# Hybridization involving s, p and d-Orbitals

October 6, 2020

- The process of mixing of the atomic orbitals is called
**hybridization.** - There are different types of hybridization like
*sp*^{2}, sp^{3}, sp, sp^{3}d, sp^{3}d^{2}, sp^{3}d, dsp^{2},d^{2}sp^{3.} - The atoms of third period elements have vacant
orbitals*d**.* - Since, the energy of
orbitals is comparable to the energy of the occupied*3d*and*3s*orbitals (the energy difference is very small), hence quite often*3p*orbitals also get involved in the bonding process.*d* - The paired electrons in the
or*3s*orbitals can be promoted to the vacant*3p*orbitals, thereby, opening up the possibilities of hybridization of*3d*,*s*and*p**d.* - the different types of hybridization involving only
,**s**and*p*orbitals are:**d**

**i) The sp ^{3}d hybridization: Trigonal bi-pyramidal hybridization**

- In
hybridization, one*sp*^{3}d, three*s*and one*p*orbital (*d*) of the same energy level (e.g., n=3) are involved in hybridization.*d*_{z}^{2} - These five orbitals combine to give five hybrid orbitals.
- These hybrid orbitals designated as
orbitals, are oriented towards the corners of trigonal bi-pyramid.*sp*^{3}d - Three orbitals forming a plane are directed towards the corner of the equilateral triangle (angle of inclination = 120
^{0}), while the other two are oriented at right angles to the plane of the triangle, above and below it. - A typical example of this type of hybridization is the formation of
**PCl**_{5}molecule. - The ground state outer electronic configuration of phosphorous is
*3s*^{2}*3p*_{x}^{1}3p_{y}^{1}3p_{z}^{1 }3d^{0}*.* - The promotion of one
electron to*3s*level gives the outer configuration*3d**,**3s*^{1}*3p*_{x}^{1}3p_{y}^{1}3p_{z}^{1 }3d^{1}*.* - These five orbitals hybridize to give five new
orbitals*sp*^{3}d*.* - Each hybrid orbital is singly occupied.
- One of the
orbitals of chlorine is singly occupied.*p* - These singly occupied
orbitals of Cl form five P-Cl bonds due to the overlap with five*p*orbitals.*sp*^{3}d - This gives a trigonal by-pyramidal geometry
**for PCl**_{5}molecule.

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**ii) The sp ^{3}d^{2} hybridization : Octahedral hybridization**

- The orbitals which take part in
hybridization are*sp*^{3}d^{2}. E.g.,*ns, n(p*_{x}, p_{y}, p_{z}), n(d_{z}^{2}, d_{x}^{2}_{-y}^{2})*3s, 3(p*_{x}, p_{y}, p_{z}), 3(d_{z}^{2}, d_{x}^{2}_{-y}^{2}). - In this case, the total number of atomic orbitals which hybridize is six.
- The six hybridized orbitals are directed towards the corner of an octahedron, and are inclined at an angel of 90
^{0}to each other. - The
hybridization is observed for sulphur in*sp*^{3}d^{2}**SF**._{6}

- The ground state outer electronic configuration of sulphur is
*3s*^{2}3p_{x}^{2 }3p_{y}^{1 }3p_{z}^{1}3d^{0}. - Promotion of two electrons, one each from
and*3s*o two*3p*_{x}t(*3d*,*d*_{z}^{2}) orbitals, followed by hybridization yields six*d*_{x}^{2}_{-y}^{2}hybrid orbitals.*sp*^{3}d^{2} - These six hybrid orbitals are directed towards the corner of a regular octahedron.
- Each hybrid orbital is occupied by one electron.
- These hybrid orbitals overlap with the half-filled
orbitals of F atoms to form six*p***S-F bonds**. This gives an octahedral geometry for**SF**_{6}.

**iii) The d ^{2}sp^{3} hybridization: Octahedral hybridization**

- The
hybridization involves two*d*^{2}sp^{3}orbitals belonging to (n-1)**d**^{th}energy level and theand*s*orbitals belonging to nth energy level, i.e.,*p**(n-1) (d*_{z}^{2}, d_{x}^{2}_{-y}^{2}), ns, n(p_{x}, p_{y}, p_{z}). - For example
*: 3d*_{z}^{2}, 3d_{x}^{2}_{-y}^{2}, 4s, 4(p_{x}, p_{y}, p_{z}) orbitals. - A typical example of this type of hybridization is the formation of a complex ion,
**[Co(NH**_{3})_{6}]^{3+}.

- The outer configuration
**of Co**is*3s*^{2}3p^{6}3d^{7}4s^{2}. - Therefore, the configuration of Co
^{3+}is*3s*^{2 }3p^{6}3d^{6}4s^{0}4p^{0}*.* - When the reaction takes place, the six electrons in
orbitals pair up. This makes two*3d*-orbitals available.*d* - These two
orbitals combine with one*3d*and three*4s*orbitals to form six*4p*hybrid orbitals.*d*^{2}sp^{3} - These six orbitals are oriented along the six corners of a regular octahedron.

**iv) The dsp ^{2} hybridization : Square planar hybridization**

- In this type of hybridization, one
orbital belonging to*d***(n-1)**energy level, and one^{th}and*s***two**orbitals belonging to the*p***nth**energy level, viz.,*(n-1) d*_{x}^{2}_{-y}^{2}, ns, np_{x}, np_{y}, np_{z}. - For example:
*3 d*are involved._{x}^{2}_{-y}^{2}, 4s, 4p_{x}, 4p_{y}, 4p_{z} - A typical example of this kind of hybridization is the formation of complex
**ion [Ni(CN)**_{4}]^{2-}.

- Nickel (atomic no. = 28) has the outer configuration
The electronic configuration of Ni*3s*^{2}, 3p^{6}, 4s^{2}, 3d^{8}.^{2+}is.*3s*^{2}3p^{6}3d^{8} - During the reaction, these eight
electrons pair up to occupy four of the*d*orbitals. This leaves one of the*d*orbitals vacant.*3d* - This
orbital combines with one 4*3d*and two*s*orbitals to give four*4p*hybrid orbitals.*dsp*^{2} - Thee four hybrid orbitals are directed towards the four corners of a square in the
plane.*xy* - This gives a
**square planar**geometry to the complex ion**[Ni(CN)**_{4}]^{2-}.

**References: **

i) https://www.masterorganicchemistry.com/2018/01/16/a-hybridization-shortcut/

ii) https://www2.chem.wisc.edu/content/hybridization#

iii) https://www.organicchemistrytutor.com/topic/hybridization/