Two elements of p-block can form following halides XF$_3$ and YF$_3$. XF$_3$ can act as Lewis acid while YF$_3$ can act as Lewis base. Then hybridization of ‘X’ and ‘Y’ in XF$_3$ and YF$_3$ is respectively.
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In Lewis acid-base theory, Lewis acids tend to have empty orbitals for accepting electron pairs, while Lewis bases have lone pairs available for donation.
Step 1: Establish the trend.
Generally, ionization energy increases across a period. However, Nitrogen ($2s^2 2p^3$) has a stable half-filled p-orbital configuration. This makes it harder to remove an electron from Nitrogen than from Oxygen ($2s^2 2p^4$), creating an anomaly in the trend.
Step 2: Assign values.
- Carbon (lowest Z, no anomaly): Lowest value \(\rightarrow\) 800.6.
- Boron (actually B<C usually, but here B is 1086? Wait, B<Be and B<C. IE order: B<C<O<N).
Let's match the order:
Boron (800.6), Carbon (1086.5), Oxygen (1313.9), Nitrogen (1402.3).
*(Note: Real values are approx B-800, C-1086, O-1314, N-1402).*
Step 3: Identify Nitrogen.
Nitrogen has the highest ionization energy among these four due to the extra stability. Therefore, the highest value corresponds to Nitrogen.
\[ \text{IE}_1(\text{N}) = 1402.3\,\text{kJ/mol} \]
