Consider the following reactions
$$\begin{aligned}\mathrm{Na_2B_4O_7} & \xrightarrow{\Delta} 2X + Y \\[6pt]\mathrm{CuSO_4} + Y &\xrightarrow{\text{Non-luminous flame}} Z + \mathrm{SO_3}\\[6pt]2Z + 2X + \mathrm{C}&\xrightarrow{\text{Luminous flame}} 2Q +\mathrm{Na_2B_4O_7} + \mathrm{CO}\end{aligned}$$
The oxidation states of Cu in Z and Q, respectively are:

The given reactions are analyzed step by step to determine the oxidation states of copper in compounds Z and Q.

Reaction 1: $$\mathrm{Na_2B_4O_7} \xrightarrow{\Delta} 2X + Y$$

Borax ($$\mathrm{Na_2B_4O_7}$$) decomposes on heating to form sodium metaborate and boron trioxide. The balanced equation is:

$$\mathrm{Na_2B_4O_7} \xrightarrow{\Delta} 2\mathrm{NaBO_2} + \mathrm{B_2O_3}$$

Thus, $$X = \mathrm{NaBO_2}$$ and $$Y = \mathrm{B_2O_3}$$.

Reaction 2: $$\mathrm{CuSO_4} + Y \xrightarrow{\text{Non-luminous flame}} Z + \mathrm{SO_3}$$

Substituting $$Y = \mathrm{B_2O_3}$$, the reaction is:

$$\mathrm{CuSO_4} + \mathrm{B_2O_3} \xrightarrow{\text{Non-luminous flame}} Z + \mathrm{SO_3}$$

In the borax bead test, copper sulfate reacts with boron trioxide to form copper metaborate. The balanced equation is:

$$\mathrm{CuSO_4} + \mathrm{B_2O_3} \to \mathrm{Cu(BO_2)_2} + \mathrm{SO_3}$$

Thus, $$Z = \mathrm{Cu(BO_2)_2}$$ (copper metaborate).

To find the oxidation state of Cu in Z:

Let the oxidation state of Cu be $$x$$. In $$\mathrm{Cu(BO_2)_2}$$, boron has an oxidation state of +3 (since it is in metaborate), and oxygen has an oxidation state of -2. The sum of oxidation states is zero:

$$x + 2 \times [\text{oxidation state of B} + 2 \times (\text{oxidation state of O})] = 0$$

$$x + 2 \times [+3 + 2 \times (-2)] = 0$$

$$x + 2 \times [3 - 4] = 0$$

$$x + 2 \times (-1) = 0$$

$$x - 2 = 0$$

$$x = +2$$

So, the oxidation state of Cu in Z is +2.

Reaction 3: $$2Z + 2X + \mathrm{C} \xrightarrow{\text{Luminous flame}} 2Q + \mathrm{Na_2B_4O_7} + \mathrm{CO}$$

Substituting $$Z = \mathrm{Cu(BO_2)_2}$$ and $$X = \mathrm{NaBO_2}$$:

$$2\mathrm{Cu(BO_2)_2} + 2\mathrm{NaBO_2} + \mathrm{C} \to 2Q + \mathrm{Na_2B_4O_7} + \mathrm{CO}$$

In the reducing luminous flame, copper is reduced. Balancing the equation:

Left side atoms: Cu = 2, B = (2 × 2) + (2 × 1) = 6, O = (2 × 4) + (2 × 2) = 12, Na = 2, C = 1.

Right side atoms: $$\mathrm{Na_2B_4O_7}$$ has Na=2, B=4, O=7; $$\mathrm{CO}$$ has C=1, O=1; so total O so far = 7 + 1 = 8, B=4. The remaining atoms: Cu=2, B=2 (since 6-4=2), O=4 (since 12-8=4) must be in 2Q.

Thus, 2Q contains 2Cu, 2B, and 4O, so Q is $$\mathrm{CuBO_2}$$ (copper metaborate).

To find the oxidation state of Cu in Q:

Let the oxidation state of Cu be $$y$$. In $$\mathrm{CuBO_2}$$, boron has an oxidation state of +3, and oxygen has an oxidation state of -2. The sum of oxidation states is zero:

$$y + \text{oxidation state of B} + 2 \times (\text{oxidation state of O}) = 0$$

$$y + 3 + 2 \times (-2) = 0$$

$$y + 3 - 4 = 0$$

$$y - 1 = 0$$

$$y = +1$$

So, the oxidation state of Cu in Q is +1.

Therefore, the oxidation states of Cu in Z and Q are +2 and +1, respectively.

The correct option is B. +2 and +1.