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Topic: Question related to bond dissociation enthalpies and bond order.  (Read 3098 times)

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Offline earthnation112

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The following is a past exam paper question:

(a)   Using bond order, compare the following bond dissociation enthalpies and explain the differences: O2 (494 kJ mol-1), O2+ (644 kJ mol-1), N2 (942 kJ mol-1) and N2+ (841 kJ mol-1).

Answer:
•   Make O2, O2+, N2 and N2+ molecular orbital diagrams.
•   Predict the bond order for these using formula.
•   Mention HOMO and LUMO and also why the bond order has increased or decreased mentioning the placement of electrons in bonding and anti-bonding orbitals.
•   Mention the relationship between bond order and bond enthalpy. Which is, bond enthalpy increases as bond order increases. As a higher bond order means the bonds are shorter making it harder for these bonds to be broken (bond dissociation).

Is my answer sufficient for a first year undergraduate studying chemistry? Any more information which I can mention?

Offline mikasaur

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Re: Question related to bond dissociation enthalpies and bond order.
« Reply #1 on: June 29, 2016, 01:37:01 AM »
Pretty well-thought-out to me. I think that's a pretty sufficient answer for a frosh undergrad course.
Or you could, you know, Google it.

Offline earthnation112

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Re: Question related to bond dissociation enthalpies and bond order.
« Reply #2 on: June 29, 2016, 09:58:51 AM »
Pretty well-thought-out to me. I think that's a pretty sufficient answer for a frosh undergrad course.

Cheers thank you for the input.

Offline Enthalpy

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Re: Question related to bond dissociation enthalpies and bond order.
« Reply #3 on: June 29, 2016, 02:47:23 PM »
To my eyes, the interesting part of the question is why O2+ is easier to break than O2. That the question also puts N2 suggests that an explanation for this is awaited in your answer.

So, at least to my eyes:
- Double-check the data
- Show the accupied antibonding orbitals of O2 and of O2+
- Try to relate it with the dissociation energy.
- And compare with N2 which resembles the expectation more.

Offline earthnation112

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Re: Question related to bond dissociation enthalpies and bond order.
« Reply #4 on: June 29, 2016, 03:25:43 PM »
To my eyes, the interesting part of the question is why O2+ is easier to break than O2. That the question also puts N2 suggests that an explanation for this is awaited in your answer.

So, at least to my eyes:
- Double-check the data
- Show the accupied antibonding orbitals of O2 and of O2+
- Try to relate it with the dissociation energy.
- And compare with N2 which resembles the expectation more.

Wouldn’t I be covered by for example comparing firstly O2 and O2+ by using the bond order formula and mentioning they bond order and why there is a change between the two. For example, mentioning where the extra electrons go and mentioning if it goes to a bonding or anti bonding orbital.

Also doing the same of N2 and N2+.

Also when comparing the bond orders mentioning the higher the bond order the more stable a molecule is, which in turns makes it harder to dissociate as it has a shorter bond length. So the molecule with the highest bond order from the ones mentioned is N2 and it is the hardest to dissociate.



Offline Enthalpy

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Re: Question related to bond dissociation enthalpies and bond order.
« Reply #5 on: June 30, 2016, 07:04:43 AM »
Well, try it and compare with the data.

To my eyes, just bonding and antibonding fits the given data nicely: increase up to N2, then decrease.

I've found interatomic distances: O2+ 132pm versus O2 121pm. Deduction?

Also, a triple bond may bring the nuclei closer than a double and a single, but each added bond is increasingly weaker nevertheless, if you think at H3C-CH3, H2C=CH2 and HC≡CH, so mind misleading ideas. Breaking at C≡C takes less than 1.5× than C=C which takes less than 2× C-C.

O, I see. The question wanted you to use the bond order.
« Last Edit: June 30, 2016, 08:01:25 AM by Enthalpy »

Offline earthnation112

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Re: Question related to bond dissociation enthalpies and bond order.
« Reply #6 on: June 30, 2016, 01:42:05 PM »
Thank you for the help  :)

Offline mjc123

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Re: Question related to bond dissociation enthalpies and bond order.
« Reply #7 on: July 01, 2016, 04:44:08 AM »
Quote
I've found interatomic distances: O2+ 132pm versus O2 121pm. Deduction?
Deduction: you've been looking in the wrong place. It should have been obvious to you from the bond order.
O2+: 1.116 Å (http://webbook.nist.gov/cgi/cbook.cgi?ID=C12185078&Mask=1000#Diatomic)
O2-: 1.35 Å (http://webbook.nist.gov/cgi/cbook.cgi?Formula=O2-&Units=SI&cDI=on)
Quote
Also, a triple bond may bring the nuclei closer than a double and a single, but each added bond is increasingly weaker nevertheless
Not necessarily; compare N2H4, N2H2 and N2.

Offline earthnation112

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Re: Question related to bond dissociation enthalpies and bond order.
« Reply #8 on: July 01, 2016, 02:48:50 PM »
Getting a bit confused here, is the following generally right:

Bond enthalpy increases as bond order increases.
Bond length decreases as bond order increases
.

The above quotation was taken from “Inorganic Chemistry (5th edition)” by Shriver & Atkins

Offline mjc123

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Re: Question related to bond dissociation enthalpies and bond order.
« Reply #9 on: July 04, 2016, 06:00:23 AM »
Yes

Offline Enthalpy

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Re: Question related to bond dissociation enthalpies and bond order.
« Reply #10 on: July 04, 2016, 06:53:00 PM »
Quote
I've found interatomic distances: O2+ 132pm versus O2 121pm. Deduction?
Deduction: you've been looking in the wrong place. It should have been obvious to you from the bond order.
O2+: 1.116 Å (http://webbook.nist.gov/cgi/cbook.cgi?ID=C12185078&Mask=1000#Diatomic)
O2-: 1.35 Å (http://webbook.nist.gov/cgi/cbook.cgi?Formula=O2-&Units=SI&cDI=on)
Thank you!

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