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Among the species A: \[CrC{{l}_{3}}\]; B: \[CuS\]; C: \[AlC{{l}_{3}}\]; D: \[ZnC{{l}_{2}}\]; which will be soluble in excess of NaOH?
(a)A,C and D
(b)C and D only
(c)B and C only
(d)A and D only

Last updated date: 13th Jun 2024
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Hint: Some of the inorganic compounds do not react with NaOH. Some when reacted with NaOH can produce inflammable gas. More polar metal salts can dissolve in excess of NaOH. d-block metal salts are more polar salts.

Complete step by step answer:
Chromium chloride when react with NaOH it gives chromium hydroxide and the salt NaCl.
When \[CrC{{l}_{3}}\] reacts with excess NaOH to give a complex and the salt NaCl.
\[CrC{{l}_{3}}+6NaOH\to Na[Cr{{(OH)}_{6}}]+3NaCl\]
Aluminium chloride reacts with NaOH to give aluminium hydroxide and salt.
When excess NaOH reacts with \[AlC{{l}_{3}}\] will give \[NaAl{{O}_{2}}\] and water. The reaction is given as
\[AlC{{l}_{3}}+3NaOH\to NaAl{{O}_{2}}+2{{H}_{2}}O\]
When \[ZnC{{l}_{2}}\] reacts with excess NaOH, it gives zinc hydroxide and a salt NaCl.
\[ZnC{{l}_{2}}+2NaOH\to Zn{{(OH)}_{2}}+2NaCl\]
Only CuS does not react with NaOH to give any hydroxide or complex. This is because copper itself is highly unreactive. It does not dissolve in NaOH until it is highly concentrated. Also CuS is less polar than the other options above.
Thus, only \[CrC{{l}_{3}}\] , \[AlC{{l}_{3}}\] and \[ZnC{{l}_{2}}\] react with excess of NaOH.

The correct answer for the question is option (a).

Additional Information:
 Sodium hydroxide or caustic soda has the ability to dissolve metals and convert them into their hydroxides. It is an ionic compound with a melting point of 604.4 degrees Fahrenheit. It dissolves extremely well in water. It reacts with acids to neutralise the solution and form salt and water. It is a highly strong base and its ions are sodium ion and hydroxide ion. It is mainly used as a cleaning agent.

In the question excess of NaOH is mentioned. So mostly compounds in excess of NaOH give complexes or oxide or hydroxide. This is due because NaOH is a polar solvent and is a strong base.