H2o2 Bonding And Lone Pairs

I never know how to answer this blazon 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 cantankerous doagram format, i dont go it why is the answer 104.5. I wouldve worked out theres 4 electron pairs just idk whatto practise from then onwards, how exercise i know theres 2 lp on central cantlet oxygen

Oxygen'due south octet is 2 bonds and two lone pairs.
Hydrogen only forms i bail.

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

Therefore, the angle is 109.5 - v = 104.five°, just as in water!

(Original post by zoae)
Oxygen's octet is two bonds and ii lone pairs.
Hydrogen only forms one bond.

The basic construction must be H-O-O-H, with 2 lone pairs on each oxygen. Any of the two oxygens is surrounded past iv centres of negative accuse (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.v - 5 = 104.5°, just every bit in water!

thank u, sorry i forgot to mention the whole molecule, information technology was a diagram of h202 and i had to piece of work H-O-Obut im still unsure how this is tetrahedral? do i have to commencement look at the whole molecule and work the overall shape? how do u decide u starting time from 109.5? tetrahedral is iv bond pairs, no lone pairs

(Original mail by usernamenew)
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 await 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 cantankerous diagram
You lot take to piece of work this out get-go. You kinda take to know that the atoms in hydrogen peroxide are ordered like H O O H (H-H-O-O wouldn't actually work), so draw the atoms in that club with some space in between the atoms. Next you lot can presume that each two atoms that are adjacent to each other are join by at least a single bond (could be a double or triple, but we do single equally a starting point), and so depict a dot and a cross betwixt each atom pair (I call up unlike coloured crosses would be helpful, one color for each cantlet's electrons):

H•

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

Both H'southward apply all their electrons in bonding already, and then now let's look at the O's. Each O has used 1 electron for each bail information technology'south made so far, and O has 6 valence electrons, then they each still need

4 electrons left to add . At the moment, O has 4 electrons around them (like the red O has 2 cerise electrons, one black and one bluish). If nosotros put the the four 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 :lol: )
___ ¨ _. ¨
Structure around each O atom
VSEPR is used to discover the shape of the electron pairs (solitary or bonding) around a central atom. With a molecule like H₂O₂, you don't have 1 central cantlet (like you do with BF₃ for example), then you have to cull one centre, discover the shape effectually that middle, then exercise the other centres.

And so let's use find the shape around one of the O atoms (which volition help detect the H-O-O angle). Let's only use the red O:

It has 4 electron pairs effectually it. According to VSEPR theory, this means that its base of operations structure is tetrahedral. This is only considering this would be the structure where the electron pairs are furthest apart from each other, resulting in the everyman energy possible for the molecule. The base angle, i.due east. the angle when all electron pairs and bonding pairs, would be 109.5°. Information technology's this detail angle considering if y'all have an actual tetrahedron with an imaginary betoken right in the middle, information technology'south the bending between that signal and 2 other vertices.
Specifically, it has two bonding and 2 lone pairs of electrons. Each lone pair reduces the angle betwixt the bonding pairs (here, the H-O and O-O bonding pairs) by virtually two.5° (lone electron pairs are more than repulsive than bonding electron pairs), so the angle between the bonding pairs (the H- O -O bending) is 104.five°.
Finally, y'all're correct: the shape effectually the O atom can't be "tetrahedral" equally at that place are only 2 atoms around information technology. If an atom has two atoms (ii bonding pairs) and 2 alone pairs, this is labelled as a bent or non-linear structure.

Hope that helps

Last edited past Professor L; 3 years agone

(Original mail by zoae)
Oxygen's octet is 2 bonds and ii solitary pairs.
Hydrogen only forms i bond.

The bones structure must exist H-O-O-H, with 2 solitary pairs on each oxygen. Whatsoever of the two oxygens is surrounded by four centres of negative charge (two pairs shared with each hydrogen, two lonely pairs). Maximum repulsion is achieved with a tetrahedral organisation (angle = 109.5°), but each solitary pair reduces the angle by 2.five° approx.

Therefore, the angle is 109.five - 5 = 104.5°, just every bit in h2o!

The pair geometry is that of a terahedral since it has 4 areas of elctron areas of density. However the molecules geometry, since it simply has 2 bonded pair is similar to a linear molecule, is 104.5 giving it a bent five-shape since the alone pairs take a larger repulsion give, and the two bonds (once more like a linear) on either side are accommodate in a aptitude shape in-order to minimize the energy acquired by repulsion. I hope I could answer your question in due time, any queries delight let me know. Goodluck in the future, take intendance.

(Original post by Professor L)
I endeavour to testify step-by-step how to figure out how the molecule looks like:

Dot and cross diagram
You have to piece of 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 depict the atoms in that club with some space in between the atoms. Next you tin can assume that each two atoms that are adjacent to each other are bring together by at least a single bond (could exist a double or triple, only nosotros do single as a starting point), so describe a dot and a cantankerous between each atom pair (I think different coloured crosses would be helpful, ane colour for each atom'southward electrons):

H•

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

Both H's use all their electrons in bonding already, so at present allow'due south await at the O's. Each O has used 1 electron for each bond it'south fabricated then far, and O has half dozen valence electrons, and then they each still need

4 electrons left to add . At the moment, O has four electrons effectually them (similar the ruby-red O has two cherry electrons, one blackness 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 lot have to add them equally two pairs of electrons, so at that place will be two alone pairs effectually each O.

H•

•Ö• •Ö••H (I dunno how well that came out :lol: )
___ ¨ _. ¨
Structure around each O atom
VSEPR is used to detect the shape of the electron pairs (lone or bonding) around a primal cantlet. With a molecule similar H₂O_ii, yous don't have 1 fundamental atom (like y'all do with BF₃ for instance), so you lot have to cull one centre, find the shape around that centre, then practise the other centres.

So permit'southward use detect the shape around one of the O atoms (which volition assistance find the H-O-O angle). Let's just utilise the crimson O:

It has iv electron pairs around it. According to VSEPR theory, this ways that its base of operations structure is tetrahedral. This is simply because this would be the construction where the electron pairs are furthest apart from each other, resulting in the lowest free energy possible for the molecule. The base angle, i.e. the angle when all electron pairs and bonding pairs, would be 109.5°. Information technology'south this particular bending because if you have an actual tetrahedron with an imaginary betoken right in the eye, it'due south the angle between that point and ii other vertices.
Specifically, it has 2 bonding and 2 lone pairs of electrons. Each lone pair reduces the angle between the bonding pairs (hither, the H-O and O-O bonding pairs) past nearly 2.5° (lonely electron pairs are more repulsive than bonding electron pairs), so the angle between the bonding pairs (the H- O -O bending) is 104.5°.
Finally, you're right: the shape effectually the O atom can't exist "tetrahedral" as there are just ii atoms around it. If an atom has 2 atoms (2 bonding pairs) and 2 lone pairs, this is labelled equally a bent or non-linear construction.

Hope that helps

give thanks u so much, that really helped!!

(Original post past η γνώση)
The pair geometry is that of a terahedral since it has 4 areas of elctron areas of density. Nevertheless the molecules geometry, since information technology but has two bonded pair is like to a linear molecule, is 104.5 giving it a bent five-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, whatsoever queries delight let me know. Goodluck in the future, accept care.

give thanks u!