Question

The amide anion \(\mathrm{NH}_{2}^{-}\) is a very strong base. On the basis of molecular orbital theory, would you expect \(\mathrm{NH}_{2}^{-}\) to be linear or bent?

Short answer

The \(\mathrm{NH}_{2}^{-}\) anion is expected to have a bent structure.

Step-by-step solution

Understand Molecular Orbitals and Electron Distribution

A molecular orbital is a region in a molecule where there's a high probability of finding an electron. In an \(\mathrm{NH}_{2}^{-}\) molecule, there's a total of 8 valence electrons. Nitrogen atom has 5 valence electrons and each Hydrogen atom contributes one, the extra electron makes it eight.

Apply Molecular Orbital Theory to \(\mathrm{NH}_{2}^{-}\) structure

The \(\mathrm{NH}_{2}^{-}\) molecule forms when one extra electron is added to the Nitrogen atom, making it an anion. This extra electron will go to the non-bonding orbital of nitrogen. Nitrogen atom forms two bond pairs with the two Hydrogen atoms, and it has one non-bonding electron pair.

Determine the Structure of \(\mathrm{NH}_{2}^{-}\)

Non-bonding electron pairs require more space than bonding pairs. Therefore, the \(\mathrm{NH}_{2}^{-}\) anion prefers a shape that minimizes electron pair repulsion and ensures maximum stability. In this case, it's a 'bent' or 'V-shaped' geometry, not linear.

Key concepts

Amide Anion \( \mathrm{NH}_{2}^{-} \)

The amide anion, \( \mathrm{NH}_{2}^{-} \), is an important ion in chemistry known for its strong basic properties. It consists of one nitrogen atom and two hydrogen atoms, along with an additional electron that gives it a negative charge. This extra electron is crucial because it significantly affects the molecule's shape and reactivity. In chemical terms, the presence of a negative charge indicates that the amide anion has an additional electron compared to its neutral form. This additional electron occupies a non-bonding molecular orbital, altering the standard shape you might expect in its absence. Given the molecular orbital theory, which helps in predicting the structure of molecules, the electron configuration greatly influences the amide anion's geometry.

Valence Electrons

Valence electrons are the outermost electrons of an atom and play a critical role in chemical bonding. For the amide anion \( \mathrm{NH}_{2}^{-} \), understanding the number and distribution of valence electrons helps predict its shape and reactivity. Let's break down the contributions:

• Nitrogen, being from the fifth group of the periodic table, has 5 valence electrons.
• Each hydrogen atom contributes 1 valence electron.
• The extra electron (accounting for the negative charge) adds 1 more.

Combining these, the total is 8 valence electrons in \( \mathrm{NH}_{2}^{-} \). These electrons arrange themselves to minimize energy and pair repulsion effects, leading to a stable configuration for the molecule.

Electron Pair Repulsion

Electron pair repulsion is a key concept in molecular geometry, dictating how atomic and electron groups arrange themselves in space. The fundamental idea is that electron pairs, being like-charges, repel each other and distribute around a central atom to minimize these repulsions.For \( \mathrm{NH}_{2}^{-} \), the nitrogen atom has two bonding pairs from the hydrogen atoms and one non-bonding or lone pair of electrons. This lone pair is especially significant as it often occupies more space than a bonding pair, due to the absence of a second atom that would help to constrain it.Therefore, this repulsion causes the molecule to adopt a bent or V-shaped structure, rather than a linear one. This bent shape, achieved by maximizing the distance between these electron pairs, helps to stabilize the molecule by minimizing repulsion and achieving a lower energy state.