Answer This Gen chimstry
Started by IsRaEl21
Well they are all isoelectronic, however, what leads you to believe the ionization energies are all equal?
They are all isoelectronic, so isn't ionization energy based on size?
Na+ would be the smallest, Na+ has the highest ionization energy.
Correct me if I'm wrong.
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Nevermind...I don't agree.
flouride is much larger than it's neutral counterpart, and the neutral F is larger than the corresponding noble gas
Thus, the ionization energy would be lower than Ne
I think if it wasn't a typo, then Ne would have the higher ionization energy considering it has a higher Z(eff). But if it actually IS Na+ instead of Na-, then I agree with everyone in Na+ having the largest IE
Ok, first the Na is Na+ Or Na-? , if is Na- , it has the LOWEST Ionization E and if it is Na+ it has the Highest Ionization E!!!
But lets assume Na- and looking for answer, "Ne" will be the highest one if we go for "Na-"
GOING ACROSS IONIZATION ENERGY INCREASES AND GOING FROM BOTTOM TO UP IONIZATION ENERGY INCREASES... IT WOULD SEEM "Ne" IS THE ANSWER BUT IF YOU HAVE "O 2-" AND "F -" AND "Na +" IT WOULD PUT ALL THE IONS IN THE SAME SPOT AS "Ne" MAKEING ALL EQUALL IN IONIZATION ENERGY
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i believe it's Na+
Ok, after your correction the answer is "Na+". I assume you know why the answer is Na+, I know many of you guys don't like my detail explanation. but let me give you a small ouote for that.
all your Ions and Ne have 10 electrons, in Na+ these 10 electrons forcing by 11 protons ( more abserbation), in Ne by 10 P in F- by 9 P and in O-2 by 8 P.
Hopefully it helped.
104.5 or in some books you can see 105 compare it to H2O.
"O" and "S'' are in same group and hybridazation for both will be SP3
hope it helped
That was a trick question.
Hybridization cannot explain the bond angle of hydrogen sulfide.
The bond angle is 92.2. (I was taught it was 90 degrees) Wikipedia says 92.2
However, the point is....is that hybridization isn't needed to explain the bond angle.
The p orbitals don't undergo hybridization, thus, maintain their original spatial configurations (90 degrees...x,y,z)
So...yeah
Just thought it was a nice question.
🙁 i suck at bond angles huh!
Well, they all have the same electronic configuration, which is a full octet, but the effective nuclear charge is going to be different because of the atomic number. Therefore, I'd venture to say that Ne is the highest ionization energy, because it has the highest atomic number.
Yes, Valence Shell Electron Pair Repulsion theory is just a model to explain the observed phenomena.
However, like most models...it doesn't always work
VSEPR works very well to explain covalent bonding for periods 1 and 2.
But this is ALWAYS the case. Just something to keep in mind.
In the case of hydrogen sulfide, the atomic P orbitals simply overlap with the S orbital of hydrogen. Thus, maintaining the original spatial orientation of the atomic orbitals.
Applying hybridization can't explain the observed bond angle of hydrogen sulfide
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If you want more info...go here. The website gives a good review of hybridization. For hydrogen sulfide specifically, go the end of the page. FINAL REMARKS ABOUT HYBRID ORBITALS.
http://www.chem1.com/acad/webtext/chembond/cb07.html
http://www.chem1.com/acad/webtext/chembond/cb07.html
Perhaps the bonding electron pairs repel each other ever so slightly enough to make it 92
When I first learned it...I just learned it as 90. But yeah...that would be my guess...slight repulsion
"oxygen atom in water molecule has two lone pairs; due to small size of oxy , the lone pair-lone pair and the bond-pair bond-pair repulsion is large and as a result tend to push the molecule apart ; however in H2S, the sulphur atom is greater in size and the lone pair tend to be far apart, thereby considerably reducing the repulsions and also the bond angle. that is the reason"
Are you saying that atomic size alone can account for the deviance from 109.5 as predicted by hybridization/VSEPR theory?
Bond angles usually explained by VSEPR which explains the stereochemistry of an atom in a molecule is determined primarily by the repulsive interactions among all the electron pairs in its valence shell.
But in this case, we have to take the hybridization theory into account. H2S is hardly hybridized. The H-S bond from using the p-orbitals, unlike 0-H using sp3 hybridized
The reason is that the large size of S allows the electron pairs to be far from each other result in repulsion's energy is not very large, and thus the energy need not be minimized.
Oxygen is smaller. Therefore, hybridization have to be made (result in more stable molecule)
Sorry, I see what you mean, but your reply is a bit confusing. I agree w/ you that hybridization is not necessary to explain hybridization.
My response was that the bond angles are the result of the overlapping atomic orbitals. Since this is the case, the spatial orientation of the atomic orbitals remain the same and can explain the 90degree bond angle.
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