Answer:
<em>the </em><em>two </em><em>elements</em><em> </em><em>are </em><em>in </em><em>the</em><em> same</em><em> </em><em>period</em><em>,</em><em> with</em><em> </em><em>element </em><em>R </em><em>the </em><em>first</em><em> </em><em>element</em><em> </em><em>in </em><em>the</em><em> </em><em>period</em><em> </em><em>and </em><em>element </em><em>Q </em><em>the </em><em>last</em><em> </em><em>element</em>
The oxidation number of iodine is 5 in Mg(IO3)2 which can be calculated as
Mg(IO3)2
MgI2O6
As we know that
Mg has +2
O has -2
So,
(+2) + 2I + 6 (-2)=0
2 + 2I - 12 =0
10+ 2I =0
10 = 2I
I =5
The products for the complete combustion of a hydrocarbon in excess air is carbon dioxide and water. Any hydrocarbon when reacted with oxygen will always yield the said products. Incomplete combustion, on the other hand, yields carbon monoxide and water.
The molecule is stable and can exist even though the number of valence electrons around central atom in the molecule are less than 8.
<h3>Is BF3 molecule stable or not?</h3>
BF3 molecule is a stable molecule because all the electrons present in the outermost shell of boron are covalently bonded with fluorine. Boron in BF3, three bonds is the maximum possible because boron only has 3 electrons to share.
So we can conclude that the molecule is stable and can exist even though the number of valence electrons around central atom in the molecule are less than 8.
Learn more about molecule here: brainly.com/question/26044300
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