1) Explain Hybridization concept of Pauling.
2) Explain how to determine the Hybridization in a given molecule.
3) Give some examples.
Answers
Concept:
Hybridization concept was invented to explain molecular structures of compounds more correctly than earlier theories. It was given by Linus Pauling and Slater. Different Orbitals (s, p, d, f) of same atom (central atom in the molecule) have different energies. They combine together and form equal number of hybrid orbitals all having the same energy and shape. Their orientation is different in space. We can explain structure, bond angles and behavior of compounds better with this Molecular Orbital Theory (MOT).
Hybrid orbitals form ONLY Sigma bonds between central atom and atoms it bonds with.
How to determine the type:
H = (V + M - C + A) / 2
H = num of orbitals involved in hybridization (2 to 7).
H is the number of sigma bonds in the molecule (or given anion or cation).
H = 2 => sp; 3 => sp2; 4 => sp3; 5 => sp3d; 6 => sp3d2; 7 =>sp3d3.
V = num of electrons in the outermost orbit of the central atom.
M = num of monovalent atoms around the central atom.
C = +ve charge on the Cation (if given group is a cation).
A = -ve Charge on the Anion (if group is an anion).
Examples
Ammonia: NH3. V = 5. M = 3, C=0. A=0. So H = 4 => sp3 Hybridization.
Exceptions:
There are exceptions to prediction of molecular shapes with this formula. PH3 does not follow it. It does not form hybridized orbitals. Only pure 3p orbitals form bonds. Bond angle is 93.5 deg. In NH3 there is sp3 hybridization. Lone pair repulsion results in bond angle of 107 deg. The reason is the bigger size of Phosphorous. Due to increased P-H bond distance in PH3, steric interaction is less . So the need for hybridization is less.
Examples:
NH4+ : V=5. M=4. C=1. A=0. So H = 4. So sp3.
SF6 : V = 6. M = 6, C = 0. A = 0. H = 6. So sp3d2.
PF6- : V=5. M=6. C=0. A=1. H=6. Sp3d2
H2S : V=6. M=2. C=0. A=0. H = 4. Sp3.
CH4: V=4. M=4. C=0. A=0. H=4. Sp3.
H2O: V=6. M=2. C=0. A=0. H=4. Sp3.
HHCO : C is central atom, bonds are with H, H & O.
V=4. M=3. C=0. A=0. H = 3.5. 3 Sigma bonds.
One Pi bond (due to 0.5) between C and O, as C=O is a double bond.
PCl5 : V=5. M=5. C=0.A=0. H = 5. Sp3d.
I Cl2- : V=7. M=2. C=0. A=1. H = 5. Sp3d.
H2 CO3: C is central atom. H-O-(C=O)-O-H. So V=4. M=3. C=0. A=0.
H = 3.5. So sp2. 0.5 => A pi bond between C and O.
H3O+ : sp3 . V=6. M=3. C=1.A=0. H=4. Sp3.
C S2 : sp2. C is central.
P Br3: sp3. P is central.
SF4: sp3d. S is central.
Be Cl2 : sp. Beryllium is central.
Xe F4 : sp3 d2. Xe is central.
Xe O F4 : sp3 d2. Xe is central.
H3 PO2 : sp3. P is central.
Table of molecules according to Hybridization:
SP: Be F2, Hg Cl2, Be Cl2, CO, C2H2, CO2, Cd Cl2, Zn Cl2, CO.
SP2: BF3, SO3, BCl3, Al Cl3, SO2, HNO3, C2H4, Pb Cl2, C6H6, SO3, H2CO3, Sn Cl2
SP3: CH4, NH3, H2O, NH4+, BF4-, H2SO4, HClO4,
PCl3, NCl3, AsCl3, HClO3, ICl2+
ICl, XeO3, SCl2, OF2, ICl2+, H2O
Tetrahedral shape.
Sp3d: PF5, SF4, Br F3, PCl5, Sb Cl5, Cl F3, Br F3, Xe F2, I Cl2-
Trigonal bipyramidal shape. d Orbital in the bond is dz^2.
Sp3d2: SF6, Cl F5, Xe F4, PF6-, AlF6^-2, SiF6^-2, PF6^-1, IF5, Br F5,
XeOF4, BrF4^-, I Cl4^-.
Octahedral shape. Two d orbitals are dx^2-y^2, dz^2.
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dsp2 : Shape is square planar. D orbital in the bond is dx^2-y^2.
d3 s : dxy, dxz, dyz orbitals participate in the bond.
dsp3 : Square bipyramidal shape.
dx^2-y^2 orbital participates in the bond.
d2sp3 : Octahedral shape.
The two d orbitals in the bond are dx^2-y^2 & dz^2.
Answer:
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