Which process could be used to calculate the bond energy for the covalent bond X-Y by dividing

its ∆H by n?
A XYn(g) → X(g) + nY(g)
B 2XYn (g) → 2XYn-1(g) + Y2(g)
C Y(g) + XYn-1(g) → XYn(g)
D nXY(g) → nX(g) + n/2 Y2(g)

Remember what bond energy is,it is the energy released when a covalent bond of a gaseous molecule is broken to give gaseous atoms and not molecules or compounds or molecules. So in A only is this happening in the rest of the options ∆H not only contains the bond breaking energy but also bond formation so the resulting ∆H for B,C and D is lesser than the value of X----Y bond energy !

The correct process for calculating the bond energy for the covalent bond X-Y by dividing its ∆H by n is option A: XYn(g) → X(g) + nY(g). This process involves breaking one mole of XYn molecules into one mole of isolated X atoms and n moles of isolated Y atoms. The enthalpy change (∆H) for this reaction represents the bond energy of the X-Y bond. Therefore, dividing ∆H by n gives the bond energy per bond.

To calculate the bond energy for the covalent bond X-Y by dividing its ∆H (enthalpy change) by n, we need to look for a process that would give us the desired equation:

X-Y (bond) → X + nY

Among the given choices, the only option that matches this desired equation is option B:

2XYn (g) → 2XYn-1(g) + Y2(g)

In this equation, we start with 2XYn molecules and break one X-Y bond, resulting in 2XYn-1 molecules and one Y2 molecule. This matches the desired equation X-Y (bond) → X + nY, where n=1.

Therefore, the correct answer is option B: 2XYn (g) → 2XYn-1(g) + Y2(g)