The standard heat of combustion is the amount of heat released when one mole of a substance undergoes complete combustion. It is usually represented by the symbol ΔHc. The standard heat of formation is the amount of heat absorbed or released when one mole of a compound is formed from its elements in their standard states. It is usually represented by the symbol ΔHf.

- Calculation of ΔHc for carbon and sulphur:

The standard heat of combustion of carbon is -393.3 kJ/mol, which means that 393.3 kJ of heat is released when one mole of carbon undergoes complete combustion. The equation for the combustion of carbon can be written as:

C(s) + O2(g) → CO2(g)

Similarly, the standard heat of combustion of sulphur is -293.72 kJ/mol, which means that 293.72 kJ of heat is released when one mole of sulphur undergoes complete combustion. The equation for the combustion of sulphur can be written as:

S(s) + O2(g) → SO2(g)

- Calculation of ΔHc for carbon disulphide:

To calculate the standard heat of combustion of carbon disulphide, we need to use the standard heat of formation values of carbon and sulphur. The equation for the combustion of carbon disulphide can be written as:

CS2(l) + 3O2(g) → CO2(g) + 2SO2(g)

From the equation, we can see that one mole of carbon disulphide produces one mole of carbon dioxide and two moles of sulphur dioxide. Using the standard heat of formation values, we can calculate the standard heat of combustion of carbon disulphide as follows:

ΔHc(CS2) = [2ΔHf(SO2) + ΔHf(CO2)] - ΔHf(CS2)

ΔHc(CS2) = [2(-293.72 kJ/mol) + (-393.3 kJ/mol)] - ΔHf(CS2)

ΔHc(CS2) = -980.74 kJ/mol - ΔHf(CS2)

From the given information, we know that the standard heat of combustion of carbon disulphide is -1108.76 kJ/mol. Therefore, we can solve for the standard heat of formation of carbon disulphide as follows:

-1108.76 kJ/mol = -980.74 kJ/mol - ΔHf(CS2)

ΔHf(CS2) = -1108.76 kJ/mol + 980.74 kJ/mol

ΔHf(CS2) = -128.02 kJ/mol

Therefore, the correct answer is option B: -128.02 kJ.