Question

Boron is unable to form \(\mathrm{BF}_{6}^{3-}\) ions due to (a) non-availability of \(d\)-orbitals (b) small size of boron atom (c) non-metallic nature (d) less reactivity towards halogens.

Short answer

Boron is unable to form \(\mathrm{BF}_{6}^{3-}\) ions due to the non-availability of d-orbitals to expand its valence shell.

Step-by-step solution

Understanding Boron's electronic configuration

Analyze the electronic configuration of boron to understand if it is capable of expanding its valence shell. Boron, with an atomic number of 5, has the electronic configuration of 1s2 2s2 2p1. This indicates that boron has only three valence electrons in the second shell, where no d-orbitals are present.

Considering the ability of boron to expand its octet

Evaluate if boron can expand its octet to accommodate additional electrons. Boron does not have access to d-orbitals in the second energy level, hence it cannot expand its octet beyond the three valence electrons present in the 2s and 2p orbitals.

Eliminating incorrect options

Rule out options (b), 'small size of boron atom', (c), 'non-metallic nature', and (d), 'less reactivity towards halogens', as they do not directly explain the inability of boron to form \(\mathrm{BF}_{6}^{3-}\) ions. The 'small size' and 'non-metallic nature' do not stop boron from forming compounds generally, and 'less reactivity towards halogens' is not relevant to the ability to form a specific ion.

Identifying the correct reason

Identify the correct reason based on the elimination in the previous step and understanding of boron's electronic configuration. Boron cannot form compounds with an expanded octet because it lacks the necessary d-orbitals to accommodate additional electrons that would be involved in forming \(\mathrm{BF}_{6}^{3-}\) ions.

Key concepts

Electronic Configuration of Boron

Understanding the electronic configuration of boron is essential when determining its chemical behavior. Boron, which sits in group 13 of the periodic table, has an atomic number of 5. This number tells us that boron has five electrons. To understand where these electrons are located, we break down the electronic configuration into energy levels and subshells.

The electronic configuration of boron can be represented as \(1s^2 2s^2 2p^1\). This configuration shows that boron has two electrons filling the 1s subshell, two electrons in the 2s subshell, and just one electron in the 2p subshell. The electrons filling these subshells follow the aufbau principle, which guides the order in which electrons populate the orbitals. According to this principle, the lowest energy orbitals are filled first, which is why the 1s subshell, the innermost level, is filled before the 2s.

Furthermore, this particular arrangement is crucial for predicting boron's chemical bonding and reactivity as it highlights that boron can form three covalent bonds by using its three valence electrons. However, without access to an additional energy subshell, such as the d-orbitals, boron's ability to form certain types of compounds is limited, which we will explore further in the discussion of octet expansion.

Boron's Valence Electrons

Moving on from understanding the electronic configuration of boron, it is equally important to discuss valence electrons because they play a pivotal role in chemical bonding.

What are Valence Electrons?

Valence electrons are the electrons located in the outermost shell of an atom. These are the electrons involved in forming bonds with other atoms.

As previously mentioned, boron has three valence electrons located in the second shell (2s and 2p orbitals). A common characteristic of group 13 elements is this trio of valence electrons, which dictates the typical trivalent chemistry we see with boron compounds. For instance, in forming trihalides, such as boron trifluoride \(BF_3\), boron uses these three valence electrons to form three covalent bonds.

However, with only three valence electrons, boron does not achieve an octet naturally and remains short of the eight electrons typically associated with a completely filled valence shell. This limitation impacts boron's ability to form more complex ions, such as \(BF_{6}^{3-}\), as we'll discuss in the context of the octet rule and d-orbitals.

D-Orbitals and Octet Expansion

The concept of octet expansion is closely linked to the ability of an element to utilize d-orbitals in bonding. The octet rule states that atoms tend to form compounds in ways that give them eight valence electrons, thereby achieving a noble gas-like electron configuration. For many elements, especially those in the second period of the periodic table, this is a guiding principle for chemical compound formation.

However, the octet rule is not universal and can be expanded upon for elements that have access to d-orbitals, which are found in the third and higher periods. Such elements can accommodate more than eight electrons by utilizing these d-orbitals. In contrast, as we've observed with boron, it is located in the second period where d-orbitals are not available. Consequently, it is unable to employ d-orbitals for bonding, which restricts it from exceeding the octet limit.

This limitation explains why boron cannot form the ion \(BF_{6}^{3-}\). To construct this ion, boron would need to bond with six fluorine atoms, which would demand the ability to expand its octet to hold more than three bonds. Without available d-orbitals, boron cannot do this, which is the correct answer to the exercise - non-availability of d-orbitals (option a). This fundamental aspect of boron chemistry highlights the influence of electronic configuration in determining the types of compounds an element can form.