A) \[B<Be<C<O<N\]
B) \[B<Be<C<N<O\]
C) \[Be<B<C<N<O\]
D) \[Be<B<C<O<N\]
Correct Answer: A
Solution :
| Ionisation potential: The amount of energy required to remove an electron from the outer-most orbit of a gaseous atom is known as ionisation potential. Elements having half-filled or completely filled orbitals are more stable than partially filled orbitals. |
| In a period from left to right ionisation potential decreases as the atomic number increases. The given elements (Be, B, C, N, O) are present in II period as |
| \[\xrightarrow[\text{Ionisation potential increases}]{Be\,\,\,\,\,\,\,\,\,B\,\,\,\,\,\,\,\,C\,\,\,\,\,\,\,N\,\,\,\,\,\,\,O}\] |
| But in case of Be and B, Be has higher ionisation potential due to stable configuration. |
| \[_{4}Be=1{{s}^{2}},\,2{{s}^{2}}\] |
| \[\] Stable configuration |
| \[_{5}B=1{{s}^{2}},2{{s}^{2}},2{{p}^{1}}\] |
| \[\] Unstable configuration |
| In the same way in case of N and O, N has higher LP. than O due to stable configuration |
| \[_{7}N=1{{s}^{2}},2{{s}^{2}}2{{p}^{3}}\] |
| \[\,\,\text{Stable}\,\text{configuration}\] |
| \[_{8}O=1{{s}^{2}},2{{s}^{2}}2{{p}^{4}}\] |
| \[\,\,\text{Unstable}\,\,\text{configuration}\] |
| So, the correct order of increasing IP will be: |
| \[B<Be<C<O<N\] |
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