Question

Consider the following unbalanced reactions and choose the statements that are true from the options given below.
$\begin{aligned} & Zn+hot\text{ }conc.{{H}_{2}}S{{O}_{4}}\to G+R+X \\\ & Zn+conc.NaOH\to T+Q \\\ & G+{{H}_{2}}S+N{{H}_{4}}OH\to Z(precipitate)+X+Y \\\ \end{aligned}$
A. Bond order of $Q$ is 1 in its ground state
B. The oxidation state of $Zn$ in $T$ is +1
C. $R$ is a V-shaped molecule
D. $Z$ is dirty white in colour

Answer 1

Think about what the products of the reactions given will be before trying to determine whether the statements are true or not. Consider the conditions in which the reaction is occurring for the accurate product.

Complete step by step solution:
To solve this question, we will first have to just consider all the products of the reaction in a random order and we can only assign the variables to the products after careful analysis of the other reactions given. Once the variables are assigned to the products, we can verify their properties and determine whether the given options are true or not. We will look at the reactions one by one in the beginning.
- Reaction 1
$Zn+hot\text{ }conc.{{H}_{2}}S{{O}_{4}}\to G+R+X$
When a metal like zinc reacts with hot and concentrated sulphuric acid, it forms a sulphate with the formation of water and the evolution of a gas with a pungent odour. The products formed are zinc sulphide, sulphur dioxide and water. We will not assign the products to the variables right now but the reaction is:
$Zn+2{{H}_{2}}S{{O}_{4(conc)}}\to ZnS{{O}_{4}}+S{{O}_{2}}+2{{H}_{2}}O$
- Reaction 2
$Zn+conc.NaOH\to T+Q$
The reactivity of zinc is very less than that of sodium, so it cannot directly replace sodium and form a hydroxide compound. Instead, it forms sodium zincate and causes the evolution of hydrogen gas as the consequence of the reaction. So, the overall reaction will be:
$Zn+2NaOH\to N{{a}_{2}}Zn{{O}_{2}}+{{H}_{2}}$
Here, we can assign the variables as one of the options explicitly states that the compound $T$ contains zinc. Hence, $Q$ will be ${{H}_{2}}$.
- Reaction 3
$G+{{H}_{2}}S+N{{H}_{4}}OH\to Z(precipitate)+X+Y$
We will now have to assign the variables as some of the variables are repeated form the first reaction and we cannot determine the products of this reaction without the products of the first reaction. Out of the three products, sulphur dioxide and water do not react with hydrogen sulphide and ammonium hydroxide to give a precipitate so they are not the variable $G$. Only zinc sulphate can react with these reactants to form a precipitate. Hence, we can say that $G$ is $ZnS{{O}_{4}}$.
When zinc sulphate reacts with hydrogen sulphide and ammonium hydroxide it gives zinc sulphide as a major product and precipitates. It also produces the salt ammonium sulphate and water as the byproducts. The reaction is as follows:
$ZnS{{O}_{4}}+{{H}_{2}}S+2N{{H}_{4}}OH\to ZnS+{{(N{{H}_{4}})}_{2}}S{{O}_{4}}+2{{H}_{2}}O$
The only common product among reaction 1 and reaction 3 is water, so we can say that $X$ is ${{H}_{2}}O$. Now we can safely conclude that $R$ is $S{{O}_{2}}$.
Now, let us look at the statements given in the options.
- For option A
The bond order of $Q$ or ${{H}_{2}}$ is asked. We know that the bond order of any molecule is the difference between the number of electrons in the bonding and antibonding orbitals divided by 2. Hydrogen does not contain any electrons in its anti-bonding orbital and 2 in its bonding orbital. Hence, the bond order of ${{H}_{2}}$ in its ground state is 1 and this option is true.
- For option B
We need to consider the oxidation state of $Zn$ in $N{{a}_{2}}Zn{{O}_{2}}$ to answer this. We know that the fixed oxidation states of $Na$ and $O$ are +1 and -2. Since this is a neutral molecule, the oxidation state of $Zn$ turns out to be +2 and this option is false.
- For option C
For this option we need to know the geometry of the $S{{O}_{2}}$ molecule. The central sulphur atom has 1 lone pair present and this causes the general shape of the molecule to appear like a V if only the nuclei of the atoms are considered. So, this option is true.
- For option D
For this option, the colour of the precipitate in the third reaction needs to be considered. We know that the colour of the compound $ZnS$ is dirty white, so this is true.

Hence, the correct options for this question are A, C, and D.

Note: Remember that in the first reaction, the fact that zinc is reacting with hot and concentrated sulphuric acid needs to be taken into consideration. If zinc reacts with aqueous ${{H}_{2}}S{{O}_{4}}$ then only 2 products are produced, zinc sulphate and hydrogen gas. Not taking this into consideration will cause unnecessary confusion.