d and f Block Elements — Page 10 | JEE Chemistry

Q91

The electronic configuration for Neodymium is: [Atomic Number for Neodymium 60]

A

[\mathrm{Xe}] 4 \mathrm{f}^4 6 \mathrm{~s}^2

B

[\mathrm{Xe}] 5 \mathrm{f}^4 7 \mathrm{~s}^2

C

[\mathrm{Xe}] 4 \mathrm{f}^6 6 \mathrm{~s}^2

D

[\mathrm{Xe}] 4 \mathrm{f}^1 5 \mathrm{~d}^1 6 \mathrm{~s}^2

Correct Answer

Option A

Solution

The electronic configuration of an element is determined by adding electrons to atomic orbitals following the Aufbau principle, the Pauli exclusion principle, and Hund's rule.

For atoms with atomic numbers higher than that of xenon (Xe, atomic number 54), the electrons start filling the outer orbitals following the xenon core.

In this case, we're looking at neodymium (Nd), which has an atomic number of 60.

To find the electronic configuration of neodymium, we first note the configuration of xenon, which serves as the core, and then add the additional electrons beyond xenon to the relevant orbitals.

Neodymium has 60 - 54 = 6 electrons more than xenon.

These additional electrons will fill the 4f and 6s orbitals.

Based on its position in the periodic table, neodymium’s electrons start filling the 4f orbitals before the 5d.

Since the 6s orbital is at a lower energy level than the 4f orbital, it gets filled before the 4f.

Neodymium will have 2 electrons in the 6s orbital, and the remaining 4 electrons will go into the 4f orbital.

Therefore, the correct electronic configuration for neodymium is:

[\mathrm{Xe}] 4 \mathrm{f}^4 6 \mathrm{~s}^2

So, Option A is the correct answer.

Q92

Given below are two statements : Statement (I) : In the Lanthanoids, the formation

\mathrm{Ce}^{+4}

is favoured by its noble gas configuration. Statement (II) :

\mathrm{Ce}^{+4}

is a strong oxidant reverting to the common +3 state. In the light of the above statements, choose the most appropriate answer from the options given below :

A Both Statement I and Statement II are false

B Statement I is true but Statement II is false

C Both Statement I and Statement II are true

D Statement I is false but Statement II is true

Correct Answer

Option C

Solution

Statement (1) is true,

\mathrm{Ce}^{+4}

has noble gas electronic configuration. Statement (2) is also true due to high reduction potential for

\mathrm{Ce}^{4+} / \mathrm{Ce}^{3+}(+1.74 \mathrm{~V})

, and stability of

\mathrm{Ce}^{3+}, \mathrm{Ce}^{4+}

acts as strong oxidizing agent.

Q93

Choose the correct option having all the elements with

\mathrm{d}^{10}

electronic configuration from the following :

A

{ }^{46} \mathrm{Pd},{ }^{28} \mathrm{Ni},{ }^{26} \mathrm{Fe},{ }^{24} \mathrm{Cr}

B

{ }^{29} \mathrm{Cu},{ }^{30} \mathrm{Zn},{ }^{48} \mathrm{Cd},{ }^{47} \mathrm{Ag}

C

{ }^{27} \mathrm{Co},{ }^{28} \mathrm{Ni},{ }^{26} \mathrm{Fe},{ }^{24} \mathrm{Cr}

D

{ }^{28} \mathrm{Ni},{ }^{24} \mathrm{Cr},{ }^{26} \mathrm{Fe},{ }^{29} \mathrm{Cu}

Correct Answer

Option B

Solution

\begin{aligned} & {[\mathrm{Cr}]=[\mathrm{Ar}] 4 \mathrm{~s}^1 3 \mathrm{~d}^5} \\ & {[\mathrm{Cd}]=[\mathrm{Kr}] 5 \mathrm{~s}^2 4 \mathrm{~d}^{10}} \\ & {[\mathrm{Cu}]=[\mathrm{Ar}] 4 \mathrm{~s}^1 3 \mathrm{~d}^{10}} \\ & {[\mathrm{Ag}]=[\mathrm{Kr}] 5 \mathrm{~s}^1 4 \mathrm{~d}^{10}} \\ & {[\mathrm{Zn}]=[\mathrm{Ar}] 4 \mathrm{~s}^2 3 \mathrm{~d}^{10}} \end{aligned}

Q94

Choose the correct statements from the following A.

\mathrm{Mn}_2 \mathrm{O}_7

is an oil at room temperature B.

\mathrm{V}_2 \mathrm{O}_4

reacts with acid to give

\mathrm{VO}_2{ }^{2+}

C.

\mathrm{CrO}

is a basic oxide D.

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

does not react with acid Choose the correct answer from the options given below :

A A, B and D only

B A, B and C only

C A and C only

D B and C only

Correct Answer

Option C

Solution

(A)

\mathrm{Mn}_2 \mathrm{O}_7

is green oil at room temperature. (B)

\mathrm{V}_2 \mathrm{O}_4

dissolve in acids to give

\mathrm{VO}^{2+}

salts. (C)

\mathrm{CrO}

is basic oxide (D)

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

is amphoteric it reacts with acid as well as base.

Q95

Identify correct statements from below: A. The chromate ion is square planar. B. Dichromates are generally prepared from chromates. C. The green manganate ion is diamagnetic. D. Dark green coloured

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

disproportionates in a neutral or acidic medium to give permanganate. E. With increasing oxidation number of transition metal, ionic character of the oxides decreases. Choose the correct answer from the options given below:

A B, D, E only

B A, B, C only

C A, D, E only

D B, C, D only

Correct Answer

Option A

Solution

A.

\mathrm{CrO}_4{ }^{2-}

is tetrahedral B.

2 \mathrm{Na}_2 \mathrm{CrO}_4+2 \mathrm{H}^{+} \rightarrow \mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7+2 \mathrm{Na}^{+}+\mathrm{H}_2 \mathrm{O}

C.

As per NCERT, green manganate is paramagnetic with 1 unpaired electron.

D.

Statement is correct E.

Statement is correct

Q96

The purest from of commercial iron is :

A pig iron

B wrought iron

C cast iron

D scrap iron and pig iron

Correct Answer

Option B

Solution

Wrought iron is purest from of commercial iron.

Q97

Which of the following statements are correct about

\mathrm{Zn}, \mathrm{Cd}

and

\mathrm{Hg}

? A. They exhibit high enthalpy of atomization as the d-subshell is full. B.

\mathrm{Zn}

and

\mathrm{Cd}

do not show variable oxidation state while

\mathrm{Hg}

shows

+\mathrm{I}

and

+\mathrm{II}

. C. Compounds of

\mathrm{Zn}, \mathrm{Cd}

and

\mathrm{Hg}

are paramagnetic in nature. D.

\mathrm{Zn}, \mathrm{Cd}

and

\mathrm{Hg}

are called soft metals. Choose the most appropriate from the options given below:

A C, D only

B B, C only

C A, D only

D B, D only

Correct Answer

Option D

Solution

(A)

\mathrm{Zn}, \mathrm{Cd}, \mathrm{Hg}

exhibit lowest enthalpy of atomization in respective transition series. (C) Compounds of

\mathrm{Zn}, \mathrm{Cd}

and

\mathrm{Hg}

are diamagnetic in nature.

Q98

The correct IUPAC name of

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

is

A Potassium tetraoxidomanganate (VI)

B Dipotassium tetraoxidomanganate (VII)

C Potassium tetraoxopermanganate (VI)

D Potassium tetraoxidomanganese (VI)

Correct Answer

Option A

Solution

\begin{aligned} & \mathrm{K}_2 \mathrm{MnO}_4 \\ & 2+\mathrm{x}-8=0 \\ & \Rightarrow \mathrm{x}=+6 \end{aligned}

\mathrm{O.S.}

of

\mathrm{Mn}=+6

IUPAC Name $=$ Potassium tetraoxidomanganate(VI)

Q99

The orange colour of

\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7

and purple colour of

\mathrm{KMnO}_4

is due to

A Charge transfer transition in both.

B

\mathrm{d} \rightarrow \mathrm{d}

transitions in

\mathrm{KMnO}_4

and charge transfer transitions in

\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7

.

C

\mathrm{d} \rightarrow \mathrm{d}

transitions in both

D

\mathrm{d} \rightarrow \mathrm{d}

transitions in

\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7

and charge transfer transitions in

\mathrm{KMnO}_4

.

Correct Answer

Option A

Solution

\left.\begin{array}{l} \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7 \rightarrow \mathrm{Cr}^{+6} \rightarrow \text{ No d - d transition } \\ \mathrm{KMnO}_4 \rightarrow \mathrm{Mn}^{7+} \rightarrow \text{ No d - d transition } \end{array}\right\} \text{ Charge transfer }

Q100

Diamagnetic Lanthanoid ions are :

A

\mathrm{La}^{3+} \& \mathrm{~Ce}^{4+}

B

\mathrm{Nd}^{3+} \& \mathrm{~Ce}^{4+}

C

\mathrm{Lu}^{3+} \& \mathrm{~Eu}^{3+}

D

\mathrm{Nd}^{3+} \& \mathrm{~Eu}^{3+}

Correct Answer

Option A

Solution

\mathrm{Ce}:[\mathrm{Xe}] 4 \mathrm{f}^1 5 \mathrm{~d}^1 6 \mathrm{~s}^2 ; \mathrm{Ce}^{4+}

diamagnetic

\mathrm{La}:[\mathrm{Xe}] 4 \mathrm{f}^0 5 \mathrm{~d}^1 6 \mathrm{~s}^2 ; \mathrm{La}^{3+}

diamagnetic