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d and f Block Elements

JEE Chemistry · 278 questions · Page 11 of 28 · Click an option or "Show Solution" to reveal answer

Q101
The electronic configuration of Einsteinium is : (Given atomic number of Einsteinium =99=99)
A [Rn]5f136 d∘7 s2[\mathrm{Rn}] 5 \mathrm{f}^{13} 6 \mathrm{~d}^{\circ} 7 \mathrm{~s}^2
B [Rn]5f126 d∘7 s2[\mathrm{Rn}] 5 \mathrm{f}^{12} 6 \mathrm{~d}^{\circ} 7 \mathrm{~s}^2
C [Rn]5f116 d∘7 s2[\mathrm{Rn}] 5 \mathrm{f}^{11} 6 \mathrm{~d}^{\circ} 7 \mathrm{~s}^2
D [Rn]5f106 d∘7 s2[\mathrm{Rn}] 5 \mathrm{f}^{10} 6 \mathrm{~d}^{\circ} 7 \mathrm{~s}^2
Correct Answer
Option C
Solution

The correct option for the electronic configuration of Einsteinium (atomic number 99) is: Option C: [Rn] 5f11 6d0 7s2 Here's why: Atomic Number and Electrons: Einsteinium has an atomic number of 99, which means it has 99 protons in its nucleus.

Since atoms are neutral, it also has 99 electrons surrounding the nucleus.

Electron Shells: Electrons fill up energy shells around the nucleus.

The order of filling follows the Aufbau principle, which generally fills lower energy shells before moving to higher ones.

Noble Gas Core Notation: The notation [Rn] represents the filled electron configuration of Radon (atomic number 86).

This is a shorthand to avoid writing out the entire configuration of the inner shells, as they are not involved in the chemical behavior of Einsteinium.

Filling the f-orbital: Einsteinium is an actinide element, located in the f-block of the periodic table.

The 5f subshell fills up after the 6s subshell.

Aufbau Principle: According to the Aufbau principle, the 5f subshell can hold up to 14 electrons.

In Einsteinium, 11 electrons fill the 5f subshell (5f11).

Empty d-orbital: The 6d subshell has a higher energy level than the 5f subshell.

Since the 5f subshell isn't completely filled, the 6d subshell remains empty (6d0).

Outermost Electrons: The remaining two electrons fill the outermost 7s subshell (7s2).

Therefore, the complete electronic configuration of Einsteinium is: [Rn] 5f11 6d0 7s2

Q102
Give below are two statements : Statement I : The higher oxidation states are more stable down the group among transition elements unlike p-block elements. Statement II : Copper can not liberate hydrogen from weak acids. In the light of the above statements, choose the correct answer from the options given below :
A Statement I is true but Statement II is false
B Both Statement I and Statement II are 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

To evaluate the given statements, let's analyze each statement in the context of chemical principles: Statement I: The higher oxidation states are more stable down the group among transition elements unlike p-block elements.

This statement is true.

In transition elements, as we move down the group, the electrons are being added to the n-1 d orbitals, where 'n' represents the principal quantum number.

The ability to exhibit higher oxidation states stems from the fact that not only are the s electrons of the outermost shell used in bonding, but the d electrons are also involved.

Moreover, down the group, the effective nuclear charge decreases, and the shielding effect increases, so electrons from both s and d orbitals can be easily used for bonding, leading to stable higher oxidation states.

This trend is opposite in the case of p-block elements, where the stability of higher oxidation states generally decreases down the group due to the inert pair effect, which makes it hard to remove electrons from the s orbital as we move down the group.

Statement II: Copper can not liberate hydrogen from weak acids.

This statement is also true.

Copper (Cu) is less reactive and lies below hydrogen in the electrochemical series, meaning it has a higher reduction potential than hydrogen.

For a metal to displace hydrogen from an acid, it must have a lower reduction potential than hydrogen.

This is not the case with copper.

Therefore, copper does not react with weak acids (like acetic acid) to liberate hydrogen gas.

This behavior can be explained through the electrochemical series, where metals placed above hydrogen are capable of displacing hydrogen from acids (exemplified by metals such as zinc or magnesium), whereas copper does not have this ability due to its relative placement in the series.

Based on the analysis, Option C is the correct answer: Both Statement I and Statement II are true.

Q103
The element which shows only one oxidation state other than its elemental form is :
A Nickel
B Titanium
C Cobalt
D Scandium
Correct Answer
Option D
Solution

The element among the given options that shows only one oxidation state other than its elemental form is Scandium.

Explanation: Nickel (Ni) can have multiple oxidation states, including +2 and +3, with +2 being the most common.

Titanium (Ti) also shows various oxidation states, most commonly +4 and +2, due to its ability to lose electrons from both its 3d and 4s subshells.

Cobalt (Co) commonly exists in the +2 and +3 oxidation states, among others, due to electron transitions in its 3d orbitals.

Scandium (Sc), however, primarily exhibits a +3 oxidation state in its compounds (excluding its 0 state in the elemental form).

This is because scandium has an electronic configuration of [Ar]3d14s2, and it loses all three outer electrons to achieve a stable configuration, resulting in a +3 oxidation state.

It does not commonly exhibit other oxidation states because removing more than three electrons would require removing electrons from the very stable noble gas core configuration ([Ar]), which is energetically unfavorable.

Therefore, the correct answer is Option D: Scandium.

Q104
Iron (III) catalyses the reaction between iodide and persulphate ions, in which A. Fe3+\mathrm{Fe}^{3+} oxidises the iodide ion B. Fe3+\mathrm{Fe}^{3+} oxidises the persulphate ion C. Fe2+\mathrm{Fe}^{2+} reduces the iodide ion D. Fe2+\mathrm{Fe}^{2+} reduces the persulphate ion Choose the most appropriate answer from the options given below:
A B and C only
B A only
C A and D only
D B only
Correct Answer
Option C
Solution

To determine the correct answer to this question, let's analyze the catalysis mechanism involving Iron (III) in the reaction between iodide (

I−\mathrm{I^-}

) and persulfate (

S2O82−\mathrm{S_2O_8^{2-}}

) ions. Iron (III), or

Fe3+\mathrm{Fe}^{3+}

, can act as a catalyst by undergoing redox reactions, cycling between

Fe2+\mathrm{Fe}^{2+}

and

Fe3+\mathrm{Fe}^{3+}

. Here’s how it typically works: 1.

Fe3+\mathrm{Fe}^{3+}

can oxidize iodide ions to iodine. The reaction will proceed as follows:

2Fe3++2I−→2Fe2++I2\mathrm{2Fe^{3+} + 2I^- \rightarrow 2Fe^{2+} + I_2}

2.

Fe2+\mathrm{Fe}^{2+}

, now formed, can be oxidized back to

Fe3+\mathrm{Fe}^{3+}

by reducing persulfate ions. The reaction will be:

2Fe2++S2O82−→2Fe3++2SO42−\mathrm{2Fe^{2+} + S_2O_8^{2-} \rightarrow 2Fe^{3+} + 2SO_4^{2-}}

Based on this mechanism,

Fe3+\mathrm{Fe}^{3+}

is responsible for oxidizing the iodide ions, and

Fe2+\mathrm{Fe}^{2+}

is responsible for reducing the persulfate ions. Thus, we can conclude: A.

Fe3+\mathrm{Fe}^{3+}

oxidises the iodide ion: This statement is correct. B.

Fe3+\mathrm{Fe}^{3+}

oxidises the persulphate ion: This statement is incorrect because

Fe3+\mathrm{Fe}^{3+}

does not oxidize persulfate ions. C.

Fe2+\mathrm{Fe}^{2+}

reduces the iodide ion: This statement is incorrect because

Fe2+\mathrm{Fe}^{2+}

does not reduce iodide ions. D.

Fe2+\mathrm{Fe}^{2+}

reduces the persulphate ion: This statement is correct.

Therefore, the most appropriate answer is: Option C: A and D only

Q105
The metal that shows highest and maximum number of oxidation state is :
A Co
B Mn
C Fe
D Ti
Correct Answer
Option B
Solution

The metal that shows the highest and maximum number of oxidation states among the options given is manganese (Mn), which is option B.

Manganese exhibits a wide range of oxidation states from +2 to +7, which is more diverse than the other metals mentioned.

To further detail: Co (Cobalt) typically exhibits oxidation states of +2 and +3, with +2 being the most common.

Although it can theoretically show +1 and +4 in very specific compounds, these are not as stable or as common as the states displayed by manganese.

Mn (Manganese) shows a remarkable series of oxidation states from +2 to +7 in its compounds, making it the element with the highest number of stable oxidation states among the options listed.

For example, MnO (in the +2 state), Mn2O3 (in the +3 state), MnO2 (in the +4 state), Mn2O7 (in the +7 state), and so forth.

Fe (Iron) commonly exhibits +2 and +3 oxidation states, such as in FeSO4 (in the +2 state) and Fe2O3 (in the +3 state).

Though there are compounds with Fe in the +6 oxidation state, like potassium ferrate (K2FeO4), these are not as common or stable as the lower oxidation states.

Ti (Titanium) shows oxidation states of +2, +3, and +4.

Among these, the +4 oxidation state (as in TiO2) is the most stable and common.

Though it offers a variety of states, it doesn't reach the variety or the maximum oxidation state shown by manganese.

Therefore, the correct answer is Option B: Mn (Manganese), which can show the highest and maximum number of oxidation states among the options provided.

Q106
While preparing crystals of Mohr's salt, dil H2SO4\mathrm{H}_2 \mathrm{SO}_4 is added to a mixture of ferrous sulphate and ammonium sulphate, before dissolving this mixture in water, dil H2SO4\mathrm{H_2SO_4} is added here to :
A prevent the hydrolysis of ferrous sulphate
B prevent the hydrolysis of ammonium sulphate
C increase the rate of formation of crystals
D make the medium strongly acidic
Correct Answer
Option A
Solution

While preparing crystal of Mohr's salt, dil.

H2SO4\mathrm{H}_2 \mathrm{SO}_4

is added to prevent hydrolysis of ferrous sulphate.

Q107
The metal that can be purified economically by fractional distillation method is :
A Fe
B Zn
C Cu
D Ni
Correct Answer
Option B
Solution

Zinc can be purified economically by fractional distillation.

Fractional distillation process utilises the boiling point difference between metal and that of impurity.

Using this process, crude zinc containing Cd, Fe and Pb as impurities can be refined.

Q108
Consider two chemical reactions (A) and (B) that take place during metallurgical process: (A) ZnCO3(s)⟶ΔZnO(s)+CO2(g)ZnC{O_{3(s)}}\overset{\Delta}\longrightarrow Zn{O_{(s)}} + C{O_{2(g)}} (B) 2ZnS(s)+3O2(g)⟶Δ2ZnO(s)+2SO2(g)2Zn{S_{(s)}} + 3{O_{2(g)}}\overset{\Delta}\longrightarrow 2Zn{O_{(s)}} + 2S{O_{2(g)}} The correct option of names given to them respectively is :
A (A) is calcination and (B) is roasting
B Both (A) and (B) are producing same product so both are roasting
C Both (A) and (B) are producing same product so both are calcination
D (A) is roasting and (B) is calcination
Correct Answer
Option A
Solution

(A)

ZnCO3(s)⟶ΔZnO(s)+CO2(g)ZnC{O_{3(s)}}\overset{\Delta}\longrightarrow Zn{O_{(s)}} + C{O_{2(g)}}

Heating in absence of oxygen in calcination. (B)

2ZnS(s)+3O2(g)⟶Δ2ZnO(s)+2SO2(g)2Zn{S_{(s)}} + 3{O_{2(g)}}\overset{\Delta}\longrightarrow 2Zn{O_{(s)}} + 2S{O_{2(g)}}

Heating in presence of oxygen in roasting. Hence, (A) is calcination while (B) in roasting.

Q109
Sulphide ion is soft base and its ores are common for metals. (a) Pb (b) Al (c) Ag (d) Mg
A (a) and (c) only
B (a) and (d) only
C (a) and (b) only
D (c) and (d) only
Correct Answer
Option A
Solution

Pb and Ag commonly exist in the form of sulphide ore like Pbs (galena) and Ag2S (Argentite) 'Al' is mainly found in the form of oxide ore whereas 'Mg' is found in the form of halide ore.

Q110
In the leaching of alumina from bauxite, the ore expected to leach out in the process by reacting with NaOH is :
A TiO2
B Fe2O3
C ZnO
D SiO2
Correct Answer
Option D
Solution

In bauxite impurities of Fe2O3, TiO2 and SiO2 are present, Fe2O3 and TiO2 are basic oxides therefore does not reacts with or dissolve in NaOH whereas SiO2 is acidic oxide it gets dissolve in NaOH, hence leach out

SiO2+2NaOH→Na2SiO3(aq.)+H2OSi{O_2} + 2NaOH \to N{a_2}Si{O_3}(aq.) + {H_2}O
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