Question

The correct order regarding the electronegativity of hybrid orbitals of carbon is

A. $$sp > s{p^2} < s{p^3}$$
B. $$sp > s{p^2} > s{p^3}$$  
C. $$sp < s{p^2} > s{p^3}$$
D. $$sp < s{p^2} < s{p^3}$$
Answer :   $$sp > s{p^2} > s{p^3}$$
Solution :
The correct order regarding the electronegativity of hybrid orbitals of carbon is $$sp > s{p^2} > s{p^3}$$   because in $$sp,s{p^2}$$  and $$s{p^3}$$  hybrid orbitals, $$s$$ - orbital character is 50%, 33.3% and 25% respectively. Due to higher $$s$$ - character electron attracting tendency, i.e. electronegativity increases.

Releted MCQ Question on
Organic Chemistry >> General Organic Chemistry

Releted Question 1

The bond order of individual carbon-carbon bonds in benzene is

A. one
B. two
C. between one and two
D. one and two alternately
Releted Question 2

Molecule in which the distance between the two adjacent carbon atoms is largest is

A. Ethane
B. Ethene
C. Ethyne
D. Benzene
Releted Question 3

Among the following, the compound that can be most readily sulphonated is

A. benzene
B. nitrobenzene
C. toluene
D. chlorobenzene
Releted Question 4

The compound 1, 2-butadiene has

A. only $$sp$$   hybridized carbon atoms
B. only $$s{p^2}$$ hybridized carbon atoms
C. both $$sp$$  and $$s{p^2}$$ hybridized carbon atoms
D. $$sp$$ , $$s{p^2}$$ and $$s{p^3}$$ hybridized carbon atoms

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