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CHXII08:THE D- & F-BLOCK ELEMENTS

321743 Read the Statement - A and Statement - B carefully to mark the correct options given below
Statement A :
In the Lanthanoids, the formation of ${Ce}^{+ 4}$ is favoured by its noble gas configuration.
Statement B :
${Ce}^{+ 4}$ is a strong oxidant reverting to the common +3 state.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

Explanation:

Both statements are true. As ${Ce}^{+ 3}$ is more stable than ${Ce}^{+ 4}, {Ce}^{+ 4}$ has tendency to accept an electron and thus, it behaves as an strong oxidising agent. So, the correct option is (3).

JEE Main - 2024

CHXII08:THE D- & F-BLOCK ELEMENTS

321744 Which among the following lanthanoids forms most stable +2 oxidation state?

1

Eu(Z  =  63)

2

Pm(Z  =  61)

3

Ce(Z  =  58)

4

Gd(Z  =  64)

Explanation:

$\begin{matrix} Ion & Valence shell \\ \begin{matrix} electronic \\ configuration \end{matrix} \\ E u^{2 +} & 4 f^{7} \\ P m^{2 +} & 4 f^{5} \\ C e^{2 +} & 4 f^{2} \\ G d^{2 +} & 4 f^{7} 5 d^{1} \end{matrix}$
Due to stable half filled 4 f orbitals, ${Eu}^{2 +}$ will be most stable, among given.

MHTCET - 2021

CHXII08:THE D- & F-BLOCK ELEMENTS

321745 What is the number of unpaired electrons present in f-orbital at +3 oxidation state of $Lu (Z = 71)$ ?

1 7

2 2

3 4

4 0

Explanation:

$_{71} Lu : [Xe] 4 f^{14} 5 d^{1} 6 s^{2}$
${Lu}^{3 +} : [Xe] 4 f^{14}$ (fulfilled f-orbitals)
Hence, number of unpaired electrons in f-orbital of ${Lu}^{3 +}$ is zero.

MHTCET - 2021

CHXII08:THE D- & F-BLOCK ELEMENTS

321746 Which of the following Lanthanoid ion is more stable in +2 oxidation state in aqueous solution?

1

{Eu}^{2 +}

2

{Ce}^{2 +}

3

{Lu}^{3 +}

4

{Fe}^{2 +}

CHXII08:THE D- & F-BLOCK ELEMENTS

321743 Read the Statement - A and Statement - B carefully to mark the correct options given below
Statement A :
In the Lanthanoids, the formation of ${Ce}^{+ 4}$ is favoured by its noble gas configuration.
Statement B :
${Ce}^{+ 4}$ is a strong oxidant reverting to the common +3 state.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

Explanation:

Both statements are true. As ${Ce}^{+ 3}$ is more stable than ${Ce}^{+ 4}, {Ce}^{+ 4}$ has tendency to accept an electron and thus, it behaves as an strong oxidising agent. So, the correct option is (3).

JEE Main - 2024

CHXII08:THE D- & F-BLOCK ELEMENTS

321744 Which among the following lanthanoids forms most stable +2 oxidation state?

1

Eu(Z  =  63)

2

Pm(Z  =  61)

3

Ce(Z  =  58)

4

Gd(Z  =  64)

Explanation:

$\begin{matrix} Ion & Valence shell \\ \begin{matrix} electronic \\ configuration \end{matrix} \\ E u^{2 +} & 4 f^{7} \\ P m^{2 +} & 4 f^{5} \\ C e^{2 +} & 4 f^{2} \\ G d^{2 +} & 4 f^{7} 5 d^{1} \end{matrix}$
Due to stable half filled 4 f orbitals, ${Eu}^{2 +}$ will be most stable, among given.

MHTCET - 2021

CHXII08:THE D- & F-BLOCK ELEMENTS

321745 What is the number of unpaired electrons present in f-orbital at +3 oxidation state of $Lu (Z = 71)$ ?

1 7

2 2

3 4

4 0

Explanation:

$_{71} Lu : [Xe] 4 f^{14} 5 d^{1} 6 s^{2}$
${Lu}^{3 +} : [Xe] 4 f^{14}$ (fulfilled f-orbitals)
Hence, number of unpaired electrons in f-orbital of ${Lu}^{3 +}$ is zero.

MHTCET - 2021

CHXII08:THE D- & F-BLOCK ELEMENTS

321746 Which of the following Lanthanoid ion is more stable in +2 oxidation state in aqueous solution?

1

{Eu}^{2 +}

2

{Ce}^{2 +}

3

{Lu}^{3 +}

4

{Fe}^{2 +}

CHXII08:THE D- & F-BLOCK ELEMENTS

321743 Read the Statement - A and Statement - B carefully to mark the correct options given below
Statement A :
In the Lanthanoids, the formation of ${Ce}^{+ 4}$ is favoured by its noble gas configuration.
Statement B :
${Ce}^{+ 4}$ is a strong oxidant reverting to the common +3 state.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

Explanation:

Both statements are true. As ${Ce}^{+ 3}$ is more stable than ${Ce}^{+ 4}, {Ce}^{+ 4}$ has tendency to accept an electron and thus, it behaves as an strong oxidising agent. So, the correct option is (3).

JEE Main - 2024

CHXII08:THE D- & F-BLOCK ELEMENTS

321744 Which among the following lanthanoids forms most stable +2 oxidation state?

1

Eu(Z  =  63)

2

Pm(Z  =  61)

3

Ce(Z  =  58)

4

Gd(Z  =  64)

Explanation:

$\begin{matrix} Ion & Valence shell \\ \begin{matrix} electronic \\ configuration \end{matrix} \\ E u^{2 +} & 4 f^{7} \\ P m^{2 +} & 4 f^{5} \\ C e^{2 +} & 4 f^{2} \\ G d^{2 +} & 4 f^{7} 5 d^{1} \end{matrix}$
Due to stable half filled 4 f orbitals, ${Eu}^{2 +}$ will be most stable, among given.

MHTCET - 2021

CHXII08:THE D- & F-BLOCK ELEMENTS

321745 What is the number of unpaired electrons present in f-orbital at +3 oxidation state of $Lu (Z = 71)$ ?

1 7

2 2

3 4

4 0

Explanation:

$_{71} Lu : [Xe] 4 f^{14} 5 d^{1} 6 s^{2}$
${Lu}^{3 +} : [Xe] 4 f^{14}$ (fulfilled f-orbitals)
Hence, number of unpaired electrons in f-orbital of ${Lu}^{3 +}$ is zero.

MHTCET - 2021

CHXII08:THE D- & F-BLOCK ELEMENTS

321746 Which of the following Lanthanoid ion is more stable in +2 oxidation state in aqueous solution?

1

{Eu}^{2 +}

2

{Ce}^{2 +}

3

{Lu}^{3 +}

4

{Fe}^{2 +}

CHXII08:THE D- & F-BLOCK ELEMENTS

321743 Read the Statement - A and Statement - B carefully to mark the correct options given below
Statement A :
In the Lanthanoids, the formation of ${Ce}^{+ 4}$ is favoured by its noble gas configuration.
Statement B :
${Ce}^{+ 4}$ is a strong oxidant reverting to the common +3 state.

1 Statement A is correct but Statement B is incorrect.

2 Statement A is incorrect but Statement B is correct.

3 Both statements are correct.

4 Both Statements are incorrect.

Explanation:

Both statements are true. As ${Ce}^{+ 3}$ is more stable than ${Ce}^{+ 4}, {Ce}^{+ 4}$ has tendency to accept an electron and thus, it behaves as an strong oxidising agent. So, the correct option is (3).

JEE Main - 2024

CHXII08:THE D- & F-BLOCK ELEMENTS

321744 Which among the following lanthanoids forms most stable +2 oxidation state?

1

Eu(Z  =  63)

2

Pm(Z  =  61)

3

Ce(Z  =  58)

4

Gd(Z  =  64)

Explanation:

$\begin{matrix} Ion & Valence shell \\ \begin{matrix} electronic \\ configuration \end{matrix} \\ E u^{2 +} & 4 f^{7} \\ P m^{2 +} & 4 f^{5} \\ C e^{2 +} & 4 f^{2} \\ G d^{2 +} & 4 f^{7} 5 d^{1} \end{matrix}$
Due to stable half filled 4 f orbitals, ${Eu}^{2 +}$ will be most stable, among given.

MHTCET - 2021

CHXII08:THE D- & F-BLOCK ELEMENTS

321745 What is the number of unpaired electrons present in f-orbital at +3 oxidation state of $Lu (Z = 71)$ ?

1 7

2 2

3 4

4 0

Explanation:

$_{71} Lu : [Xe] 4 f^{14} 5 d^{1} 6 s^{2}$
${Lu}^{3 +} : [Xe] 4 f^{14}$ (fulfilled f-orbitals)
Hence, number of unpaired electrons in f-orbital of ${Lu}^{3 +}$ is zero.

MHTCET - 2021

CHXII08:THE D- & F-BLOCK ELEMENTS

321746 Which of the following Lanthanoid ion is more stable in +2 oxidation state in aqueous solution?

1

{Eu}^{2 +}

2

{Ce}^{2 +}

3

{Lu}^{3 +}

4

{Fe}^{2 +}

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