The first ionization potential of $N a$ is 5.1 eV. The value of electron gain enthalpy of $N a^{+}$ would 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
$N a \to N a^{+} + e^{-}, I . P . = + 5.1 e V$

$N a^{+} + e^{-} \to N a, △ H_{e g} = - 5.1 e V$

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

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118

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