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Question

What is the value of electron affinity of $$Na^+$$ if $$IE_1$$ of Na = 5.1 eV ?


A
0 eV
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B
-3.52 eV
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C
-5.1 eV
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D
-2.84 eV
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Solution

The correct option is C -5.1 eV
$$e^-$$ affinity is opposite to $$IE$$, i.e., $$e^-$$ affinity is energy released/required during addition of $$e^-$$ whereas $$IE$$ is the energy required/released during removal of $$e^-$$. 

Thus, $$EA=-IE$$

$$\therefore EA (Na^+)=-5.1eV$$

Hence, the correct option is $$C$$

Chemistry

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