The first ionisation potential of Na is 5.1 eV. The value of electron gain enthalpy of $N a^{+}$ will be

-2.55 eV

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-5.1 eV

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-10.2 eV

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+2.55 eV

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Solution

The correct option is B
-5.1 eV

$N a \to N a^{+} + e^{-} (F i r s t I . E) N a^{+} + e^{-} \to N a$
Electron gain enthalpy of $N a^{+}$ is reverse of (IE)
Because reaction is reverse so
$Δ H$ (eq) =-5.1 eV

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The first ionization potential of Na is 5.1 eV. The value of electron gain enthalpy of Na⁺ will be:

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