Boron can't form which one of the following anions?
(a) BF6^3-
(b) BH4^-
(c) B(OH)4^-
(d) BO2^-

Q: Why does Boron form $BF_4^-$ if it only has 3 valence electrons?

While Boron has 3 valence electrons to form 3 covalent bonds (like in $BF_3$), it still has one empty $2p$ orbital. A fluoride ion ($F^-$) can donate a lone pair into this empty orbital, forming a coordinate covalent bond and creating $BF_4^-$.

Q: What is the maximum covalency of any 2nd-period element?

The maximum covalency for any 2nd-period element (like C, N, O, F) is strictly 4. This is because the 2nd shell only contains one $2s$ and three $2p$ orbitals, making a total of 4 orbitals available for bonding.

Q: Why can Aluminum form $AlF_6^{3-}$ but Boron cannot form $BF_6^{3-}$?

Aluminum is in the 3rd period, meaning its valence shell is $n=3$. The 3rd shell contains $3s$, $3p$, and empty $3d$ orbitals. Aluminum can use these empty $3d$ orbitals to expand its octet and accommodate 6 bonds, whereas Boron has no $d$-orbitals.

Q: Does Boron have an incomplete octet in $BF_3$?

Yes! In $BF_3$, Boron only has 6 electrons in its valence shell (3 pairs from the 3 bonds). It is an electron-deficient molecule and acts as a Lewis acid, eagerly waiting to accept a lone pair to complete its octet.

Q: Is the fourth bond in $BF_4^-$ different from the other three?

Initially, the fourth bond is a coordinate covalent bond (both electrons come from the $F^-$ ion). However, once formed, all four B-F bonds in $BF_4^-$ are identical and indistinguishable due to $sp^3$ hybridization.

Question

Boron can't form which one of the following anions?

Options:

(A) $BF_6^{3-}$

(B) $BH_4^-$

(C) $B(OH)_4^-$

(D) $BO_2^-$

Sigma Percentile · Accepted Answer

### The Limits of Bonding: Why Boron Can't Expand Its Octet

When we study the $p$-block elements, one of the most fascinating concepts is how atoms decide the maximum number of bonds they can form. This property, known as **covalency**, is strictly governed by the availability of orbitals in the atom's valence shell. Let's dive into the quantum mechanics of Boron to understand why it behaves the way it does.

### The Ground State and Excitation

Boron has an atomic number of $Z=5$. In its ground state, its electronic configuration is $1s^2 2s^2 2p^1$. At first glance, it only has one unpaired electron in the $2p$ subshell. However, atoms rarely bond in their ground state.

When Boron approaches other atoms to form bonds, it absorbs a small amount of energy to excite one of its $2s$ electrons into an empty $2p$ orbital. The new excited state configuration becomes $1s^2 2s^1 2p^2$. Now, Boron has exactly three unpaired electrons, allowing it to form three covalent bonds, as seen in molecules like $BF_3$ or $BCl_3$.

### The Magic of the Empty Orbital

Even after forming three bonds, if we look closely at the $2p$ subshell, we notice something critical: **there is still one completely empty $2p$ orbital**.

This empty orbital acts as a perfect docking station for a lone pair of electrons. A Lewis base, such as a fluoride ion ($F^-$) or a hydroxide ion ($OH^-$), can donate its lone pair into this empty orbital, forming a coordinate covalent bond. This allows Boron to form a fourth bond, reaching a maximum covalency of 4. This is exactly how anions like $BH_4^-$, $B(OH)_4^-$, and $BF_4^-$ are formed.

### The Hard Limit: No d-orbitals

So, if Boron can form 4 bonds, why can't it form 5 or 6? The answer lies in its position on the periodic table. Boron is a **second-period element** ($n=2$). According to the rules of quantum mechanics, the second principal quantum shell only contains $s$ and $p$ subshells. **There are no $2d$ orbitals in existence.**

Because Boron has exactly four orbitals in its valence shell (one $2s$ and three $2p$), it is physically impossible for it to accommodate more than 8 electrons (4 pairs). Therefore, its covalency is strictly capped at 4.

When we look at the option $BF_6^{3-}$, it requires Boron to form 6 bonds. Since Boron lacks the $d$-orbitals necessary to expand its octet, the formation of $BF_6^{3-}$ is completely impossible. In contrast, elements in the third period, like Aluminum, possess vacant $3d$ orbitals and can easily form octahedral complexes like $AlF_6^{3-}$. Always keep an eye on the period number when dealing with maximum covalency!

Animated Solution for Chemistry - s and p-Block Elements: Boron can't form which one of the following anions?

Select Answer:

Visualized Solution

$B (Z = 5)$

Excited State

Maximum Covalency

Absence of $d$ -orbitals

Analyzing $B F_{6}^{3 -}$

The Way Forward

The Sigma Insight: Group 13 Elements

The Limits of Bonding

The Ground State and Excitation

The Magic of the Empty Orbital

The Hard Limit

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