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 I.P. than 'O' due to stable configuration \[\begin{align} & _{7}N=1{{s}^{2}},2{{s}^{2}}2{{p}^{3}} \\ & \,Stable\,configuration \\ \end{align}\] \[\begin{align} & _{8}O=1{{s}^{2}},2{{s}^{2}}2{{p}^{4}} \\ & \,Unstable\,\,configuration \\ \end{align}\] So, the correct order of increasing IP will be: B < Be < C < O < N.
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