What will be the order of 1^{st} ionisation e | vedclass

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(A) The $1^{st}$ ionisation energy is the energy required to remove the most loosely bound electron from an isolated gaseous atom.In a group of the pe

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$Li > Na > K$

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(A) The $1^{st}$ ionisation energy is the energy required to remove the most loosely bound electron from an isolated gaseous atom.In a group of the periodic table,as we move from top to bottom,the atomic size increases and the valence electrons are further away from the nucleus,resulting in a decrease in ionisation energy.For the alkali metals (Group $1$),the order of ionisation energy is $Li > Na > K > Rb > Cs$.Therefore,the correct order for $Li$,$Na$,and $K$ is $Li > Na > K$.

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

What will be the order of $1^{st}$ ionisation energy for the alkali metals $Li$,$Na$,and $K$?

Similar Questions

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:

The first ionisation energies of alkaline earth metals are higher than those of the alkali metals. This is because

Which of the following elements has the lowest $IE_1$ value?

In the given graph,which point represents an alkali metal with the least atomic number?

Which electrons are removed to determine the ionization enthalpy of beryllium and boron,respectively?

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