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Consider the bent shape of $\ce{H2O}$ with a bond angle of $104°$. From my understanding, this is due to the lone pair repulsion between the 2 pairs of lone pair electrons. Hence, it adopts a tetrahedral structure, with a smaller bond angle due to the 2 lone pairs. We know this from VESPR.

However consider the following diagram

enter image description here

What is inherently wrong with it? To me, it seems that $\ce{H2O}$ could adopt this linear shape without violating any rules. As its symmetrical, the repulsions should essentially cancel each other out, leaving this linear shape. However, obviously this structure is not possible. So what is wrong with it? Why can't the lone pairs be situated on opposite sides as such?

Furthermore, wouldn't placing the lone pairs on opposite sides maximise this distance between them. Hence shouldn't the above be favourable?

Poutnik
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Quin Gardiner Bax
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1 Answers1

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You mentioned that it adopts a tetrahedral. Take a look at this image of a tetrahedral: enter image description here

Let's take a look at some possible positions the lone pairs (denoted by purple) can be. enter image description hereenter image description here enter image description here

As you can see, no matter which two places you put the lone pairs, the structure will be identical (they are all the same, just rotated).

So, while it may look like it based on the lewis structure, the lone pairs cannot be on "opposite sides," they will always be next to each other.

jb0
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