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Which of the following is incorrect?

A$\mathrm{IE}_1$ of $\mathrm{Li} < \mathrm{IE}_1$ of Be
B$\mathrm{IE}_1$ of $\mathrm{Be} < \mathrm{IE}_1$ of B
C$\mathrm{IE}_1$ of $\mathrm{Li} > \mathrm{IE}_1$ of Na
D$\mathrm{IE}_1$ of $\mathrm{He} > \mathrm{IE}_1$ of Ne
Answer & Solution
Correct answer: B. $\mathrm{IE}_1$ of $\mathrm{Be} < \mathrm{IE}_1$ of B
Across a period, first ionization energy generally increases, but there are known exceptions due to subshell stability. In period $2$, $\mathrm{Be}$ has configuration $1s^2 2s^2$ and $\mathrm{B}$ has configuration $1s^2 2s^2 2p^1$. Removing a $2p$ electron from $\mathrm{B}$ is easier than removing a filled $2s$ electron from $\mathrm{Be}$, so $$\mathrm{IE}_1(\mathrm{Be}) > \mathrm{IE}_1(\mathrm{B}).$$ Therefore statement $\mathrm{Be} < \mathrm{B}$ is incorrect. Also, $$\mathrm{IE}_1(\mathrm{Li}) < \mathrm{IE}_1(\mathrm{Be})$$ is correct, $$\mathrm{IE}_1(\mathrm{Li}) > \mathrm{IE}_1(\mathrm{Na})$$ is correct because ionization energy decreases down the group, and $$\mathrm{IE}_1(\mathrm{He}) > \mathrm{IE}_1(\mathrm{Ne})$$ is also correct. After checking all options, the incorrect statement is $\text{(B)}$.
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