QQuestionChemistry
QuestionChemistry
Draw the Lewis structure and predict the geometry for OF^2.
Tetrahedral
Bent
Linear
T-shaped
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Answer
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Step 1:: Determine the total number of valence electrons in the OF^2 molecule.
The valence electrons of oxygen (O) is 6, and each fluorine (F) atom has 7 valence electrons. Since there are two fluorine atoms in OF^2, we have: Total valence electrons = Valence electrons of O + 2 * Valence electrons of F = 6 + 2 * 7 = 6 + 14 = 20 valence electrons
Step 2:: Draw the Lewis structure for OF^2.
Step 3:
Place the central atom, oxygen, and then distribute the remaining valence electrons around the oxygen and fluorine atoms. O: ____F - O - F____ : :
Step 4:
Distribute the remaining valence electrons as lone pairs on the oxygen and fluorine atoms to satisfy the octet rule. O: ____F : : O : F____ : : : : : : : :
Step 5:
Add the remaining valence electrons to the oxygen atom as a lone pair. O: ____F : : O : F____ : : e- : : e- : : e- : : e-
Step 6:: Predict the molecular geometry for OF^2.
Step 7:
Count the regions of electron density around the central atom (oxygen). There are 4 regions of electron density (2 bonding pairs and 2 lone pairs).
Step 8:
Refer to the AXN system, where A is the central atom, X is the number of bonding pairs, and N is the number of lone pairs. Here, A = O, X = 2, and N = 2.
Step 9:
Using the AXN system, we can predict the molecular geometry for OF^2. With 2 bonding pairs and 2 lone pairs, the molecular geometry is bent or V-shaped.
Final Answer
O: ____F : : O : F____ : : e- : : e- : : e- : : e- The molecular geometry for OF^2 is bent or V-shaped.
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