1012 , \(\text{kJ mol}^{-1}\)
1402 , \( \text{kJ mol}^{-1}\)
834 , \( \text{kJ mol}^{-1}\)
947 , \( \text{kJ mol}^{-1}\)
Step 1: Understand the question.
We are given the first ionisation enthalpy values for group 15 elements (N, P, As, Sb, Bi) and need to identify the correct value for the element that is the main component of the apatite family.
Step 2: Identify the element related to the apatite family.
Apatite is a group of phosphate minerals with the general formula \( \text{Ca}_5(\text{PO}_4)_3(\text{F, Cl, OH}) \). The key element forming these compounds is phosphorus (P), as it is present in the form of phosphate ions \( \text{PO}_4^{3-} \).
Hence, the element in question is phosphorus (P).
Step 3: Recall the first ionisation enthalpy trend for group 15 elements.
The group 15 elements are: N, P, As, Sb, Bi.
The first ionisation enthalpy decreases down the group due to increasing atomic size and shielding effect.
Approximate values (in kJ mol⁻¹):
- N: 1402
- P: 1012
- As: 947
- Sb: 834
- Bi: 703
Step 4: Identify the correct value.
For phosphorus (the main component of the apatite family), the first ionisation enthalpy is 1012 kJ mol⁻¹.
Final Answer:
\[ \boxed{1012 \, \text{kJ mol}^{-1}} \]
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The reaction : \(A_2 \rightleftharpoons 2A\)

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The reaction : \(A_2 \rightleftharpoons 2A\)
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