# BOND ANGle of h-o-o in h2o2

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#1
I never know how to answer this type of questionwhen they ask for the speicifix bond angle of a specific bond and not of the whole molecule, idk how to draw the H O O in cross doagram format, i dont get it why is the answer 104.5. I wouldve worked out theres 4 electron pairs but idk whatto do from then onwards, how do i know theres 2 lp on central atom oxygen
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1 year ago
#2
Oxygen's octet is two bonds and two lone pairs.
Hydrogen only forms one bond.

The basic structure must be H-O-O-H, with two lone pairs on each oxygen. Any of the two oxygens is surrounded by four centres of negative charge (two pairs shared with each hydrogen, two lone pairs). Maximum repulsion is achieved with a tetrahedral arrangement (angle = 109.5°), but each lone pair reduces the angle by 2.5° approx.

Therefore, the angle is 109.5 - 5 = 104.5°, just as in water!
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#3
(Original post by zoae)
Oxygen's octet is two bonds and two lone pairs.
Hydrogen only forms one bond.

The basic structure must be H-O-O-H, with two lone pairs on each oxygen. Any of the two oxygens is surrounded by four centres of negative charge (two pairs shared with each hydrogen, two lone pairs). Maximum repulsion is achieved with a tetrahedral arrangement (angle = 109.5°), but each lone pair reduces the angle by 2.5° approx.

Therefore, the angle is 109.5 - 5 = 104.5°, just as in water!

thank u, sorry i forgot to mention the whole molecule, it was a diagram of h202 and i had to work H-O-Obut im still unsure how this is tetrahedral? do i have to first look at the whole molecule and work the overall shape? how do u decide u start from 109.5? tetrahedral is 4 bond pairs, no lone pairs
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1 year ago
#4
thank u, sorry i forgot to mention the whole molecule, it was a diagram of h202 and i had to work H-O-Obut im still unsure how this is tetrahedral? do i have to first look at the whole molecule and work the overall shape? how do u decide u start from 109.5? tetrahedral is 4 bond pairs, no lone pairs
I try to show step-by-step how to figure out how the molecule looks like:

Dot and cross diagram
You have to work this out first. You kinda have to know that the atoms in hydrogen peroxide are ordered like H O O H (H-H-O-O wouldn't really work), so draw the atoms in that order with some space in between the atoms. Next you can assume that each 2 atoms that are next to each other are join by at least a single bond (could be a double or triple, but we do single as a starting point), so draw a dot and a cross between each atom pair (I think different coloured crosses would be helpful, one colour for each atom's electrons):

H••O••O••H (I'd draw the electrons like ":" though)

Both H's use all their electrons in bonding already, so now let's look at the O's. Each O has used 1 electron for each bond it's made so far, and O has 6 valence electrons, so they each still need 4 electrons left to add. At the moment, O has 4 electrons around them (like the red O has 2 red electrons, one black and one blue). If we put the the 4 electrons left around each O, that would give each O its desired octet of 8 electrons. You have to add them as two pairs of electrons, so there will be 2 lone pairs around each O.

H••Ö••Ö••H (I dunno how well that came out )
___¨ _.¨
Structure around each O atom
VSEPR is used to find the shape of the electron pairs (lone or bonding) around a central atom. With a molecule like H2O2, you don't have 1 central atom (like you do with BF3 for example), so you have to choose one centre, find the shape around that centre, then do the other centres.

So let's use find the shape around one of the O atoms (which will help find the H-O-O angle). Let's just use the red O:
• It has 4 electron pairs around it. According to VSEPR theory, this means that its base structure is tetrahedral. This is only because this would be the structure where the electron pairs are furthest apart from each other, resulting in the lowest energy possible for the molecule. The base angle, i.e. the angle when all electron pairs and bonding pairs, would be 109.5°. It's this particular angle because if you have an actual tetrahedron with an imaginary point right in the middle, it's the angle between that point and 2 other vertices.
• Specifically, it has 2 bonding and 2 lone pairs of electrons. Each lone pair reduces the angle between the bonding pairs (here, the H-O and O-O bonding pairs) by about 2.5° (lone electron pairs are more repulsive than bonding electron pairs), so the angle between the bonding pairs (the H-O-O angle) is 104.5°.
• Finally, you're right: the shape around the O atom can't be "tetrahedral" as there are only 2 atoms around it. If an atom has 2 atoms (2 bonding pairs) and 2 lone pairs, this is labelled as a bent or non-linear structure.

Hope that helps Last edited by Professor L; 1 year ago
1
1 year ago
#5
(Original post by zoae)
Oxygen's octet is two bonds and two lone pairs.
Hydrogen only forms one bond.

The basic structure must be H-O-O-H, with two lone pairs on each oxygen. Any of the two oxygens is surrounded by four centres of negative charge (two pairs shared with each hydrogen, two lone pairs). Maximum repulsion is achieved with a tetrahedral arrangement (angle = 109.5°), but each lone pair reduces the angle by 2.5° approx.

Therefore, the angle is 109.5 - 5 = 104.5°, just as in water!
The pair geometry is that of a terahedral since it has 4 areas of elctron areas of density. However the molecules geometry, since it only has 2 bonded pair is similar to a linear molecule, is 104.5 giving it a bent v-shape since the lone pairs have a larger repulsion give, and the 2 bonds (again like a linear) on either side are arrange in a bent shape in-order to minimize the energy caused by repulsion. I hope I could answer your question in due time, any queries please let me know. Goodluck in the future, take care.
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#6
(Original post by Professor L)
I try to show step-by-step how to figure out how the molecule looks like:

Dot and cross diagram
You have to work this out first. You kinda have to know that the atoms in hydrogen peroxide are ordered like H O O H (H-H-O-O wouldn't really work), so draw the atoms in that order with some space in between the atoms. Next you can assume that each 2 atoms that are next to each other are join by at least a single bond (could be a double or triple, but we do single as a starting point), so draw a dot and a cross between each atom pair (I think different coloured crosses would be helpful, one colour for each atom's electrons):

H••O••O••H (I'd draw the electrons like ":" though)

Both H's use all their electrons in bonding already, so now let's look at the O's. Each O has used 1 electron for each bond it's made so far, and O has 6 valence electrons, so they each still need 4 electrons left to add. At the moment, O has 4 electrons around them (like the red O has 2 red electrons, one black and one blue). If we put the the 4 electrons left around each O, that would give each O its desired octet of 8 electrons. You have to add them as two pairs of electrons, so there will be 2 lone pairs around each O.

H••Ö••Ö••H (I dunno how well that came out )
___¨ _.¨
Structure around each O atom
VSEPR is used to find the shape of the electron pairs (lone or bonding) around a central atom. With a molecule like H2O2, you don't have 1 central atom (like you do with BF3 for example), so you have to choose one centre, find the shape around that centre, then do the other centres.

So let's use find the shape around one of the O atoms (which will help find the H-O-O angle). Let's just use the red O:
• It has 4 electron pairs around it. According to VSEPR theory, this means that its base structure is tetrahedral. This is only because this would be the structure where the electron pairs are furthest apart from each other, resulting in the lowest energy possible for the molecule. The base angle, i.e. the angle when all electron pairs and bonding pairs, would be 109.5°. It's this particular angle because if you have an actual tetrahedron with an imaginary point right in the middle, it's the angle between that point and 2 other vertices.
• Specifically, it has 2 bonding and 2 lone pairs of electrons. Each lone pair reduces the angle between the bonding pairs (here, the H-O and O-O bonding pairs) by about 2.5° (lone electron pairs are more repulsive than bonding electron pairs), so the angle between the bonding pairs (the H-O-O angle) is 104.5°.
• Finally, you're right: the shape around the O atom can't be "tetrahedral" as there are only 2 atoms around it. If an atom has 2 atoms (2 bonding pairs) and 2 lone pairs, this is labelled as a bent or non-linear structure.

Hope that helps thank u so much, that really helped!!
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#7
(Original post by η γνώση)
The pair geometry is that of a terahedral since it has 4 areas of elctron areas of density. However the molecules geometry, since it only has 2 bonded pair is similar to a linear molecule, is 104.5 giving it a bent v-shape since the lone pairs have a larger repulsion give, and the 2 bonds (again like a linear) on either side are arrange in a bent shape in-order to minimize the energy caused by repulsion. I hope I could answer your question in due time, any queries please let me know. Goodluck in the future, take care.
thank u!
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