JEE MAIN - Chemistry (2023 - 11th April Morning Shift - No. 6)

For elements $$\mathrm{B}, \mathrm{C}, \mathrm{N}, \mathrm{Li}, \mathrm{Be}, \mathrm{O}$$ and $$\mathrm{F}$$, the correct order of first ionization enthalpy is
$$\mathrm{Li}<\mathrm{Be}<\mathrm{B}<\mathrm{C}<\mathrm{O}<\mathrm{N}<\mathrm{F}$$
$$\mathrm{B}>\mathrm{Li}>\mathrm{Be}>\mathrm{C}>\mathrm{N}>\mathrm{O}>\mathrm{F}$$
$$\mathrm{Li}<\mathrm{Be}<\mathrm{B}<\mathrm{C}<\mathrm{N}<\mathrm{O}<\mathrm{F}$$
$$\mathrm{Li}<\mathrm{B}<\mathrm{Be}<\mathrm{C}<\mathrm{O}<\mathrm{N}<\mathrm{F}$$

Explanation

The ionization energy of an atom or molecule describes the minimum amount of energy required to remove an electron from the atom or molecule in the gaseous state. It generally increases across a period (from left to right) on the periodic table because the number of protons is increasing, making the nucleus more positively charged and therefore more strongly attracting the negatively charged electrons.

However, there are exceptions to this trend when half-filled or fully-filled sub-shells are present, which are particularly stable configurations. The increased stability of these configurations makes it harder to remove an electron, increasing the ionization energy.

The order of first ionization enthalpy for these elements, considering the above factors, should be:

Li < B < Be < C < O < N < F

This is because:

  • Lithium (Li) has the lowest nuclear charge in this group and thus the lowest ionization energy.

  • Beryllium (Be) has a fully filled 2s subshell, which increases its ionization energy above that of boron (B).

  • Carbon (C) has more protons than boron, increasing its ionization energy.

  • Nitrogen (N) has a half-filled 2p subshell, which increases its ionization energy above that of oxygen (O).

  • Fluorine (F) has the highest nuclear charge in this group and thus the highest ionization energy.

So the correct answer is: $$\mathrm{Li}<\mathrm{B}<\mathrm{Be}<\mathrm{C}<\mathrm{O}<\mathrm{N}<\mathrm{F}$$

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