The $EN$ (Electronegativity) of element $A$ is $E_1$ and its $IP$ (Ionization Potential) is $E_2$. Hence,its $EA$ (Electron Affinity) will be:

  • A
    $2E_1 - E_2$
  • B
    $E_1 - E_2$
  • C
    $E_1 - 2E_2$
  • D
    $(E_1 + E_2) / 2$

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