Answer
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Hint: To recall the definition of the reducing agent which says that those reagents in a chemical reaction which themselves oxidizes or loses the electrons in a redox reaction.
Complete step by step solution:
Given in the question: Inorganic compound is a strong reducing agent which on hydrolysis gives a white turbidity and the compound A also reduces iodine and gives chromyl chloride test and it forms white precipitate with NaOH and reduces auric chloride to produce purple of cassius. The reactions involved are mentioned below:
The compound A is $SnC{{l}_{2}}$ , and the hydrolysis of $SnC{{l}_{2}}$is mentioned below:
$SnC{{l}_{2}}+{{H}_{2}}O\to \underset{(B)}{\mathop{Sn(OH)Cl}}\,+HCl$
And the compound B formed is $\underset{(B)}{\mathop{Sn(OH)Cl}}\,$ Now the compound a forms white precipitate when reacts with excess NaOH. The reaction is mentioned below:
$SnC{{l}_{2}}+NaOH\to Sn(\underset{(C)}{\mathop{OH{{)}_{2}}}}\,+HCl$
The compound C $Sn(\underset{(C)}{\mathop{OH{{)}_{2}}}}\,$when dissolved in excess of NaOH it gets dissolved. The reaction is mentioned below.
$Sn(\underset{(C)}{\mathop{OH{{)}_{2}}}}\,+2NaOH\to N{{a}_{2}}Sn{{O}_{2}}+2{{H}_{2}}O$
The compound (A) reduces auric chloride to produce purple of cassius the reaction is mentioned below.\[3SnC{{l}_{3}}+2AuC{{l}_{3}}\to 3SnC{{l}_{4}}+2Au\]
And the compound (A) also reduces iodine and gives chromyl chloride test the reaction is mentioned below:
\[3SnC{{l}_{2}}+2HCl+{{I}_{2}}\to SnC{{l}_{4}}+2HI\]
Thus the product (C) contains 2 OH groups.
Note: It is important to know about the basic terms mentioned in the question as Hydrolysis. It is a chemical reaction in which a molecule of water breaks down one or more chemical bonds and leads to formation of new compounds.
Complete step by step solution:
Given in the question: Inorganic compound is a strong reducing agent which on hydrolysis gives a white turbidity and the compound A also reduces iodine and gives chromyl chloride test and it forms white precipitate with NaOH and reduces auric chloride to produce purple of cassius. The reactions involved are mentioned below:
The compound A is $SnC{{l}_{2}}$ , and the hydrolysis of $SnC{{l}_{2}}$is mentioned below:
$SnC{{l}_{2}}+{{H}_{2}}O\to \underset{(B)}{\mathop{Sn(OH)Cl}}\,+HCl$
And the compound B formed is $\underset{(B)}{\mathop{Sn(OH)Cl}}\,$ Now the compound a forms white precipitate when reacts with excess NaOH. The reaction is mentioned below:
$SnC{{l}_{2}}+NaOH\to Sn(\underset{(C)}{\mathop{OH{{)}_{2}}}}\,+HCl$
The compound C $Sn(\underset{(C)}{\mathop{OH{{)}_{2}}}}\,$when dissolved in excess of NaOH it gets dissolved. The reaction is mentioned below.
$Sn(\underset{(C)}{\mathop{OH{{)}_{2}}}}\,+2NaOH\to N{{a}_{2}}Sn{{O}_{2}}+2{{H}_{2}}O$
The compound (A) reduces auric chloride to produce purple of cassius the reaction is mentioned below.\[3SnC{{l}_{3}}+2AuC{{l}_{3}}\to 3SnC{{l}_{4}}+2Au\]
And the compound (A) also reduces iodine and gives chromyl chloride test the reaction is mentioned below:
\[3SnC{{l}_{2}}+2HCl+{{I}_{2}}\to SnC{{l}_{4}}+2HI\]
Thus the product (C) contains 2 OH groups.
Note: It is important to know about the basic terms mentioned in the question as Hydrolysis. It is a chemical reaction in which a molecule of water breaks down one or more chemical bonds and leads to formation of new compounds.
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