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Question

We consider formation of NaCl(s) using Born-Haber cycle which involves following steps:
(i) Na(s)12CI2NaCI(s)ΔH=Q
(II) Na(s)Na(g)ΔH=S
(III) Na(g)Na+(g)+eΔH=I
(IV) 12CI2(g)CI(g)ΔH=D2
(v) CI(g)+eCI(g)ΔH=?
(vI) Na+(g)+CI(g)NaCI(s)ΔH=U
Can you derive value of (EA) of CI(g) atom.

A
S+I+D/2
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B
Q+USID/2
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C
I+QU
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D
S+Q
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Solution

The correct option is C Q+USID/2
(V) involves addition of electron to CI(g)
CI(g)+eCI(g) EA=x
Energy released is (EA)
Adding (ii) to (vi) gives
Na(s)12CI2(g)NaCI(s)
Which is also the step (i), hence
S+I+D2+x(EA)U=Q
hence, x is calculated .

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