Question

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

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

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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