HELP: MODELING MOLECULES!!!!!!!!!!!!!!!!

Chemistry

1 answer:

garik1379 [7]3 years ago

8 0

<span>Sulfur Hexachloride SCl6 So now we count the number of valence electrons each has by seeing what column it's in, (1-8) not counting the columns of the transition metals. Since Sulfur is in the 6th and Chlorine is in the 7th, and there are 6 chlorines, we can add up all their valence electrons: 6*1+7*6=48 valence electrons. But remember that electrons come in pairs, either in bonds or as lone pairs. So I usually divide the valence electron number by 2 and just think about placing pairs. It's up to you, but I think it's convenient since we can count "1" in our mind each time we place a bond or a electron pair. So we need to place 24 pairs/bonds. So we can guess that sulfur is a central atom and draw out a bond from sulfur to each chlorine. Since Sulfur is in the 3rd row it can use d-orbitals to break the octet rule. So when we bond all the chlorines onto sulfur we get:
(see the figure)

and
</span><span>So we made 6 bonds, that means we used up 12 electrons, so if you're counting (AND YOU SHOULD BE!) you have 36 electrons or simply 18 electron pairs left to place. Now let's give chlorine a neutral charge.</span>

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1 . $Al_2O_3$

2. $TiO_2$

Explanation:

The more stable the ionic compound, the more is it lattice energy.

The more the charge on the cation and the anion, the greater is the lattice energy.
The less the size of the cation and the anion, the greater is the lattice energy.

Scandium oxide ( $Sc_2O_3$ ) is an oxide in which $Sc^{3+}$ behaves as cation and $O^{2-}$ behaves as anion.

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2. $TiO_2$

This is because the charge on the cation $Ti^{4+}$ is greater than that of $Sc^{3+}$ and also the size of the cation $Ti^{4+}$ is smaller than $Sc^{3+}$ . Thus, this corresponds to higher lattice energy.