d and f Block Elements — Page 9 | JEE Chemistry

Q81

Which halogen is known to cause the reaction given below :

2 \mathrm{Cu}^{2+}+4 \mathrm{X}^{-} \rightarrow \mathrm{Cu}_{2} \mathrm{X}_{2}(\mathrm{s})+\mathrm{X}_{2}

A All halogens

B Only Iodine

C Only Bromine

D Only Chlorine

Correct Answer

Option B

Solution

The given reaction describes the formation of a copper(II) halide and the corresponding diatomic halogen molecule.

Not all halogens will cause this reaction to occur.

This reaction is known to occur with iodine (I) because the formation of $\mathrm{I}_{2}$ (a diatomic iodine molecule) is easier due to the relatively low bond energy of the $\mathrm{I}-\mathrm{I}$ bond.

On the other hand, the bond energies of $\mathrm{Cl}-\mathrm{Cl}$ and $\mathrm{Br}-\mathrm{Br}$ are higher, making it more difficult for these diatomic halogen molecules to form.

Thus, these halogens would not typically cause this reaction to occur.

Therefore, the correct answer is: Option B : Only Iodine.

Q82

In chromyl chloride, the number of d-electrons present on chromium is same as in (Given at no. of

\mathrm{Ti}: 22, \mathrm{~V}: 23, \mathrm{Cr}: 24, \mathrm{Mn}: 25, \mathrm{Fe}: 26

)

A Mn (VII)

B Ti (III)

C Fe (III)

D V (IV)

Correct Answer

Option A

Solution

When a mixture containing chloride ion is heated with $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ and concentrated $\mathrm{H}_2 \mathrm{SO}_{4^{\prime}}$ deep orange-red fumes of chromyl chloride $\left(\mathrm{CrO}_2 \mathrm{Cl}_2\right)$ are formed

\begin{aligned} & \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+4 \mathrm{NaCl}+6 \mathrm{H}_2 \mathrm{SO}_4 \rightarrow 2 \mathrm{KHSO}_4+4 \mathrm{NaHSO}_4 +2 \mathrm{CrO}_2 \mathrm{Cl}_2 \uparrow+3 \mathrm{H}_2 \mathrm{O} \end{aligned}

Orange-red fumes So in this case, $\mathrm{X}$ in $\mathrm{CrO}_2 \mathrm{Cl}_2$ .

Oxidation state of $\mathrm{Cl}=-1, \mathrm{O}=-2, \mathrm{Cr}=x$

x+2 \times(-2)+2 \times(-1)=0 \Rightarrow x=+6

Q83

Strong reducing and oxidizing agents among the following, respectively, are :

A

\mathrm{Ce}^{4+}

and

\mathrm{Eu}^{2+}

B

\mathrm{Eu}^{2+}

and

\mathrm{Ce}^{4+}

C

\mathrm{Ce}^{4+}

and

\mathrm{Tb}^{4+}

D

\mathrm{Ce}^{3+}

and

\mathrm{Ce}^{4+}

Correct Answer

Option B

Solution

Strong reducing agents are those that readily donate electrons, and strong oxidizing agents are those that readily accept electrons.

$\mathrm{Ce}^{4+}$ : It can accept an electron to form $\mathrm{Ce}^{3+}$ , so it is a strong oxidizing agent.

$\mathrm{Eu}^{2+}$ : It can donate an electron to form $\mathrm{Eu}^{3+}$ , so it is a strong reducing agent.

So the correct answer is: $\mathrm{Eu}^{2+}$ and $\mathrm{Ce}^{4+}$ .

Q84

Element not present in Nessler's reagent is :

A I

B

\mathrm{N}

C

\mathrm{K}

D

\mathrm{Hg}

Correct Answer

Option B

Solution

Nessler's reagent is a 2% solution of potassium tetraiodomercurate(II) in aqueous potassium hydroxide.

Its chemical formula is

\mathrm{K}_2[\mathrm{HgI}_4]

.

The elements present in it are Potassium (K), Mercury (Hg), and Iodine (I).

Nitrogen (N) is not present in Nessler's reagent.

Q85

The pair that has similar atomic radii is :

A Mo and W

B Mn and Fe

C Ti and Hf

D Sc and Ni

Correct Answer

Option A

Solution

Electron configuration of Mo(Molybdenum) = [Kr] 4d5 5s1.

So it belongs to 4d series.

Electron configuration of W(Tungsten) = [Xe] 4f14 5d4 6s2.

So it belongs to 5d series.

Size of 3d < 4d = 5d.

Due to lanthanoid contraction size of 4d and 5d series elements are almost same.

Q86

The correct order of the first ionization enthalpies is :

A Ti < Mn < Zn < Ni

B Zn < Ni < Mn < Ti

C Mn < Ti < Zn < Ni

D Ti < Mn < Ni < Zn

Correct Answer

Option D

Solution

In a period, from left to right ionization enthalpy increses.

So in period 4, correct order for those 4 elements is Ti < Mn < Ni < Zn Exceptions in period 4 are (1) V(Vanadium) < Ti(Titanium) (2) Ni(Titanium) < Co(Cobalt) (3) Ga(Gallium) < Zn(Zinc) (4) Se(Selenium) < As(Arsenic)

Q87

During the reaction of permanganate with thiosulphate, the change in oxidation of manganese occurs by value of 3. Identify which of the below medium will favour the reaction.

A aqueous neutral

B both aqueous acidic and faintly alkaline

C both aqueous acidic and neutral

D aqueous acidic

Correct Answer

Option A

Solution

In a neutral or weakly alkaline solution, permanganate ions are reduced to MnO₂, with a change in oxidation state from +7 to +4, which is a change of 3 units.

The balanced redox reaction is : 3MnO₄⁻ + 5S₂O₃²⁻ + 2H₂O → 3MnO₂ + 5SO₄²⁻ + 10OH⁻

Q88

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : In aqueous solutions

\mathrm{Cr}^{2+}

is reducing while

\mathrm{Mn}^{3+}

is oxidising in nature. Reason (R) : Extra stability to half filled electronic configuration is observed than incompletely filled electronic configuration. In the light of the above statements, choose the most appropriate answer from the options given below :

A (A) is true but (R) is false

B Both (A) and (R) are true and (R) is the correct explanation of (A)

C Both (A) and (R) are true but (R) is not the correct explanation of (A)

D (A) is false but (R) is true

Correct Answer

Option B

Solution

Let us analyze both the Assertion (A) and the Reason (R) to determine the correct answer.

Assertion (A): In aqueous solutions, $\mathrm{Cr}^{2+}$ is reducing while $\mathrm{Mn}^{3+}$ is oxidising in nature.

This assertion is true.

The $\mathrm{Cr}^{2+}$ ion has a tendency to be oxidized to $\mathrm{Cr}^{3+}$ because $\mathrm{Cr}^{3+}$ has a more stable electronic configuration.

Therefore, $\mathrm{Cr}^{2+}$ acts as a reducing agent.

On the other hand, $\mathrm{Mn}^{3+}$ tends to stabilize by being reduced to $\mathrm{Mn}^{2+}$ , which is a more stable electronic configuration due to having a half-filled d5 subshell ( $\mathrm{Mn}^{2+}$ : [Ar] 3d5).

Thus, $\mathrm{Mn}^{3+}$ acts as an oxidizing agent.

Reason (R): Extra stability to half filled electronic configuration is observed than incompletely filled electronic configuration.

This reason is also true.

Half-filled and fully filled electron configurations, like those found in $\mathrm{Mn}^{2+}$ and $\mathrm{Cr}^{0}$ respectively, are particularly stable due to symmetry and exchange energy considerations.

This is a reason why certain ions like $\mathrm{Mn}^{3+}$ want to be reduced to $\mathrm{Mn}^{2+}$ , and why $\mathrm{Cr}^{0}$ (with a [Ar] 3d5 4s1 configuration) is not as common as $\mathrm{Cr}^{3+}$ (which has a [Ar] 3d3 configuration, and while not half-filled, is favored due to crystal field stabilization energy considerations in octahedral complexes).

If we evaluate these statements together, we see that Assertion (A) is directly related to the electronic configurations and their stability, as asserted by Reason (R).

The stability of half-filled and full orbitals contributes to the observed redox behavior of $\mathrm{Cr}^{2+}$ and $\mathrm{Mn}^{3+}$ in aqueous solutions.

Therefore, the correct answer is Option B: Both (A) and (R) are true and (R) is the correct explanation of (A).

Q89

The refining method used when the metal and the impurities have low and high melting temperatures, respectively, is :

A liquation

B vapour phase refining

C distillation

D zone refining

Correct Answer

Option A

Solution

Liquation method is used when the melting point of metal is less compare to the melting point of the associated impurity.

Q90

Which of the following electronic configuration would be associated with the highest magnetic moment?

A

[\mathrm{Ar}] 3 \mathrm{~d}^7

B

[\mathrm{Ar}] 3 \mathrm{~d}^8

C

[\mathrm{Ar}] 3 \mathrm{~d}^3

D

[\mathrm{Ar}] 3 \mathrm{~d}^6

Correct Answer

Option D

Solution

The magnetic moment of an atom or ion with unpaired electrons is given by the formula

\mu = \sqrt{n(n+2)}

Bohr magnetons (BM), where

n

is the number of unpaired electrons.

The magnetic moment depends on the number of unpaired electrons: more unpaired electrons lead to a higher magnetic moment.

Looking at the given options with respect to their electron configurations and calculating their respective magnetic moments based on their unpaired electrons:

[\mathrm{Ar}] 3 \mathrm{~d}^7

has 3 unpaired electrons, yielding a magnetic moment of

\sqrt{15}

BM.

[\mathrm{Ar}] 3 \mathrm{~d}^8

has 2 unpaired electrons, for a magnetic moment of

\sqrt{8}

BM.

[\mathrm{Ar}] 3 \mathrm{~d}^3

also has 3 unpaired electrons, similar to

3 \mathrm{~d}^7

, so its magnetic moment is likewise

\sqrt{15}

BM.Finally,

[\mathrm{Ar}] 3 \mathrm{~d}^6

has 4 unpaired electrons, resulting in the highest magnetic moment among the options,

\sqrt{24}

BM. Hence, the electron configuration associated with the highest magnetic moment is

[\mathrm{Ar}] 3 \mathrm{~d}^6

.