Question

Outermost electronic configurations of four elements A,B,C,D are given below:
(A) 3s² (B) s²3p¹ (C) 3s²3p³ (D) 3s²3p⁴
The correct order of fist ionization enthalpy for them is:

• A. (A)< (B)< (C) < (D)
• B. (B)< (A)< (D)< (C)
• C. (B) <(D)< (A)< (C)
• D. (B)< (A)< (C) < (D)

Answer 1

The Correct Option is B

To determine the correct order of first ionization enthalpy for the elements A, B, C, and D based on their outermost electronic configurations, we need to understand the concept of ionization enthalpy and compare these configurations. Ionization enthalpy is the energy required to remove an electron from an atom or ion in the gaseous state. Generally, for elements in the same period, ionization enthalpy increases across a period due to increasing nuclear charge and decreasing atomic radius.

The given electronic configurations are for elements in the third period: 

• (A) 3s² - This configuration corresponds to Magnesium (Mg)
• (B) s²3p¹ - This configuration corresponds to Aluminium (Al)
• (C) 3s²3p³ - This configuration corresponds to Phosphorus (P)
• (D) 3s²3p⁴ - This configuration corresponds to Sulfur (S)

The order of first ionization enthalpy is determined based on their position in the periodic table and the stability provided by half-filled and fully filled orbitals:

1. Element (B) Aluminium (Al) has a lower ionization enthalpy than Magnesium (A) because it starts filling the 3p orbital, which is less stable compared to a filled 3s orbital.
2. Element (A) Magnesium (Mg) has a higher ionization enthalpy than Aluminium due to a completely filled 3s orbital providing greater stability.
3. Element (D) Sulfur (S) has higher ionization enthalpy than Magnesium (A) because it is further along the period and has an increased nuclear charge.
4. Element (C) Phosphorus (P) has the highest ionization enthalpy due to a half-filled 3p orbital, which is more stable compared to the partially filled p orbitals.

Thus, the correct order of first ionization enthalpy is:
\((B) < (A) < (D) < (C)\)