The first ionization enthalpies of Mg and Al | vedclass

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Question

(D) Magnesium $(Mg)$ has the atomic number $12$ and electronic configuration $[Ne] 3s^2$. Since the $3s$ orbital is fully filled,it is more stable and

Vedclass · Accepted Answer

$IE_1(Mg) = 737.7 \ kJ \cdot mol^{-1}, IE_1(Al) = 577.5 \ kJ \cdot mol^{-1}$

Vedclass · Answer

(D) Magnesium $(Mg)$ has the atomic number $12$ and electronic configuration $[Ne] 3s^2$. Since the $3s$ orbital is fully filled,it is more stable and requires more energy to remove an electron.Aluminium $(Al)$ has the atomic number $13$ and electronic configuration $[Ne] 3s^2 3p^1$. The electron is removed from the $3p$ orbital,which is easier to remove than an electron from a fully filled $3s$ orbital.Therefore,the first ionization enthalpy of $Mg$ $(737.7 \ kJ \cdot mol^{-1})$ is higher than that of $Al$ $(577.5 \ kJ \cdot mol^{-1})$.

Element	$IE_1 + IE_2$ $(kJ/mol)$	$IE_3 + IE_4$ $(kJ/mol)$
$P$	$2.45$	$8.82$
$Q$	$2.85$	$6.11$

The first ionization enthalpies of $Mg$ and $Al$ can be expected to be

Similar Questions

The second ionization energy is maximum for

The correct order of $C$,$N$,$O$ and $F$ in terms of second ionisation potential is

If energy absorbed for conversion of $M_{(g)}$ into $M_{(g)}^{+}$ for $Li, Be, B, C$ and $N$ is $a, b, c, d$ and $e$ (in $k\,cal/mol$) respectively,then the correct order will be:

The sum of $IE_1 + IE_2$ and $IE_3 + IE_4$ for elements $P$ and $Q$ are given below:

Element $IE_1 + IE_2$ $(kJ/mol)$ $IE_3 + IE_4$ $(kJ/mol)$

$P$ $2.45$ $8.82$

$Q$ $2.85$ $6.11$

Then,according to the given information,the incorrect statement$(s)$ is/are:

Which of the following statements is incorrect?

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