Given below are two statements. One is labell | vedclass

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(A) The electronic configuration of $N$ $(Z=7)$ is $1s^2 2s^2 2p^3$,which has a stable half-filled $p$-orbital.The electronic configuration of $O$ $(Z

Vedclass · Accepted Answer

Both $A$ and $R$ are correct and $R$ is the correct explanation of $A$.

Vedclass · Answer

(A) The electronic configuration of $N$ $(Z=7)$ is $1s^2 2s^2 2p^3$,which has a stable half-filled $p$-orbital.The electronic configuration of $O$ $(Z=8)$ is $1s^2 2s^2 2p^4$.Due to the half-filled stability of the $2p$ subshell in nitrogen,it requires more energy to remove an electron compared to oxygen.In oxygen,the $2p^4$ configuration means that two electrons must occupy the same $2p$ orbital,leading to increased inter-electronic repulsion,which makes it easier to remove an electron.Thus,both Assertion $A$ and Reason $R$ are correct,and $R$ is the correct explanation for $A$.

Element	$IE_1$	$IE_2$	$IE_3$
$P$	$495.8$	$4562$	$6910$
$Q$	$737.7$	$1451$	$7733$
$R$	$577.5$	$1817$	$2745$

Similar Questions

Amongst the elements with the following electronic configurations,which one of them may have the highest ionisation energy?

Element having highest $I.P.$ value is

Which of the following orders of ionization energy is correct?

The first three ionisation energies (in $kJ/mol$) of three representative elements are given below:

Element $IE_1$ $IE_2$ $IE_3$

$P$ $495.8$ $4562$ $6910$

$Q$ $737.7$ $1451$ $7733$

$R$ $577.5$ $1817$ $2745$

Which of the following options is incorrect?

The correct order of $IE_2$ is

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