Why Some Elements Cannot Form MF6 Ions (The Covalency Limit)
Chemistry · Periodic Classification Of Properties · NEET
Boron cannot form the BF6 3- (an MF6 type) ion because it is a second-period element with only four valence orbitals (2s and 2p) and no d-orbitals. This caps its maximum covalency at 4, so it can only reach BF4 -, never BF6 3-. Aluminium, gallium and indium sit below boron and have empty d-orbitals, so they can expand their covalency to 6. Memory hook: "No d, no six" - the first element of a group is stuck at covalency 4.
Boron (period 2) has only four valence orbitals and no d-orbital, capping its covalency at 4 - so BF6 3- is impossible. Al, Ga and In have empty d-orbitals and reach covalency 6, forming MF6 3- ions.
Your doubts, answered
Why can boron not form the BF6 3- ion?
Boron is in the second period. Its outer shell (n=2) has only the 2s and 2p orbitals, which is four orbitals in total. Four orbitals can hold at most four bonds, so boron's maximum covalency is 4. To make BF6 3- boron would need to form 6 bonds, which needs 6 orbitals. There is no 2d orbital in nature (d-orbitals start only at n=3), so boron simply runs out of orbitals. It stops at BF4 -.
What is 'covalency' and how is it different from oxidation state?
Covalency is the number of shared electron-pair bonds an atom actually forms - it counts bonds. Oxidation state is a charge bookkeeping number and can be different. For example, in AlF6 3- aluminium has oxidation state +3 but covalency 6 (it forms 6 Al-F bonds). NEET often mixes these up in one option to trap you, so read the word carefully.
Why do Al, Ga and In form MF6 3- but boron does not?
Al, Ga and In are below boron in Group 13, so they are in period 3 or higher. From n=3 onward, empty d-orbitals become available (3d for Al). These extra d-orbitals let the atom form more than 4 bonds. So aluminium can use up to six orbitals and make AlF6 3- (an octahedral ion). Boron has no such d-orbital, so it is left behind at covalency 4.
Is this the same idea as 'boron cannot expand its octet'?
Yes, they are two names for the same fact. 'Expanding the octet' means holding more than 8 electrons around the central atom, which needs d-orbitals. Boron has no accessible d-orbitals, so it cannot expand its octet and cannot go beyond covalency 4. That is why BF6 3- is impossible but AlF6 3- is fine.
Does this rule apply only to boron, or to all first-period-of-a-group elements?
It applies to the first member of every s- and p-block group (Li, Be, B, C, N, O, F). All of them are in period 2 and have only four valence orbitals (2s + 2p), so their maximum covalency is 4. This is one reason the first element of each group shows anomalous (different) behaviour compared to the elements below it. This is a favourite NEET topic.
⚠️ The NEET trap ✗ Picking Al, Ga or In as the element that cannot form MF6 3-, thinking the biggest or smallest atom is the odd one out. ✓ Boron (B) is the correct answer. It is the only second-period element in the list, so it has no d-orbitals and a maximum covalency of 4, which blocks BF6 3-. 🧠 See MF6 3-? Ask 'which one has NO d-orbital?' The period-2 first member (here boron) is always the loser.
Real NEET questions
NEET 2018
Which one of the following elements is unable to form MF6 3- ion?
A · B ✓
B · Al
C · Ga
D · In
Solution: Boron is a second-period element. Its valence shell (n=2) has only the 2s and three 2p orbitals - four orbitals total - and there is no 2d orbital. So boron's maximum covalency is 4; it can form BF4 - but not BF6 3-. Aluminium, gallium and indium lie below boron and have empty d-orbitals available, so they can expand their covalency to 6 and form MF6 3- ions. Therefore the element that cannot form MF6 3- is boron, option A.
NEET 2022 / 2026
Identify the incorrect statement (option about covalency shown):
A · Largest and smallest among Mg, Mg2+, Al, Al3+ are Al and Mg2+ ✓
B · IUPAC name of Z=107 is Unnilseptium
C · Li-Mg similarity is the diagonal relationship
D · Oxidation state and covalency of Al in [AlCl(H2O)5]2+ are 3 and 6
Solution: This question tests whether you can separate oxidation state from covalency. In [AlCl(H2O)5]2+, aluminium forms 6 bonds (1 Al-Cl + 5 Al-O), so its covalency is 6, while its oxidation state is +3 - so option D is correct and not the answer. Aluminium can reach covalency 6 because it has accessible 3d-orbitals, exactly the property boron lacks. The incorrect statement is A (the largest species is neutral Mg, and the smallest is Al3+, not as stated).
Solved Periodic Classification Of Properties NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
No. Boron has only four valence orbitals (2s and 2p) and no d-orbital, so its maximum covalency is 4. The most fluorine it can hold is in BF4 -, never BF6 3-.
Why do d-orbitals matter here?
To form 6 bonds an atom needs 6 orbitals. Beyond the four s and p orbitals, the extra orbitals must come from the d-subshell. d-orbitals exist only from n=3 onward, so period-2 elements like boron cannot use them.
What is the maximum covalency of second-period elements?
Four. Every period-2 element (Li, Be, B, C, N, O, F) has only four valence orbitals, so none of them can go beyond covalency 4. This is part of the anomalous behaviour of the first element of each group.
Does aluminium really form AlF6 3-?
Yes. Aluminium has empty 3d-orbitals, so it can expand its covalency to 6 and form the octahedral AlF6 3- (hexafluoroaluminate) ion, found in cryolite Na3AlF6.
Why is this asked in NEET?
It links three high-yield ideas: anomalous behaviour of second-period elements, the covalency limit, and the difference between covalency and oxidation state. NEET 2018 asked it directly and 2022/2026 tested it inside a covalency-vs-oxidation-state trap.
