Boron cannot form which one of the following anions?

Boron is in Period 2 of the periodic table. Hence:

• Its valence shell is the second shell (n = 2).
• The maximum set of valence orbitals available is one 2s orbital and three 2p orbitals (total 4).
• There are no 2d orbitals of comparable energy, so boron cannot expand its octet beyond 8 electrons and cannot attain a coordination number greater than 4.

Now examine each anion:

Case A: $$\text{BH}_4^-$$

This is the tetrahydroborate ion. Boron uses its four valence orbitals to form four $$\sigma$$ bonds with H, giving a tetrahedral $$sp^3$$ arrangement. Coordination number = 4, which is possible. So this anion is stable and known (e.g. in $$\text{NaBH}_4$$).

Case B: $$\text{B(OH)}_4^-$$

Here boron again forms four $$\sigma$$ bonds (one to each OH group). Coordination number = 4, within the octet limit. The anion exists in alkaline solutions of boric acid. Hence it can be formed.

Case C: $$\text{BO}_2^-$$

This is metaborate. Boron is three-coordinate here (one double-bond equivalent to O and two single B-O bonds aggregated in polymeric forms). Coordination number = 3, easily accommodated. The ion exists in many borate salts such as $$\text{NaBO}_2\cdot3\text{H}_2\text{O}$$.

Case D: $$\text{BF}_6^{3-}$$

To surround boron with six fluoride ligands, a coordination number = 6 is required. Achieving a hexacoordinate environment needs six hybrid orbitals (usually $$sp^3d^2$$), which in turn demand the participation of d-orbitals. Because boron belongs to the second period, there are no 2d orbitals of suitable energy, so it cannot expand its octet to accommodate six F ligands. Consequently such anion cannot be formed.

Therefore boron is unable to form $$\text{BF}_6^{3-}$$, while all the other listed anions are known.

Option D which is: $$\text{BF}_6^{3-}$$