EN(mulliken scale) of the element (A) is $E_{1}$ and IP is $E_{2}$ . Hence, EA will be:

2 E_{1} - E_{2}

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E_{1} - E_{2}

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E_{1} - 2 E_{2}

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(E_{1} + E_{2}) / 2

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Solution

The correct option is B $2 E_{1} - E_{2}$
As we know,
$E N = \frac{(I P) + (E A)}{2}$

$E A = 2 E N - I P$

$E A = 2 E_{1} - E_{2}$

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No/2 atoms of $X_{(g)}$ are converted into $X_{(g)^{+}}^{+}$ by energy $E_{1}$ , No/2 atoms of $X_{g}$ are converted into $X_{(g)}^{-}$ by energy $E_{2}$ . Hence ionization potential and electron affinity of $X_{(g)}$ per atom are

N_o/2 atoms of X(g) are converted into X⁺(g) by energy E₁. N_o/2 atoms of X(g) are converted into X^-(g) by energy E₂. Hence the ionisation potential and electron affinity of X(g) per atom are:

A. 2E₁/N_o., 2(E₂-E₁)/N_o

B. 2E₁/N_o.,. 2E₂/N_o

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