the structure of `SF_6` is octahedral. what is the structure of `SF_4`

Options

A. Trigonal bipyramidal
B. pyramidal
C. square planar
D. trigonal planar

Answer

To determine the structure of sulfur tetrafluoride (SF₄), we will follow these steps:

### Step 1: Determine the Valence Electrons
Sulfur (S) is in group 16 of the periodic table and has 6 valence electrons. Each fluorine (F) atom has 7 valence electrons, and since there are 4 fluorine atoms, they contribute a total of \(4 \times 7 = 28\) valence electrons. Therefore, the total number of valence electrons in SF₄ is:
\[ 6 \text{ (from S)} + 28 \text{ (from 4 F)} = 34 \text{ valence electrons} \]

### Step 2: Determine the Hybridization
To find the hybridization of sulfur in SF₄, we can use the formula:
\[ \text{Hybridization} = \frac{1}{2} \left( \text{Number of valence electrons} + \text{Number of monovalent atoms} - \text{Charge on cation} + \text{Charge on anion} \right) \]
Here, the number of monovalent atoms (the 4 fluorine atoms) is 4, and there are no charges on the molecule. Thus, we have:
\[ \text{Hybridization} = \frac{1}{2} \left( 6 + 4 + 0 + 0 \right) = \frac{10}{2} = 5 \]
This indicates that the hybridization of sulfur in SF₄ is \(sp^3d\).

### Step 3: Determine the Molecular Geometry
With \(sp^3d\) hybridization, the ideal geometry is trigonal bipyramidal. However, since there are 4 bond pairs (from the 4 fluorine atoms) and 1 lone pair, the molecular geometry will be affected by the presence of the lone pair.

### Step 4: Identify the Shape
In a trigonal bipyramidal arrangement, the lone pair will occupy one of the equatorial positions to minimize repulsion. Therefore, the actual shape of SF₄ will be distorted from the ideal trigonal bipyramidal shape due to the lone pair, resulting in a seesaw shape.

### Conclusion
The structure of SF₄ is **seesaw** due to the presence of one lone pair and four bond pairs.

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