The electronic configurations of elements A, | vedclass

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(A) The elements $A, B$ and $C$ are Lithium $(Li)$,Sodium $(Na)$ and Potassium $(K)$ respectively,which belong to Group $1$ of the periodic table.As w

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$A > B > C$

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(A) The elements $A, B$ and $C$ are Lithium $(Li)$,Sodium $(Na)$ and Potassium $(K)$ respectively,which belong to Group $1$ of the periodic table.As we move down the group,the atomic size increases due to the addition of new shells.Ionization potential is inversely proportional to the atomic size.Therefore,as the atomic size increases from $A$ to $C$,the energy required to remove the valence electron decreases.The correct order of first ionization potential is $A > B > C$.

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

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 $C$?

Similar Questions

The first ionisation potential is maximum for

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 elements has the lowest ionisation potential?

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What is the correct order of the first ionization energy for $Li, Be, B,$ and $Na$?

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