d and f Block Elements — Page 20 | JEE Chemistry

Q191

The dark purple colour of

\mathrm{KMnO}_{4}

disappears in the titration with oxalic acid in acidic medium. The overall change in the oxidation number of manganese in the reaction is :

A 5

B 1

C 7

D 2

Correct Answer

Option A

Solution

$2 \overset{+7}{\mathrm{KMnO}_{4}}+5 \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}+3 \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow$

\mathrm{K}_{2} \mathrm{SO}_{4}+2 \overset{+2}{\mathrm{MnSO}_{4}}+10 \mathrm{CO}_{2}+8 \mathrm{H}_{2} \mathrm{O}

Change is oxidation state $M n$ is 5.

Q192

The total number of

\mathrm{Mn}=\mathrm{O}

bonds in

\mathrm{Mn}_{2} \mathrm{O}_{7}

is __________.

A 4

B 5

C 6

D 3

Correct Answer

Option C

Solution

Structure of $\mathrm{Mn}_{2} \mathrm{O}_{7}$ is as : $\therefore$ There are total $6\, \mathrm{M}=\mathrm{O}$ bonds are present in $\mathrm{Mn}_{2} \mathrm{O}_{7}$ compound.

Q193

In neutral or alkaline solution,

\mathrm{MnO}_{4}^{-}

oxidises thiosulphate to :

A

\mathrm{S}_{2} \mathrm{O}_{7}^{2-}

B

\mathrm{S}_{2} \mathrm{O}_{8}^{2-}

C

\mathrm{SO}_{3}^{2-}

D

\mathrm{SO}_{4}^{2-}

Correct Answer

Option D

Solution

$8 \mathrm{MnO}_{4}^{-}+3 \mathrm{~S}_{2} \mathrm{O}_{3}^{2-}+ \mathrm{H}_{2} \mathrm{O}\rightarrow 8 \mathrm{MnO}_{2}+6 \mathrm{SO}_{4}^{2-}+$ $2 \mathrm{OH}^{-}$

Q194

The reaction of zinc with excess of aqueous alkali, evolves hydrogen gas and gives :

A

\mathrm{Zn}(\mathrm{OH})_{2}

B

\mathrm{ZnO}

C

\left[\mathrm{Zn}(\mathrm{OH})_{4}\right]^{2-}

D

\left[\mathrm{ZnO}_{2}\right]^{2-}

Correct Answer

Option C

Solution

Zinc dissolves in excess of aqueous alkali.

\mathrm{Zn}+2 \mathrm{OH}^{-}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow\left[\mathrm{Zn}(\mathrm{OH})_{4}\right]^{2-}+\mathrm{H}_{2} \uparrow

(Tetrahydroxozincate(II) ion) However, this reaction in NCERT is given as

\mathrm{Zn}+2 \mathrm{NaOH} \longrightarrow \mathrm{Na}_{2} \mathrm{ZnO}_{2}+\mathrm{H}_{2} \uparrow

\mathrm{ZnO}_{2}^{2-}

is anhydrous form of

\left[\mathrm{Zn}(\mathrm{OH})_{4}\right]^{2-}

.

\mathrm{ZnO}_{2}^{2-}+2 \mathrm{H}_{2} \mathrm{O} \rightleftharpoons\left[\mathrm{Zn}(\mathrm{OH})_{4}\right]^{2-}

So in aqueous medium best answer of this question is $\left[\mathrm{Zn}(\mathrm{OH})_{4}\right]^{2-}$ .

Q195

In following pairs, the one in which both transition metal ions are colourless is :

A

\mathrm{Sc}^{3+}, \mathrm{Zn}^{2+}

B

\mathrm{Ti}^{4+}, \mathrm{Cu}^{2+}

C

\mathrm{V}^{2+}, \mathrm{Ti}^{3+}

D

\mathrm{Zn}^{2+}, \mathrm{Mn}^{2+}

Correct Answer

Option A

Solution

$\mathrm{Sc}^{+3}$ and $\mathrm{Zn}^{+2}$ are colourless as they contain no unpaired electron.

Whereas the transition metal ions $\mathrm{Cu}^{+2}, \mathrm{Ti}^{+3}, \mathrm{~V}^{+2}$ and $\mathrm{Mn}^{+2}$ are coloured as they contain unpaired electrons.

The unpaired electron from lower energy $d$ orbital gets excited to a higher energy $d$ orbital on absorbing light of frequency which lies in visible region.

The colour complementary to light absorbed is observed.

Q196

\mathrm{KMnO}_4

oxidises

\mathrm{I}^{-}

in acidic and neutral/faintly alkaline solutions, respectively, to :

A

\mathrm{IO}_3^{-} ~\&~ \mathrm{IO}_3^{-}

B

\mathrm{I}_2 ~\&~ \mathrm{I}_2

C

\mathrm{I}_2 ~\&~ \mathrm{IO}_3^{-}

D

\mathrm{IO}_3^{-} ~\&~ \mathrm{I}_2

Correct Answer

Option C

Solution

Potassium permanganate ( $\mathrm{KMnO}_4$ ) is a strong oxidizing agent that can oxidize iodide ions ( $\mathrm{I}^{-}$ ) to different products depending on the pH of the solution.

In acidic solutions, $\mathrm{KMnO}_4$ oxidizes $\mathrm{I}^{-}$ to molecular iodine ( $\mathrm{I}_2$ ) according to the following equation:

2\mathrm{MnO}_4^{-} + 16\mathrm{H}^{+} + 10\mathrm{I}^{-} \rightarrow 2\mathrm{Mn}^{2+} + 8\mathrm{H}_2\mathrm{O} + 5\mathrm{I}_2

In neutral or faintly alkaline solutions, the oxidation product is different because the $\mathrm{I}^{-}$ is oxidized to iodate ion ( $\mathrm{IO}_3^{-}$ ).

The reaction in a neutral or slightly alkaline solution is as follows:

2\mathrm{MnO}_4^{-} + 2\mathrm{H}_2\mathrm{O} + 10\mathrm{I}^{-} \rightarrow 2\mathrm{MnO}_2 + 4\mathrm{OH}^{-} + 5\mathrm{IO}_3^{-}

So according to the reactions above, in acidic solutions, $\mathrm{I}^{-}$ is oxidized to $\mathrm{I}_2$ while in neutral or faintly alkaline solutions, $\mathrm{I}^{-}$ is oxidized to $\mathrm{IO}_3^{-}$ .

Therefore, the correct answer is: Option C: $\mathrm{I}_2 ~\&~ \mathrm{IO}_3^{-}$

Q197

During the qualitative analysis of

\mathrm{SO}_{3}^{2-}

using dilute

\mathrm{H}_{2} \mathrm{SO}_{4}, \mathrm{SO}_{2}

gas is evolved which turns

\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}

solution (acidified with dilute

\mathrm{H}_{2} \mathrm{SO}_{4}

) :

A blue

B black

C red

D green

Correct Answer

Option D

Solution

\mathrm{S{O_2} + C{r_2}O_7^{2 - }\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{}} \mathop {C{r^{3 + }}}\limits_{(green)} + SO_4^{2 - }}

Q198

Given below are two statements : Statement I : Nickel is being used as the catalyst for producing syn gas and edible fats. Statement II : Silicon forms both electron rich and electron deficient hydrides. In the light of the above statements, choose the most appropriate answer from the options given below :

A Statement I is correct but statement II is incorrect

B Both the statements I and II are correct

C Both the statements I and II are incorrect

D Statement I is incorrect but statement II is correct

Correct Answer

Option A

Solution

Statement I is correct, Statement II is incorrect as Si forms electron precise hydride.

Q199

The set of correct statements is : (i) Manganese exhibits +7 oxidation state in its oxide. (ii) Ruthenium and Osmium exhibit +8 oxidation in their oxides. (iii) Sc shows +4 oxidation state which is oxidizing in nature. (iv) Cr shows oxidising nature in +6 oxidation state.

A (ii), (iii) and (iv)

B (ii) and (iii)

C (i) and (iii)

D (i), (ii) and (iv)

Correct Answer

Option D

Solution

(i)

\mathrm{Mn_2O_7\to Mn}

in (+7) oxidation state (ii) It is also correct (iii) Sc only shows +3 oxidation state (iv)

\mathrm{Cr^{+6}}

is oxidising in nature

Q200

\mathrm{KMnO}_4

decomposes on heating at

513 \mathrm{~K}

to form

\mathrm{O}_2

along with

A

\mathrm{K}_2 \mathrm{MnO}_4 ~\& \mathrm{~Mn}

B

\mathrm{MnO}_2 ~\& \mathrm{~K}_2 \mathrm{O}_2

C

\mathrm{K}_2 \mathrm{MnO}_4 ~\& \mathrm{~MnO}_2

D

\mathrm{Mn} ~\& \mathrm{~KO}_2

Correct Answer

Option C

Solution

\mathrm{KMnO}_4 \xrightarrow{\Delta} \mathrm{K}_2 \mathrm{MnO}_4+\mathrm{MnO}_2+\mathrm{O}_2