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  • Ionization potential and electron affinity of fluorine are respectively.

Ionization potential and electron affinity of fluorine are \mathrm{17.42 \: and \: 3.45 eV} respectively. The electronegativity of fluorine on Mulliken scale and Pauling scale is m&p respectively, then \mathrm{(m + p)/ 1.011 \: is:}
 

Option: 1

13


Option: 2

14


Option: 3

15


Option: 4

16


Answers (1)

best_answer

According to Mulliken's Scale:
Electronegativity \mathrm{= m = (I.E+E.A)/2 = (17.42+3.45)/2 = 10.435}
Thus, the value of m is \mathrm{10.435.}

According to Pauling's Scale:
Electronegativity\mathrm{=p=\left ( I.E+E.A \right )/\left ( 2\times 2.8 \right )=\left ( 17.42+3.45 \right )\5.6=3.73}

Thus, Value of \mathrm{ p\: is \: 3.73}
\mathrm{\therefore (m + p)/1.011 = (10.435+3.73)/1.011 = 14}

Hence option 2 is correct.

Posted by

Pankaj Sanodiya

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