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Predicting the Hybridization of Simple Molecules

  • Page ID
    35873
  • Prof. Linus Pauling (1931) first developed the Hybridization state theory in order to explain the structure of molecules such as methane (CH4).1This concept was developed for simple chemical systems but this one applied more widely later on and from today’s point of view it is considered an operative empirical for excusing the structures of organic and inorganic compounds along with their related problems. An innovative method proposed for the determination of hybridization state on time economic ground 2,3,4.

    Prediction of sp, sp2, sp3 Hybridization state

    We Know, hybridization is nothing but the mixing of orbital’s in different ratio to form some newly synthesized orbitals called hybrid orbitals. The mixing pattern is as follows:

    s + p (1:1) - sp hybrid orbital; s + p (1:2) - sp2 hybrid orbital ; s + p (1:3) - sp3 hybrid orbital

    Formula used for the determination of sp, sp2 and sp3 hybridization state:                        

    Power on the Hybridization state of the central atom = (Total no of σ bonds around each central atom -1)

    All single (-) bonds are σ bond, in double bond (=) there is one σ and 1π, in triple bond (≡) there is one σ and 2π. In addition to these each lone pair (LP) and Co-ordinate bond can be treated as one σ bond subsequently.

    Eg.:

    a. In NH3: central atom N is surrounded by three N-H single bonds i.e. three sigma (σ) bonds and one lone pair (LP) i.e. one additional σ bond. So, in NH3 there is a total of four σ bonds [3 bond pairs (BPs) + 1 lone pair (LP)] around central atom N. Therefore, in this case power of the hybridization state of N = 4-1 = 3 i.e. hybridization state = sp3.  

    b. In H2O: central atom O is surrounded by two O-H single bonds i.e. two sigma (σ) bonds and two lone pairs i.e. two additional σ bonds. So, altogether in H2O there are four σ bonds (2 bond pairs + 2 lone pairs) around central atom O, So, in this case power of the hybridization state of O = 4-1 =3 i.e. hybridization state of O in H2O = sp3

    c. In H3BO3:- B has 3σ bonds (3BPs but no LPs) and oxygen has 4σ bonds (2BPs & 2LPs) so, in this case power of the hybridization state of B = 3-1 = 2 i.e. B is sp2 hybridized in H3BO3. On the other hand, power of the hybridization state of O = 4-1= 3 i.e. hybridization state of O in H3BO3 is sp3.

    d. In I-Cl: I and Cl both  have 4σ bonds and 3LPs, so, in this case power of the hybridization state of both I and Cl =  4 - 1 = 3 i.e. hybridization state of I and Cl both are sp3.

    e. In CH2=CH2: each carbon is attached with 2 C-H single bonds (2 σ bonds) and one C=C bond (1σ bond), so, altogether there are 3 sigma bonds. So, in this case, power of the hybridization state of both C = 3-1 = 2 i.e. hybridization state of both C’s are sp2

    Prediction of sp3d, sp3d2, and sp3d3 Hybridization States

    In case of sp3d, sp3d2 and sp3d3 hybridization state there is a common term sp3 for which 4 sigma bonds are responsible. So, in addition to 4 sigma bonds, for each additional sigma, added one d orbital gradually as follows:-

    5σ bonds = 4σ bonds + 1 additional σ bond = sp3d hybridization

    6σ bonds = 4σ bonds + 2 additional σ bonds = sp3d2 hybridization

    7σ bonds = 4σ bonds + 3 additional σ bonds = sp3d3 hybridization

    Eg:-

    a.IF4+I has 7 e-s in its outermost shell, so, in this case, subtract one e- from 7 i.e. 7 – 1 = 6. So, out of 6 electrons, 4 electrons form 4 I-F bonds i.e. 4 sigma bonds and there is one LP. So, altogether there are 5 σ bonds. So, 5σ bonds = 4 σ bonds + 1 additional σ bond = sp3d hybridization

    b.IF7: 7 I-F single bonds i.e. 7σ bonds = 4σ bonds + 3 additional σ bonds = sp3d3 hybridization.

    c.ICl2-: I has 7 e-s in its outermost shell, so, in this case, add one e- with 7(overall charge on the compound) i.e. 07+1= 08. So, out of 08 electrons, 02 electrons form 02 I-Cl bonds i.e. 02 sigma bonds and there is 03 LPs. So, altogether there are 05σ bonds. So, 5σ bonds = 04 σ bonds + 01 additional σ bond = sp3d hybridization.

    d. XeF4: Xe, an inert gas, consider 8 e-s in its outermost shell, 04 of which form 04 Xe-F sigma bonds and there is two LPs, i.e. altogether there is 06 σ bonds = 04 σ bonds  + 02 additional σ bonds = sp3d2 hybridization.

    In case of determination of the hybridization state by using the above method, one must have a clear idea about the outermost electrons of different family members in the periodic table as follows:

                     Family                               Outermost electrons

    Nitrogen family                                  05

    Oxygen family                                    06

    Halogen family                                   07

    Inert gas family                                   08

    In case of cationic species you must remove requisite electron / electrons from the outermost orbit of the central atom and incase of anionic species you must add requisite electron with the outermost electrons of the central atom. Examples have been explored in Table 1.

    Table 1: Total number of σ bonds and Hybridization State

    Total number of sigma (σ) bonds

    Nature of Hybridization State

    Examples

    2

    sp

    BeCl2, HgCl2,C2H2,CO2,CO,CdCl2, ZnCl2 etc.

    3

    sp2

    BCl3, AlCl3,C2H4,C6H6,SO2,SO3,HNO3,

    H2CO3,SnCl2, PbCl2 etc.

    4

    sp3

    NH4+, BF4-, H2SO4, HClO4,PCl3, NCl3, AsCl3, HClO3,ICl2+,OF2,HClO2,SCl2,HClO, ICl, XeO3 etc.

    5

    sp3d

    PCl5, SbCl5, SF4, ClF3, BrF3, XeF2, ICl2- etc.

    6

    sp3d2

    SF6, AlF63-, SiF62-, PF6-, IF5, BrF5, XeOF4, XeF4, BrF4-, ICl4- etc.

    7

    sp3d3

    IF7, XeF6 etc.

    References

    1. L. Pauling, J. Am. Chem. Soc., 1931, 53(4), 1367-1400, doi:10.1021/ja01355a027
    2. Arijit Das,‘New Innovative Methods for Prediction of Hybridization State in a  Very Short Time’, IJAR, 2013, 3(07), 594, ISSN-2249-555X
    3. Arijit Das, ‘Simple Thinking Makes Chemistry Metabolic And Interesting- A Review Article’, IOSR-JAC, 2013, 6(4), 8-15, e-ISSN: 2278-5736, doi:10.9790/5736-0640815
    4. Arijit Das, R.Sanjeev and V.Jagannadham, “Innovative And Time Economic Pedagogical Views In Chemical Education – A Review Article”, World Journal of Chemical Education, 2014, 2(3), 29-38, Science and Education Publishing , USA, DOI:10.12691/wjce-2-3-1

    Contributor

    • Dr. Arijit Das, Ph.D. (Inorganic Chemistry), MACS ( Invited,USA ), SFICS, MISC, MIAFS (India), Assistant Professor, Department of Chemistry, Ramthakur College, Agartala, Tripura(W), Tripura, India, Pin-799003.