The values of Delta _iH_1, Delta _iH_2, Delta | vedclass

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Question

(B) The successive ionization energies are given as: $\Delta _iH_1 = 7.5 \ eV$,$\Delta _iH_2 = 25.6 \ eV$,$\Delta _iH_3 = 48.6 \ eV$,and $\Delta _iH_4

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$1s^2 \ 2s^2 \ 2p^6 \ 3s^2 \ 3p^1$

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(B) The successive ionization energies are given as: $\Delta _iH_1 = 7.5 \ eV$,$\Delta _iH_2 = 25.6 \ eV$,$\Delta _iH_3 = 48.6 \ eV$,and $\Delta _iH_4 = 170.6 \ eV$. $A$ large jump in ionization energy indicates the removal of an electron from a stable,fully-filled or half-filled shell (noble gas configuration). Here,the jump from $\Delta _iH_3$ $(48.6 \ eV)$ to $\Delta _iH_4$ $(170.6 \ eV)$ is very large,which suggests that the atom has $3$ valence electrons. Among the given options,the configuration $1s^2 \ 2s^2 \ 2p^6 \ 3s^2 \ 3p^1$ (Aluminum,$Z=13$) has $3$ valence electrons in the $3s$ and $3p$ orbitals. Therefore,the correct electronic configuration is $1s^2 \ 2s^2 \ 2p^6 \ 3s^2 \ 3p^1$.

Element	Ionization energy $1^{st}$ $(kJ/mol)$	Ionization energy $2^{nd}$ $(kJ/mol)$
$X$	$495$	$4563$
$Y$	$731$	$1450$

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.

Similar Questions

The element having the greatest difference between its first and second ionization energies is:

Consider the elements $Ne$,$Na$,and $Mg$. The element with the highest first ionization enthalpy and the element with the lowest second ionization enthalpy,respectively,are:

Which of the following sets is correct regarding the ionization potential?

Which of the following represents the correct order of ionization enthalpy?
$(i) Be^{+} > Be$ $(ii) Be > Be^{+}$ $(iii) C > Be$ $(iv) B > Be$

Identify the elements $X$ and $Y$ using the ionisation energy values given below :

Element Ionization energy $1^{st}$ $(kJ/mol)$ Ionization energy $2^{nd}$ $(kJ/mol)$

$X$ $495$ $4563$

$Y$ $731$ $1450$

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