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Question

(C) The successive ionization enthalpy values are given as: $\Delta_i H_1 = 899 \ kJ/mol$,$\Delta_i H_2 = 1757 \ kJ/mol$,$\Delta_i H_3 = 14847 \ kJ/mo

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Alkaline earth metal group

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(C) The successive ionization enthalpy values are given as: $\Delta_i H_1 = 899 \ kJ/mol$,$\Delta_i H_2 = 1757 \ kJ/mol$,$\Delta_i H_3 = 14847 \ kJ/mol$,and $\Delta_i H_4 = 17948 \ kJ/mol$. We observe a very large jump in energy between the second and third ionization enthalpies ($\Delta_i H_2$ to $\Delta_i H_3$). This indicates that the element has $2$ valence electrons in its outermost shell,as removing the third electron requires breaking a stable noble gas configuration. Elements with $2$ valence electrons belong to Group $2$,which is the alkaline earth metal group.

List-$I$ ($IE_1, IE_2, IE_3$ in $kJ \ mol^{-1}$)	List-$II$ (Element)
$A$. $1510$	$1$. $H$
$B$. $495, 6500, 10200$	$2$. $Li$
$C$. $840, 1630, 13100$	$3$. $Be$
$D$. $600, 2050, 3100$	$4$. $B$

The successive ionization enthalpy values for an unknown element are $\Delta_i H_1 = 899 \ kJ/mol$,$\Delta_i H_2 = 1757 \ kJ/mol$,$\Delta_i H_3 = 14847 \ kJ/mol$,and $\Delta_i H_4 = 17948 \ kJ/mol$. To which group of the periodic table does this element belong?

Similar Questions

Match List-$I$ with List-$II$ with correct code:

List-$I$ ($IE_1, IE_2, IE_3$ in $kJ \ mol^{-1}$) List-$II$ (Element)

$A$. $1510$ $1$. $H$

$B$. $495, 6500, 10200$ $2$. $Li$

$C$. $840, 1630, 13100$ $3$. $Be$

$D$. $600, 2050, 3100$ $4$. $B$

The electronic configurations of elements $A, B$ and $C$ are $[He] 2s^1$,$[Ne] 3s^1$ and $[Ar] 4s^1$ respectively. Which one of the following orders is correct for the first ionization potentials (in $kJ \ mol^{-1}$) of $A, B$ and $C$?

The second ionization enthalpy of an element is always higher than its first ionization enthalpy because

The correct order of first ionization potential $(IP)$ is:

In a group of elements,as the atomic number increases,which of the following is $NOT$ observed?

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