Which of the following orders is correct for | vedclass

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(B) The second ionisation energy $(IE_{2})$ is the energy required to remove an electron from a monovalent cation. The electronic configurations of th

Vedclass · Accepted Answer

$C < N < F < O$ $(2^{nd} \ I.E.)$

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(B) The second ionisation energy $(IE_{2})$ is the energy required to remove an electron from a monovalent cation. The electronic configurations of the corresponding cations are:$C^{+} (2s^{2} 2p^{1})$$N^{+} (2s^{2} 2p^{2})$$O^{+} (2s^{2} 2p^{3})$$F^{+} (2s^{2} 2p^{4})$Due to the stable half-filled $2p^{3}$ configuration of $O^{+}$,it has a higher $IE_{2}$ than $F^{+}$. The correct order of $IE_{2}$ is $C K (19) > Cl (17) > S (16)$,the $IE$ order should be $Ca^{2+} > K^{+} > Cl^{-} > S^{2-}$. Thus,option $A$ is incorrect. Re-evaluating $B$: $C^{+} (2s^{2} 2p^{1})$,$N^{+} (2s^{2} 2p^{2})$,$F^{+} (2s^{2} 2p^{4})$,$O^{+} (2s^{2} 2p^{3})$. $O^{+}$ is more stable than $F^{+}$,so $IE_{2}$ of $O > F$. The correct order is $C < N < F < O$.

Which of the following orders is correct for the property mentioned in brackets?

Similar Questions

Which of the following electronic configurations is associated with the biggest jump between the second and third ionization energies?

Given below are two statements:
Statement $I$: The first ionization energy of $Pb$ is greater than that of $Sn$.
Statement $II$: The first ionization energy of $Ge$ is greater than that of $Si$.
In the light of the above statements,choose the correct answer from the options given below:

Which of the following elements will have the lowest first ionisation energy?

Which of the following has the least ionization potential?

Assertion $(A)$: The first ionisation energy of $Be$ is greater than that of $B$.
Reason $(R)$: $2p$ orbital has lower energy than $2s$ orbital.

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