The electronegativity of fluorine from the following data isEH−H=104.2kcalmol−1,EF−F=36.6kcalmol−1,EH−F=134.6  kcalmol −1(Round off to nearest integer)

We know that electronegativity of H = 2.1. Since bond energy is given in   k cal mol −, so :XF−XH=0.208BEH−F−BEH−H×BEF−F1/21/2XF−2.1=0.208134.6−{104.2×36.6}1/21/2XF=2.1+0.208134.6−(3813.72)1/2 i.e., 61.761/2=2.1+0.208×(72.84)1/2=2.1+(0.208×8.53)=2.1+1.77=3.87