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QuestionChemistry

Draw the Lewis structure and predict the geometry for OF^2. Tetrahedral Bent Linear T-shaped
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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.