You are given Avogadro's no . Of X atoms . If half of the atoms of X transfer one electron to the other half of X atoms 409kJ must be added . If these X− ions are subsequently converted to X+, and additional 733kJ must be added . Calculate IP and EA of X in eV Uses (1eV=1.602×10−19J and Na=6.023×1023) .

Question

You are given Avogadro's no . Of X atoms . If half of the atoms of X transfer one electron to the other half of X atoms 409kJ must be added . If these X− ions are subsequently converted to X+, and additional 733kJ must be added . Calculate IP and EA of X in eV Uses (1eV=1.602×10−19J and Na​=6.023×1023) .

Filo · Accepted Answer

Solution: Give X→X++e,ΔH=IP1​aeV  X=e→X−,ΔH=EA1​=−beV  If 12N​ atoms of X lose electrons which are atlen up by remaining 2Z​ of X to giveX− then  a×2N​−b×2N​=1.602×1019409×103​eV.  or a−b=1.602×10−19×6.023×1023409×103×2​  ∴a−b=8.477  Now , N/2 of X− lose two electrons to giev X+  X−→X=e,ΔH=+EA1​=+b  X→X++e,Δ=+IP1​=+a  ∴a×2N​+b×2N​=1.602×10−19733×103​eV  or a+b=1.602×1019×6.023×1023733×103×20​  a+b=15.194  ∴a=11.7835eV and b=3,358eV.

Question Text	You are given Avogadro's no . Of $X$ atoms . If half of the atoms of $X$ transfer one electron to the other half of $X$ atoms $409 k J$ must be added . If these $X^{-}$ ions are subsequently converted to $X^{+}$ , and additional $733 k J$ must be added . Calculate $I P$ and $E A$ of X in $e V$ Uses $(1 e V = 1.602 \times 10^{- 19} J$ and $N_{a} = 6.023 \times 10^{23})$ .
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