X→N2⋅∆Y→H2OZColourless Gas ↓CuSO4 TBlue Colour Then, substances Y and Tare :

The reaction is,3Mg→N2,ΔMg3N2H2O⟶Mg(OH)2+NH3(colourless gas)→CuSo4CuSO4.4NH3Thus the Y=MgO and T=CuSO4⋅4NH3.

X→N2⋅∆Y→H2OZColourless Gas ↓CuSO4 TBlue Colour Then, substances Y and Tare :

The reaction is,3Mg→N2,ΔMg3N2H2O⟶Mg(OH)2+NH3(colourless gas)→CuSo4CuSO4.4NH3Thus the Y=MgO and T=CuSO4⋅4NH3.

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$X \overset{N_{2} \cdot ∆}{\to} Y \overset{H_{2} O}{\to} \underset{Colourless Gas}{Z} ↓ {CuSO}_{4}$
$\underset{Blue Colour}{T}$

Then, substances Y and Tare :

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$Y = {Mg}_{3} N_{2}$ and $T = {CuSO}_{4} \cdot 5 H_{2} O$

$Y = Mg {({NO}_{3})}_{2}$ and $T = CuO$

$Y = {MgO}_{}$ and $T = {CuSO}_{4} \cdot 4 {NH}_{3}$

$Y = {Mg}_{3} N_{2}$ and $T = {CuSO}_{4} \cdot 4 {NH}_{3}$

answer is B.

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Detailed Solution

The reaction is,

$3 Mg \overset{N_{2}, Δ}{\to} {Mg}_{3} N_{2}$

$H_{2} O ⟶ Mg (OH)_{2} + {NH}_{3}$ (colourless gas)

$\overset{{CuSo}_{4}}{\to} {CuSO}_{4} . 4 {NH}_{3}$

Thus the $Y = {MgO}_{}$ and $T = {CuSO}_{4} \cdot 4 {NH}_{3}$ .

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X→N2⋅∆Y→H2OZColourless Gas ↓CuSO4 TBlue Colour Then, substances Y and Tare :

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$X \overset{N_{2} \cdot ∆}{\to} Y \overset{H_{2} O}{\to} \underset{Colourless Gas}{Z} ↓ {CuSO}_{4}$
$\underset{Blue Colour}{T}$

Then, substances Y and Tare :