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Ionization Enthalpy

If Mg → M+g...

Question

If $M (g) \to M^{+} (g) + e^{-}; Δ H = 100 e V M (g) \to M^{2 +} (g) + 2 e^{-}; Δ H = 250 e V$
Which is/are correct statement(s)?
( $I E =$ ionization energy, $Δ H_{g e} =$ electron gain enthalpy)

I E_{1}

M (g)

is 100 eV

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I E_{1}

M^{+} (g)

is 150 eV

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I E_{2}

M (g)

is 250 eV

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Δ H_{g e}

M^{2 +} (g)

is -150 eV

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Solution

The correct option is D $Δ H_{g e}$ of $M^{2 +} (g)$ is -150 eV
$I E_{1} f o r M^{+} i s a c t u a l l y I E_{2}$

$M^{+} (g) \to M^{+ 2} (g) + e^{-}$

$I E_{2} = (250 - 100) e V = 150 e V$
Since, $I E = - E A$
Hence,
$Δ H_{g e} (M^{2 +}) = - I E_{2} = - 150 e V$

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Similar questions

M (g) \to M^{+} (g) + e^{-}; Δ H = 100 e V M (g) \to M^{2 +} (g) + 2 e^{-}; Δ H = 250 e V

I E =

Δ H_{g e} =

M (g) \to M^{+} (g) + e^{-}; Δ H = 100 e V M (g) \to M^{2 +} (g) + 2 e^{-}; Δ H = 250 e V

I E =

Δ H_{g e} =

M (g) \to M^{+} (g) + e^{-}; Δ H = 90 e V

M (g) \to M^{2 +} (g) + 2 e^{-}; Δ H = 260 e V

M^{2 +} (g)

M (g) \to M^{+} (g) + e^{-}, Δ H = 100 e V, M (g) \to M^{2 +} (g) + 2 e^{-}, Δ H = 250 e V

M (g) \to M^{+} (g) + e^{-}; Δ H = 100 e V

M (g) \to M^{2 +} (g) + 2 e^{-}; Δ H = 250 e V

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