Home
Class 12
CHEMISTRY
Zn^(2+) ions are....... and..........

`Zn^(2+)` ions are....... and.......

Text Solution

AI Generated Solution

To answer the question regarding the properties of \( \text{Zn}^{2+} \) ions, we can follow these steps: ### Step-by-Step Solution: 1. **Identify the Oxidation State of Zinc:** - Zinc (Zn) typically has an oxidation state of +2 in its ionic form, which is represented as \( \text{Zn}^{2+} \). 2. **Determine the Electron Configuration of Zinc:** ...
Promotional Banner

Topper's Solved these Questions

  • d- AND- f- BLOCK ELEMENTS

    ICSE|Exercise EXERCISE (PART-I)(OBJECTIVE QUESTIONS)(THE CORRECT ALTERNATIVE FROM THE CHOICES GIVEN : )|27 Videos
  • d- AND- f- BLOCK ELEMENTS

    ICSE|Exercise EXERCISE (PART-I)(CORRECT THE FOLLOWING STATEMENTS BY CHANGING THE UNDERLINE PART OF THE SENTENCE : )|15 Videos
  • COORDINATION COMPOUNDS

    ICSE|Exercise ISC EXAMINATION QUESTION (PART- II DESCRIPTIVE QUESTIONS)|39 Videos
  • DISTINCTION BETWEEN PAIRS OF COMPOUNDS

    ICSE|Exercise QUESTIONS |155 Videos

Similar Questions

Explore conceptually related problems

In the zinc blende structure, zinc ions occupy alternate tetrahedral voids and S^(2-) ions exist as ccp the radii of Zn^(2+) and S^(2-) ions are 0.83Å and 1.74 Å respectively.The edge length of the ZnS unit cell is :

In the sphalerite (ZnS) structure S^(2-) ions form a face-centred cubic lattice. Then Zn^(2+) ions are present on the body diagonals at

In Zinc blende structure, Zn^(+2) ions are present in alternate tetrahedral voids and S^(-2) in ccp. The coordination number of Zn^(+2) and S^(-2) are respectively

Pb^(2+), Cu^(2+), Zn^(2+) and Ni^(2+) ions are present in a given acidic solution. On passing hydrogen sulphide gas through this solution, the available precipitate will contain

Why Zn^(2+) ions are colourless while Ni^(2+) ions are green and Cu^(2+) ions are blue in colour?

A: Zn (II) salts are diamagnetic R : Zn^(2+) ion has one unpaired electron.

Statement-I: In the Daniel cell, if concentration of Cu^(2+) and Zn^(2+) ions are doubled the emf of the cell will not change. Because Statement-II: If the concentration of ions in contact with the metals is doubled, the electrode potential is doubled.

A solution containing both Zn^(2+) and Mn^(2+) ions at a concentration of 0.01M is saturated with H_(2)S . What is pH at which MnS will form a ppt ? Under these conditions what will be the concentration of Zn^(2+) ions remaining in the solution ? Given K_(sp) of ZnS is 10^(-22) and K_(sp) of MnS is 5.6 xx 10^(-16), K_(1) xx K_(2) of H_(2)S = 1.10 xx 10^(-21) .

Assertion (A) : In a Daniell cell, if the concentration of Cu^(2+) and Zn^(2+) ions are doubled, the EMF of the cell will be doubled. Reason (R) : If the concentration of ions in contact with metals is doubled, the electrode potential is doubled. (a)If both (A) and (R) are correct, and (R) is the correct explanation of (A) . (b)If both (A) and (R) are correct, but (R) is not the correct explanation of (A) . (c)If (A) is correct, but (R) is incorrect. (d)If (A) is incorrect, (R) is correct.

What products are formed ? When : (i)Disodium hydrogen phosphate is added to magnesium sulphate in presence of ammonium chloride and aqueous ammonia. (ii)A solution containing Zn^(2+) ions is poured in an aqueous ammonia. (iii) Bi(NO_(3))_(3) solution is mixed with KI and then resulting precipitate is heated with water. (iv)Disodium hydrogen phosphate is boiled with concentrated HNO_(3) and ammonium molybdate reagent.