An inorganic compound (A) is a strong reducing agent. Its hydrolysis in water gives a white turbidity (B). An aqueous solution of. (A) Gives a white precipitate (C) with NaOH solution which is soluble in excess of NaOH. (A) reduces auric chloride to produce purple of cassius, (A) also reduces iodine and gives chromyl chloride test. Find a total number of OH groups present in one molecule of product.
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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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