Which of these arrangement of elements is in order of increasing ionization energy?

Analyze the question: We need to arrange B, C, N and F in order of increasing first ionization energy.

Concept: Ionization energy is the energy needed to remove the outermost electron from a gaseous atom. Across a period (left to right), nuclear charge increases while the atomic radius decreases. The outer electrons are held more tightly, so ionization energy generally increases.

There are small exceptions where a half-filled or fully filled subshell is extra stable.

Solution: B, C, N and F are all in Period 2.

Electron configurations:

• B: 1s² 2s² 2p¹
• C: 1s² 2s² 2p²
• N: 1s² 2s² 2p³ (half-filled p subshell, extra stable)
• F: 1s² 2s² 2p⁵

Approximate first ionization energies (kJ/mol):

• B: 801
• C: 1086
• N: 1402
• F: 1681

Order of increasing ionization energy: B < C < N < F

The notable exception in this period is between Be and B, and between N and O. Neither affects this set, because O is not included. N is still higher than C, and F is higher than N.

Checking the options:

• A: puts F before N, which is wrong.
• B: Boron, Carbon, Nitrogen, Fluorine, which is correct.
• C: puts C before B, which is wrong.
• D and E: wrong order (D is the decreasing order).

Answer: B

Tip: Across a period, ionization energy rises from left to right, so for Period 2 elements just follow the order in the periodic table: Li, Be, B, C, N, O, F, Ne. Remember only the two dips (Be to B, and N to O). Here, with no dip involved, the order is simply the order of atomic number.