The Periodic Table Chapter 8

The Periodic Table Chapter 8
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Electron Configurations of Cations and Anions
Of Representative Elements Na [Ne]3s1 Na+ [Ne] Atoms lose electrons so that cation has a noble-gas outer electron configuration. Ca [Ar]4s2 Ca2+ [Ar] Al [Ne]3s23p1 Al3+ [Ne] H 1s1 H− 1s2 or [He] Atoms gain electrons so that anion has a noble-gas outer electron configuration. F 1s22s22p5 F− 1s22s22p6 or [Ne] O 1s22s22p4 O2− 1s22s22p6 or [Ne] N 1s22s22p3 N3− 1s22s22p6 or [Ne]

Isoelectronic: have the same number of electrons, and hence the same ground-state electron configuration Na+: [Ne] Al3+: [Ne] F-: 1s22s22p6 or [Ne] O2-: 1s22s22p6 or [Ne] N3-: 1s22s22p6 or [Ne] Na+, Al3+, F−, O2−, and N3− are all isoelectronic with Ne

Electron Configurations of Cations of Transition Metals
When a cation is formed from an atom of a transition metal, electrons are always removed first from the ns orbital and then from the (n – 1)d orbitals. Fe: [Ar]4s23d6 Mn: [Ar]4s23d5 Fe2+: [Ar]4s03d6 or [Ar]3d6 Mn2+: [Ar]4s03d5 or [Ar]3d5 Fe3+: [Ar]4s03d5 or [Ar]3d5

Atomic Radii covalent radius metallic radius

8.2 Referring to a periodic table, arrange the following atoms in order of increasing atomic radius: P, Si, N.

8.2 Strategy What are the trends in atomic radii in a periodic group and in a particular period? Which of the preceding elements are in the same group? in the same period? Solution From Figure 8.2 we see that N and P are in the same group (Group 5A). Therefore, the radius of N is smaller than that of P (atomic radius increases as we go down a group). Both Si and P are in the third period, and Si is to the left of P. Therefore, the radius of P is smaller than that of Si (atomic radius decreases as we move from left to right across a period). Thus, the order of increasing radius is N < P < Si

Cation is always smaller than atom from which it is formed.
Anion is always larger than atom from which it is formed.

Ionization energy is the minimum energy (kJ/mol) required to remove an electron from a gaseous atom in its ground state. I1 + X (g) X+(g) + e- I1 first ionization energy I2 + X+(g) X2+(g) + e- I2 second ionization energy I3 + X2+(g) X3+(g) + e- I3 third ionization energy I1 < I2 < I3

General Trends in First Ionization Energies
Increasing First Ionization Energy Increasing First Ionization Energy

8.4 Which atom should have a smaller first ionization energy: oxygen or sulfur? Which atom should have a higher second ionization energy: lithium or beryllium?

8.4 Strategy First ionization energy decreases as we go down a group because the outermost electron is farther away from the nucleus and feels less attraction. Removal of the outermost electron requires less energy if it is shielded by a filled inner shell. Solution (a) Oxygen and sulfur are members of Group 6A. They have the same valence electron configuration (ns2np4), but the 3p electron in sulfur is farther from the nucleus and experiences less nuclear attraction than the 2p electron in oxygen. Thus, we predict that sulfur should have a smaller first ionization energy.

8.4 (b) The electron configurations of Li and Be are 1s22s1 and 1s22s2, respectively. The second ionization energy is the minimum energy required to remove an electron from a gaseous unipositive ion in its ground state. For the second ionization process, we write Because 1s electrons shield 2s electrons much more effectively than they shield each other, we predict that it should be easier to remove a 2s electron from Be+ than to remove a 1s electron from Li+.

Electron affinity is the negative of the energy change that occurs when an electron is accepted by an atom in the gaseous state to form an anion. X (g) + e X-(g) F (g) + e F-(g) DH = -328 kJ/mol EA = +328 kJ/mol O (g) + e O-(g) DH = -141 kJ/mol EA = +141 kJ/mol

Variation of Electron Affinity With Atomic Number (H – Ba)

8.5 Why are the electron affinities of the alkaline earth metals, shown in Table 8.3, either negative or small positive values?

8.5 Strategy What are the electron configurations of alkaline earth metals? Would the added electron to such an atom be held strongly by the nucleus? Solution The valence electron configuration of the alkaline earth metals is ns2, where n is the highest principal quantum number. For the process where M denotes a member of the Group 2A family, the extra electron must enter the np subshell, which is effectively shielded by the two ns electrons (the ns electrons are more penetrating than the np electrons) and the inner electrons. Consequently, alkaline earth metals have little tendency to pick up an extra electron.

Chemical Bonding I: Basic Concepts
Chapter 9 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Lewis Dot Symbols for the Representative Elements &
Noble Gases

The Ionic Bond Ionic bond: the electrostatic force that holds ions together in an ionic compound. Li+ F - Li + F 1s2 1s22s22p6 1s22s1 1s22s22p5 [He] [Ne] Li Li+ + e- LiF e- + F - F - Li+ + Li+

Why should two atoms share electrons?
A covalent bond is a chemical bond in which two or more electrons are shared by two atoms. Why should two atoms share electrons? 7e- 7e- 8e- 8e- F F + F Lewis structure of F2 lone pairs F single covalent bond single covalent bond F

Lewis structure of water
single covalent bonds 2e- 8e- 2e- H + O + H O H or Double bond – two atoms share two pairs of electrons 8e- 8e- 8e- O C or O C double bonds Triple bond – two atoms share three pairs of electrons 8e- N 8e- or N triple bond

Lengths of Covalent Bonds
Bond Lengths Triple bond < Double Bond < Single Bond

Polar covalent bond or polar bond is a covalent bond with greater electron density around one of the two atoms H F electron rich region electron poor region e- poor e- rich F H d+ d-

Electron Affinity - measurable, Cl is highest
Electronegativity is the ability of an atom to attract toward itself the electrons in a chemical bond. Electron Affinity - measurable, Cl is highest X (g) + e X-(g) Electronegativity - relative, F is highest

The Electronegativities of Common Elements

9.2 Classify the following bonds as ionic, polar covalent, or covalent: the bond in HCl the bond in KF the CC bond in H3CCH3

9.2 Strategy We follow the 2.0 rule of electronegativity difference and look up the values in Figure 9.5. Solution The electronegativity difference between H and Cl is 0.9, which is appreciable but not large enough (by the 2.0 rule) to qualify HCl as an ionic compound. Therefore, the bond between H and Cl is polar covalent. The electronegativity difference between K and F is 3.2, which is well above the 2.0 mark; therefore, the bond between K and F is ionic. The two C atoms are identical in every respect—they are bonded to each other and each is bonded to three other H atoms. Therefore, the bond between them is purely covalent.

Writing Lewis Structures
Draw skeletal structure of compound showing what atoms are bonded to each other. Put least electronegative element in the center. Count total number of valence e−. Add 1 for each negative charge. Subtract 1 for each positive charge. Draw a single covalent bond between the central and each surrounding atom. Complete an octet for all atoms except hydrogen, beginning with the surrounding atoms. If central atom has fewer than eight electrons, form double and triple bonds on central atom as needed.

NF3 is a colorless, odorless, unreactive gas.
9.3 Write the Lewis structure for nitrogen trifluoride (NF3) in which all three F atoms are bonded to the N atom. NF3 is a colorless, odorless, unreactive gas.

9.3 Solution We follow the preceding procedure for writing Lewis structures. Step 1: The N atom is less electronegative than F, so the skeletal structure of NF3 is Step 2: The outer-shell electron configurations of N and F are 2s22p3 and 2s22p5, respectively. Thus, there are 5 + (3 × 7), or 26, valence electrons to account for in NF3.

9.3 Step 3: We draw a single covalent bond between N and each F, and complete the octets for the F atoms. We place the remaining two electrons on N: Because this structure satisfies the octet rule for all the atoms, step 4 is not required. Check Count the valence electrons in NF3 (in bonds and in lone pairs). The result is 26, the same as the total number of valence electrons on three F atoms (3 × 7 = 21) and one N atom (5).

HNO3 is a strong electrolyte.
9.4 Write the Lewis structure for nitric acid (HNO3) in which the three O atoms are bonded to the central N atom and the ionizable H atom is bonded to one of the O atoms. HNO3 is a strong electrolyte.

9.4 Solution We follow the procedure already outlined for writing Lewis structures. Step 1: The skeletal structure of HNO3 is Step 2: The outer-shell electron configurations of N, O, and H are 2s22p3, 2s22p4, and 1s1, respectively. Thus, there are 5 + (3 × 6) + 1, or 24, valence electrons to account for in HNO3.

9.4 Step 3: We draw a single covalent bond between N and each of the three O atoms and between one O atom and the H atom. Then we fill in electrons to comply with the octet rule for the O atoms: Step 4: We see that this structure satisfies the octet rule for all the O atoms but not for the N atom. The N atom has only six electrons. Therefore, we move a lone pair from one of the end O atoms to form another bond with N.

9.4 Now the octet rule is also satisfied for the N atom:
Check Make sure that all the atoms (except H) satisfy the octet rule. Count the valence electrons in HNO3 (in bonds and in lone pairs). The result is 24, the same as the total number of valence electrons on three O atoms (3 × 6 = 18), one N atom (5), and one H atom (1).

9.5 Write the Lewis structure for the carbonate ion ( ).

9.5 Solution We follow the preceding procedure for writing Lewis structures and note that this is an anion with two negative charges. Step 1: We can deduce the skeletal structure of the carbonate ion by recognizing that C is less electronegative than O. Therefore, it is most likely to occupy a central position as follows:

9.5 Step 2: The outer-shell electron configurations of C and O are 2s22p2 and 2s22p4, respectively, and the ion itself has two negative charges. Thus, the total number of electrons is 4 + (3 × 6) + 2, or 24. Step 3: We draw a single covalent bond between C and each O and comply with the octet rule for the O atoms: This structure shows all 24 electrons.

9.5 Step 4: Although the octet rule is satisfied for the O atoms, it is not for the C atom. Therefore, we move a lone pair from one of the O atoms to form another bond with C. Now the octet rule is also satisfied for the C atom: Check Make sure that all the atoms satisfy the octet rule. Count the valence electrons in (in chemical bonds and in lone pairs). The result is 24, the same as the total number of valence electrons on three O atoms (3 × 6 = 18), one C atom (4), and two negative charges (2).

A resonance structure is one of two or more Lewis structures for a single molecule that cannot be represented accurately by only one Lewis structure. O + - O + -

Exceptions to the Octet Rule
The Incomplete Octet Be – 2e- 2H – 2x1e- 4e- BeH2 H Be B – 3e- 3F – 3x7e- 24e- 3 single bonds (3x2) = 6 9 lone pairs (9x2) = 18 Total = 24 F B BF3

Exceptions to the Octet Rule
Odd-Electron Molecules N – 5e- O – 6e- 11e- NO N O The Expanded Octet (central atom with principal quantum number n > 2) S F S – 6e- 6F – 42e- 48e- 6 single bonds (6x2) = 12 18 lone pairs (18x2) = 36 Total = 48 SF6

PF5 is a reactive gaseous compound.
9.10 Draw the Lewis structure for phosphorus pentafluoride (PF5), in which all five F atoms are bonded to the central P atom. PF5 is a reactive gaseous compound.

9.10 Strategy Note that P is a third-period element. We follow the procedures given in Examples 9.5 and 9.6 to draw the Lewis structure and calculate formal charges. Solution The outer-shell electron configurations for P and F are 3s23p3 and 2s22p5, respectively, and so the total number of valence electrons is 5 + (5 × 7), or 40. Phosphorus, like sulfur, is a third-period element, and therefore it can have an expanded octet.

9.10 The Lewis structure of PF5 is
Note that there are no formal charges on the P and F atoms. Check Although the octet rule is satisfied for the F atoms, there are 10 valence electrons around the P atom, giving it an expanded octet.

Single bond < Double bond < Triple bond
The enthalpy change required to break a particular bond in one mole of gaseous molecules is the bond enthalpy. Bond Enthalpy H2 (g) H (g) + DHo = kJ Cl2 (g) Cl (g) + DHo = kJ HCl (g) H (g) + Cl (g) DHo = kJ O2 (g) O (g) + DHo = kJ O N2 (g) N (g) + DHo = kJ N Bond Enthalpies Single bond < Double bond < Triple bond

Chemical Bonding II: Molecular Geometry and Hybridization of Atomic Orbitals
Chapter 10 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Valence shell electron pair repulsion (VSEPR) model:
Predict the geometry of the molecule from the electrostatic repulsions between the electron (bonding and nonbonding) pairs. Class # of atoms bonded to central atom # lone pairs on central atom Arrangement of electron pairs Molecular Geometry AB2 2 linear linear B

0 lone pairs on central atom
Cl Be 2 atoms bonded to central atom

Arrangement of electron pairs
VSEPR Class # of atoms bonded to central atom # lone pairs on central atom Arrangement of electron pairs Molecular Geometry AB2 2 linear linear trigonal planar trigonal planar AB3 3

VSEPR Class # of atoms bonded to central atom # lone pairs on central atom Arrangement of electron pairs Molecular Geometry AB2 2 linear linear AB3 3 trigonal planar AB4 4 tetrahedral tetrahedral

VSEPR Class # of atoms bonded to central atom # lone pairs on central atom Arrangement of electron pairs Molecular Geometry AB2 2 linear linear AB3 3 trigonal planar AB4 4 tetrahedral tetrahedral trigonal bipyramidal trigonal bipyramidal AB5 5

VSEPR Class # of atoms bonded to central atom # lone pairs on central atom Arrangement of electron pairs Molecular Geometry AB2 2 linear linear AB3 3 trigonal planar AB4 4 tetrahedral tetrahedral AB5 5 trigonal bipyramidal AB6 6 octahedral octahedral

> bonding-pair vs. bonding- pair repulsion lone-pair vs. lone-pair
lone-pair vs. bonding- >

10.1 Use the VSEPR model to predict the geometry of the following molecules and ions: AsH3 OF2 C2H4

10.1 Strategy The sequence of steps in determining molecular geometry is as follows: Solution The Lewis structure of AsH3 is There are four electron pairs around the central atom; therefore, the electron pair arrangement is tetrahedral (see Table 10.1).

10.1 Recall that the geometry of a molecule is determined only by the arrangement of atoms (in this case the As and H atoms). Thus, removing the lone pair leaves us with three bonding pairs and a trigonal pyramidal geometry, like NH3. We cannot predict the HAsH angle accurately, but we know that it is less than 109.5° because the repulsion of the bonding electron pairs in the As—H bonds by the lone pair on As is greater than the repulsion between the bonding pairs. (b) The Lewis structure of OF2 is There are four electron pairs around the central atom; therefore, the electron pair arrangement is tetrahedral.

10.1 Recall that the geometry of a molecule is determined only by the arrangement of atoms (in this case the O and F atoms). Thus, removing the two lone pairs leaves us with two bonding pairs and a bent geometry, like H2O. We cannot predict the FOF angle accurately, but we know that it must be less than 109.5° because the repulsion of the bonding electron pairs in the O−F bonds by the lone pairs on O is greater than the repulsion between the bonding pairs. (c) The Lewis structure of is

10.1 There are four electron pairs around the central atom; therefore, the electron pair arrangement is tetrahedral. Because there are no lone pairs present, the arrangement of the bonding pairs is the same as the electron pair arrangement. Therefore, has a tetrahedral geometry and the ClAlCl angles are all 109.5°. (d) The Lewis structure of is There are five electron pairs around the central I atom; therefore, the electron pair arrangement is trigonal bipyramidal. Of the five electron pairs, three are lone pairs and two are bonding pairs.

10.1 Recall that the lone pairs preferentially occupy the equatorial positions in a trigonal bipyramid (see Table 10.2). Thus, removing the lone pairs leaves us with a linear geometry for , that is, all three I atoms lie in a straight line. (e) The Lewis structure of C2H4 is The C=C bond is treated as though it were a single bond in the VSEPR model. Because there are three electron pairs around each C atom and there are no lone pairs present, the arrangement around each C atom has a trigonal planar shape like BF3, discussed earlier.

10.1 Thus, the predicted bond angles in C2H4 are all 120°. Comment
(1) The ion is one of the few structures for which the bond angle (180°) can be predicted accurately even though the central atom contains lone pairs. (2) In C2H4, all six atoms lie in the same plane. The overall planar geometry is not predicted by the VSEPR model, but we will see why the molecule prefers to be planar later. In reality, the angles are close, but not equal, to 120° because the bonds are not all equivalent.

10.2 Predict whether each of the following molecules has a dipole moment: BrCl BF3 (trigonal planar) CH2Cl2 (tetrahedral)

10.2 Solution (a) Because bromine chloride is diatomic, it has a linear geometry. Chlorine is more electronegative than bromine (see Figure 9.5), so BrCl is polar with chlorine at the negative end Thus, the molecule does have a dipole moment. In fact, all diatomic molecules containing different elements possess a dipole moment.

10.2 (b) Because fluorine is more electronegative than boron, each B−F bond in BF3 (boron trifluoride) is polar and the three bond moments are equal. However, the symmetry of a trigonal planar shape means that the three bond moments exactly cancel one another: An analogy is an object that is pulled in the directions shown by the three bond moments. If the forces are equal, the object will not move. Consequently, BF3 has no dipole moment; it is a nonpolar molecule.

10.2 (c) The Lewis structure of CH2Cl2 (methylene chloride) is
This molecule is similar to CH4 in that it has an overall tetrahedral shape. However, because not all the bonds are identical, there are three different bond angles: HCH, HCCl, and ClCCl. These bond angles are close to, but not equal to, 109.5°.

10.2 Because chlorine is more electronegative than carbon, which is more electronegative than hydrogen, the bond moments do not cancel and the molecule possesses a dipole moment: Thus, CH2Cl2 is a polar molecule.

Hybridization – mixing of two or more atomic orbitals to form a new set of hybrid orbitals
Mix at least 2 nonequivalent atomic orbitals (e.g. s and p). Hybrid orbitals have very different shape from original atomic orbitals. Number of hybrid orbitals is equal to number of pure atomic orbitals used in the hybridization process. Covalent bonds are formed by: Overlap of hybrid orbitals with atomic orbitals Overlap of hybrid orbitals with other hybrid orbitals

Formation of sp3 Hybrid Orbitals

Formation of Covalent Bonds in CH4

sp3-Hybridized N Atom in NH3
Predict correct bond angle

Formation of sp Hybrid Orbitals

Formation of sp2 Hybrid Orbitals

How do I predict the hybridization of the central atom?
Draw the Lewis structure of the molecule. Count the number of lone pairs AND the number of atoms bonded to the central atom # of Lone Pairs + # of Bonded Atoms Hybridization Examples 2 sp BeCl2 3 sp2 BF3 4 sp3 CH4, NH3, H2O 5 sp3d PCl5 6 sp3d2 SF6

Sigma (s) and Pi Bonds (p)
1 sigma bond Single bond Double bond 1 sigma bond and 1 pi bond Triple bond 1 sigma bond and 2 pi bonds

The Periodic Table Chapter 8

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