Answer
Verified
429.6k+ views
Hint: The inclination of a molecule to pull in holding electrons towards itself is called its electronegativity.
The more noteworthy the s-character of the half and half orbitals, the more prominent is the electronegativity on the grounds that a s orbital holds electrons all the more firmly to the core
Regarding Electronegativity: \[sp{\text{ }} > {\text{ }}s{p^2} > {\text{ }}s{p^3}\]
Complete step by step answer:
Electronegativity increments with the s-character of the half and half orbital utilized for making covalent bond.
It could be clarified by taking a straightforward illustration of three least difficult hydrocarbons
Ethane \[{C_2}{H_6}\]→ where \[C\] uses- \[s{p^3}\] hybrid orbital with \[25\% \] s-Character
Ethene \[{C_2}{H_4}\]→ where \[C\] uses-\[s{p^2}\] hybrid orbital with \[33.33\% \] s-Character
Ethyne \[{C_2}{H_2}\]→ where \[C\] uses-\[sp\] hybrid orbital with half s-Character
S orbital being the orbital nearer to the \[C\]-atom the electron pair of the bond shaped by the orbital having higher s-character is pulled in addition towards \[C\]-particle indicating higher electronegativity .
Along these lines the request for EN of Carbon
\[Ethyne > Ethene > Ethane\]
The electronegativity of carbon relies upon its hybridization state. Carbons that are \[s{p^2}\] - hybridized are fairly more electronegative than \[s{p^3}\] - hybridized carbons; sp-hybridized carbons are much more electronegative by another \[0.2\] units. This implies that \[s{p^3}-{\text{ }}s{p^2}\] carbon–carbon bonds have a slight bond dipole.
Bond-strength impacts and electronegativity impacts. The request can't be a bond-strength impact, since bond qualities increment with expanding s character. That is, the strength of a \[C\left( {sp} \right)--{\text{ }}H\] bond is more prominent than that of a \[C\left( {s{p^2}} \right)--{\text{ }}H\] bond, which is more noteworthy than that of a \[\;C\left( {s{p^3}} \right)--{\text{ }}H\] bond. Hence, this should be an electronegativity impact. This corrosiveness request, at that point, is reliable with expanding electronegativity of carbons with expanding s character in their hybridizations.
Note:
Three kinds of hybridisation − \[sp\], \[s{p^2}\] and \[s{p^3}\], are found in carbon particles relying on the number of atoms connected to the carbon.
A \[sp\] hybrid orbital has half s-character and half p-character; a \[s{p^2}\] hybrid orbital has \[33.33\% \]s-character and \[66.66\% \] p-character; a \[s{p^3}\] hybrid orbital has \[25\% \] s-character and \[75\% \] p-character.
The more noteworthy the s-character of the half and half orbitals, the more prominent is the electronegativity on the grounds that a s orbital holds electrons all the more firmly to the core
Regarding Electronegativity: \[sp{\text{ }} > {\text{ }}s{p^2} > {\text{ }}s{p^3}\]
Complete step by step answer:
Electronegativity increments with the s-character of the half and half orbital utilized for making covalent bond.
It could be clarified by taking a straightforward illustration of three least difficult hydrocarbons
Ethane \[{C_2}{H_6}\]→ where \[C\] uses- \[s{p^3}\] hybrid orbital with \[25\% \] s-Character
Ethene \[{C_2}{H_4}\]→ where \[C\] uses-\[s{p^2}\] hybrid orbital with \[33.33\% \] s-Character
Ethyne \[{C_2}{H_2}\]→ where \[C\] uses-\[sp\] hybrid orbital with half s-Character
S orbital being the orbital nearer to the \[C\]-atom the electron pair of the bond shaped by the orbital having higher s-character is pulled in addition towards \[C\]-particle indicating higher electronegativity .
Along these lines the request for EN of Carbon
\[Ethyne > Ethene > Ethane\]
The electronegativity of carbon relies upon its hybridization state. Carbons that are \[s{p^2}\] - hybridized are fairly more electronegative than \[s{p^3}\] - hybridized carbons; sp-hybridized carbons are much more electronegative by another \[0.2\] units. This implies that \[s{p^3}-{\text{ }}s{p^2}\] carbon–carbon bonds have a slight bond dipole.
Bond-strength impacts and electronegativity impacts. The request can't be a bond-strength impact, since bond qualities increment with expanding s character. That is, the strength of a \[C\left( {sp} \right)--{\text{ }}H\] bond is more prominent than that of a \[C\left( {s{p^2}} \right)--{\text{ }}H\] bond, which is more noteworthy than that of a \[\;C\left( {s{p^3}} \right)--{\text{ }}H\] bond. Hence, this should be an electronegativity impact. This corrosiveness request, at that point, is reliable with expanding electronegativity of carbons with expanding s character in their hybridizations.
Note:
Three kinds of hybridisation − \[sp\], \[s{p^2}\] and \[s{p^3}\], are found in carbon particles relying on the number of atoms connected to the carbon.
A \[sp\] hybrid orbital has half s-character and half p-character; a \[s{p^2}\] hybrid orbital has \[33.33\% \]s-character and \[66.66\% \] p-character; a \[s{p^3}\] hybrid orbital has \[25\% \] s-character and \[75\% \] p-character.
Recently Updated Pages
Fill in the blanks with suitable prepositions Break class 10 english CBSE
Fill in the blanks with suitable articles Tribune is class 10 english CBSE
Rearrange the following words and phrases to form a class 10 english CBSE
Select the opposite of the given word Permit aGive class 10 english CBSE
Fill in the blank with the most appropriate option class 10 english CBSE
Some places have oneline notices Which option is a class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How do you graph the function fx 4x class 9 maths CBSE
When was Karauli Praja Mandal established 11934 21936 class 10 social science CBSE
Which are the Top 10 Largest Countries of the World?
What is the definite integral of zero a constant b class 12 maths CBSE
Why is steel more elastic than rubber class 11 physics CBSE
Distinguish between the following Ferrous and nonferrous class 9 social science CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE