Given below are two statements.
Statement I: Colourless cupric metaborate is reduced to cuprous meta

Q: Why is the luminous flame considered reducing?

A luminous flame burns with a yellow color because it contains unburnt carbon particles due to incomplete combustion. These carbon particles act as strong reducing agents, readily reducing metal oxides to lower oxidation states or even to the metallic state.

Q: Why is $Cu^+$ colourless while $Cu^{2+}$ is blue?

Color in transition metal complexes usually arises from d-d electron transitions. $Cu^+$ has a $3d^{10}$ electron configuration, meaning all its d-orbitals are completely filled. With no empty spaces for electrons to transition into, it cannot absorb visible light, making it colourless. $Cu^{2+}$, on the other hand, is $3d^9$, allowing for d-d transitions that absorb red/orange light and transmit blue.

Q: What exactly is the 'glassy bead' made of?

When borax ($Na_2B_4O_7 \cdot 10H_2O$) is heated strongly, it loses its water of crystallization and melts into a transparent, glass-like bead. This bead is a mixture of sodium metaborate ($NaBO_2$) and boric anhydride ($B_2O_3$).

Q: Can we get pure metallic copper in this test?

Yes! If you heat the copper bead in a highly reducing flame (rich in carbon) for a longer time, the cuprous metaborate can be further reduced to form red, opaque metallic copper.

Q: Why does copper sulphate react with boric anhydride?

In the hot flame, copper sulphate decomposes to form copper oxide ($CuO$) and sulfur trioxide ($SO_3$). The basic copper oxide then reacts with the acidic boric anhydride ($B_2O_3$) to form the salt, cupric metaborate ($Cu(BO_2)_2$).

Question

Given below are two statements.
Statement I: Colourless cupric metaborate is reduced to cuprous metaborate in a luminous flame.
Statement II: Cuprous metaborate is obtained by heating boric anhydride and copper sulphate in a non-luminous flame.
In the light of the above statements, choose the most appropriate answer from the options given below.

Options:

(A) Both statement I and statement II are true.

(B) Both statement I and statement II are false.

(C) Statement I is true but statement II is false.

(D) Statement I is false but statement II is true.

Sigma Percentile · Accepted Answer

### The Magic of the Borax Bead Test

Imagine you are a chemical detective, and your only tools are a Bunsen burner, a platinum wire, and a pinch of white powder. Welcome to the world of qualitative inorganic analysis! The Borax Bead Test is one of the most elegant and visually stunning methods used to identify transition metals.

Transition metals are famous for their vibrant colors, which arise from the intricate dance of electrons within their partially filled d-orbitals. By subjecting these metals to different thermal environments—specifically, oxidizing and reducing flames—we can force them to change their oxidation states, resulting in a spectacular shift in colors.

### The Anatomy of a Flame

Before we dive into the chemistry, we must understand our primary tool: the flame. A standard Bunsen burner flame has two distinct zones that act as chemical reagents:

**1. The Non-Luminous Flame (Oxidizing Zone):** This is the hot, blue, outer part of the flame where combustion is complete. Because there is an excess of oxygen, this flame acts as an **oxidizing agent**, pushing metals into their higher oxidation states.

**2. The Luminous Flame (Reducing Zone):** This is the cooler, yellow, inner part of the flame. The yellow color comes from glowing, unburnt carbon particles due to incomplete combustion. These carbon particles are hungry for oxygen and act as powerful **reducing agents**, forcing metals into their lower oxidation states.

### The Chemistry of the Bead

When we heat Borax ($Na_2B_4O_7 \cdot 10H_2O$) on a platinum wire loop, it first swells up as it loses its water of crystallization. Upon further heating, it melts into a clear, transparent, glass-like bead.

This glassy bead is a mixture of sodium metaborate and boric anhydride:
$$Na_2B_4O_7 \xrightarrow{\Delta} 2NaBO_2 + B_2O_3$$

The magic happens when we touch this hot bead to a tiny speck of a transition metal salt and put it back into the flame. The acidic boric anhydride ($B_2O_3$) reacts with the basic metal oxides to form colored metal metaborates.

### Analyzing Statement I: The Reducing Flame

Let's look at the first statement: *"Colourless cupric metaborate is reduced to cuprous metaborate in a luminous flame."*

There is a massive trap here! The statement correctly identifies that reduction happens in the luminous flame, but it completely botches the colors.

Cupric metaborate ($Cu(BO_2)_2$) contains copper in the $+2$ oxidation state ($Cu^{2+}$). The $Cu^{2+}$ ion has a $3d^9$ electron configuration. Because it has an unpaired electron, it can undergo d-d transitions, absorbing red light and appearing **beautifully blue**. It is absolutely not colourless!

When we place this blue bead into the luminous (reducing) flame, the unburnt carbon particles strip oxygen away, reducing the $Cu^{2+}$ to $Cu^+$:
$$2Cu(BO_2)_2 + 2NaBO_2 + C \xrightarrow{	ext{Luminous flame}} 2CuBO_2 + Na_2B_4O_7 + CO$$

The resulting product is cuprous metaborate ($CuBO_2$). Copper is now in the $+1$ oxidation state ($Cu^+$), which has a completely filled $3d^{10}$ configuration. With no room for d-d electron transitions, it cannot absorb visible light, making it **colourless**.

Because the statement falsely claims that cupric metaborate is colourless, **Statement I is entirely false.**

### Analyzing Statement II: The Oxidizing Flame

Now, let's dissect the second statement: *"Cuprous metaborate is obtained by heating boric anhydride and copper sulphate in a non-luminous flame."*

Remember our flame anatomy? The non-luminous flame is an **oxidizing flame**. It favors the formation of higher oxidation states.

When copper sulphate ($CuSO_4$) is heated with boric anhydride ($B_2O_3$) in this oxygen-rich environment, it forms cupric metaborate ($Cu(BO_2)_2$), where copper sits comfortably in its higher $+2$ oxidation state:
$$CuSO_4 + B_2O_3 \xrightarrow{	ext{Non-luminous flame}} Cu(BO_2)_2 + SO_3$$

As we established earlier, cupric metaborate is blue. The statement claims that cuprous metaborate (the $+1$ state) is formed, which contradicts the oxidizing nature of the non-luminous flame.

Therefore, **Statement II is also completely false.**

### The Final Verdict

Both statements attempt to trick you by swapping the colors and the oxidation states associated with the respective flames. In the Borax Bead Test, the oxidizing flame yields the blue cupric metaborate, while the reducing flame yields the colourless cuprous metaborate (or even red metallic copper if reduced further).

Since both statements are factually incorrect, the right choice is that both Statement I and Statement II are false.

Select Answer:

Visualized Solution

Borax Bead Test

Analyzing Statement I

Reaction in Luminous Flame

Analyzing Statement II

Reaction in Non-Luminous Flame

Final Conclusion

The Sigma Insight: d-block Elements

The Magic of the Borax Bead Test

The Anatomy of a Flame

The Chemistry of the Bead

Analyzing Statement I

Analyzing Statement II

The Final Verdict

Similar Questions

Which one of the following statements is correct?

Excess of $KI$ reacts with $CuSO_{4}$ solution and then $Na_{2} S_{2} O_{3}$ solution is added to it. Which of the following statements is incorrect for this reaction?

The incorrect statement(s) among (1)-(3) is (are) 1. W(VI) is more stable than Cr(VI). 2. in the presence of HCl, permanganate titrations provide satisfactory results. 3. some lanthanoid oxides can be used as phosphors.

Fusion of $MnO_{2}$ with $KOH$ in presence of $O_{2}$ produces a salt $W$ . Alkaline solution of $W$ upon eletrolytic oxidation yields another salt $X$ . The manganese containing ions present in $W$ and $X$ , respectively, are $Y$ and $Z$ . Correct statement(s) is (are)

A red solid is insoluble in water. However, it becomes soluble if some KI is added to water. Heating the red solid in a test tube results in liberation of some violet coloured fumes and droplets of a metal appear on the cooler parts of the test tube. The red solid is

If $Fe^{3 +}$ and $Cr^{3 +}$ both are present in group III of qualitative analysis, then distinction can be made by

Potassium permanganate on heating at 513 K gives a product which is

In the given chemical reaction, colours of the $Fe^{2 +}$ and $Fe^{3 +}$ ions, are respectively $5 Fe^{2 +} + MnO_{4}^{-} + 8 H^{+} ⟶ Mn^{2 +} + 4 H_{2} O + 5 Fe^{3 +}$

Select Answer:

Visualized Solution

Borax Bead Test

Analyzing Statement I

Reaction in Luminous Flame

Analyzing Statement II

Reaction in Non-Luminous Flame

Final Conclusion

The Sigma Insight: d-block Elements

The Magic of the Borax Bead Test

The Anatomy of a Flame

The Chemistry of the Bead

Analyzing Statement I

Analyzing Statement II

The Final Verdict

Similar Questions

Which one of the following statements is correct?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Excess of KI reacts with CuSO4​ solution and then Na2​S2​O3​ solution is added to it. Which of the following statements is incorrect for this reaction?

The incorrect statement(s) among (1)-(3) is (are) 1. W(VI) is more stable than Cr(VI). 2. in the presence of HCl, permanganate titrations provide satisfactory results. 3. some lanthanoid oxides can be used as phosphors.

A red solid is insoluble in water. However, it becomes soluble if some KI is added to water. Heating the red solid in a test tube results in liberation of some violet coloured fumes and droplets of a metal appear on the cooler parts of the test tube. The red solid is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If Fe3+ and Cr3+ both are present in group III of qualitative analysis, then distinction can be made by

Potassium permanganate on heating at 513 K gives a product which is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In the given chemical reaction, colours of the Fe2+ and Fe3+ ions, are respectively 5Fe2++MnO4−​+8H+⟶Mn2++4H2​O+5Fe3+

Excess of $KI$ reacts with $CuSO_{4}$ solution and then $Na_{2} S_{2} O_{3}$ solution is added to it. Which of the following statements is incorrect for this reaction?

If $Fe^{3 +}$ and $Cr^{3 +}$ both are present in group III of qualitative analysis, then distinction can be made by

In the given chemical reaction, colours of the $Fe^{2 +}$ and $Fe^{3 +}$ ions, are respectively $5 Fe^{2 +} + MnO_{4}^{-} + 8 H^{+} ⟶ Mn^{2 +} + 4 H_{2} O + 5 Fe^{3 +}$