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Understanding Hybridization, Molecular Polarity, and Atomic Trends in Chemistry

This review focuses on the concepts of hybridization, molecular shapes, and polarity in various chemical compounds. It covers the hybridization of carbon atoms in compounds like CH4, C2H2, C2H4, and C2H6, as well as more complex structures like XeF4 and SF6. The review also predicts the polarity of molecules such as PCl3, SO3, and IF, providing insights on how molecular geometry and lone pairs influence molecular characteristics. Additionally, it addresses trends in atomic size, ionization energy, and electronegativity to facilitate understanding chemical behavior.

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Understanding Hybridization, Molecular Polarity, and Atomic Trends in Chemistry

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  1. Unit 7-9 Test Review APC MM 12-13

  2. 14 lone pairs missing • See picture • See picture • 7 pi bonds (7 double bonds)

  3. 8 lone pairs (green circled) • 4 pi bonds (4 double bonds) 3. Shapes aren’t the same. Carbon lacks a lone pair so its shape is trigonal planar. Nitrogen has a lone pair so it is trigonal pyramidal.

  4. Tell the hybridization for a carbon in the following… 1.   CH4 – sp3 2.   C2H2 – sp 3.   C2H4 – sp2 4.   C2H6 – sp3

  5. Predict whether the following molecules will be polar (have a dipole moment) or nonpolar (have a zero dipole moment.) 1.   PCl3 – polar 2.   SO3 – nonpolar 3.   SF6 – nonpolar

  6. Tell the hybridization for the following… 1.   XeF4– sp3d2 2.   XeF2– sp3d 3.   SF6– sp3d2 4.   SF4– sp3d

  7. Chlorines are 180° to each other in the para isomer which will cancel out the dipoles.

  8. Chlorines are in the opposite directions which will cancel out each other

  9. 1.   IF– polar 2.   CS2 – nonpolar 3.   IF5 – polar

  10. Linear (3), trigonal planar, tetrahedral, trigonal bipyramidal, octahedral, square planar • Linear (2)/dumbbell, bent, trigonal pyramidal, seesaw, t-shaped, square pyramidal • Polar • Nonpolar

  11. (614 kJ/mol + 146 kJ/mol) – (2•358 kJ/mol +348 kJ/mol) = -304 kJ/mol

  12. Although I3- is known, F3- does not form. Explain your answer. Fluorine doesn’t have an available d-orbital to put extra electrons in like Iodine does because Fluorine is in the 2nd period and Iodine is in the 5th period of the PT.

  13. Pick the element with the larger value depending on the trend presented. 1.   atomic size: Rb or Sr 2.   ionization energy: Cl or Br 3.   electronegativiy: As or S 4.   cation size: Ca2+ or K+ 5. anion size: O2- or F-

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