hybridisation notes & tricks for neet & jee

Hybridisation notes:

Formula for hybridisation:

-->H=1/2 ( V + M - C + A )

V = Number of valence electrons

M = Monovalent atoms bonded to central atom

C = Cationic charge

A = Anionic charge

For H=2->sp

H=3->sp2

H=4->sp3

H=5->sp3d

H=6->sp3d2

H=7->sp3d3

Eg:BeCl2

The electronic configuration of 'Be' in ground state is 1s(2) 2s(2)

Since there are no unpaired electrons, it undergoes excitation by giving one of its 2s electron to 2p orbital.

Thus in the excited state, the electronic configuration of Be is 1s(2) 2s(1)2p(1).

‘Be’ has 2 valence electrons,so V=2.

bonds attatched to ‘Be’ are 2 ‘Cl’atoms,M=2.

cationic charge=0,C=0.

anionic charge=0,A=0.

H=1/2(2+2–0+0)=>H=2.

therefore,the hybridisation for BeCl2 is sp hybridisation.

example 2 -:

1. CH4

So CH4 has central atom carbon which has four valence electrons and it has 4 monovalent hydrogen atom. As there is no plus or minus charge on compound so

0.5(4+4)= 4

2) SO4^2-

The central atom is sulphur with 6 valence electrons. It has no monovalent atom. The total Anion charge is 2

So 0.5(6+2) = 4

Now match the number you get with:-

Sp- 2

Sp2-3

Sp3-4

Sp3d-5

Sp3d2- 6 and so on

The above two species hence have sp3 hybridisation.

Note:-

1. Monovalent atoms means hydrogen, alkali metals and group 17 elements
2. If charge on compound is -2 you have to take 2 and never take the sign. Only the number. Remember the number of positive charge is to be subtracted in formula and vice versa
3. Rarely in 2–3 compound the formula won't work

Hybridisation for a molecule is given by:

1/2(V+H−C+A)

Where,

V = Number of valance electrons in central atom

H = Number of surrounding monovalent atoms

C = Cationic charge

A = Anionic charge

If this expression is 2 then it's a sp hybridised, 3 then sp2 hybridised and so on.

but there is some exceptions in this formula:

1. For example, PCl3 and POCl3. For PCl3 you would get 0.5×(5+3)=4 and for POCl3 you would get 0.5×(5+3)=4. The reason the O made no difference is because phosphorus uses two of its electrons to bond with both of which are nothing but valence electrons of phosphorus, only present as a lone pair in one case and a sigma bond in the other.
2. Also, this formula does not work when you have triple bonds, say for HCN.
3. 

You can find number of lone pair also

Another method that you can follow-

NO2 is sp2 hybridised, in classical sense.

Generally, Single electron orbitals are unhybridised, pure p -orbitals, like in methyl free radical.

However, Single electron orbital is said to be hybridised, when central atom is bonded to highly electronegative atoms/groups. It is because of decreased electron density at central atom which then attracts odd electron-orbital density thereby size of odd-electron orbital is decreased. Hence odd-electron orbital it acquires some “s” character and become hybridised as a result.

for Similar reason, ClO2, ClO3 & CF3 are sp3 hybridised.

Bond angle in NO2 is 115 degree. and here both N-O bonds are equivalent due to resonance.

(NO2)-, [nitrite ion or nitro ion] is sp2 hybridised as well, here lone pair orbital is hybridised. remember here,

• (a) both N-O bonds are equivalent due to resonance
• (b) Double bonds behave similar to lone pair, with respect to repulsions, in effect. (remember, SO2 has bond angle approx 120 degree)

Here, bond angle in (NO2)- is 115 degree.

if we consider, NO2+(nitronium ion), it is sp hybridised(bond angle 180 degree).

note—[NO2+ has non-equivalent resonating structures]

[related example- Acylium cations(RCO+) are linear, sp hybridised, triply bonded resonating structure is more stable due to complete octet of all atoms

The sp3d3 hybridization has a pentagonal bipyramidal geometry i.e., five bonds in a plane, one bond above the plane and one below it.

So to achieve such geometry we need to choose orbitals in such a way such that five of them are in one plane and two in planes perpendicular to it

We have one s orbital – s

three p orbitals – px py pz

and five d orbitals – dxy dyz dxz dx2-y2 dz2

The only possible combination in which 5 are in a plane and 2 above it is

s px py pz dxy dx2-y2 dz2

in which s px py dxy dx2-y2 are in xy plane and pz and dz2 perpendicular to it.

So the d orbitals being used are dxy dx2-y2 dz2

sp3d3 hybridisation :

The mixing of one s, three p and three d- atomic orbitals to form seven equivalent sp3d3 hybrid orbitals of equal energy. This hybridization is known as sp3d3 hybridization.

It's orbitals which take part in sp3d3 hybridisation are ss, px, py, dxy, dx2-y2 are in xy plane and pz and dz2 perpendicular to it.

• Seven sp3d3 hybrid orbitals are directed towards the corners of a pentagonal bipyramid.
• These are not equivalent hybrid orbitals because five of them are directed towards the corners of a regular pentagon while the remaining two are directed above and below the plane.
• The geometry is pentagonal bipyramidal and bond angle is 72 degree and 90 degree.

When in complex coordination compound if inner d orbital takes parts in hybridisation.

Then Hybridisation will be d2sp3.

Coordination compound have d2sp3 hybridisation in two case-

1.When 2 inner d orbital are vacant.(d1,d2,d3)

2.when there is sfl(strong field ligand) pr in compound which force them to pair and vacant the 2 d orbital…(d4,d5,d6,d7)

➡️Sp3d2 hybridisation takes place when inner d orbital have d8,d9,d10 configuration in pr of any type of ligand sfl or wfl.

➡️Sp3d2 hybridisation will also assign to compound when there is d4,d5,d6,d7 and ligand is wfl.

I.e- when ligand is weak.(sp3d2) [Cof6]3-

I.e-sfl +d8——[Ni(NH3)6]2+

The shapes and bond angles also depend on the presence of lone pairs on the central atom. These lone pairs (LP) tend to distort the geometry of the molecule and also decrease the standard bond angles in the regular geometry.

Therefore the molecules can be classified broadly into regular and irregular(have LP):

a) Regular- you can check the bond angles and geometry from the table given below

b) Irregular

note that these molecules have one or multiple LP which distorts the geometry and reduces the bond angles which have different values which depend on the electronegativity of the central atom (so the values of bond angles are not mentioned here)

sp_____linear
~180°
sp2____trigonal-
(planar)
~120°
sp3____tetrahedral
~109°5'
dsp2___square-(planar)
~90°
sp3d___trigonal-
(bipyramidal)
~90°;120°
dsp3___square-
(pyramidal)
~90°
sp3d2/d2sp3___octahedral
~90°