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Q.

EN of the element (A) is E1 and IP is E2. Hence
EA will be

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a

E1-E2

b

E1-2E2

c

2E1-E2

d

E1+E2/2

answer is A.

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Detailed Solution

We know that Mulliken scale considers the electronegativity of an element as the average of the ionization potential and the electron affinity of that element. 

EN=(IP)+(EA)2

Now, According to question

EA=2EN-IP

EA=2E1-E2

Hence, option(a) is the answer.


 

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