Give five examples of structures with this formula (c6h12). at least one should contain a ring, and at least one should contain
a double bond.
1 answer:
Answer: -
Following are five examples of structures with the chemical formula C₆H₁₂
Compound A is Hexene.
Compound B is 2-Hexene.
Compound C is 3-Hexene.
Compound D is Cyclohexane.
Compound E is Methylcyclopentane.
As we can see Hexene, 2- Hexene and 3-Hexene all have double bonds.
Cyclohexane and Methylcyclopentane contains a ring.
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The question is incomplete, here is the complete question:
Problem page in each of the molecules drawn below one chemical bond is colored red. Decide whether this bond is likely to be polar or not. if the bond is likely to be polar, write down the chemical symbol for the atom which will have more negative charge.
The image is attached below.
<u>Answer:</u>
<u>For carbon dioxide molecule:</u> The bond is considered as polar and the elecvtronegative atom is oxygen.
<u>For water molecule:</u> The bond is considered as polar and the elecvtronegative atom is oxygen.
<u>Explanation:</u>
There are two types of covalent bonds:
- <u>Polar covalent bond:</u> This bond is formed when difference in electronegativity between the atoms is present. When atoms of different elements combine, it results in the formation of polar covalent bond. <u>For Example:</u>
etc..
- <u>Non-polar covalent bond:</u> This bond is formed when there is no difference in electronegativity between the atoms. When atoms of the same element combine, it results in the formation of non-polar covalent bond. <u>For Example:</u>
etc..
<u>In carbon dioxide molecule:</u>
The given bond is present between C and O atom.
Electronegativity value of C = 2.5
Electronegativity value of O = 3.5
Electronegativity difference = (3.5 - 2.5) = 1
As, electronegativity difference is present. So, the bond is considered as polar and the elecvtronegative atom is oxygen.
<u>In water molecule:</u>
The given bond is present between H and O atom.
Electronegativity value of H = 2.1
Electronegativity value of O = 3.5
Electronegativity difference = (3.5 - 2.1) = 1.4
As, electronegativity difference is present. So, the bond is considered as polar and the elecvtronegative atom is oxygen.
