Question

Which of the following impart green colour to the burner flame? (a) \(\mathrm{Na}(\mathrm{OMe})\) (b) \(\mathrm{B}(\mathrm{OMe})_{3}\) (c) \(\mathrm{Sn}(\mathrm{OH})_{2}\) (d) \(\mathrm{Al}(\mathrm{OPR})_{3}\)

Short answer

(b) \(\mathrm{B}(\mathrm{OMe})_{3}\) imparts a green color to the flame.

Step-by-step solution

Understanding Flame Colors

When compounds are heated in a flame, they emit light at characteristic wavelengths, which are visible as different colors. Different elements impart different colors to a flame due to the excitement of their electrons. Our task is to find which compound imparts a green color.

Analyzing Element Contributions

Each compound contains different elements that might contribute to the flame color. Among the given options, compounds containing Boron (B) are known to impart a green color to a flame.

Identifying the Correct Compound

Looking at the options, (a) \(\mathrm{Na}(\mathrm{OMe})\), (b) \(\mathrm{B}(\mathrm{OMe})_{3}\), (c) \(\mathrm{Sn}(\mathrm{OH})_{2}\), (d) \(\mathrm{Al}(\mathrm{OPR})_{3}\), option (b) contains Boron, which is known for imparting a green color to a burner flame.

Key concepts

Flame Color

Every time you see a flame, you're witnessing a fascinating array of light and colors. The color of a flame depends on different elements that are being heated. When a substance burns, the heat energy boosts the electrons to higher energy levels. As these electrons return to their ground state, they emit light at specific wavelengths. This light is what we perceive as flame color.

When you heat a metal or compound in a flame, you'll observe a range of colors based on its metallic content. For example:

• Sodium (Na) gives off a bright yellow flame.
• Copper (Cu) burns with a green or blue flame.
• Potassium (K) emits a lilac or light purple flame.

This phenomenon is known as the flame test and it's a simple method often used in chemistry to detect the presence of certain metal ions based on these characteristic colors.

Boron Compounds

Boron is fascinating because of the unique color it imparts when burned. Compounds containing boron often exhibit a glowing green color in a flame. That's why in the original exercise, compound (b) \(\mathrm{B}(\mathrm{OMe})_{3}\), which contains boron, was noted to provide this distinctive green hue.

The unique green flame color of boron compounds is due to their specific electron configurations and the interplay of energy levels. When boron compounds such as boric acid or borate salts are heated, the energy excites the electrons to higher energy levels. As the electrons return to their original state, they release energy in the form of green light, which we see as the signature color of boron.

Electron Excitation

Electron excitation is a pivotal concept to understand why certain compounds emit light at different colors. When a compound is heated, the thermal energy excites the electrons, moving them from their original energy levels to higher ones.

This movement is temporary, and as electrons return to their stable states, they release energy in the form of light. The characteristics of this light depend on the energy difference between the excited and ground states. Each element has a unique set of energy levels, leading to different wavelengths of light being emitted.

• Larger energy differences lead to shorter wavelengths (towards violet).
• Smaller energy differences result in longer wavelengths (towards red).

This is why we see a variety of colors in a flame test, from red through green and blue, depending on the electrons' transitions in the atoms of different compounds. It's also a principle that finds use in applications like fireworks and spectroscopic analysis.