In: Chemistry
Identify the molecular geometry for each of the following triatomic molecules using the VSEPR model. Then, determine which molecules would be expected to have at least one normal mode that is infrared active. Explain your reasoning
a. NO2
b. PCl2 +
c. BrF2 -
d. N3 -
molecular geometry |
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a. NO2 |
nonlinear, |
NO2 there is an essential number of electrons yet a non-basic number of electron sets. Since any orbital can suit 0, 1, or 2 electrons, 21/2 electron sets must be set into three orbitals. In this way the geometry is based upon a trigonal-planar course of action of electron sets. Since the solitary match orbital is just half filled, it requests less space, and the O-N-O point opens out a bit (to 134.1°) from the perfect trigonal edge of 120°. Expansion of one electron to NO2 gives the nitrite anion NO2-. This last electron finishes the half-possessed solitary combine orbital and this filling of the orbital causes it to round out, thus shut the O-N-O bond point to 115°. |
NO2 is nonlinear, the number of vibrational modes is 3N–6, where N is the number of atoms. modes = 3*3–6 = 3 modes. |
b. PCl2 + |
Linear |
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c. BrF2 - |
linear molecule |
F-Br-F This will have 22 electrons (7 from every iota + 1 for the - charge) Start by giving both F iotas a full octet. Counting the sets in the bonds, that take 16, abandoning you with 6 addiitonal electrons. Those will make up 3 unshared sets of electrons on the Br molecule. That will give you a straight particle with the geometry of electrons/bonds around the Br iotas in a trigonal bipyramid game plan. The bond edge ought to be 180 degrees, a the atom ought to have no net dipole minute. |
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d. N3 - |
linear |
there are no lone pairs on the central atom to begin with. |