Rank the following in order of increasing bond ang...

Question

Rank the following in order of increasing bond angles: OF2, SnF2, XeF2(1) OF2 < XeF2 < SnF2(2) OF2 < SnF2 < XeF2(3) SnF2 < OF2 < XeF2(4) XeF2 < OF2 < SnF2(5) XeF2 < SnF2 < OF2

Jules Bruno · Accepted Answer

We’re being asked to rank the following in order of increasing bond angles: OF2, SnF2, XeF2. Since we don’t know the Lewis structure for each, we need to do the following steps:Step 1: Determine the central atom in this molecule.Step 2: Calculate the total number of valence electrons present.Step 3: Draw the Lewis structure for the molecule.Step 4: Determine the number of electron groups around the indicated atom.Step 5: Determine the electron geometry and bond angle using this:Electron Regions          Electronic Geometry          Bond Angles2                                      linear                                    180˚3                                      trigonal planar                      120˚4                                      tetrahedral                            109.5˚5                                      trigonal bipyramidal              90˚, 120˚, and 180˚6                                      octahedral                            90˚ and 180˚[readmore]For OF2:Step 1: Oxygen is less electronegative than fluorine so oxygen is the central atom.Step 2: The total number of valence electrons present in OF2 is:            Group             Valence ElectronsO         6A                   1 × 6 e– = 6 e–F         7A                    2 × 7 e– = 14 e–                              Total:  20 valence e–Step 3: Halogens, when not the central atom, form only 1 bond, which means the fluorine atoms each form 1 bond. Oxygen prefers to form 2 bonds. The incomplete Lewis structure of OF2 is:A bond corresponds to 2 electrons shared between two atoms so we currently have 2 × 2 e– = 4 val. e–, which means we’re still lacking 16 e–. We can add lone pairs to the atoms. The complete Lewis structure is:Step 4: Oxygen is surrounded by 4 electron groups: 2 bonds and 2 lone pairs.Step 5: Since there are 4 electron groups around oxygen, the electron geometry is tetrahedral, whose ideal bond angles are 109.5˚. Recall that lone pairs compress the bond angle, causing a deviation from the ideal bond angle.Therefore, the bond angles in OF2 is 109.5˚.For SnF2:Step 1: Tin is less electronegative than fluorine so tin is the central atom.Step 2: The total number of valence electrons present in SnF2 is:            Group             Valence ElectronsSn       4A                   1 ×  4 e– = 4 e–F         7A                    2 × 7 e– = 14 e–                              Total:  18 valence e–Step 3: Halogens, when not the central atom, form only 1 bond, which means the fluorine atoms each form 1 bond. The incomplete Lewis structure of SnF2 is:A bond corresponds to 2 electrons shared between two atoms so we currently have 2 × 2 e– = 4 val. e–, which means we’re still lacking 14 e–. We can add lone pairs to the atoms. The complete Lewis structure is:Step 4: Tin is surrounded by 3 electron groups: 2 bonds and 1 lone pairs.Step 5: Since there are 3 electron groups around tin, the electron geometry is trigonal planar, whose ideal bond angles are 120˚. Recall that lone pairs compress the bond angle, causing a deviation from the ideal bond angle.Therefore, the bond angles in SnF2 is 120˚.For XeF2:Step 1: Xenon is less electronegative than fluorine so oxygen is the central atom.Step 2: The total number of valence electrons present in XeF2 is:            Group             Valence ElectronsXe        8A                   1 × 8 e– = 8 e–F         7A                    2 × 7 e– = 14 e–                              Total:  22 valence e–Step 3: Halogens, when not the central atom, form only 1 bond, which means the fluorine atoms each form 1 bond. The incomplete Lewis structure of XeF2 is:A bond corresponds to 2 electrons shared between two atoms so we currently have 2 × 2 e– = 4 val. e–, which means we’re still lacking 16 e–. We can add lone pairs to the atoms. The complete Lewis structure is:Step 4: Xenon is surrounded by 5 electron groups: 2 bonds and 3 lone pair.Step 5: Since there are 5 electron groups around iodine, the electron geometry is trigonal bipyramidal, whose ideal bond angles are 90˚, 120˚, and 180˚. Recall that lone pairs compress the bond angle, causing a deviation from the ideal bond angle.Therefore, the bond angles in XeF2 (F-Xe-F bond) is 180˚.Ranking the in order of increasing bond angles: (2) OF2 < SnF2 < XeF2.

Problem: Rank the following in order of increasing bond angles: OF2, SnF2, XeF2(1) OF2 < XeF2 < SnF2(2) OF2 < SnF2 < XeF2(3) SnF2 < OF2 < XeF2(4) XeF2 < OF2 < SnF2(5) XeF2 < SnF2 < OF2

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Problem: Rank the following in order of increasing bond angles: OF2, SnF2, XeF2(1) OF2 &lt; XeF2 &lt; SnF2(2) OF2 &lt; SnF2 &lt; XeF2(3) SnF2 &lt; OF2 &lt; XeF2(4) XeF2 &lt; OF2 &lt; SnF2(5) XeF2 &lt; SnF2 &lt; OF2

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Problem: Rank the following in order of increasing bond angles: OF2, SnF2, XeF2(1) OF2 < XeF2 < SnF2(2) OF2 < SnF2 < XeF2(3) SnF2 < OF2 < XeF2(4) XeF2 < OF2 < SnF2(5) XeF2 < SnF2 < OF2