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Electronegativity

Which of the ...

Question

Which of the following equation represents first enthalpy of ionization?

A

{H g}_{(s)} \to {H g}_{(g)}^{+} + e^{-}

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B

{H g}_{(l)} \to {H g}_{(g)}^{+} + e^{-}

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C

{H g}_{(g)} \to {H g}_{(g)}^{+} + e^{-}

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D

{H g}_{(s)} \to {H g}_{(g)^{2 +}}^{+} + e^{-}

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Solution

The correct option is D ${H g}_{(g)} \to {H g}_{(g)}^{+} + e^{-}$
Solution:- (C) ${H g}_{(g)} ⟶ {H g}^{+}_{(g)} + e^{-}$
The first ionisation energy is the energy required to remove one mole of the most loosely held electrons from one mole of gaseous atoms to produce 1 mole of gaseous ions each with a charge of $+ 1$ .
In the reaction, ${H g}_{(g)} ⟶ {H g}^{+}_{(g)} + e^{-}$ , 1 mole of $H g$ in gaseous state is producing 1 mole of ${H g}^{+}$ ion in gaseous state, hence the equation represents first enthalpy of ionization.

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Similar questions

H g_{2}^{2 +} + 2 e^{-} \to 2 H g ⟹ E^{0} = 0.789 V

H g^{2 +} + 2 e^{-} \to H g ⟹ E^{0} = 0.854 V

Calculate the equilibrium constant for

H g_{2}^{2 +} \to H g + H g^{2 +}

Q. The equilibrium constant for the following reaction :

H g^{2 +} + H g \to H g_{2}^{2 +}

E_{(H g | H g^{2 +})}^{⊖} = - 0.788 V; E_{(H g_{2}^{2 +} | 2 H g^{2 +})}^{⊖} = - 0.92 V

K_{e q} = 2.98 \times 10^{x}

x

E_{A g^{+} / A g}^{o} = 0.80 V, E_{M g^{2 +} / M g}^{o} = - 2.37 V, E_{C u^{2 +} / C u}^{o} = 0.34 V, E_{H g^{2 +} / H g}^{o} = 0.79 V

.
Which of the following statement is correct?

K_{c}

H g^{2 +} + H g ⇌ H g_{2}^{2 +}

E_{2 H g / H g^{2 +}}^{o} = - 0.788 V

E_{H g_{2}^{2 +} / 2 H g^{2 +}}^{o} = - 0.920 V

Write answer as nearest integer after dividing by 100.

Given:
$E_{A g^{+} / A g}^{0} = 0.80 V$ ,
$E_{M g^{2 +} / M g}^{0} = - 2.37 V$ ,
$E_{C u^{2 +} / C u}^{0} = 0.34 V$ ,
$E_{H g^{2 +} / H g}^{0} = 0.79 V$
Which of the following statements is/are correct ?

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