Problem 113 - A Complete Resource Book in Chemistry for JEE Main

Question

Specify the coordination geometry and hybridization of \(\mathrm{N}\) and \(\mathrm{B}\) atoms in a \(1: 1\) complex of \(\mathrm{BF}_{3}\) and \(\mathrm{NH}_{3}\) (a) \(\mathrm{N}\) : tetrahedral, sp ; B: tetrahedral, sp \(^{3}\) (b) N: pyramidal, sp'; B: pyramidal, sp \(^{3}\) (c) N: pyramidal, sp \(^{3} ;\) B: planar, sp \(^{2}\) (d) \(\mathrm{N}\) : pyramidal, \(\mathrm{sp}^{3} ; \mathrm{B}:\) tetrahedral, \(\mathrm{sp}^{3}\)

Step-by-Step Solution

Answer

The coordination geometry and hybridization are (d): N: pyramidal, sp³; B: tetrahedral, sp³.

Step 1: Identify the initial molecules

Firstly, examine the simple structures of \( \mathrm{BF}_3 \) and \( \mathrm{NH}_3 \). \( \mathrm{BF}_3 \) is a planar molecule with \( \mathrm{B} \) being \( \mathrm{sp}^2 \) hybridized. \( \mathrm{NH}_3 \) is a pyramidal molecule with \( \mathrm{N} \) in an \( \mathrm{sp}^3 \) hybridization state.

Step 2: Coordinate Bond Formation

In the formation of a \(1:1\) complex, \( \mathrm{BF}_3 \) acts as a Lewis acid and \( \mathrm{NH}_3 \) acts as a Lewis base. The \( \mathrm{NH}_3 \) donates its lone pair to the \( \mathrm{B} \) atom of \( \mathrm{BF}_3 \), resulting in a coordinate covalent bond.

Step 3: Determine the Coordination Geometry and Hybridization of \( \mathrm{N}\)

The geometry around \( \mathrm{N} \) in \( \mathrm{NH}_3 \) remains pyramidal even after the formation of the complex because it maintains its \( \mathrm{sp}^3 \) hybridization with the formation of four sigma bonds including the lone pair.

Step 4: Determine the Coordination Geometry and Hybridization of \( \mathrm{B}\)

Upon accepting the lone pair from \( \mathrm{NH}_3 \), \( \mathrm{B} \) in \( \mathrm{BF}_3 \) gains an additional bond and transitions to a tetrahedral geometry. To accommodate four regions of electron density, \( \mathrm{B} \) becomes \( \mathrm{sp}^3 \) hybridized.

Step 5: Match the solution with the options provided

From the analysis, \( \mathrm{N} \) maintains a pyramidal shape with \( \mathrm{sp}^3 \) hybridization, and \( \mathrm{B} \), after the lone pair donation, becomes tetrahedral and \( \mathrm{sp}^3 \) hybridized. Compare these results to the given options: (a), (b), (c), and (d). The correct option is (d).

Key Concepts

Coordination Geometry

In coordination chemistry, coordination geometry describes the spatial arrangement of atoms around a central atom in a complex. It plays a crucial role in understanding how molecules are structured in space. For instance, in our example of the complex formation between \( \mathrm{BF}_3 \) and \( \mathrm{NH}_3 \):

• The \( \mathrm{BF}_3 \) begins as a planar molecule with a trigonal planar coordination around boron.
• When the lone pair from \( \mathrm{NH}_3 \) coordinates with boron, this arrangement changes.
• After coordination, \( \mathrm{B} \) becomes tetrahedral due to the addition of a fourth region of electron density.
• Meanwhile, the \( \mathrm{N} \) atom in \( \mathrm{NH}_3 \) retains its original pyramidal structure.

Understanding this helps in predicting molecular shapes and bond angles, which can affect properties like polarity and reactivity.

Hybridization Theory

Hybridization theory is essential in determining the geometry and bonding in coordination compounds like the \( 1:1 \) complex of \( \mathrm{BF}_3 \) and \( \mathrm{NH}_3 \). This theory describes how atomic orbitals mix to form new hybrid orbitals that can accommodate bonding pairs of electrons.

• Before donation of the lone pair, \( \mathrm{B} \) in \( \mathrm{BF}_3 \) is \( \mathrm{sp}^2 \) hybridized, corresponding to its planar geometry.
• Upon donation of the lone pair by \( \mathrm{NH}_3 \), the \( \mathrm{sp}^2 \) hybridization changes to \( \mathrm{sp}^3 \), allowing tetrahedral geometry.
• \( \mathrm{N} \), initially \( \mathrm{sp}^3 \) hybridized in \( \mathrm{NH}_3 \), maintains this hybridization as its tetrahedral electron-pair geometry isn’t altered, just one position is filled with a lone pair giving a pyramidal shape.

Recognizing these hybridization changes is pivotal in predicting the behavior and reactivity of complexes.

Lewis Acid-Base Reaction

The interaction between \( \mathrm{BF}_3 \) and \( \mathrm{NH}_3 \) in this example is a classic Lewis acid-base reaction, where bond formation occurs through electron pair donation from a base to an acid.

• \( \mathrm{BF}_3 \) acts as a Lewis acid, which is a compound that can accept a pair of electrons due to the empty orbital on \( \mathrm{B} \).
• \( \mathrm{NH}_3 \) functions as a Lewis base, characterized by the presence of a lone pair on nitrogen available for donation.
• When the lone pair from \( \mathrm{NH}_3 \) is donated to \( \mathrm{B} \) in \( \mathrm{BF}_3 \), a coordinate covalent bond is formed.

This fundamental reaction underpins much of coordination chemistry and explains the formation of complex species in solutions.

Complex Formation

Complex formation is a process where molecules or ions form a new, stable entity. In the \( 1:1 \) complex of \( \mathrm{BF}_3 \) and \( \mathrm{NH}_3 \), the formation of the complex involves the transfer of a lone electron pair, resulting in a stable structure.

• The \( \mathrm{BF}_3 \) molecule alone doesn’t complete an octet and hence seeks additional electron pairs.
• The \( \mathrm{NH}_3 \) molecule, with its available lone pair, is the perfect donor to complete the electron requirement of \( \mathrm{B} \).
• Once the complex is formed, \( \mathrm{B} \) now completes its oble gas configuration—a key to stabilizing the molecular structure.
• This process is central to many reactions in chemistry, particularly in catalysis where complexation enhances reactivity.

By understanding how complexes form, students can better grasp why certain molecules come together and how they behave in chemical reactions.