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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: earthnation112 on June 28, 2016, 04:10:04 PM

Title: Question related to bond dissociation enthalpies and bond order.
Post by: earthnation112 on June 28, 2016, 04:10:04 PM: 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?
Title: Re: Question related to bond dissociation enthalpies and bond order.
Post by: mikasaur 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.
Title: Re: Question related to bond dissociation enthalpies and bond order.
Post by: earthnation112 on June 29, 2016, 09:58:51 AM: Quote from: mikasaur 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.

Cheers thank you for the input.
Title: Re: Question related to bond dissociation enthalpies and bond order.
Post by: Enthalpy 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.
Title: Re: Question related to bond dissociation enthalpies and bond order.
Post by: earthnation112 on June 29, 2016, 03:25:43 PM: Quote from: Enthalpy 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.

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.
Title: Re: Question related to bond dissociation enthalpies and bond order.
Post by: Enthalpy 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.
Title: Re: Question related to bond dissociation enthalpies and bond order.
Post by: earthnation112 on June 30, 2016, 01:42:05 PM: Thank you for the help :)
Title: Re: Question related to bond dissociation enthalpies and bond order.
Post by: mjc123 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.
Title: Re: Question related to bond dissociation enthalpies and bond order.
Post by: earthnation112 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
Title: Re: Question related to bond dissociation enthalpies and bond order.
Post by: mjc123 on July 04, 2016, 06:00:23 AM: Yes
Title: Re: Question related to bond dissociation enthalpies and bond order.
Post by: Enthalpy on July 04, 2016, 06:53:00 PM: Quote from: mjc123 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)
Thank you!