Match List-I with List-II.

List-I (Salt)
A. LiCl
B. NaCl
C. RbCl
D. CsCl

List-II (Flame colour wav

Q: Why do only some metals show a color in the flame test?

Metals with low ionization energies (like alkali and alkaline earth metals) have valence electrons that can be easily excited by the relatively low thermal energy of a Bunsen flame. Transition metals often require much higher energies to excite their electrons, so they don't always show a visible flame color.

Q: Does the anion (like Chloride or Nitrate) affect the flame color?

No, the color is strictly characteristic of the metal cation. However, chlorides are most commonly used in flame tests because they are highly volatile, meaning they vaporize easily in the flame, producing a clearer and brighter color.

Q: Why does Cesium have the shortest wavelength if it's at the bottom of the group?

While ionization energy decreases down the group, the specific electronic transition that emits visible light in Cesium ($6p \rightarrow 6s$) involves a larger energy gap compared to the prominent visible transitions in the other metals, resulting in a higher energy (blue) photon with a shorter wavelength.

Q: What is the flame color of Potassium?

Potassium imparts a pale violet or 'Lilac' color to the flame. It is often masked by the intense yellow of Sodium impurities, so it is sometimes viewed through cobalt blue glass to filter out the yellow light.

Q: Do I need to memorize the exact nanometer values for JEE?

Not necessarily. If you remember the colors (Crimson, Yellow, Red-Violet, Blue) and their relative order in the VIBGYOR spectrum, you can easily match them to the given numerical wavelengths by sorting them from shortest (Blue) to longest (Red).

Question

Match List-I with List-II.

\begin{array}{ll} 	extbf{List-I (Salt)} & 	extbf{List-II (Flame colour wavelength)} \ 	ext{A. LiCl} & 	ext{1. } 455.5 	ext{ nm} \ 	ext{B. NaCl} & 	ext{2. } 670.8 	ext{ nm} \ 	ext{C. RbCl} & 	ext{3. } 780.0 	ext{ nm} \ 	ext{D. CsCl} & 	ext{4. } 589.2 	ext{ nm} \end{array}

Choose the correct answer from the options given below.

Options:

(A) A $ightarrow$ 4, B $ightarrow$ 2, C $ightarrow$ 3, D $ightarrow$ 1

(B) A $ightarrow$ 2, B $ightarrow$ 1, C $ightarrow$ 4, D $ightarrow$ 3

(C) A $ightarrow$ 1, B $ightarrow$ 4, C $ightarrow$ 2, D $ightarrow$ 3

(D) A $ightarrow$ 2, B $ightarrow$ 4, C $ightarrow$ 3, D $ightarrow$ 1

Sigma Percentile · Accepted Answer

### The Magic of the Flame Test

Have you ever wondered why fireworks burst into brilliant reds, yellows, and blues? The secret lies in the atomic structure of the elements used, and the **flame test** is the perfect laboratory demonstration of this phenomenon.

Alkali metals, which belong to Group 1 of the periodic table, are famous for their highly reactive nature. But they also have another defining characteristic: they possess very low ionization enthalpies. This means their outermost valence electron is loosely bound to the nucleus.

### The Physics of Emission

When an alkali metal salt (like a chloride) is introduced into a hot Bunsen burner flame, the heat energy is absorbed by the atoms. This thermal energy is sufficient to excite the loosely bound valence electron from its ground state to a higher, unstable energy level.

Because the excited state is unstable, the electron quickly drops back down to its original ground state. As it falls, it must release the extra energy it absorbed. It does this by emitting a photon of light. The energy of this photon ($E$) is directly related to its wavelength ($\lambda$) by the famous Planck-Einstein relation:

$$E = \frac{hc}{\lambda}$$

Since the energy gaps between the atomic orbitals are unique for every element, the wavelength of the emitted light is also unique. This is why different metals produce different, characteristic colors!

### Decoding the Colors

Let's break down the specific metals given in our problem:

**1. Lithium (Li):** Lithium imparts a beautiful **Crimson Red** color to the flame. In the visible spectrum, red light has a relatively low energy and a longer wavelength. The characteristic wavelength for Lithium is exactly **$670.8 	ext{ nm}$**.

**2. Sodium (Na):** If you've ever seen the warm glow of old street lamps, you've seen Sodium in action. It produces an intense, persistent **Yellow** flame. The wavelength corresponding to this yellow emission is **$589.2 	ext{ nm}$**.

**3. Rubidium (Rb):** The name Rubidium is derived from the Latin word *rubidus*, meaning deepest red. True to its name, it imparts a **Red-Violet** color to the flame. This corresponds to the longest wavelength among our options, which is **$780.0 	ext{ nm}$**.

**4. Cesium (Cs):** Cesium gets its name from the Latin word *caesius*, meaning sky blue. It burns with a brilliant **Blue** flame. Blue light is highly energetic, which means it has a much shorter wavelength. For Cesium, this is **$455.5 	ext{ nm}$**.

### The Final Match

By pairing the metals with their characteristic wavelengths, we get:
- $	ext{LiCl}$ (A) $ightarrow$ $670.8 	ext{ nm}$ (2)
- $	ext{NaCl}$ (B) $ightarrow$ $589.2 	ext{ nm}$ (4)
- $	ext{RbCl}$ (C) $ightarrow$ $780.0 	ext{ nm}$ (3)
- $	ext{CsCl}$ (D) $ightarrow$ $455.5 	ext{ nm}$ (1)

This perfectly aligns with option (d). Memorizing these characteristic colors is a surefire way to secure easy marks in inorganic chemistry!

Select Answer:

Visualized Solution

\text{The Flame Test}

\text{Mechanism of Emission}

\text{Lithium (Li)}

\text{Sodium (Na)}

\text{Rubidium (Rb)}

\text{Cesium (Cs)}

\text{Final Match}

\text{The Way Forward}

The Sigma Insight: Alkali Metals

The Magic of the Flame Test

The Physics of Emission

Decoding the Colors

The Final Match

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