d and f Block Elements

JEE Chemistry · 278 questions · Page 13 of 28 · Click an option or "Show Solution" to reveal answer

Q121
Identify the element for which electronic configuration in +3 oxidation state is [Ar]3d5 :
A Ru
B Mn
C Co
D Fe
Correct Answer
Option D
Solution

Mn(25) = [Ar]3d54s2 Mn+3 = [Ar]3d44s0 Ru-belongs to 4d transition series Co (27) = [Ar]3d74s2 Co+3 = [Ar]3d64s0 Fe(26) = [Ar]3d64s2 Fe+3 = [Ar]3d54s0

Q122
Which of the following acts as a strong reducing agent? (Atomic number: Ce=58,Eu=63,Gd=64,Lu=71\mathrm{Ce}=58, \mathrm{Eu}=63, \mathrm{Gd}=64, \mathrm{Lu}=71)
A Eu2+\mathrm{Eu}^{2+}
B Gd3+\mathrm{Gd}^{3+}
C Lu3+\mathrm{Lu}^{3+}
D Ce4+\mathrm{Ce}^{4+}
Correct Answer
Option A
Solution

Eu2+\mathrm{Eu}^{2+} is a strong reducing agent changing to the common +3 state while all other are provided at their higher oxidation states.

Q123
The first transition series metal ' MM ' has the highest enthalpy of atomisation in its series. One of its aquated ion (Mn+)\left(\mathrm{M}^{\mathrm{n}+}\right) exists in green colour. The nature of the oxide formed by the above Mn+\mathrm{M}^{\mathrm{n}+} ion is:
A basic
B neutral
C amphoteric
D acidic
Correct Answer
Option A
Solution

In the 3d transition series, Vanadium (V) has the highest enthalpy of atomization.

The ion V3+\mathrm{V}^{3+} is known to exhibit a green color in its aqueous form.

The oxide formed by the V3+\mathrm{V}^{3+} ion is V2O3\mathrm{V}_2 \mathrm{O}_3, which is classified as a basic oxide.

Q124
Which one of the following ores is best concentrated by froth floatation method?
A Siderite
B Galena
C Malachite
D Magnetite
Correct Answer
Option B
Solution

Froth floatation method is mainly applicable for sulphide ores.

(1)(1)

Malachite ore :

Cu(OH)2,CuCO3Cu{\left( {OH} \right)_2},\,\,CuC{O_3}
(2)(2)

Magnetite ore :

Fe3O4F{e_3}{O_4}
(3)(3)

Siderite ore :

FeCO3FeC{O_3}
(4)(4)

Galena ore :

PbSPbS

(Sulphide Ore)

Q125
Given below are two statements: Statement I : CrO3\mathrm{CrO}_3 is a stronger oxidizing agent than MoO3\mathrm{MoO}_3 Statement II : Cr(VI)\mathrm{Cr}(\mathrm{VI}) is more stable than Mo(VI)\mathrm{Mo}(\mathrm{VI}) 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 true
C Both Statement I and Statement II are false
D Statement I is false but Statement II is true
Correct Answer
Option A
Solution

Statement I is true while Statement II is false.

Cr(VI) is less stable than Mo(VI).

Consequently, CrO₃ more readily reduces to Cr³⁺ compared to MoO₃, exhibiting a stronger oxidizing nature.

Q126
Which one of the following ores is best concentrated by froth – floatation method?
A Magnetite
B Malachite
C Galena
D Cassiterite
Correct Answer
Option C
Solution

NOTE : Galena is PbS and thus purified by froth floatation method.

Froath floatation method is used to concentrate sulphide ores.

This method is based on the preferential wetting properties with the froathing agent and water.

Q127
During the process of electrolytic refining of copper, some metals present as impurity settle as ‘anode mud’ These are
A Sn and Ag
B Pb and Zn
C Ag and Au
D Fe and Ni
Correct Answer
Option C
Solution

During the process of electrolytic refining Ag and Au are obtained as anode mud.

Q128
The idea of froth floatation method came from a person X and this method is related to the process Y of ores. X and Y, respectively, are :
A washer woman and concentration
B fisher woman and concentration
C wisher man and reduction
D fisher man and reduction
Correct Answer
Option A
Solution

Froth floatation is a method of concentration and it was discovered by a washer women.

Q129
The pair that does NOT require calcination is -
A Fe2O3 and CaCO3⋅MgCO3
B ZnCO3 and CaO
C ZnO and Fe2O3⋅xH2O
D ZnO and MgO
Correct Answer
Option D
Solution

In calcination, we convert a compound or ore into oxide.

In option (D) you can see there is ZnO and MgO, as they are already in oxide form so no calcination required for then.

Q130
In the Hall-Heroult process, aluminium is formed at the cathode. The cathode is made out of -
A Copper
B Platinum
C Carbon
D Pure aluminium
Correct Answer
Option C
Solution

In the Hall-Heroult process the cathode is made of carbon.

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