According to VSEPR model, the shape of `[XeOF_(5)]^(-)` is
According to VSEPR model, the shape of `[XeOF_(5)]^(-)` is
A
Octahedral
B
Triagonal bipyramidal
C
Square pyramidali
D
pentagonal monopyramidal
Text Solution
AI Generated Solution
The correct Answer is:
To determine the shape of the ion \([XeOF_5]^-\) according to the VSEPR (Valence Shell Electron Pair Repulsion) model, we can follow these steps:
### Step 1: Count the Valence Electrons
- **Xenon (Xe)** has 8 valence electrons.
- **Oxygen (O)** has 6 valence electrons.
- Each **Fluorine (F)** has 7 valence electrons, and there are 5 fluorine atoms, contributing \(5 \times 7 = 35\) electrons.
- The overall charge of the ion is -1, which means we add 1 more electron.
**Total Valence Electrons Calculation:**
\[
8 \, (Xe) + 6 \, (O) + 35 \, (5F) + 1 \, (charge) = 50 \, \text{valence electrons}
\]
### Step 2: Determine the Number of Bonding and Lone Pairs
- Each bond represents a pair of electrons. Since we have 5 fluorine atoms and 1 oxygen atom bonded to xenon, that accounts for 6 bonding pairs.
- The total number of pairs of electrons is \(50 \div 2 = 25\) pairs.
- Since we have 6 bond pairs, the remaining pairs are:
\[
25 - 6 = 19 \, \text{lone pairs}
\]
However, this calculation is incorrect as we need to consider the lone pairs more carefully.
### Step 3: Correctly Identify Lone Pairs
- We have 6 bond pairs, and we can see that there is 1 lone pair left from the total of 25 pairs:
\[
1 \, \text{lone pair}
\]
### Step 4: Determine the Hybridization
- The hybridization can be determined by the number of bonding pairs and lone pairs. We have 6 bond pairs and 1 lone pair, which gives us a total of 7 regions of electron density.
- The hybridization corresponding to 7 regions is \(sp^3d^3\).
### Step 5: Determine the Geometry
- The geometry for \(sp^3d^3\) hybridization is typically octahedral, but due to the presence of a lone pair, the shape will adjust.
- The presence of a lone pair will distort the octahedral shape into a pentagonal bipyramidal shape.
### Conclusion
Thus, the shape of \([XeOF_5]^-\) according to the VSEPR model is **pentagonal monopyramidal**.
### Final Answer
**The shape of \([XeOF_5]^-\) is pentagonal monopyramidal.**
---
To determine the shape of the ion \([XeOF_5]^-\) according to the VSEPR (Valence Shell Electron Pair Repulsion) model, we can follow these steps:
### Step 1: Count the Valence Electrons
- **Xenon (Xe)** has 8 valence electrons.
- **Oxygen (O)** has 6 valence electrons.
- Each **Fluorine (F)** has 7 valence electrons, and there are 5 fluorine atoms, contributing \(5 \times 7 = 35\) electrons.
- The overall charge of the ion is -1, which means we add 1 more electron.
...
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