a)

(b) P atoms, trigonal pyramidal; Due south atoms, aptitude, with two lonely pairs; Cl atoms, trigonal pyramidal; (c) Hybridization about P, South, and Cl is, in all cases, sp ³; (d) Oxidation states P +1, [latex]\text{S} - 1\frac{1}{3}[/latex], Cl +5, O –2. Formal charges: P 0; S 0; Cl +ii: O –1

13.

Phosphorus and nitrogen can class sp ³ hybrids to form iii bonds and concord 1 lonely pair in PF_three and NF_iii, respectively. However, nitrogen has no valence d orbitals, and so it cannot form a set of sp ³ d hybrid orbitals to bind five fluorine atoms in NF₅. Phosphorus has d orbitals and can demark v fluorine atoms with sp ³ d hybrid orbitals in PF₅.

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