JEE MAIN - Chemistry (2025 - 29th January Evening Shift - No. 13)

First ionisation enthalpy values of first four group 15 elements are given below. Choose the correct value for the element that is a main component of apatite family :
834 kJ mol−1
947 kJ mol−1
1402 kJ mol−1
1012 kJ mol−1

Explanation

Apalite family - A group of phosphate minerals that includes fluorapatite, chlorapatite, and hychoxyapatite. Chemical formula $$C{a_{10}}{(P{O_4})_6}{(OH,F,Cl)_2}$$

$$\left. \matrix{ Fluorapatite \to C{a_{10}}{(P{O_4})_6}{F_2} \hfill \cr Chlorapatite \to C{a_{10}}{(P{O_4})_6}C{l_2} \hfill \cr Hydroxyapatite \to C{a_{10}}{(P{O_4})_6}{(OH)_2} \hfill \cr} \right\}\matrix{ {Main\,components:} \cr {Calcium\,and\,phosphate} \cr }$$

So, the element considered is phosphorus.

The first four group 15 elements are Nitrogen (N), Phosphorus (P), Arsenic (As) and Antimony (Sb). The outer configuration of phosphorus is half filled and hence stable $$({s^2}{p^3})$$.

For a stable configuration, ionization enthalpy is high. Ionization enthalpy - Energy required to remove an electron from the gaseous atom.

Nitrogen has the highest first ionization enthalpy in group 15.

Down the group, ionization enthalpy decreases.

Down the group, atomic size increases, valence electrons are less attracted to the nucleus. And here, electron removal will become easier. So, ionization enthalpy is also decreases.

The highest value of ionization enthalpy 1402 KJ mol$^{-1}$ corresponds to nitrogen element.

So, the first ionization enthalpy of the element that is a main component of apatite family is option 4) 1012 kJ mol$^{-1}$.

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