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 $Al$ $(577.5 \ kJ \cdot mol^{-1})$.

$B$	$801$
$Al$	$577$
$Ga$	$579$
$In$	$558$
$Tl$	$589$

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

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The values of $\Delta _iH_1, \Delta _iH_2, \Delta _iH_3$,and $\Delta _iH_4$ for an atom are $7.5 \ eV, 25.6 \ eV, 48.6 \ eV$,and $170.6 \ eV$ respectively. Determine the electronic configuration of the atom.

The element having the highest first ionization enthalpy is

The first ionization enthalpy values (in $kJ\,mol^{-1}$) of group $13$ elements are :

$B$ $801$

$Al$ $577$

$Ga$ $579$

$In$ $558$

$Tl$ $589$

How would you explain this deviation from the general trend?

What is second ionization enthalpy?

In the long form of the periodic table,the elements having the lowest ionisation potentials are present in:

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