# (solved)Question 3.16 of NCERT Class XI Chemistry Chapter 3

Among the second period elements the actual ionization enthalpies are in the order Li < B < Be < C < O < N < F < Ne. Explain why (i) Be has higher ∆i H than B (ii) O has lower ∆i H than N and F?
(Rev. 27-Mar-2023)

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### Question 3.16NCERT Class XI Chemistry

Among the second period elements the actual ionization enthalpies are in the order Li < B < Be < C < O < N < F < Ne. Explain why (i) Be has higher ∆i H than B (ii) O has lower ∆i H than N and F?

### Solution in Detail(video solution below this)

$\displaystyle \underline{\underline{\text{the Expected Order}}}$

We know that as we move from left to right in a period (i) the new electrons get added to the same principal shell and (ii) the atomic radii continue to decrease regularly.

The decreasing atomic radius, along with the fact that the shielding due to inner principal shells remains same, suggests that the outer electrons become more strongly attracted to the nucleus and, therefore, we expect the ionization enthalpy to increase continuously from Li to F.

$\displaystyle \underline{\underline{\text{why Be and B break the trend?}}}$

When it comes to trends in energy, we have to look very finely on the orbitals rather than the orbits!

Electronic configuration of Be is: $\displaystyle 1s^22s^2$

Electronic configuration of B is: $\displaystyle 1s^22s^22p^1$

the electron removed in the case of B is $\displaystyle 2p^1$ whereas in the case of Be it is $\displaystyle 2s^2$

Point 1: both the electrons belong to same n = 2, but different orbitals

Point 2: the 2p-electron of B is shielded by the 2s-electrons, whereas the 2s electron of Be is shielded by 1s electrons. Hence, the 2p electron of B has less penetration to the nucleus, and can therefore be easily removed!

$\displaystyle \underline{\underline{\text{why N and O break the trend?}}}$

Electronic configuration of N is: $\displaystyle 1s^22s^22p^3$

Electronic configuration of O is: $\displaystyle 1s^22s^22p^4$

Point 1: One pair of the 2p electrons of oxygen share a common 2p orbital due to which they suffer repulsion, and one of them can be easily removed.

Point 2: The p-orbitals of N are half-filled, have the same spin, and have an exchange energy stabilization due to which more energy is needed to remove an electron.

Point 3: The p-orbitals of O can become half-filled and symmetrical by removal of one electron. The exchange energy stabilization may be the reason due to which less energy is needed to remove one of the paired electrons.

$\displaystyle \underline{\underline{\text{Is the reasoning correct?}}}$

The reasoning seems correct if we observe the exceptions in the second ionization enthalpies.

The exceptions of first IE move one step to the right when it comes to second IE. Instead of B the exception is shown by C, and instead of O the exception is shown by its successor florine F.

Take this example: B+ with $\displaystyle 1s^22s^2$ takes the place of Be, whereas C+ takes the place of B with $\displaystyle 1s^22s^22p^1$, which, justifiably, makes C+ as the exception for the second IE!

Similar shift to the right is observed for 3rd IE. Hence, our reasoning is correct.

### Video Explanation

Please watch this youtube video for a quick explanation of the solution: