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Polar Molecular Compounds

Polar Molecular Compounds. SWBAT: Identify Polar and Nonpolar Molecules. Periodic Table Trends. Atomic Radii. Electronegativity. Ability of an atom to attract electrons when bonded with another atom. Half the distance between the nuclei of two adjacent atoms. Atomic Radii Trend.

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Polar Molecular Compounds

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  1. Polar Molecular Compounds SWBAT: Identify Polar and Nonpolar Molecules

  2. Periodic Table Trends Atomic Radii Electronegativity Ability of an atom to attract electrons when bonded with another atom • Half the distance between the nuclei of two adjacent atoms

  3. Atomic Radii Trend

  4. Atomic Radii Trend

  5. Electronegativity Trend • Electronegativity is highest within the nonmetals with Fluorine having the highest. Why are noble gases excluded?

  6. Electronegativity Trend

  7. Three Types of Bonds: non-polar, polar, ionic Use Electronegativity (EN) values to determine type of bond EN value: H = 2.1 C = 2.5 N = 3.0 O = 3.5 F = 4.0 P = 2.1 S = 2.5 Cl = 3.0 Br = 2.8 I = 2.5 • Every element has a specific EN value • We will focus on the elements of organic chemistry C,H,N,O,P,S and Halogens • Do not need to memorize

  8. Electronegativity Differenceand type of bond

  9. 0 - .4 = nonpolar .5 – 1.7 = polar 1.8 and greater = ionic Rating of Covalent Bonds: • Example 1: What type of bond is H-Cl? • EN: H = 2.1 Cl: 3.0 • Subtract ENs from each other, larger – smaller • 3.0 – 2.1 = 0.9 compare to rating • Redraw compound with partial charges: δ+ & δ- δ+ δ- H – Cl Dipole Moment 0 - .4 = nonpolar .5 – 1.7 = polar 1.8 and greater = ionic *Polar Covalent

  10. Rating of Covalent Bonds: 0 - .4 = nonpolar .5 – 1.7 = polar 1.8 and greater = ionic Example 1: What type of bond is H-Br? EN: H = 2.1 Br: 2.8 Subtract ENs from each other, larger – smaller 2.8 – 2.1 = 0.7 compare to rating Redraw compound with partial charges: δ+ & δ- δ+ δ- H – Br Dipole Moment 0 - .4 = nonpolar .5 – 1.7 = polar 1.8 and greater = ionic *Polar Covalent

  11. Rating of Covalent Bonds: 0 - .4 = nonpolar .5 – 1.7 = polar 1.8 and greater = ionic Example 2: What type of bond is C-H? EN: H = 2.1 C: 2.5 Subtract ENs from each other, larger – smaller 2.5 – 2.1 = 0.4 compare to rating Less than 0.5 = no partial charges, no dipole arrow C – H 0 - .4 = nonpolar .5 – 1.7 = polar 1.8 and greater = ionic *Nonpolar Covalent

  12. Tips • Bonds between different elements always have some kind of dipole moment regardless of polarity rating • Draw the dipole arrow of Br – Pd (use just their position in the PT) Br – Pd • Bonds between same elements never have a dipole moment regardless of individual EN • No dipole between F – F or H – H

  13. Practice Problems • Using electronegativity difference, classify each bond as ionic, polar covalent, or nonpolar covalent. • Show the partial charge for any polar covalent bonds. • K – F EN values: K = 0.8 & F = 4.0   • O – O EN values: O = 3.5 • I – Cl EN values: I = 2.5, &Cl = 3.0

  14. Practice Problems • Using electronegativity difference, classify each bond as ionic, polar covalent, or nonpolar covalent. • Show the partial charge for any polar covalent bonds. • K – F EN values: K = 0.8 & F = 4.0 • 4.0 - 0.8 = 3.2 - ionic bond • O – O EN values: O = 3.5 • 3.5 - 3.5 = 0 - nonpolar covalent bond • I – Cl EN values: I = 2.5 & Cl = 3.0 • 3.0 - 2.5 = 0.5 - polar covalent bond δ+ δ- I – Cl

  15. Practice Problems • The bonds between the following pairs of elements are covalent.  Arrange them according to polarity, listing the most polar first.a.  H – Cl b.  H – C c.  H – Fd.  H – O e.  H – H f.  S – Cl

  16. Homework Problems • 4.50 Describe the end in EN going down a group. • 4.52 Approximately what EN difference would you expect for a polar covalent bond? • 4.54 Predict whether each of the following bonds is ionic, polar covalent or nonpolar covalent: • A. Si-O B. K-Cl C. S-F D. P-Br E. Li-O F. N-P • 4.56 For each of the following bonds, indicate the positive end with δ+ and the negative end with δ-. Write an arrow to show the dipole for each. • A. Si-Br B. Se-F C. Br-F D. N-H E. N-P

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