A) \[Be>B>C>N>F\]
B) \[N>F>C>B>Be\]
C) \[F>N>C>Be>B\]
D) \[N>F>B>C>Be\]
E) \[F>C>N>B>Be\]
Correct Answer: C
Solution :
In general, as we move from left to right in a period, the ionization enthalpy increases with increasing atomic numbers. But the ionization enthalpy of B is lower than that of Be. This is because the electronic configuration of \[(1{{s}^{2}}\text{ }2{{s}^{2}}\text{ }2{{p}^{1}})\]is less stable than that of Be \[(1{{s}^{2}}\text{ }2{{s}^{2}})\]which has completely filled orbitals. As a result, the 2p-electron of B is not strongly attracted by the nucleus as the 2s-electron of Be. Thus, the increasing order of first ionization enthalpies of elements is \[F>N>C>Be>B\]
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