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EXAMPLE 10.1

Solution:. SKILLBUILDER 10.1. Writing Lewis Structures for Elements. Write a Lewis structure for Mg. FOR MORE PRACTICE. Example 10.12; Problems 27, 28. EXAMPLE 10.1. Writing Lewis Structures for Elements. Write a Lewis structure for phosphorus.

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EXAMPLE 10.1

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  1. Solution: SKILLBUILDER 10.1 Writing Lewis Structures for Elements Write a Lewis structure for Mg. FOR MORE PRACTICE Example 10.12; Problems 27, 28. EXAMPLE 10.1 Writing Lewis Structures for Elements Write a Lewis structure for phosphorus. Since phosphorus is in Group 5A in the periodic table, it has 5 valence electrons. Represent these as five dots surrounding the symbol for phosphorus.

  2. Solution: SKILLBUILDER 10.2 Writing Ionic Lewis Structures Write a Lewis structure for the compound NaBr. FOR MORE PRACTICE Example 10.13; Problems 39, 40. EXAMPLE 10.2 Writing Ionic Lewis Structures Write a Lewis structure for the compound MgO. Draw the Lewis structures of magnesium and oxygen by drawing two dots around the symbol for magnesium and six dots around the symbol for oxygen. In MgO, magnesium loses its 2 valence electrons, forming a 2+ charge, and oxygen gains 2 electrons, forming a 2– charge and acquiring an octet.

  3. Solution: The formula is therefore CaCl2. FOR MORE PRACTICE Example 10.14; Problems 41, 42, 43, 44. SKILLBUILDER 10.3 Using Lewis Theory to Predict the Chemical Formula of an Ionic Compound Use Lewis theory to predict the formula for the compound that forms between magnesium and nitrogen. EXAMPLE 10.3 Using Lewis Theory to Predict the Chemical Formula of an Ionic Compound Use Lewis theory to predict the formula for the compound that forms between calcium and chlorine. Draw the Lewis structures of calcium and chlorine by drawing two dots around the symbol for calcium and seven dots around the symbol for chlorine. Calcium must lose its 2 valence electrons (to effectively get an octet in its previous principal shell), while chlorine needs to gain only 1 electron to get an octet. Consequently, the compound that forms between Ca and Cl must have two chlorine atoms to every one calcium atom.

  4. Solution: Following the symmetry guideline, we write: Total number of electrons for Lewis structure = Total number of electrons for Lewis structure = Writing Lewis Structures for Covalent Compounds EXAMPLE 10.4 EXAMPLE 10.5 Write a Lewis structure for CO2. Write a Lewis structure for CCl4. 1.Write the correct skeletal structure for the molecule. Solution: Following the symmetry guideline, we write: O C O 2.Calculate the total number of electrons for the Lewis structure by summing the valence electrons of each atom in the molecule.

  5. Bonding electrons first. (4 of 16 electrons used) Lone pairs on terminal atoms next. (16 of 16 electrons used) Bonding electrons first. (8 of 32 electrons used) Lone pairs on terminal atoms next. (32 of 32 electrons used) Move lone pairs from the oxygen atoms to bonding regions to form double bonds. Writing Lewis Structures for Covalent Compounds Continued EXAMPLE 10.4 EXAMPLE 10.5 3.Distribute the electrons among the atoms, giving octets (or duets for hydrogen) to as many atoms as possible. Begin with the bonding electrons, and then proceed to lone pairs on terminal atoms, and finally to lone pairs on the central atom. 4.If any atoms lack an octet, form double or triple bonds as necessary to give them octets. Since all of the atoms have octets, the Lewis structure is complete.

  6. FOR MORE PRACTICE Example 10.15; Problems 49, 50, 51, 52, 53, 54. Writing Lewis Structures for Covalent Compounds Continued EXAMPLE 10.4 EXAMPLE 10.5 SKILLBUILDER 10.4 SKILLBUILDER 10.5 Write a Lewis structure for CO. Write a Lewis structure for H2CO.

  7. Solution: EXAMPLE 10.6 Writing Lewis Structures for Polyatomic Ions Write the Lewis structure for the NH4+ ion. Begin by writing the skeletal structure. Since hydrogen atoms must be terminal, and following the guideline of symmetry, put the nitrogen atom in the middle surrounded by four hydrogen atoms. Calculate the total number of electrons for the Lewis structure by summing the number of valence electrons for each atom and subtracting 1 for the positive charge. Next, place 2 electrons between each pair of atoms. Since the nitrogen atom has an octet and since all of the hydrogen atoms have duets, the placement of electrons is complete. Write the entire Lewis structure in brackets and write the charge of the ion in the upper right corner.

  8. SKILLBUILDER 10.6 Writing Lewis Structures for Polyatomic Ions Write a Lewis structure for the ClO– ion. FOR MORE PRACTICE Problems 57bcd, 58abc, 59, 60. EXAMPLE 10.6 Writing Lewis Structures for Polyatomic Ions Continued

  9. EXAMPLE 10.7 Writing Resonance Structures Write a Lewis structure for the NO3– ions. Include resonance structures. Begin by writing the skeletal structure. Using the guideline of symmetry, make the three oxygen atoms terminal. Solution: O O N O Sum the valence electrons (adding 1 electron to account for the –1 charge) to determine the total number of electrons in the Lewis structure. Place 2 electrons between each pair of atoms. Distribute the remaining electrons, first to terminal atoms. Since there are no electrons remaining to complete the octet of the central atom, form a double bond by moving a lone pair from one of the oxygen atoms into the bonding region with nitrogen. Enclose the structure in brackets and write the charge at the upper right.

  10. SKILLBUILDER 10.7 Writing Resonance Structures Write a Lewis structure for the NO2- ion. Include resonance structures. FOR MORE PRACTICE Example 10.16; Problems 57, 58, 59, 60. EXAMPLE 10.7 Writing Resonance Structures Continued Notice that you could have formed the double bond with either of the other two oxygen atoms. Since the three Lewis structures are equally correct, write the three structures as resonance structures.

  11. Solution: PCl3 has 26 electrons. Solution: [NO3]– has 24 electrons. Three of the four electron groups around P are bonding groups, and one is a lone pair. All three of the electron groups around N are bonding groups. Predicting Geometry Using VSEPR EXAMPLE 10.8 EXAMPLE 10.9 Predict the electron and molecular geometry of PCl3. Predict the electron and molecular geometry of the [NO3]- ion. 1. Draw a Lewis structure for the molecule. 2. Determine the total number of electron groups around the central atom. Lone pairs, single bonds, double bonds, and triple bonds each count as one group. The central atom (P) has four electron groups. The central atom (N) has three electron groups (the double bond counts as one group). 3. Determine the number of bonding groups and the number of lone pairs around the central atom. These should sum to the result from Step 2. Bonding groups include single bonds, double bonds, and triple bonds.

  12. SKILLBUILDER 10.8 SKILLBUILDER 10.9 Predict the molecular geometry of ClNO (N is the central atom). Predict the molecular geometry of The SO32– ion. FOR MORE PRACTICE Example 10.17; Problems 67, 68, 71, 72, 75, 76. Predicting Geometry Using VSEPR Continued EXAMPLE 10.8 EXAMPLE 10.9 4. Use Table 10.1 to determine the electron geometry and molecular geometry The electron geometry is tetrahedral (four electron groups), and the molecular geometry—the shape of the molecule—is trigonal pyramidal (four electron groups, three bonding groups, and one lone pair). The electron geometry is trigonal planar (three electron groups), and the molecular geometry—the shape of the molecule—is trigonal planar (three electron groups, three bonding groups, and no lone pairs).

  13. EXAMPLE 10.10 Classifying Bonds as Pure Covalent, Polar Covalent, or Ionic Determine whether the bond formed between each of the following pairs of atoms is pure covalent, polar covalent, or ionic. (a) Sr and F (b) N and Cl (c) N and O Solution: (a) From Figure 10.2, we find the electronegativity of Sr (1.0) and of F (4.0). The electronegativity difference (ΔEN) is: ΔEN = 4.0 – 1.0 = 3.0 (b) From Figure 10.2, we find the electronegativity of N (3.0) and of Cl (3.0). The electronegativity difference (ΔEN) is: ΔEN = 3.0 – 3.0 = 0 Using Table 10.2, we classify this bond as pure covalent. (c) From Figure 10.2, we find the electronegativity of N (3.0) and of O (3.5). The electronegativity difference (ΔEN) is: ΔEN = 3.0 – 3.0 = 0 Using Table 10.2, we classify this bond as polar covalent.

  14. SKILLBUILDER 10.10 Classifying Bonds as Pure Covalent, Polar Covalent, or Ionic Determine whether the bond formed between each of the following pairs of atoms is pure covalent, polar covalent, or ionic. (a) I and I (b) Cs and Br (c) P and O FOR MORE PRACTICE Problems 83, 84. EXAMPLE 10.10 Classifying Bonds as Pure Covalent, Polar Covalent, or Ionic Continued

  15. Solution: SKILLBUILDER 10.11 Determining Whether a Molecule Is Polar Determine whether CH4 is polar. FOR MORE PRACTICE Example 10.18; Problems 91, 92, 93, 94. EXAMPLE 10.11 Determining Whether a Molecule Is Polar Determine whether NH3 is polar. Begin by drawing the Lewis structure of NH3. Since N and H have different electronegativities, the bonds are polar The geometry of NH3 is trigonal pyramidal (four electron groups, three bonding groups, one lone pair). Draw a three dimensional picture of NH3 and imagine each bond as a rope that is being pulled. The pulls of the ropes do not cancel and the molecule is polar.

  16. Solution: Since S is in Group 6A, it has 6 valence electrons. We draw these as dots surrounding its symbol, S. EXAMPLE 10.12 Lewis Structures for Elements What is the Lewis structure of sulfur?

  17. Solution: EXAMPLE 10.13 Writing Lewis Structures for Ionic Compounds Write a Lewis structure for lithium bromide.

  18. Solution: The Lewis structures of K and S are: Potassium must lose 1 electron and sulfur must gain 2. Consequently, we need two potassium atoms to every sulfur atom. The Lewis structure is: The correct formula is K2S. EXAMPLE 10.14 Using Lewis Theory to Predict the Chemical Formula of an Ionic Compound Use Lewis theory to predict the formula for the compound that forms between potassium and sulfur.

  19. Solution: S C S EXAMPLE 10.15 Writing Lewis Structures for Covalent Compounds Write a Lewis structure for CS2.

  20. Solution: We can write a Lewis structure for SeO2 by following the steps for writing covalent Lewis structures. We find that we can write two equally correct structures, so we draw them both as resonance structures. EXAMPLE 10.16 Writing Resonance Structures Write resonance structures for SeO2.

  21. Solution: The Lewis structure for SeO2 (as we saw in Example 10.16) is composed of the following two resonance structures. Either of the resonance structures will give the same geometry. Total number of electron groups = 3 Number of bonding groups = 2 Number of lone pairs = 1 Electron geometry = Trigonal planar Molecular geometry = Bent EXAMPLE 10.17 Predicting the Shapes of Molecules Predict the geometry of SeO2.

  22. Solution: Se and O are nonmetals with different electronegativities (2.4 for Se and 3.5 for O). Therefore, the Se–O bonds are polar. As we saw in Example 10.17, the geometry of SeO2 is bent. The polar bonds do not cancel but rather sum to give a net dipole moment. Therefore the molecule is polar. EXAMPLE 10.18 Determining Whether a Molecule Is Polar Determine whether is SeO2 polar.

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