JEE Mains · Chemistry · STD 11 - 3. Classification of elements and periodicity in properties
The first ionisation potential of \(Na\) is \(5.1 \,eV.\) The value of electron gain enthalpy of \(Na^+\) will be : ............... \(\mathrm{eV}\)
- A \(-2.55\)
- B \(-5.1\)
- C \(-10.2\)
- D \(+2.55\)
Answer & Solution
Correct Answer
(B) \(-5.1\)
Step-by-step Solution
Detailed explanation
\(\mathrm{Na} \rightarrow \mathrm{Na}^{+}+\mathrm{e}^{-} \quad \mathrm{IE}=5.1 \mathrm{eV} \ldots \ldots\) \(\mathrm{Na}^{+}+\mathrm{e}^{-} \rightarrow \mathrm{Na}\) Second reaction is inverse of first reaction, therefore electron goes enthalapy of \(N a^{+}\) is reverse of…
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