Pls explain the structure b and c how can here n is sp2 ?

c is not sp2 hybridised

it will be sp2 as nitrogen is double bonded

and double bonded nitrogen is sp2 hybridised

but N not in conjugation so the free l.p also count as singma so total sigma 4 and it will be sp3

but it has formed pi bond also

so we can say sp2

its answer given in the answer key is abcd please check

because i have already done this assignment

double bonded nitrogen atom is also sp2

but and is given wrong

concepts never wrong ....according to concept

no of sigmabond ...+ localised l.p = hybridisation

wait I will send u the pic

i agree concept never go wrong but in the goc lecture i have done same example in which it is said to be sp2 because it already formed pi bond

seee this ....

just wait

ok

please correct in your option b in your notes as boron have not empty d orbital it has empty p orbital

sry ...I will correct it

tell me clearly that c is sp2 or sp3

please check in the above two examples

please check

did you get it please tell me

yes ur correct here also 3 sigma possible another l.p which I m counting is not possible

now you understood completely or have any doubt

can u give me ans of 9.10 and 13 no question of this assignment pls ..

can u ..?

mesomeric always get priority over all other electronic effects in comparison of carbocation stability

but in q 10 all the c+ attached to +m group

no please check that in q10 in a and b carbocation is one carbon away from the lone pair containing groups for mesomeric it should be adjacent or dirctly attached to lp containing groups as in option b and d

sorry as in option c and d

in b no mesomeric is present

yes I apply this in case of b their is a free l.p so I told u that b is sp3

B is sp2 hybridized Simun

how can u explain by draw it

B has only 3 valence electron. So only 3 hybrid orbital will be there.

So I have no copy pen Now. Please understand.

sir the N in c is sp2 or sp3 ..?

sir c also 3 sigma so it is also sp2

but the N in B is also a localised l.p also

4 sigma so sp3 in the B

In option B Boron has empty p orbital. If any lone pair is adjacent to p orbital then lone pair is also used for delocalization.

ohhhh I don't know about this so tqq ...☺☺

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