Question

shapes of hybridization

1.
$no {}{3} {}^{ - }$
2.CO3—​

Answer 1

First we have to find the total electron pair of carbon,

Total electron pair = 1/2[ V+M -C + A]

V = total no of valence electron of atom under consideration

M= total no. Of monovalent atom.

C = total cationic charge.

A = total anionic charge.

Therefore, for carbon in CO3 2-,

Total electron pair = 1/2[4+0–0+2]

Total electron pair= 3

And for 3 electron pair , there exist “ sp2” hybridization.

Answer 2

Answer:

As we know formula,

Hybridization= 1/2 (valence electron of central atom+ No. of surrounding monovalent atom- cationic charge+ anionic charge)

there is a assumption that O is not considered in monovalent atom:

H=1/2(5+0–0+1) = 3

i.e. sp2

and another way for hybridization of NO3- is, The central nitrogen atom is bonded to three oxygen atoms, and has no lone pairs. If you were to draw the Lewis dot structure, one of the nitrogen-oxygen bonds would be a double bond, and the other two would be single bonds, but the three bonds can resonate with each other, so actually it's more like they're all 1 1/3 bonds. Since it's a symmetric atom with three substituents, the geometry around the central atom will be trigonal planar, and the hybridization will be sp2.

Each of nitrogen's three sp2 orbitals overlaps with the s orbital of one of the oxygen atoms. The left-over p orbital on the nitrogen is involved in a resonating double bond with the three oxygen atoms, and it basically ends up sharing three pairs of electrons between four orbitals (the p orbital of the nitrogen, and one p orbital of each of the oxygen atoms). Each oxygen atom has two more p orbitals left, which each hold a lone pair of electrons. Except I think that those lone pairs would be involved in resonance as well, so it would end up as nine pairs of electrons shared between ten orbitals, but I'm not certain of that last part.