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

- 10.2 eV

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+2.55 eV

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- 2.55 eV

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- 5.1 eV

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Solution

The correct option is D
- 5.1 eV

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Similar questions

N a

5.1 e V

N a^{+}

N a^{+}

I E_{1}

N a = 5.1 e V

{N a}^{+}

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N A = 5.1 e V

N a^{+}

118

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