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Which one of the following order is correct for the first ionisation energies of the elements?
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The correct answer is:
B < Be < O < N
First ionisation energy increases in a period. Thus, the first IE of the elements of the second period should be as follows
$\mathrm{Be} < \mathrm{B} < \mathrm{N} < \mathrm{O}$
But in practice, the elements do not follow the above order. The first IE of these elements is
$\mathrm{B} < \mathrm{Be} < \mathrm{O} < \mathrm{N}$
The lower IE of $\mathrm{B}$ than that of $\mathrm{Be}$ is because in $\mathrm{B}$ $\left(1 s^2, 2 s^2 2 p^1\right)$, electron is to be removed from $2 p$ which is easy while in $\operatorname{Be}\left(1 s^2, 2 s^2\right)$, electron is to be removed from $2 s$ which is difficult. The low IE of $\mathrm{O}$ than that of $\mathrm{N}$ is because of the half-filled $2 p$ orbitals in $\mathrm{N}\left(1 s^2, 2 s^2 2 p^3\right)$.
$\mathrm{Be} < \mathrm{B} < \mathrm{N} < \mathrm{O}$
But in practice, the elements do not follow the above order. The first IE of these elements is
$\mathrm{B} < \mathrm{Be} < \mathrm{O} < \mathrm{N}$
The lower IE of $\mathrm{B}$ than that of $\mathrm{Be}$ is because in $\mathrm{B}$ $\left(1 s^2, 2 s^2 2 p^1\right)$, electron is to be removed from $2 p$ which is easy while in $\operatorname{Be}\left(1 s^2, 2 s^2\right)$, electron is to be removed from $2 s$ which is difficult. The low IE of $\mathrm{O}$ than that of $\mathrm{N}$ is because of the half-filled $2 p$ orbitals in $\mathrm{N}\left(1 s^2, 2 s^2 2 p^3\right)$.
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