The first and second ionisation enthalpies of | vedclass

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Question

(A) The $IE$ values indicate that the metal belongs to the $I^{st}$ group because the second $IE$ is very high,suggesting the presence of only one val

Vedclass · Accepted Answer

$1$ and $0.5$

Vedclass · Answer

(A) The $IE$ values indicate that the metal belongs to the $I^{st}$ group because the second $IE$ is very high,suggesting the presence of only one valence electron.Therefore,the metal hydroxide will be of the type $MOH$.The reaction with $HCl$ is:$MOH + HCl \rightarrow MCl + H_{2}O$From the stoichiometry,$1 \; mol$ of $MOH$ reacts with $1 \; mol$ of $HCl$.The reaction with $H_{2}SO_{4}$ is:$2MOH + H_{2}SO_{4} \rightarrow M_{2}SO_{4} + 2H_{2}O$Dividing by $2$ to match $1 \; mol$ of $MOH$:$MOH + \frac{1}{2} H_{2}SO_{4} \rightarrow \frac{1}{2} M_{2}SO_{4} + H_{2}O$From the stoichiometry,$1 \; mol$ of $MOH$ reacts with $0.5 \; mol$ of $H_{2}SO_{4}$.Thus,$1$ mole of $HCl$ and $0.5$ mole of $H_{2}SO_{4}$ are required.

The first and second ionisation enthalpies of a metal are $496$ and $4560 \; kJ \; mol^{-1},$ respectively. How many moles of $HCl$ and $H_{2}SO_{4}$ respectively,will be needed to react completely with $1$ mole of the metal hydroxide?

Similar Questions

Which of the following statements is incorrect?

If first ionization enthalpies of element $X$ and $Y$ are $419 \ kJ \ mol^{-1}$ and $590 \ kJ \ mol^{-1}$,respectively and second ionization enthalpies of $X$ and $Y$ are $3069 \ kJ \ mol^{-1}$ and $1145 \ kJ \ mol^{-1}$,respectively. Then the correct statement is :-

Which of the following orders regarding ionization energy is correct?

$A$ neutral atom will have the lowest ionization potential when its electronic configuration is

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