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Ques: The strength of bonds by 2s-2s , 2p-2p , 2p-2s overlap has order
(1) s-s>s-p>p-p
(2) p-p >s-p >s-s
Explain with reason ?

Diksha Rana , 8 Years ago
Grade 12th pass
anser 5 Answers
Umakant biswal

Last Activity: 8 Years ago

@ diksha 
acc to the question , option no 1 that is s-s>s-p>p-p turns out to be the correct answer . 
as u knows s-s is head on head overlapping and hence , its the strongest . and the 2nd strongest is s-p due to intermixing of overlapping .. 
and the least strong is p-p due to axial overlapping .. 
HOPE IT CLEARS YOUR DOUBT 
IF ANY MORE , THEN LET ME KNOW 
ALL THE BEST ..
 

Saurav

Last Activity: 7 Years ago

Second option is correct.On the basis of extent of overlapping, greater the area of overlapping greater is the strength of bond.Since the area of overlapping is greatest in case of p-p bond due to its dumbbell shape, so it has greatest strength followed by s-p and then s-s.

Javid Ahmad dar

Last Activity: 5 Years ago

The extent of overlapping depends on principl quantum number ,,,1s-1s,will show much extent than 2p_2p,which show much than 2s_2p

Aashish Sharma

Last Activity: 4 Years ago

It depends if we talk about size then s-s is smaller in size hence the distance of nucleus is small from electron hence s-s bond is stronger than s-p and p-p 
If we talk about direction then p-p overlapping is directional and s-s overlapping is non directional si p-p overlapping is strong 
 
One more factor that is area p orbital have more area than s orbital hence there will be extent overlapping so p-p overlapping is strong
 

preet3shukla

Last Activity: 3 Years ago

It depend on the extent of overlapping as p-p has more extent of overlapping.It doesnot has any affect of IND(internuclear distance ) imfact all pp ss and sp have same IND.
 

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