Question

With which of the following electronic configuration of an atom has the lowest ionisation enthalpy?

• A. $1s^2 , 2s^2, 2p^5$
• B. $1s^2,2s^2 2p^3$
• C. $1s^2,2s^2 2p^6,3s^1$
• D. $1s^2 , 2s^2 2p^6$

Answer 1

The Correct Option is C

To determine which electronic configuration has the lowest ionization enthalpy, we need to understand the factors affecting ionization energy. Ionization energy is the energy required to remove an electron from an atom. It depends on several factors, including:

1. Atomic size: Larger atoms with electrons further from the nucleus have lower ionization energy.
2. Effective nuclear charge: A higher effective nuclear charge increases ionization energy as the nucleus pulls the electrons more strongly.
3. Electronic configuration: Half-filled and fully filled subshells are more stable due to symmetrical distribution, increasing ionization energy.

Let's analyze the given configurations:

• 1s^2 , 2s^2, 2p^5: This configuration belongs to fluorine. The 2p subshell is one electron short of being filled, but fluorine is highly electronegative and retains its electrons strongly.
• 1s^2,2s^2 2p^3: This configuration matches nitrogen. It has a half-filled 2p subshell, providing extra stability and higher ionization energy.
• 1s^2,2s^2 2p^6,3s^1: This is the configuration of sodium, where the single 3s electron is further away from the nucleus and easily removed, thus requiring lower ionization energy.
• 1s^2 , 2s^2 2p^6: This configuration corresponds to neon, which is a noble gas with a full outer shell, leading to very high ionization energy.

The correct answer is 1s^2,2s^2 2p^6,3s^1 (sodium) because the outermost electron in the 3s orbital experiences less nuclear attraction due to being further from the nucleus, and the subshell is neither half-filled nor fully filled, which provides less stability compared to the others.

Therefore, sodium has the lowest ionization enthalpy among the given configurations.