Question

Which of the following electronic configuration an atom has the lowest ionization enthalpy?

• A. 1s² 2s² 2p³
• B. 1s² 2s² 2p⁶ 3s¹
• C. 1s² 2s² 2p⁶
• D. 1s² 2s² 2p⁵

Answer 1

The correct option is B

1s² 2s² 2p⁶ 3s¹

• Ionization energy or enthalpy is the measure of the force of attraction between the nucleus and the valence electrons. In other words, it is the amount of energy required to remove an electron from an isolated atom or molecule
• The element which has completely filled or exactly half-filled outermost orbital with electrons have acquired extra stability. So, it has maximum ionization enthalpy.

Explanation for correct option:

(B)

• The electronic configuration of an atom is given (1s² 2s² 2p⁶ 3s¹).
• The given atom is sodium(Na) because it has a total of 11 electrons, which is an alkali metal.
• Here, the outermost subshell of Na i.e. s-subshell is partially filled.
• It has 1 valence electron. So, it has a higher tendency to lose 1 electron and complete its octet and attain the nearest noble gas configuration.
• Thus, the electronic configuration of an atom(Na) $(1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^1)$has the lowest ionization enthalpy.

Explanation for incorrect option:

(A)

• The electronic configuration of an atom is given (1s² 2s² 2p³).
• The given atom is nitrogen(N) because it has a total of 7 electrons.
• The given atom(N) has exactly a half-filled p-subshell.
• So, it acquires extra stability.
• It has a low tendency to lose electrons.
• Thus, the electronic configuration of an atom(N) (1s² 2s² 2p³) has more ionization enthalpy than Na.

(C)

• The electronic configuration of an atom is given (1s² 2s² 2p⁶).
• The given atom is neon(Ne) because it has a total of 10 electrons. It is a noble gas.
• The given atom(Ne) has a completely filled p-subshell.
• As we know that noble gas has a completely filled valence shell and it has a stable electronic configuration.
• Thus, the electronic configuration of an atom (1s² 2s² 2p⁶) has maximum ionization enthalpy.

(D)

• The electronic configuration of an atom is given (1s² 2s² 2p⁵).
• The given atom is fluorine(F)) because it has a total of 9 electrons.
• The given atom (F) has 7 valence electrons.
• So, it tends to gain 1 electron and complete its octet and attain the nearest noble gas configuration.
• Thus, the electronic configuration of an atom (1s² 2s² 2p⁵) has more ionization enthalpy.

Conclusion: Hence, option (B) is correct and 1s² 2s² 2p⁶ 3s¹ electronic configuration of an atom has the lowest ionization enthalpy