Given that δ h ∘ f [ br ( g ) ] = 111.9 kj ⋅ mol − 1 δ h ∘ f [ c ( g ) ] = 716.7 kj ⋅ mol − 1 δ h ∘ f [ cbr 4 ( g ) ] = 29.4 kj ⋅ mol − 1 calculate the average molar bond enthalpy of the carbon‑bromine bond in a cbr 4 molecule.

283.725 kJ ⋅ mol − 1

Explanation:

C(s) + 2Br2(g) ⇒ CBr4(g) , Δ H ∘ = 29.4 kJ ⋅ mol − 1

Br2(g) ⇒ Br(g) ,  Δ H ∘ = 111.9 kJ ⋅ mol − 1

C(s) ⇒ C(g) ,  Δ H ∘ = 716.7 kJ ⋅ mol − 1

4*eqn(2) + eqn(3) ⇒ 2Br2(g) + C(s) ⇒ 4 Br(g) + C(g) , Δ H ∘ = 1164.3 kJ ⋅ mol − 1

eqn(1) - eqn(4) ⇒ 4 Br(g) + C(g) ⇒ CBr4(g) , Δ H ∘ = -1134.9 kJ ⋅ mol − 1

so,

   average bond enthalpy is = 283.725 kJ ⋅ mol − 1

indium, and thallium have three electrons i think

i think its 100

i hope this !