Bond energies of `H_(2),Cl_(2)` and `HCl` are respectively `"104, 58 and 103 k cal mol"^(-1)`. Calculate Pauling's electronegativity of chlorine. 

Explanation

Average of bond energies of `H_(2) and Cl_(2)` is the calculated bond energy of `HCI= (104 +58)/(81)" k cal mol"^(-1)`
Experimental bond energy of `HCl = 103 k cal mol"^(-1)`
`Delta = "Bond (resonance) stabilisation energy" = 103 - 81 = "22k cal mol"^(-1)`
`X_(1)-X_(2) = 0.208 sqrtDelta = 0.208sqrt(22) = 0.98`
Since Pauling.s electronegativity of hydrogen is 2.05, that of chlorine `= 2.05 +0.98 = 3.03`