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(B) The difference $(IP_2 - IP_1)$ is related to the energy required to remove the second electron compared to the first. For $Mg$ $(Z=12)$,the electr

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$Mg$

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(B) The difference $(IP_2 - IP_1)$ is related to the energy required to remove the second electron compared to the first. For $Mg$ $(Z=12)$,the electronic configuration is $[Ne] 3s^2$. $IP_1$ involves removing an electron from the stable $3s^2$ orbital,while $IP_2$ involves removing an electron from the $3s^1$ orbital. For $Al$ $(Z=13)$,the configuration is $[Ne] 3s^2 3p^1$. $IP_1$ involves removing the $3p^1$ electron,and $IP_2$ involves removing a $3s^2$ electron. Generally,the jump $(IP_2 - IP_1)$ is significantly higher when the second electron is removed from a more stable or inner shell. For $Mg$,$(IP_2 - IP_1)$ is approximately $15.03 \ eV - 7.64 \ eV = 7.39 \ eV$,which is $For $Al$,$(IP_2 - IP_1)$ is approximately $18.82 \ eV - 5.98 \ eV = 12.84 \ eV$,which is $> 11 \ eV$. Thus,the condition holds for $Mg$.

$I.P.$ \text{ Trend}	Reason
$A$. $N > O$	Half-filled configuration
$B$. $Zr < Hf$	Lanthanide contraction
$C$. $Na > K$	$Z_{eff}$
$D$. $Al < Ga$	Transition contraction

For which element is the difference between the second and first ionization potential $(IP_2 - IP_1 < 11 \ eV)$ found?

Similar Questions

Ionisation energy in group $I$ elements varies in the decreasing order as:

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

Identify the incorrect match regarding Ionization Potential $(I.P.)$ trends:

$I.P.$ \text{ Trend} Reason

$A$. $N > O$ Half-filled configuration

$B$. $Zr < Hf$ Lanthanide contraction

$C$. $Na > K$ $Z_{eff}$

$D$. $Al < Ga$ Transition contraction

Which of the following has the least ionization potential?

Which of the following statements is correct regarding the ionization potential $(IP)$?

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