Question

An orange solid X on heating gives a colourless gas Y and an only green residue Z. gas Y on treatment with Mg produces a greenish-yellow solid substance…….
(A)- $M{{g}_{3}}{{N}_{2}}$
(B)- $MgO$
(C)- $M{{g}_{2}}{{O}_{3}}$
(D)- $MgC{{l}_{2}}$

Answer 1

The approach to answer this question is to follow the leads given in the question in the given sequence as
$X(orange)\xrightarrow{\Delta }Y(gas)\,+\,Z(green)$
$Y\,+\,Mg\to \,?(greenish-yellow\,solid)$

Complete step by step solution:

Let’s look at the answer:
It is given that X is an orange coloured compound and we know that salts of dichromate are orange coloured.
Therefore X is ${{(N{{H}_{4}})}_{2}}C{{r}_{2}}{{O}_{7}}$
Now, on heating ${{(N{{H}_{4}})}_{2}}C{{r}_{2}}{{O}_{7}}$, it decomposes into chromium oxide, dinitrogen gas and water vapour
The reaction is s follows:
${{(N{{H}_{4}})}_{2}}C{{r}_{2}}{{O}_{7}}\xrightarrow{\Delta }\,C{{r}_{2}}{{O}_{3}}\,+\,{{N}_{2}}\,+\,2{{H}_{2}}O$
Where, $C{{r}_{2}}{{O}_{3}}$ is the green solid and ${{N}_{2}}$is the colourless gas.
Mg on reaction with ${{N}_{2}}$ gives a special compound magnesium nitride ($M{{g}_{3}}{{N}_{2}}$)
The reaction is as follows:
$3Mg\,+\,{{N}_{2}}\,\to \,M{{g}_{3}}{{N}_{2}}$
$M{{g}_{3}}{{N}_{2}}$ is an inorganic compound. It is greenish yellow in colour at room temperature and pressure. It is used in the synthesis of borazon as a catalyst.
Hence, the answer to the given question is option (A).

Additional Information: Salts of Cr are colourful due to the presence of unpaired electrons in the d-orbital of chromium. These d-d electronic transitions impart colours to the salts of Cr.

Note: Mg is an element of the second group and of third period. It shows diagonal properties with Li. It shows many properties which are different from the other elements of the period. This is because of its small size and high ionization enthalpy. Only Mg reacts with ${{N}_{2}}$ to give $M{{g}_{3}}{{N}_{2}}$.