Consider the following conversionsi) O(g) +e- → O(g)-. ∆H1ii) F(g) +e- → F(g)-. ∆H2iii) Cl(g) +e- → Cl(g)-. ∆H3iv) Na(g) → Na(g)+.∆H4Incorrect statement is

O(g) + e- → O-(g) is exothermicF(g) + e- → F-(g) is exothermicCl(g) + e- → Cl-(g) is exothermicNa(g) → Na+(g) is endothermic.ΔH1 , ΔH2 and ΔH3 are negative where as ΔH4 is positiveThus 3rd option is incorrect.

Consider the following conversionsi) O(g) +e- → O(g)-. ∆H1ii) F(g) +e- → F(g)-. ∆H2iii) Cl(g) +e- → Cl(g)-. ∆H3iv) Na(g) → Na(g)+.∆H4Incorrect statement is

O(g) + e- → O-(g) is exothermicF(g) + e- → F-(g) is exothermicCl(g) + e- → Cl-(g) is exothermicNa(g) → Na+(g) is endothermic.ΔH1 , ΔH2 and ΔH3 are negative where as ΔH4 is positiveThus 3rd option is incorrect.

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Consider the following conversions
i) O_(g) +e^- → O_(g)^-. $∆$ H₁
ii) F_(g) +e^- → F_(g)^-. $∆$ H₂
iii) Cl_(g) +e^- → Cl_(g)^-. $∆$ H₃
iv) Na_(g) → Na_(g)⁺. $∆$ H₄
Incorrect statement is

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$∆$ H₂ is more negative than $∆$ H₁

$∆$ H₁, $∆$ H₂ and H₃ are negative while $∆$ H₄is positive

$∆$ H₁ and $∆$ H₂ is less negative than $∆$ H₃

$∆$ H₂, $∆$ H₃ are negative while $∆$ H₁is positive

answer is C.

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Detailed Solution

O(g) + e- → O^-(g) is exothermic
F(g) + e- → F^-(g) is exothermic
Cl(g) + e- → Cl^-(g) is exothermic
Na(g) → Na⁺(g) is endothermic.
ΔH₁ , ΔH₂ and ΔH₃ are negative where as ΔH₄ is positive
Thus 3^rd option is incorrect.