d and f Block Elements — Page 22 | JEE Chemistry

Q211

The amphoteric oxide among

\mathrm{V}_2 \mathrm{O}_3, \mathrm{~V}_2 \mathrm{O}_4

and

\mathrm{V}_2 \mathrm{O}_5

, upon reaction with alkali leads to formation of an oxide anion. The oxidation state of V in the oxide anion is :

A + 4

B + 5

C + 3

D + 7

Correct Answer

Option B

Solution

\begin{aligned} &\mathrm{V}_2 \mathrm{O}_5+\text{ alkali } \rightarrow \mathrm{VO}_4^{3-}\\ &\text{ In } \mathrm{VO}_4^{3-} \text{ ion, vanadium is in }+5 \text{ oxidation state } \end{aligned}

Q212

Preparation of potassium permanganate from

\mathrm{MnO}_2

involves two step process in which the 1st step is a reaction with KOH and

\mathrm{KNO}_3

to produce

A

\mathrm{K}_3 \mathrm{MnO}_4

B

\mathrm{K}_4\left[\mathrm{Mn}(\mathrm{OH})_6\right]

C

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

D

\mathrm{KMnO}_4

Correct Answer

Option C

Solution

The preparation of potassium permanganate ( $\mathrm{KMnO}_4$ ) from manganese dioxide ( $\mathrm{MnO}_2$ ) involves a two-step process.

In the first step, $\mathrm{MnO}_2$ reacts with potassium hydroxide ( $\mathrm{KOH}$ ) and potassium nitrate ( $\mathrm{KNO}_3$ ) under heating.

This reaction produces potassium manganate ( $\mathrm{K}_2 \mathrm{MnO}_4$ ) as an intermediate product.

The reaction can be depicted as follows: $\mathrm{MnO}_2 \xrightarrow[\mathrm{KNO}_3, \mathrm{\Delta}]{\mathrm{KOH}} \mathrm{K}_2 \mathrm{MnO}_4$

Q213

The metal ion whose electronic configuration is not affected by the nature of the ligand and which gives a violet colour in non-luminous flame under hot condition in borax bead test is

A

\mathrm{Cr}^{3+}

B

\mathrm{Ti}^{3+}

C

\mathrm{Ni}^{2+}

D

\mathrm{Mn}^{2+}

Correct Answer

Option C

Solution

$\mathrm{Ni}^{+2}$ gives violet coloured bead in non-luminous flame under hot conditions. $\mathrm{Ni}^{+2}$ has $\mathrm{d}^8$ configuration which does not depend on nature of ligand present in octahedral complex.

\mathrm{Ni}^{+2}: \mathrm{t}_{2 \mathrm{~g}}{ }^6 \mathrm{e}_{\mathrm{g}}{ }^2

Q214

The correct decreasing order of spin only magnetic moment values (BM) of Cu+, Cu2+, Cr2+ and Cr3+ ions is :

A Cr3+ > Cr2+ > Cu+ > Cu2+

B Cu+ > Cu2+ > Cr3+ > Cr2+

C Cr2+ > Cr3+ > Cu2+ > Cu+

D Cu2+ > Cu+ > Cr2+ > Cr3+

Correct Answer

Option C

Solution

$\mathrm{Cu}^{+}:[\mathrm{Ar}] 3 \mathrm{~d}^{10}$ , Spin only magnetic moment $=0$ B.M.

$\mathrm{Cu}^{+2}:[\mathrm{Ar}] 3 \mathrm{~d}^9$ , Spin only magnetic moment $=\sqrt{3}$ B.M.

$\mathrm{Cr}^{+2}:[\mathrm{Ar}] 3 \mathrm{~d}^4$ , Spin only magnetic moment $=\sqrt{24}$ B.M.

$\mathrm{Cr}^{+3}:[\mathrm{Ar}] 3 \mathrm{~d}^3$ , Spin only magnetic moment $=\sqrt{15}$ B.M.

Order of $\mu: \mathrm{Cr}^{+2}>\mathrm{Cr}^{+3}>\mathrm{Cu}^{+2}>\mathrm{Cu}^{+}$

Q215

Extraction of copper by smelting uses silica as an additive to remove :

A Cu2S

B FeO

C FeS

D Cu2O

Correct Answer

Option B

Solution

The gangue material in the ore is removed by adding another chemical substance called flux.

If ore has basic impurities, such as FeO, CaO, MgO, etc. suitable acid flux such as SiO2 , P2O5 , etc. are used.

For example, in the extraction of copper, ferrous oxide (FeO), a basic impurity, is removed by adding silica.

FeO (gangue) and SiO2 (flux) together form slag FeSiO3.

Q216

The metal ions that have the calculated spin-only magnetic moment value of 4.9 B.M. are : A.

\mathrm{Cr}^{2+}

B.

\mathrm{Fe}^{2+}

C.

\mathrm{Fe}^{3+}

D.

\mathrm{Co}^{2+}

E.

\mathrm{Mn}^{3+}

Choose the correct answer from the options given below:

A A, D and E Only

B A, B and E Only

C A, C and E Only

D B and E Only

Correct Answer

Option B

Solution

To determine which metal ions have a calculated spin-only magnetic moment of 4.9 Bohr Magnetons (B.M.), we can use the formula: $\text{Magnetic Moment (M.M)} = \sqrt{n(n+2)} \, \text{B.M.}$ where $n$ is the number of unpaired electrons.

Given that the magnetic moment is 4.9 B.M., we can set up the equation: $4.9 = \sqrt{n(n+2)}$ Solving this equation, we find that $n = 4$ .

Let's analyze each metal ion: (A) ${}_{24} \mathrm{Cr}^{2+}$ : Electronic configuration: $[\mathrm{Ar}] 3d^4$ .

This ion has 4 unpaired electrons.

(B) ${}_{26} \mathrm{Fe}^{2+}$ : Electronic configuration: $[\mathrm{Ar}] 3d^6$ .

This ion also has 4 unpaired electrons.

(C) ${}_{26} \mathrm{Fe}^{3+}$ : Electronic configuration: $[\mathrm{Ar}] 3d^5$ .

This ion has 5 unpaired electrons, which does not match the required $n$ value of 4.

(D) ${}_{27} \mathrm{Co}^{2+}$ : Electronic configuration: $[\mathrm{Ar}] 3d^7$ .

This ion has 3 unpaired electrons.

(E) ${}_{25} \mathrm{Mn}^{3+}$ : Electronic configuration: $[\mathrm{Ar}] 3d^4$ .

This ion has 4 unpaired electrons.

Based on the number of unpaired electrons, the metal ions with a spin-only magnetic moment of approximately 4.9 B.M. are $\mathrm{Cr}^{2+}$ , $\mathrm{Fe}^{2+}$ , and $\mathrm{Mn}^{3+}$ .

Q217

Aluminium is extracted by the electrolysis of

A bauxite

B alumina

C alumina mixed with molten cryolite

D molten cryolite

Correct Answer

Option C

Solution

Pure aluminium can be obtained by electrolysis of a mixture containing alumina, crayolite and fluorspar in the ratio

20:24:20.

The fusion temperature of this mixture is

{900^ \circ }C

and it is a good conductor of electricity.

Q218

The metal extracted by leaching with a cyanide is

A Mg

B Ag

C Cu

D Na

Correct Answer

Option B

Solution

Silver ore forms a soluble complex with

NaCN

from which silver is precipitated using scrap zinc.

A{g_2}S + 2NaCN \to

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Na\left[ {Ag{{\left( {CN} \right)}_2}} \right]\overset{{Zn}}\longrightarrow

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,N{a_2}\left[ {Zn{{\left( {CN} \right)}_4}} \right] + Ag \downarrow

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,

sodargento-cynanide(soluble)

Q219

Which of the following factors is of no significance for roasting sulphide ores to the oxides and not subjecting the sulphide ores to carbon reduction directly?

A Metal sulphides are thermodynamically more stable than CS2

B CO2 is thermodynamically more stable than CS2

C Metal sulphides are less stable than the corresponding oxides

D CO2 is more volatile than CS2

Correct Answer

Option C

Solution

NOTE : The reduction of metal sulphides by carbon reduction process is not spontaneous because

\Delta G

for such a process is positive. The reduction of metal oxide by carbon reduction process is spontaneous as

\Delta G

for such a process is negative. From this we find that on thermodynamic considerations

C{O_2}

is more stable than

C{S_2}

and the metal sulphides are more stable than corresponding oxides.

In view of above the factor listed in choice (c) is incorrect and so is of no significance.

Hence the correct answer is

(c)

Q220

Which method of purification is represented by the following equation : Ti (s) + 2I2 (g)

\overset{{523K}}\longrightarrow

TiI4 (g)

\overset{{1700K}}\longrightarrow

Ti (s) + 2I2 (g)

A zone refining

B cupellation

C Poling

D Van Arkel

Correct Answer

Option D

Solution

Van Arkel is a method in which heat treatment it used to purify metal in this process metals are converted into other metal compound for loosly coupled like as iodine to make metal iodide which are easily decomposed and give pure metal.

The process is known as Van Arkel method.