$Δ H_{f} =$ -98.2 K.Cal/mole
$S_{N a} =$ 36 K.Cal/mole
$I_{N a} =$ 118.5 K.Cal/mole
$\frac{1}{2} D_{C l_{2}} =$ 29 K.Cal/mole
$U_{N a C l} =$ -184.2 K.Cal/mole
From the data given below for $N a C l$ , the electron affinity of chlorine $[- E_{a}]$ is:

-97.5 K.cal/ mole

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-108 K.cal/mole

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-75 K.cal/mole

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-128 K.cal/mole

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Solution

The correct option is A -97.5 K.cal/ mole
From Born-Haber Cycle,
We have,
$Δ H_{f} = S_{N a} + I_{N a} + \frac{1}{2} D_{C l_{2}} + E_{a} + U_{N a C l}$
$- 98.2 = 36 + 118.5 + 29 + (- 184.2) + E_{a}$
$E_{a} = - 97.5 K . C a l / m o l e$

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ΔHf= -98.2 K.Cal/moleSNa= 36 K.Cal/moleINa= 118.5 K.Cal/mole12DCl2= 29 K.Cal/moleUNaCl= -184.2 K.Cal/moleFrom the data given below for NaCl, the electron affinity of chlorine [−Ea] is:

The correct option is A -97.5 K.cal/ moleFrom Born-Haber Cycle,We have,ΔHf=SNa+INa+12DCl2+Ea+UNaCl−98.2=36+118.5+29+(−184.2)+EaEa=−97.5K.Cal/mole

$Δ H_{f} =$ -98.2 K.Cal/mole
$S_{N a} =$ 36 K.Cal/mole
$I_{N a} =$ 118.5 K.Cal/mole
$\frac{1}{2} D_{C l_{2}} =$ 29 K.Cal/mole
$U_{N a C l} =$ -184.2 K.Cal/mole
From the data given below for $N a C l$ , the electron affinity of chlorine $[- E_{a}]$ is:

The correct option is A -97.5 K.cal/ mole
From Born-Haber Cycle,
We have,
$Δ H_{f} = S_{N a} + I_{N a} + \frac{1}{2} D_{C l_{2}} + E_{a} + U_{N a C l}$
$- 98.2 = 36 + 118.5 + 29 + (- 184.2) + E_{a}$
$E_{a} = - 97.5 K . C a l / m o l e$