Class 11th

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.

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.

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

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.

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⁻.

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

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

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (ii) is the correct answer.

The oxidation states are given below-

In NH₂OH oxidation of N is -1.

NH₄NO₃ exists as NH₄⁺.NO₃^-,    

Thus, the oxidation state of N in NH₄⁺ is -3 while in NO₃^? is +5.

In N₂H₄ oxidation of N is -2

In N₃H oxidation of N is $\frac{-1}{3}$

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (i) is the correct answer

In ionic hydrides hydrogen exists in -1 oxidation state because the hydrogen acquires negative charge in the presence of its companion.