Question

First ionisation enthalpy values of the first four group 15 elements are given below. Choose the correct value for the element that is a main component of the apatite family:

• A. 1012 , \(\text{kJ mol}^{-1}\)
• B. 1402 , \( \text{kJ mol}^{-1}\)
• C. 834 , \( \text{kJ mol}^{-1}\)
• D. 947 , \( \text{kJ mol}^{-1}\)

Hint:

The ionization enthalpy increases across a period as the effective nuclear charge increases, but it decreases down a group due to an increase in atomic size. Phosphorus is a key element in the apatite family and has a relatively moderate ionization enthalpy.

Answer 1

The Correct Option is A

The apatite family is composed of minerals that are mainly made of phosphate ions, and one of the key elements in this family is phosphorus. Phosphorus is the element in Group 15, and its ionization enthalpy is \( 1012 \, \text{kJ mol}^{-1} \). Hence, the correct value corresponds to the ionization enthalpy of phosphorus.
Thus, the correct answer is \( \boxed{(1)} \).

Answer 2

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:
\[ \boxed{1012 \, \text{kJ mol}^{-1}} \]