Question

Oxidation state of central metal of \(Z\) and \(Q\) are:

• A. \(+2\) and \(+1\)
• B. \(+1\) and \(+2\)
• C. \(+2\) and \(+2\)
• D. \(+1\) and \(+1\)

Hint:

In borax bead test: {non-luminous flame} favors higher oxidation state, while {luminous flame} favors lower oxidation state.

Answer 1

The Correct Option is A

Concept:
Borax bead test is used to identify metal ions based on color changes and oxidation states in different flames.

Heating borax produces boric oxide and sodium metaborate.
Transition metal salts form colored metaborates.
Oxidation state of metal depends on flame conditions.

Step 1: Decomposition of Borax
\[ \mathrm{Na_2B_4O_7\cdot 10H_2O} \xrightarrow{\Delta} 2\,\mathrm{NaBO_2} + \mathrm{B_2O_3} \] Thus: \[ X = \mathrm{NaBO_2}, \quad Y = \mathrm{B_2O_3} \]
Step 2: Reaction with Copper Sulphate
\[ \mathrm{CuSO_4 + NaBO_2} \xrightarrow{\text{non-luminous flame}} \mathrm{Cu(BO_2)_2} \ (Z) \] In \(Z\), copper exists as \(\mathrm{Cu^{2+}}\).
Step 3: Heating in Luminous Flame
\[ Z \xrightarrow{\text{luminous flame}} \mathrm{Cu_2O} \ (Q) \] In \(Q = \mathrm{Cu_2O}\), copper is in \(+1\) oxidation state.
Final Conclusion:
\[ \boxed{\text{Oxidation state in } Z = +2,\quad \text{Oxidation state in } Q = +1} \]