Dear V S Kirankumar 

Conduction band (CB) and valence band (VB) potentials of semiconductors are calculated using the following empirical equations:

       ECB = χ – Ee ‏– 0.5Eg                                                       (1)

       EVB = ECB + Eg                                                               (2)

Where EVB and ECB are the VB and CB potentials, respectively. Moreover, Ee is the energy of free electrons vs. hydrogen (4.5 eV). Finally, χ is the electronegativity of semiconductor and it was calculated by the following equation:

      χ = [x(A)ax(B)bx(C)c]1/(a+b+c)                                                           (3)

In which a, b, and c are the number of atoms in the compounds. For example, in the case of Cu2CoS4, the calculation procedure is as follows:

 χ calculation for Cu2CoS4

For Cu:   EIE = 745.5   kj/mol     ÷  96.48 = 7.73 eV

               EEA = 119.2   kj/mol     ÷ 96.48 = 1.23 eV

 XBi = 1/2(7.73+1.23) = 4.48

 For Co:  EIE = 760.4kj/mol     ÷   96.48  = 7.88 eV

               EEA = 64    kj/mol     ÷  96.48  =  0.66 eV

XI = 1/2(7.88+0.66) = 4.27

 For S:    EIE = 999.6   kj/mol     ÷   96.48  = 10.36 eV

              EEA =  200.4     kj/mol     ÷  96.48  = 2.08eV

XS = 1/2(10.36+2.08) = 6.22

XCu2CoS4 = (4.48×4.48×4.27×6.22×6.22×6.22×6.22)1/7 = 5.36

For more information, please refer to the attached paper.

Dear V S Kirankumar 

Conduction band (CB) and valence band (VB) potentials of semiconductors are calculated using the following empirical equations:

       ECB = χ – Ee ‏– 0.5Eg                                                       (1)

       EVB = ECB + Eg                                                               (2)

Where EVB and ECB are the VB and CB potentials, respectively. Moreover, Ee is the energy of free electrons vs. hydrogen (4.5 eV). Finally, χ is the electronegativity of semiconductor and it was calculated by the following equation:

      χ = [x(A)ax(B)bx(C)c]1/(a+b+c)                                                           (3)

In which a, b, and c are the number of atoms in the compounds. For example, in the case of Cu2CoS4, the calculation procedure is as follows:

 χ calculation for Cu2CoS4

For Cu:   EIE = 745.5   kj/mol     ÷  96.48 = 7.73 eV

               EEA = 119.2   kj/mol     ÷ 96.48 = 1.23 eV

 XBi = 1/2(7.73+1.23) = 4.48

 For Co:  EIE = 760.4kj/mol     ÷   96.48  = 7.88 eV

               EEA = 64    kj/mol     ÷  96.48  =  0.66 eV

XI = 1/2(7.88+0.66) = 4.27

 For S:    EIE = 999.6   kj/mol     ÷   96.48  = 10.36 eV

              EEA =  200.4     kj/mol     ÷  96.48  = 2.08eV

XS = 1/2(10.36+2.08) = 6.22

XCu2CoS4 = (4.48×4.48×4.27×6.22×6.22×6.22×6.22)1/7 = 5.36

For more information, please refer to the attached paper.

Thank you sir., very clear explanation It is highly helpful for my research.

 Excellent description by Aziz Habibi-Yangjeh. But sir can you please suggest me how to calculate the X value of gC3N4 and graphene/RGO like material. As we dont know the exact no of atoms present in the compound.  How can I calculate the same.

Dear Alaka Samal

The calculations for g-C3N4 is as follows:

 CB and VB for C3N4

 For C:   EIE = 1086.5   kj/mol     ÷   96.48  = 11.26 eV

              EEA =  121.8   kj/mol     ÷  96.48  = 1.26 eV

 XC = 1/2(11.26+1.26) = 6.26

 For N:

            EIE = 1402.3   kj/mol     ÷   96.48  = 14.53 eV

           EEA = 7           kj/mol     ÷  96.48  =  0.07 eV

 XN = 1/2(14.53+0.07) = 7.3

XC3N4 = (6.26×6.26×6.26×7.3×7.3×7.3×7.3)1/7 = 4.73

Eg = 2.7 eV

CBC3N4 = 4.73 - 4.5 - 2/7/2 = -1.13 eV

VBC3N4 = -1.13 + 2.7 = 1.57 eV

Aziz Habibi-Yangjeh - 

I appreciate your detailed and thorough answer for the calculations on band-edge positions of semiconducting materials.  I am hoping that you can provide one additional point of clarification.  In the example on graphitic carbon nitride (g-C3N4), I agree with all of your posted values on the ionization energy and electron affinity of the individual elements.  Similarly, I agree with your calculated electronegativities.  However, by my calculations, the seventh root of the product of the electronegativity values is not 4.73.  Rather, it is 6.8 - which makes more sense as the geometric mean of 6.26 (taken three times) and 7.3 (taken four times).  However, the value of 4.73 - which is closer to the 8th or 9th root of the product of the absolute electronegativity values - is widely used in the literature and there is no doubt that this value produces band edge positions that seem to match experimental results.

I would welcome any clarification on this point, since I have been struggling with this problem for a bit and this calculation would be extremely helpful to my work.

Hi nanba,

read this attached article it will help you.

Thank you to all 

Excellent and detailed explanation by Pro. Aziz Habibi-Yangjeh. But I have a small suggestion /correction in the calculation.

According to Mulikken's electronegativity concept the 'I' (electron ionization energy EIE) and 'E' (electron Affinity EA) which were about 2.8 times higher than Pauling's method. So it should be included in the calculation.

Therefore the the calculations has to be E+I/2*2.8*96.48 OR simply E+I/540.28

Ex: CuAl2O4 system,

For Cu: EIE = 745.5   kj/mol    

               EEA = 119.2   kj/mol    

E+I = 846.7 kj/mol 

 X = 846.7 /540.28 = 1.6

This value would closely match with the standard electronegativity value of Cu.

Similarly Al and O has to be derived and substituted in final equation as Prof. Aziz explained. You can also refer this in the book 'Concise Inorganic Chemistry by J.D. Lee, page number 162.

Thank you.

Thanks John Thurston, I really appreciate your question as I was having a similar problem of calculating electronegativity of graphitic carbon nitride.

Thank you Prof. Aziz Habibi-Yangjeh . Recently I have read this and this information will definitely help many researcher. Thank you again Prof. Aziz Habibi-Yangjeh

Thanks All. It is very useful information and discussion.

Dear Prof. Aziz Habibi-Yangjeh , Thank you for your explanation. Can you please explain about the X value of ZnO ?

Dear Dr. Sanjit Manohar Majhi

Please see the attached file.

Dear Prof Aziz Habibi-Yangjeh ,

Thank you so much for your valuable time to explain for me. It is highly appreciated.

Best regards

Sanjit

Dear Dr. Aziz habibi, your information is very helpful but I am still confused to calculate X value of my sample because I didn't get last line in calculation as follows

XC3N4 = (6.26×6.26×6.26×7.3×7.3×7.3×7.3)1/7 = 4.73

How is it possible please explain? How comes 4.73 after this calculation?

It will be really appreciated, I need this urgent.

according to the above calculation ( XC3N4 = (6.26×6.26×6.26×7.3×7.3×7.3×7.3)1/7 = 4.73 ) i think it will be 6.83 @Sonam Goyal.

Sir , is it multiplication or addition sign?

@Sonam Goyal, multiplicationa and addition signs both are there according to this is the formula, χ = [x(A)ax(B)bx(C)c]1/(a+b+c) . your material is C3N4

X (elemental electronegativity) of C is 6.26 and X of N is 7.3, so according to the formula

for C3N4,

X (electronegatvity of material)= (6.26×6.26×6.26×7.3×7.3×7.3×7.3)1/3+4 = 6.83

Dear Prof Aziz Habibi-Yangjeh ,

Thank you so much for your patient answering and valuable information, but there is still some confusing about C3N4.

Would you mind explaining more about the line XC3N4 = (6.26×6.26×6.26×7.3×7.3×7.3×7.3)1/7 = 4.73 ?

I get 6.83 instead of 4.73. I think the geometric average between 6.26 and 7.3 should not be smaller than both of them.

Thank you for your kindness.

Best regards,

Licheng Lou

Dear Professor Aziz Habibi-Yangjeh,

Thank you very much for sharing this valuable information. Indeed, I do appreciate that.

Thank you very for enlightening us. This was just a missing number of the whole puzzle for developing my mechanism

I am no longer struggling to calculate electronegativity of my semiconductors. This has really helped me alot. Thanks Prof Habib

Thank you very much

it was perfect answer.

Aziz Habibi-Yangjeh sir, please can you tell me for electronegativity of CuO

After considering these calculations for some time, I am wondering if this theory simply does not work very well for non-ionic systems where there is not a large difference in absolute electronegativity between the different components. The math works very well for classic transition metal based semiconductors such as TiO2 and ZnO. It seems to have problems (as noted in many of the comments above) with organic materials, such as g-C3N4 where the bonding would be understood to be more covalent in nature. In these cases, I would suggest that it is better to experimentally determine the band edge positions using XPS, UPS or another, similar technique.

Dear Aziz Habibi-Yangjeh ,

How to calculate VB and CB it for reduced graphene oxide.

The potentials of the conduction band (CB) and valence band (VB) of a semiconductor can be calculated using Mulliken electronegativity theory :

ECB = X- Ec - 0.5 Eg (Eq.1)

EVB = ECB + Eg (Eq.2)

Where Ec is the energy of free electrons vs. hydrogen (4.5 eV) and χ is Mulliken electronegativity of the semiconductor. X was calculated by the following equation:

χ = [x(A)^ax(B)^bx(C)^c ] ^(1/(a+b+c) ) (Eq.3)

a, b, and c are the number of atoms in the compounds.

χ calculation for CuAl2O4

Cu:

EIE (Molar ionization energies) = 7.73 eV

EEA (Electron affinity) = 1.23 eV

XCu = (7.73+1.23)1/2 = 4.48

Al:

EIE = 5.98 eV

EEA = 0.43 eV

XAl= (5.98+0.43)1/2 = 3.20

O:

EIE = 13.61 eV

EEA = 1.46 eV

XO = (13.61+1.46)1/2 = 7.54

XCuAl2O4 = (4.48×3.20×3.20×7.54×7.54×7.54×7.54)1/7 = 5.48

ECB = 5.48 – 4.5 ‏– 0.5Eg

EVB= 1.48+ 0.5Eg

Thank you for all the patiently answered questions. I'm still confused on how to calculate X for oxygen doped g-C3N4 in which case, i dont know the number of atoms of oxygen doped in it. Thank you

Dear Prof

Thank you for your patience in putting these together. It has helped me tremendously

@Touahra Fouzia. Your explanation is very helpful for ECB and EVB calculation. Tq sir.