Chemistry NCERT Exemplar Solutions Class 11th Chapter Eight
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New answer posted
4 months agoContributor-Level 10
This is a Matching Type Questions as classified in NCERT Exemplar
(i) → (d); (ii) → (e); (iii) → (c); (iv) → (a)
New answer posted
4 months agoContributor-Level 10
This is a Multiple Choice Questions as classified in NCERT Exemplar
Option (i) and (ii) are the correct answers.
The ones which will act as anodes when connected to standard hydrogen electrode as they have more negative standard reduction potential as compared to standard hydrogen electrode. The one which will act as cathodes when connected to standard hydrogen electrode as they have more positive standard reduction potential as compared to standard hydrogen electroDE
New answer posted
4 months agoContributor-Level 10
This is a Multiple Choice Questions as classified in NCERT Exemplar
Hence, option (iii) and (iv) are the correct answers.
The given reaction is as below-
P4+3OH- +3H2O→PH3 + 3H2PO2-
The above reaction is a kind of disproportionate reaction in which phosphorous is being reduced as well as oxidized whereas hydrogen remains same in +1 oxidation state.
New answer posted
4 months agoContributor-Level 10
This is a Multiple Choice Questions as classified in NCERT Exemplar
Option (ii), (iii) and (iv) are the correct answers.
Elements that are having only s-electrons in the valence shell do not show more than one oxidation state (shows only one oxidation state of +1). Hence, (b), (c) having incompletely filled d-orbital's in the outermost shell show variable oxidation states. Element with outer electronic configuration as 3d1 4S2 shows variable oxidation states of +2 and +3 and the element with outer electronic configuration as 3d24S2 shows variable oxidation states of +2, +3 and +4. P-Block elements also show va
New answer posted
4 months agoContributor-Level 10
This is a Multiple Choice Questions as classified in NCERT Exemplar
Option (iii) and (iv) is the correct answer.
The given equation is as-
Zn+2HCl→ZnCl2+H2
In this Zn has a negative E? value, which means it will undergo oxidation and will act as a reducing agent (reductant). Zn can produce H2 gas with HCl, as hydrogen has higher standard reduction potential than Zn, hydrogen will undergo reduction and will act as oxidant.
New answer posted
4 months agoContributor-Level 10
This is a Multiple Choice Questions as classified in NCERT Exemplar
Options (iii) and (iv) are the correct answers.
As, in the given reaction below-
2KClO3→2KCl+3O2
Potassium remains in same oxidation state and oxygen is being oxidized
New answer posted
4 months agoContributor-Level 10
This is a Multiple Choice Questions as classified in NCERT Exemplar
Option (iii) is the correct answer.
As, Cl, Br, I all are having -1 to +7 oxidation state. But Oxidation state of F is fixed (-1) as it is the most electronegative element and do not loose electron. Hence, it does not show disproportionate tendency.
New answer posted
4 months agoContributor-Level 10
This is a Multiple Choice Questions as classified in NCERT Exemplar
Option (iv) is the correct answer.
Disproportionate reactions are defined as the reactions in which the same substance is oxidized as well as reduced. Here, the below reaction is given as-
2NO2 + 2OH- →NO2 -+ NO3- +H2O
In this reaction, N is both oxidized as well as reduced since O.N. of N increases from +4 in NO3−? to +5 in NO2 ? and decreases from +4 in NO to +3 in NO2−.
New answer posted
4 months agoContributor-Level 10
This is a Multiple Choice Questions as classified in NCERT Exemplar
Option (iv) is the correct answer
For 3d1 4s2 can exhibit the highest oxidation state as 2+1 = +3,
For 3d34s2can exhibit the highest oxidation state as 2+3 = +5
For 3d54s1can exhibit the highest oxidation state as 1+5 = +6
For 3d54s2can exhibit the highest oxidation state as 2+5 = +7
New answer posted
4 months agoContributor-Level 10
This is a Multiple Choice Questions as classified in NCERT Exemplar
Option (i) is the correct answer
Let the oxidation number of central atom be 'y'.
Oxidation number of O = −2
CrO2−: y + 2 * (−2) = −1
y − 4 = −1 or y = +3
CrO42−? : y + 4 * (−2) = −2
y − 8 = −2 or y = +6
ClO3−: y + 3 * (−2) = −1
y − 6 = −1 or y = +5
MnO4−: y + 4 * (−2) = −1
y − 8 = −1 or y = +7
increasing order of oxidation number of the central atom is:
CrO2 – , ClO3 –, CrO4 2–, MnO4
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