Advertisements
Advertisements
प्रश्न
A violet compound of manganese (A) decomposes on heating to liberate oxygen and compounds (B) and (C) of manganese are formed. Compound (C) reacts with KOH in the presence of potassium nitrate to give compound (B). On heating compound (C) with conc. \[\ce{H2SO4}\] and \[\ce{NaCl}\], chlorine gas is liberated and a compound (D) of manganese along with other products is formed. Identify compounds A to D and also explain the reactions involved.
Advertisements
उत्तर
The compounds A, B, C and D are given as under:
A = \[\ce{KMnO4}\]
B = \[\ce{K2MnO4}\]
C = \[\ce{MnO2}\]
D = \[\ce{MnCl2}\]
The reactions are explained as under:
\[\ce{\underset{(A)}{KMnO4} ->[Δ] \underset{(B)}{K2MnO4} + \underset{(C)}{MnO2} + O2}\]
\[\ce{MnO2 + KOH + O2 -> 2K2MnO4 + 2H2O}\]
\[\ce{MnO2 + 4NaCl + 4H2SO4 -> \underset{(D)}{MnCl2} + 2NaHSO4 + 2H2O + Cl2}\]
APPEARS IN
संबंधित प्रश्न
The elements of 3d transition series are given as: Sc Ti V Cr Mn Fe Co
Answer the following: Which element has the highest m.p?
Account for the following:
Cu+ ion is unstable in aqueous solution.
|
`E_((M^(2+)/M)` |
Cr | Mn | Fe | Co | Ni | Cu |
| -0.91 | -1.18 | -0.44 | -0.28 | -0.25 | -0.34 |
From the given data of E0 values, answer the following questions :
(1) Why is `E_(((Cu^(2+))/(Cu)))` value exceptionally positive
(2) Why is `E_(((Mn^(2+))/(Mn)))` value is highly negative as compared to other elements
(3) Which is the stronger reducing agents Cr2+ or Fe2+ ? Give Reason.
In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomization of zinc is the lowest, i.e., 126 kJ mol−1. Why?
Which of the d-block elements may not be regarded as the transition elements?
What are alloys?
How would you account for the following?
Zr (Z = 40) and Hf (Z = 72) have almost identical radii.
How would you account for the following?
Transition metals and their compounds act as catalysts.
Complete and balance the following chemical equations
`Fe^(2+) + MnO_4^(-) + H^+ ->`
Explain why Mn2+ is more stable than Fe2+ towards oxidation to +3 state. (At. no. of Mn = 25, Fe = 26)
Give reason for the following:
The transition metals generally form coloured compounds.
Which among the following transition metal has the lowest melting point?
Why EΘ values for Mn, Ni and Zn are more negative than expected?
Match the properties given in Column I with the metals given in Column II.
| Column I (Property) | Column II (Metal) | |
| (i) | Element with highest second ionisation enthalpy |
(a) \[\ce{Co}\] |
| (ii) | Element with highest third ionisation enthalpy |
(b) \[\ce{Cr}\] |
| (iii) | \[\ce{M}\] in \[\ce{M(CO)6}\] is | (c) \[\ce{Cu}\] |
| (iv) | Element with highest heat of atomisation |
(d) \[\ce{Zn}\] |
| (e) \[\ce{Ni}\] |
Answer the following question:
Which element of the first transition series has highest third ionisation enthalpy?
Transition metals can act as catalysts because these can change their oxidation state. How does \[\ce{Fe(III)}\] catalyse the reaction between iodide and persulphate ions?
Read the passage given below and answer the following question.
|
Are there nuclear reactions going on in our bodies? There are nuclear reactions constantly occurring in our bodies, but there are very few of them compared to the chemical reactions, and they do not affect our bodies much. All of the physical processes that take place to keep a human body running are chemical processes. Nuclear reactions can lead to chemical damage, which the body may notice and try to fix. The nuclear reaction occurring in our bodies is radioactive decay. This is the change of a less stable nucleus to a more stable nucleus. Every atom has either a stable nucleus or an unstable nucleus, depending on how big it is and on the ratio of protons to neutrons. The ratio of neutrons to protons in a stable nucleus is thus around 1 : 1 for small nuclei (Z < 20). Nuclei with too many neutrons, too few neutrons, or that are simply too big are unstable. They eventually transform to a stable form through radioactive decay. Wherever there are atoms with unstable nuclei (radioactive atoms), there are nuclear reactions occurring naturally. The interesting thing is that there are small amounts of radioactive atoms everywhere: in your chair, in the ground, in the food you eat, and yes, in your body. The most common natural radioactive isotopes in humans are carbon-14 and potassium-40. Chemically, these isotopes behave exactly like stable carbon and potassium. For this reason, the body uses carbon-14 and potassium-40 just like it does normal carbon and potassium; building them into the different parts of the cells, without knowing that they are radioactive. In time, carbon-14 atoms decay to stable nitrogen atoms and potassium-40 atoms decay to stable calcium atoms. Chemicals in the body that relied on having a carbon-14 atom or potassium-40 atom in a certain spot will suddenly have a nitrogen or calcium atom. Such a change damages the chemical. Normally, such changes are so rare, that the body can repair the damage or filter away the damaged chemicals. The natural occurrence of carbon-14 decay in the body is the core principle behind carbon dating. As long as a person is alive and still eating, every carbon-14 atom that decays into a nitrogen atom is replaced on average with a new carbon-14 atom. But once a person dies, he stops replacing the decaying carbon-14 atoms. Slowly the carbon-14 atoms decay to nitrogen without being replaced, so that there is less and less carbon-14 in a dead body. The rate at which carbon-14 decays is constant and follows first order kinetics. It has a half-life of nearly 6000 years, so by measuring the relative amount of carbon-14 in a bone, archeologists can calculate when the person died. All living organisms consume carbon, so carbon dating can be used to date any living organism, and any object made from a living organism. Bones, wood, leather, and even paper can be accurately dated, as long as they first existed within the last 60,000 years. This is all because of the fact that nuclear reactions naturally occur in living organisms. |
Why is Carbon-14 radioactive while Carbon-12 not? (Atomic number of Carbon: 6)
Assertion (A): Transition metals have high enthalpy of atomisation.
Reason (R): Greater number of unpaired electrons in transition metals results in weak metallic bonding.
Consider the following standard electrode potential values:
\[\ce{Sn^{2+}_{ (aq)} + 2e^- -> Sn_{(s)}}\]; E0 = −0.14 V
\[\ce{Fe^{3+}_{ (aq)} + e^- -> Fe^{2+}_{ (aq)}}\]; E0 = +0.77 V
What is the cell reaction and potential for the spontaneous reaction that occurs?
In order to protect iron from corrosion, which one will you prefer as a sacrificial electrode, Ni or Zn? Why? (Given standard electrode potentials of Ni, Fe and Zn are -0.25 V, -0.44 V and -0.76 V respectively.)
