2
$\begingroup$

Why is the correct order of second ionisation potential of $\ce{Li , Be , B , C }$ such as: $\ce{Be} <\ce C <\ce B <\ce{Li} $? I understand that $\ce{Li+}$ has a stable inert gas electronic configuration and $\ce{B+}$ has one fully filled s orbital. Hence these will have a high value of ionisation potential. But on what basis do we then arrange the ionisation potentials of carbon and beryllium? What is the general steps of rules to solve such kinds of questions?

$\endgroup$

1 Answer 1

1
$\begingroup$

You are correct about $\ce{Li^+}$: the second electron to be extracted would be one of the two extremely tightly bound $\text{1s}$ electrons, which explains the very high second ionisation energy of that ion.

As for the comparison $\ce{Be^+}$ to $\ce{C^+}$, the main effect here is effective nuclear charge, $Z+1$, which in the case of $\ce{Be^+}$ is $5$ and in the case of $\ce{C^+}$ is $7$.

Equivalent electrons (same or similar orbitals) are more tightly bound to the nucleus, hence harder to extract, for higher $Z+1$. This results in a trend of higher second ionisation energies for higher $Z$, across a period.

This is a general trend across each period.

What is the general steps of rules to solve such kinds of questions?

Chemguide is a good place to start.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.