Sapling #12
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Sapling #12
An organic compound that is distilled from wood has a molar mass of 32.04g/mol. Its composition by mass is 37.5% carbon, 12.6% hydrogen, and 49.9% oxygen.
I was able to find the molecule, CH4O, and the Lewis structure, but the next part of the question asks for the hybridization of the oxygen atom. Could someone help me find the hybridized orbital?
I was able to find the molecule, CH4O, and the Lewis structure, but the next part of the question asks for the hybridization of the oxygen atom. Could someone help me find the hybridized orbital?
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Re: Sapling #12
Finding the hybridization of the oxygen is similar to finding the hybridization of the carbon. You'd look at how many regions of electron density the oxygen atom has (# of bonds + lone pairs), and use that to find the hybridization.
Hope this helps!
Hope this helps!
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Re: Sapling #12
Hello, as the person above said, the hybridization of the atoms depend on the region of electron density. From the Lewis structure, you see that both C and O have four regions of electron density, so the hybridization for the oxygen and carbon is sp3 hybridization. Hope this helps!
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Re: Sapling #12
You can find the hybridization of the oxygen by looking at how many regions of electron density it has. Because it has two bonds and two lone pairs, the oxygen has 4 regions of electron density, giving it a sp3 hybridization
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Re: Sapling #12
The hybridization of atoms depends on the number of regions of electron density. both C and O have 4 regions, therefore it has a sp3 hybridization.
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Re: Sapling #12
Because C has 4 regions of electron density (4 single bonds), and O also has 4 regions of electron density (2 single bonds and 2 lone pairs), the hybridization of both C and O is sp3.
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Re: Sapling #12
To find the hybridization of the oxygen atom you must first consider the region of electron densities on the atom. By looking at the oxygen atom, you can see it that it has two bonds and two lone pairs giving it four regions of electron density. Therefore, you can conclude that the oxygen atom has a hybridization of sp3
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Re: Sapling #12
The hybridization is the electron density around the atom. Double and triple bonds also just count as a single region of electron density.
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Re: Sapling #12
To find the hybridization of the oxygen atom in CH4O, we just have to look at how many regions of electron density are present on the oxygen atom like the bonds connected to it and lone pairs
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Re: Sapling #12
Finding hybridization involves counting the number of regions of electron density around the atom. In the carbon, it has 4 single bonds, 3 to each hydrogen and 1 to the oxygen. This results in a sp3 hybridization as there are 4 regions of electron density. Oxygen has two lone pairs of electrons, a bond to carbon, and a bond to hydrogen, making up 4 regions of electron density and thus, sp3 hybridization.
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Re: Sapling #12
To find the hybridization of a given atom, simply look at the regions of electron density around the atom! This includes bond pairs and lone pairs. This will allow you to figure out the hybridization.
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Re: Sapling #12
Hi! If you still need help with this question, the first step I did was count up all the regions of electron density and then used the hybridization guidelines to determine the answer.
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Re: Sapling #12
In order to find the hybridized orbitals of the oxygen, you should first look to see how many electron domains it has. Since there are two bonding pairs and two lone pairs, there are 4 domains of electron density, which refers to sp3.
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Re: Sapling #12
So finding the hybridization of the oxygen is, like everyone else is saying, the same idea as finding the hybridization of the carbon. So look at your structure and count all the regions of electron density that Oxygen has, and then pick the hybridization that matches that number of regions. Hope this helps :)
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Re: Sapling #12
Hi! The hybridization of the carbon and oxygen atoms depends on how many regions of electron density it has. The carbon atom has 4 regions of electron density because it has 4 single bonds. The oxygen atom has 4 regions of electron density because it has 2 single bonds and 2 lone pairs. Thus, the hybridization of both the carbon and oxygen atoms would be sp3.
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Re: Sapling #12
Since you have found the lewis structure already, look at how many bonds/lone pairs the oxygen atom has. We can see that it has 4 bonds, which means that it has a hybridization of sp^3.
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Re: Sapling #12
Hi! The carbon atom has single bonds to four other atoms via four equivalent sp^3-hybridized orbitals. The oxygen atom has single bonds to two other atoms and two lone pairs of electrons. Therefore, oxygen is also sp^3-hybridized. I hope this helps!
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Re: Sapling #12
to find the hybrized orbital you would count up all the regions of electron density that the oxygen has and use that to determine its hybridization. Since oxygen has 2 lone pairs and 2 bonds, it has 4 regions of electron density so its hybridization is sp^3
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Re: Sapling #12
I always make sure that the number of regions of electron density match up with the number of hybridized orbitals. Sometimes I try to read the hybridized orbitals as s1p1, s1p2, s1p3... etc rather than just sp or sp2. This makes it clearer when I add it up so that I don't miss that s counts as one of the orbitals.
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Re: Sapling #12
Remember that double and triple bonds don't matter in hybridization, only regions (so any type of bond counts as 1) and lone pairs are counted. Oxygen has two bonds and two lone pairs which gives it an sp3 hybridization.
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Re: Sapling #12
When you're trying to find the hybridization orbitals just look for the amount of electron dense areas. C and O in that problem have 4 areas so you get sp3 because sp3 has 4 areas of electron density, S P P P being the 4 different orbitals that can have electrons.
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Re: Sapling #12
When trying to find hybridization you should look at the number of electron dense areas (so bonds and lone pairs). In this problem carbon has 4 single bonds meaning it has 4 electron dense areas and oxygen has 2 single bonds and 2 lone pairs meaning that it also has 4 electron dense areas. This means that they have a hybridization of sp3.
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Re: Sapling #12
Because there is a total of 4 bonds for C, and 2 bonds with O and 2 lone pairs on O, the hybridization is found out to be SP3
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Re: Sapling #12
to find the hybridization for the oxygen, you have to count the bonded ATOMS (not just bonds!) and the lone pairs as well, so it would come out to 2 atoms and 2 lone pairs, which is an sp3 orbital
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Re: Sapling #12
I used this video to help me determine the hybridization! hope it helps :)
https://www.youtube.com/watch?v=4xl0BD-tMeA
https://www.youtube.com/watch?v=4xl0BD-tMeA
Re: Sapling #12
I was confused on this question too, but you have to determine the number of electron density regions. Once you know that, you can count the hybridization out. It has four electron density regions, so the hybridization would be sp3.
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Re: Sapling #12
When trying to find the hybridization orbitals look for the regions of e- density. C and O in that problem have 4 areas, u get sp3, S P P P being the 4 different orbitals that can have electrons.
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Re: Sapling #12
For this problem, we can observe from the Lewis structure that both carbon and oxygen have 4 areas of electron density, with carbon being bonded to 4 atoms while oxygen is bonded to 2 atoms and has 2 additional lone pairs. Hence, since they both have 4 regions of electron density, we can deduce that the hybridized orbital for both of them must be .
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Re: Sapling #12
This problem, we observe that both the carbon and the hydrogen have 4 places of electron densities. Electron densities are calculated by the number of bonds and lone pairs. Since it is four, it is sp^3.
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Re: Sapling #12
It depends on the electron densities, which in this case, there are 4, meaning it is sp^3
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