Bond Order

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sid bauer
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Bond Order

Postby sid bauer » Sat Jul 15, 2017 12:44 am

NEED CLARIFICATION: How do you calculate bond order? And what does it mean in terms of chemistry? Can bond order be calculated for just one atom, or does it have to be two atoms joined together?

Thank you!

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Re: Bond Order

Postby Chem_Mod » Sat Jul 15, 2017 9:17 am

The bond order indicates the number of bonds between two atoms. The simplest way to determine the bond order would be to draw the Lewis structure of the molecule and look at how many bonds form between the atoms. A single bond is a bond order of 1, a double 2, triple 3, and so on.

The significance of the bond order is that it indicates the strength and length of bonds. The higher the bond order, the more electron density there is in the bonds, the stronger the bond, and the shorter the bond.

Hope this helps!

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Re: Bond Order

Postby 604744616 » Mon Jul 17, 2017 2:18 pm

I need help on understanding hybridization!!! Can anyone explain to me the bond order, how do we go about recognizing what bond order is which from a Lewis structure? I am just confused on this whole lecture from today, if anyone can please help me and going over this that would be greatly appreciated

Wu Yuchen A1
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Re: Bond Order

Postby Wu Yuchen A1 » Tue Jul 18, 2017 2:14 pm

IN the case that Lewis structure is completely correct, you can simply count the number of bonds between two atoms and get the result as bond order of the two bonded atoms. Hybridization state depends on the number of regions of electron density. State of hybridization isn't influenced by the bond order of atoms( no matter single/double/triple bond or lone pair, it's counted as one electron density region).

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Re: Bond Order

Postby derek1d » Mon Jul 24, 2017 10:32 pm

Does counting the bonds and classifying them based upon the total number of shared electrons stand across all cases? I have come across examples where a linear structure was given and the correct hybridization turned out to be sp^3. Can someone help clarify why that would be?

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