Home
Class 12
CHEMISTRY
Assertion: Zn(OH)(2) dissolves in an exc...

Assertion: `Zn(OH)_(2)` dissolves in an excess of NaOH solution as well as `NH_(4)OH` solution.
Reason: `Zn(OH)_(2)` forms the soluble zincate salts with these alkalies.

A

If both (A) and (B) are correct and (R ) is the correct explqanation of (A)

B

If both (A) and (B) are correct but (R ) is not the correct explqanation of (A)

C

If (A) is correct ,but (R ) is incorrect

D

If (A) is incorrect ,but (R ) is correct

Text Solution

Verified by Experts

The correct Answer is:
c
Correct Reason (R )
`Zn(OH)_(2)` form the insoluble zincate salt with alkline
`Zn(OH)_(2) + 2NaOH rarr underset("insoluble sodium zincate")(Na_(2)ZnO_(2))`
Promotional Banner

Topper's Solved these Questions

  • QUALITATIVE INORGANIC SALT ANALYSIS

    CENGAGE CHEMISTRY ENGLISH|Exercise Exercises (Integer) (Naming And Terminology|15 Videos

  • QUALITATIVE INORGANIC SALT ANALYSIS

    CENGAGE CHEMISTRY ENGLISH|Exercise Exercises (Fill In The Blanks)|42 Videos

  • QUALITATIVE INORGANIC SALT ANALYSIS

    CENGAGE CHEMISTRY ENGLISH|Exercise Exercises (Single Correct) Part-D (Miscellaneous)|6 Videos

  • P-BLOCK GROUP 18 ELEMENTS - THE INERT GASES

    CENGAGE CHEMISTRY ENGLISH|Exercise Ex 5.1 (Objective)|14 Videos

  • REDUCTION AND OXIDATION REACTION OF ORGANIC COMPOUNDS

    CENGAGE CHEMISTRY ENGLISH|Exercise SUBJECTIVE TYPE|4 Videos

Similar Questions

Explore conceptually related problems

Assertion: Be(OH)_(2) dissolves in excess NaOH solution. Reason: Be(OH)_(2) is an acidic compound.

A : Zn dissolve in excess of NaOH solution so as to give H_(2) R : Zn(OH)_(2) is neutral is nature.

If group IV the ppt of Zn(OH)_(2) dissolve in excess of NaOH due to the formation of

White precipitate of Zn(OH)_(2) dissolves in:

Assertion: Zn(OH)_(2) is dissolved in both NH_(4)OH and NaOH solution Reason- NaOH and NH_(4)OH being basic can dissolve amphoteric Zn(OH)_(2) .

NH_4OH is highly soluble in

Assertion : Cu(OH))_(2) is soluble in NH_(4)OH but not in NaOH . Reason : Cu(OH)_(2) forms a soluble complex with NH_(3) .

What happen when Zn(OH)_2 is treated with excess of NaOH ?

Reaction of Zn(OH)_(2) with NaOH produces:

Assertion (A): Be(OH)_(2) is soluble in NaOH . Reason (R ): Be(OH)_(2) is amphoteric in nature.