d and f Block Elements — Page 21 | JEE Chemistry

Q201

Given below are two statement: one is labelled as Assertion

\mathbf{A}

and the other is labelled as Reason

\mathbf{R}

Assertion A:

5 \mathrm{f}

electrons can participate in bonding to a far greater extent than

4 \mathrm{f}

electrons Reason R:

5 \mathrm{f}

orbitals are not as buried as

4 \mathrm{f}

orbitals In the light of the above statements, choose the correct answer from the options given below

A Both A and R are true and R is the correct explanation of A

B Both A and R are true but R is NOT the correct explanation of A

C A is true but R is false

D A is false but R is true

Correct Answer

Option A

Solution

Assertion A: 5f electrons can participate in bonding to a far greater extent than 4f electrons This is correct.

In actinides (elements in which 5f orbitals are being filled), the 5f electrons can participate in bonding.

This contrasts with lanthanides (elements in which 4f orbitals are being filled), where the 4f electrons largely do not participate in bonding.

Reason R: 5f orbitals are not as buried as 4f orbitals This is also correct.

The term "buried" refers to the degree to which an orbital is shielded from the outside environment by other electrons.

The 5f orbitals are less shielded, or less "buried," than the 4f orbitals.

This means they are more exposed to the outside environment and can more readily interact with other atoms to form bonds.

This is the reason why 5f electrons can participate in bonding to a greater extent than 4f electrons.

Therefore, both A and R are true, and R is the correct explanation of A.

Q202

Prolonged heating is avoided during the preparation of ferrous ammonium sulphate to :

A prevent hydrolysis

B prevent breaking

C prevent oxidation

D prevent reduction

Correct Answer

Option C

Solution

Prolonged heating is avoided during the preparation of ferrous ammonium sulfate to prevent oxidation.

Ferrous ammonium sulfate is a green crystalline solid that is used as a reducing agent in various chemical reactions.

When heated, it can undergo oxidation to form ferric ammonium sulfate, which is a brown crystalline solid.

Here is a chemical equation for the oxidation of ferrous ammonium sulfate:

\begin{equation} \ce{FeSO4(NH4)2SO4.6H2O + O2 -> Fe2(SO4)3(NH4)2SO4.12H2O} \end{equation}

As you can see, the oxidation of ferrous ammonium sulfate results in the formation of ferric ammonium sulfate, which is a brown crystalline solid.

This is why prolonged heating is avoided during the preparation of ferrous ammonium sulfate.

Here are the other options and why they are incorrect: Option A: Hydrolysis is the process of a chemical compound reacting with water.

Ferrous ammonium sulfate is not hydrolyzed by water, so prolonged heating is not necessary to prevent hydrolysis.

Option B: Breaking is the process of a chemical compound being physically broken apart.

Ferrous ammonium sulfate is not broken apart by heat, so prolonged heating is not necessary to prevent breaking.

Option D: Reduction is the process of a chemical compound gaining electrons.

Ferrous ammonium sulfate is a reducing agent, so it is not reduced by heat.

Q203

Which of the following statements are correct? (A) The M

^{3+}

/M

^{2+}

reduction potential for iron is greater than manganese. (B) The higher oxidation states of first row d-block elements get stabilized by oxide ion. (C) Aqueous solution of Cr

^{2+}

can liberate hydrogen from dilute acid. (D) Magnetic moment of V

^{2+}

is observed between 4.4 - 5.2 BM. Choose the correct answer from the options given below :

A (B), (C) only

B (A), (B) only

C (A), (B), (D) only

D (C), (D) only

Correct Answer

Option A

Solution

The reduction electrode potential of $\mathrm{Mn}^{3+} / \mathrm{Mn}^{2+}$ is $+1.57 \mathrm{~V}$ while that of $\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}$ is $+0.77 \mathrm{~V}$ , hence $\mathrm{A}$ is wrong.

Higher oxidation state of smaller $d$ -block elements is stabilized (or say form compounds) with smaller anion oxide that can be explained by stearic reason hence $B$ is correct.

The oxidation electrode potential of $\mathrm{Cr}^{2+} / \mathrm{Cr}^{3+}$ is $+0.41 \mathrm{~V}$ hence it can reduce $\mathrm{H}^{+}$ and so liberate $\mathrm{H}_2$ .

The unpaired electrons in $\mathrm{V}^{2+}$ are 3 hence the magnetic moment of $\mathrm{V}^{2+}$ will be lesser than 4.4 BM.

Hence, only $B$ and $C$ are correct.

Q204

The electronic configuration of

\mathrm{Cu}(\mathrm{II})

is

3 \mathrm{~d}^9

whereas that of

\mathrm{Cu}(\mathrm{I})

is

3 \mathrm{~d}^{10}

. Which of the following is correct?

A

\mathrm{Cu}(\mathrm{II})

is more stable

B

\mathrm{Cu}

(II) is less stable

C

\mathrm{Cu}(\mathrm{I})

and

\mathrm{Cu}(\mathrm{II})

are equally stable

D Stability of

\mathrm{Cu}(\mathrm{I})

and

\mathrm{Cu}(\mathrm{II})

depends on nature of copper salts

Correct Answer

Option A

Solution

\mathrm{Cu}^{2+}

is more stable.

\mathrm{Cu}^{+} \longrightarrow \mathrm{Cu}^{2+}+\mathrm{Cu}

disproportionation reaction

Q205

The number of ions from the following that have the ability to liberate hydrogen from a dilute acid is _________.

\mathrm{Ti}^{2+}, \mathrm{Cr}^{2+} \text{ and } \mathrm{V}^{2+}

A 1

B 2

C 3

D 0

Correct Answer

Option C

Solution

To determine the ability of the ions to liberate hydrogen from a dilute acid, we need to understand the standard electrode potentials of these ions in comparison to hydrogen.

The standard electrode potential of hydrogen (

\mathrm{H}^{+}/\mathrm{H}_{2}

) is set at 0 volts.

Any metal ion with a negative standard reduction potential compared to hydrogen has the ability to liberate hydrogen gas from dilute acids because they are more likely to donate electrons to the

\mathrm{H}^{+}

ions found in acids, reducing them to

\mathrm{H}_{2}

gas. The standard reduction potentials for the ions mentioned are as follows:

\mathrm{Ti}^{2+}/\mathrm{Ti}

is more negative than that of

\mathrm{H}^{+}/\mathrm{H}_{2}

.

\mathrm{Cr}^{2+}/\mathrm{Cr}

also has a more negative standard reduction potential compared to

\mathrm{H}^{+}/\mathrm{H}_{2}

. The standard reduction potential for

\mathrm{V}^{2+}/\mathrm{V}

is similarly more negative than that of

\mathrm{H}^{+}/\mathrm{H}_{2}

. Due to their negative standard reduction potentials relative to hydrogen, all three ions (

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

) have the capacity to liberate hydrogen from a dilute acid. Therefore, the correct answer is: Option C: 3

Q206

In the context of the Hall-Heroult process for the extraction of Al, which of the following statements is false?

A Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

B Al3+ is reduced at the cathode to form Al

C Na3AlF6 serves as the electrolyte

D CO and CO2 are produced in this process

Correct Answer

Option C

Solution

In the metallurgy of aluminium purified

A{l_2}{O_3}

is mixed with

N{a_3}Al{F_6}\,\,

\,\,\,

or

\,\,\,

Ca{F_2}

which lowers the melting point of the mix and brings conductivity.

Q207

Arrange the following elements in the increasing order of number of unpaired electrons in it. (A)

\mathrm{Sc}

(B)

\mathrm{Cr}

(C)

\mathrm{V}

(D)

\mathrm{Ti}

(E)

\mathrm{Mn}

Choose the correct answer from the options given below :

A

(\mathrm{A})<(\mathrm{D})<(\mathrm{C})<(\mathrm{B})<(\mathrm{E})

B

(\mathrm{A})<(\mathrm{D})<(\mathrm{C})<(\mathrm{E})<(\mathrm{B})

C

(\mathrm{B})<(\mathrm{C})<(\mathrm{D})<(\mathrm{E})<(\mathrm{A})

D

(\mathrm{C})<(\mathrm{E})<(\mathrm{B})<(\mathrm{A})<(\mathrm{D})

Correct Answer

Option B

Solution

The number of unpaired electrons in a transition metal is determined by its electronic configuration in the context of its ground state.

The elements listed, Scandium (Sc), Chromium (Cr), Vanadium (V), Titanium (Ti), and Manganese (Mn), are transition metals that have different numbers of unpaired electrons.

Here is the electronic configuration and the corresponding number of unpaired electrons for each: Scandium (Sc): Electronic Configuration:

[\mathrm{Ar}] \, 3d^1 \, 4s^2

Unpaired Electrons: 1 Titanium (Ti): Electronic Configuration:

[\mathrm{Ar}] \, 3d^2 \, 4s^2

Unpaired Electrons: 2 Vanadium (V): Electronic Configuration:

[\mathrm{Ar}] \, 3d^3 \, 4s^2

Unpaired Electrons: 3 Chromium (Cr): Electronic Configuration:

[\mathrm{Ar}] \, 3d^5 \, 4s^1

Unpaired Electrons: 6 Manganese (Mn): Electronic Configuration:

[\mathrm{Ar}] \, 3d^5 \, 4s^2

Unpaired Electrons: 5 Arranging them in increasing order of number of unpaired electrons: $$ (\mathrm{A}) Therefore, the correct choice is Option B.

Q208

The number of element from the following that do not belong to lanthanoids is

\mathrm{Eu}, \mathrm{Cm}, \mathrm{Er}, \mathrm{Tb}, \mathrm{Yb}

and

\mathrm{Lu}

A 3

B 1

C 4

D 5

Correct Answer

Option B

Solution

To determine the number of elements from the given list that do not belong to the lanthanoids, let's first define what lanthanoids are.

The lanthanoids consist of the 15 chemical elements in the periodic table from Lanthanum ( $\mathrm{La}$ ) with atomic number 57 to Lutetium ( $\mathrm{Lu}$ ) with atomic number 71.

Thus, lanthanoids include: Lanthanum ( $\mathrm{La}$ ) Cerium ( $\mathrm{Ce}$ ) Praseodymium ( $\mathrm{Pr}$ ) Neodymium ( $\mathrm{Nd}$ ) Promethium ( $\mathrm{Pm}$ ) Samarium ( $\mathrm{Sm}$ ) Europium ( $\mathrm{Eu}$ ) Gadolinium ( $\mathrm{Gd}$ ) Terbium ( $\mathrm{Tb}$ ) Dysprosium ( $\mathrm{Dy}$ ) Holmium ( $\mathrm{Ho}$ ) Erbium ( $\mathrm{Er}$ ) Thulium ( $\mathrm{Tm}$ ) Ytterbium ( $\mathrm{Yb}$ ) Lutetium ( $\mathrm{Lu}$ ) Given the list of elements: Eu (Europium), Cm (Curium), Er (Erbium), Tb (Terbium), Yb (Ytterbium), and Lu (Lutetium), we can now check which of these are not part of the lanthanoids: $\mathrm{Eu}$ (Europium) belongs to the lanthanoids.

$\mathrm{Cm}$ (Curium) does not belong to the lanthanoids; it is an actinoid.

$\mathrm{Er}$ (Erbium) belongs to the lanthanoids.

$\mathrm{Tb}$ (Terbium) belongs to the lanthanoids.

$\mathrm{Yb}$ (Ytterbium) belongs to the lanthanoids.

$\mathrm{Lu}$ (Lutetium) belongs to the lanthanoids.

From the given list, only Cm (Curium) does not belong to the lanthanoids, which means the correct answer is: Option B: 1

Q209

The correct option with order of melting points of the pairs (Mn, Fe), (Tc, Ru) and (Re, Os) is :

A Fe < Mn, Ru < Tc and Re < Os

B Fe < Mn, Ru < Tc and Os < Re

C Mn < Fe, Tc < Ru and Re < Os

D Mn < Fe, Tc < Ru and Os < Re

Correct Answer

Option D

Solution

Melting point of the pairs $\left(M_n, F_e\right),\left(T_c, R_u\right)$ , and $\left(\mathrm{Re}_{\mathrm{e}}, \mathrm{O}_{\mathrm{s}}\right)$ : $M n, T_c, \mathrm{Re}$ are same group elements.

Also, $\mathrm{Fe}, \mathrm{Bu}, \mathrm{Os}$ , are same group elements.

$\mathrm{Mn}, \mathrm{Fe}$ Fe has a higher melting point than $M_n, M_n$ \left[\begin{array}{cc}M_n & F_e \\ T_c & R_u \\ R_e & O_s\end{array}\right] $"Melting point property is directly proportional to the strength of metallic bond, which proportional to the delocalize electrons. The more electrons are free to move around, the stronger will be the metallic bond." For Mn and$ \mathrm{Fe}, \quad\left(\mathrm{Mn}-d^5 s^2, \mathrm{Fe}-d^6 s^2\right) $.$ M_n $has a weaker metallic bond due to its electronic configuration (half filled configuration). The half filled configuration (stable) results in less delocalization of electrons and weaker attraction between atoms, requiring less energy to break apart and met. But, Fe has localized electrons and hence stronger metallic bonds, leading to higher melting point.$ T_c, R u $Tc has lower melting point than$ \mathrm{Ru}, \mathrm{Tc}_c

T c-d^5 s^2, R u-d^7 s^1

Tc has halt filled d-orbitals, which makes its electronic configuration stable.

This configuration causes the nucleus to hold the electrons tightly, which weakens the metallic bond and it results in lower melting point. $\mathrm{Re}, \mathrm{Os}_{\mathrm{s}}$ $\operatorname{Re}$ has a higher melting point than $\mathrm{Os}_{\mathrm{s}}, \mathrm{Os} The higher melting point of$ R e $is due to the strength of atomic bonding. Re trams strong metallic bonds than Us. Due to the langer size of Re , its valence elections are not tightly hooded by the nucleus and hence, the valence elections causes stronger metallic bond. As a result, melting point increases. Osmium tills weaker metallic bonds than$ \operatorname{Re} $due to its crystal arrangement and electronic Structure. Comparing Re, and OS,$ R_e $is slightly smaller in atomic radius which allows atoms to pack more efficiently than is and has stronger bonds and higher meeting point. So,$ $\begin{aligned} & M_n Correct answer: Option 4)$ \mathrm{Mn}

Q210

Lanthanoid ions with

4 \mathrm{f}^7

configuration are : (A)

\mathrm{Eu}^{2+}

(B)

\mathrm{Gd}^{3+}

(C)

\mathrm{Eu}^{3+}

(D)

\mathrm{Tb}^{3+}

(E)

\mathrm{Sm}^{2+}

Choose the correct answer from the options given below :

A (B) and (E) only

B (B) and (C) only

C (A) and (D) only

D (A) and (B) only

Correct Answer

Option D

Solution

Let's analyze the electron configurations of the given lanthanide ions.

Europium ( $\mathrm{Eu}$ ) The neutral atom of Eu (atomic number 63) typically has the configuration: $[\mathrm{Xe}]\,4f^7\,6s^2.$ For $\mathrm{Eu}^{2+}$ , two electrons are removed, usually from the $6s$ orbital, resulting in: $[\mathrm{Xe}]\,4f^7.$ Thus, $\mathrm{Eu}^{2+}$ has a $4f^7$ configuration.

For $\mathrm{Eu}^{3+}$ , three electrons are removed (the two $6s$ electrons and one $4f$ electron), giving: $[\mathrm{Xe}]\,4f^6.$ Therefore, $\mathrm{Eu}^{3+}$ does not have a $4f^7$ configuration.

Gadolinium ( $\mathrm{Gd}$ ) The neutral atom of Gd (atomic number 64) is usually represented as: $[\mathrm{Xe}]\,4f^7\,5d^1\,6s^2.$ For $\mathrm{Gd}^{3+}$ , three electrons are removed (typically the $5d^1$ and the two $6s$ electrons), resulting in: $[\mathrm{Xe}]\,4f^7.$ So, $\mathrm{Gd}^{3+}$ has a $4f^7$ configuration.

Terbium ( $\mathrm{Tb}$ ) The neutral atom of Tb (atomic number 65) typically has the configuration: $[\mathrm{Xe}]\,4f^9\,6s^2.$ For $\mathrm{Tb}^{3+}$ , removal of three electrons gives: $[\mathrm{Xe}]\,4f^8.$ Hence, $\mathrm{Tb}^{3+}$ does not have a $4f^7$ configuration.

Samarium ( $\mathrm{Sm}$ ) The neutral atom of Sm (atomic number 62) generally has: $[\mathrm{Xe}]\,4f^6\,6s^2.$ For $\mathrm{Sm}^{2+}$ , after removal of two electrons (likely from the $6s$ orbital), the configuration is: $[\mathrm{Xe}]\,4f^6.$ Therefore, $\mathrm{Sm}^{2+}$ does not have a $4f^7$ configuration.

Thus, the ions with a $4f^7$ configuration are: (A) $\mathrm{Eu}^{2+}$ (B) $\mathrm{Gd}^{3+}$ The correct answer is: Option D – (A) and (B) only.