lassify these bonds as ionic, polar covalent, or nonpolar covalent. Ionic Polar covalent Nonpolar covalent
1 answer:
The question is incomplete, complete question is:
Classify these bonds as ionic, polar covalent, or nonpolar covalent.
Na-F
P-Cl
Cl-Cl
Answer:
Na-F : The bond present is ionic bond.
PCl : The bond present is polar covalent bond.
ClCl : The bond present is non polar covalent bond.
Explanation:
1) Ionic bonds are defined as the bonds which are formed by the complete transfer of electrons from cation (positively charged ions) to anion (negatively charged ions). For Example: NaCl, etc.
Ionic compound :This compound is formed when difference in electronegativity between the atoms is very large.
2) Covalent bond is defined as the bond which is formed by the sharing of electrons between the atoms. For Example: HCl, etc.
Polar covalent compound: This compound 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. For Example: etc..
Non-polar covalent compound: This compound 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. For Example: etc.
Na-F : Two different atoms with high electronegativity difference
Ionic compound because sodium is metal and fluorine is non metal and the difference between their electronegativity is very large.The bond present is ionic bond.
PCl : Two different atoms with low electronegativity difference
Polar covalent compound because phosphorus and fluorine both are non metal and the difference between their electronegativity is not very large.The bond present is polar covalent bond.
ClCl : Two same atoms with nil electronegativity difference.
Non Polar covalent compound because both are chlorine atom (non metal) and the difference between their electronegativity is zero.The bond present is non polar covalent bond.
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Answer:
The bands are due to:
λmax = 289 nm n→π* transition (E = 12)
λmax = 182 nm π→π* transition (E=10000)
Explanation:
The two types of acetaldehyde transition are as follows:
n→π* and π→π*
From the attached diagram we have to:
ΔEn→π* < ΔEπ→π*
ΔEα(1/λ)
Thus:
λn→π* > λπ→π*
In n→π* spin forbidden, the intensity is low. Thus, the molar extinction E for n→π* is very low.
The same way, for π→π* spin allowed the intensity is high. Thus, the molar extinction coefficient E for π→π* is high too.
The bands are due to:
λmax = 289 nm n→π* transition (E = 12)
λmax = 182 nm π→π* transition (E=10000)
