MCQ Questions for Class 12 Engineering Chemistry The D-And F-Block Elements Quiz 4

Zinc chromate and lead chromate are used as:

0%
lakes
0%
lacquers
0%
mordants
0%
pigments

Explanation

Zinc Chromate (ZnCrO

$_4$ ) and Lead Chromate (PbCrO

$_4$ ) are used as pigments, paints, spray paints, artist paints , etc.

The correct statement for d-block element is:

0%
It shows magnetic property
0%
It has variable valency
0%
It has tendency of formation of colored ions
0%
It has complete d-orbitals

The atomic numbers of elements of second inner transition elements lie in the range of :

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88 to 101
0%
89 to 102
0%
90 to 103
0%
91 to 104

The chemical similarity between boron and silicon mainly due to equal value of their:

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electronegativity
0%
nuclear charge
0%
charge to (ionic radius) $^{2}$ ratio
0%
atomic volume

Beryllium resembles Aluminium in properties. This is mainly due to:

0%
equal electro negativity values of elements
0%
equal atomic volumes of the elements
0%
equal electron affinity
0%
equal nuclear charges in their atoms

The starting element in the incomplete period of long form of periodic table

0%
$Cs$
0%
$Rn$
0%
$Fr$
0%
$Th$

Elements which generally exhibit variable oxidation states and form colored ions are ___________.

0%
metalloids
0%
transition elements
0%
non-metals
0%
gases

Most of the radio active elements are present in:

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Lanthanoids
0%
Actinoids
0%
Representative elements
0%
Second transition series

The element californium belongs to a family of :

0%
actinide series
0%
alkali metal family
0%
alkaline earth family
0%
lanthanide series

Explanation

Hint: It belongs to f-block element.

Correct answer: $A$

Explanation for correct option :

In the periodic table, Actinoid element or the actinide element, any of a series of

$15$ consecutive chemical elements from actinium to lawrencium (atomic numbers

$(89-103$ ).

The atomic number of californium is

$98$

Let's write an electronic configuration for this

its electronic configuration is

$Rn^{86} 7s^2 5f^{10}$

Hence, it is an f-block element and as it is in the 7th period, it is a part of the actinide series.

Explanation for correct options :

Alkali and alkaline earth metals belong to s-block elements hence these both options are incorrect
Lanthanide series have elements $57$ to $71$ but californium has atomic number $98$ , hence it not belongs to lanthanide series.

Transition metals are often paramagnetic owing to:

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high m.p. and b.p. of transition metals
0%
presence of vacant d-orbitals
0%
presence of one or more unpaired d-electrons
0%
their less electropositive nature as compared to the elements of groups IA and IIA

Explanation

The general outer electronic configuration of d-block elements is

$(n-1)d^{1-10}\ ns^{1-2}$ . Hence, d - block elements often have one or more unpaired d - electrons.

Due to presence of these unpaired electrons, transition elements show paramagnetism.

Therefore transition elements are often paramagnetic due to presence of one or more unpaired d electrons.

Option C is correct.

A reduction in atomic size with increase in atomic number is a characteristic of elements of:

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d-block
0%
f-block
0%
radioactive series
0%
high atomic masses

Lanthanoid contraction occurs because

0%
the 4f electrons, which are gradually added, create strong shielding effect
0%
the 4f orbitals are greater in size than the 3d and 3f orbitals
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the 5f orbitals strongly penetrate into the 4f orbitals
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the poor shielding effect of 4f electrons is coupled with increased attraction between the nucleus and the added electrons

Explanation

Lanthanoid contraction occurs because the poor shielding effect of 4f electrons is coupled with increased attraction between the nucleus and the added electrons.

The atomic and ionic radii (

$M^{3+}$ ions) of Lanthanide elements decrease with increase in atomic number.

Which of the following has smallest ionic radius ?

0%
Lu $^{3+}$
0%
Nd $^{3+}$
0%
Dy $^{3+}$
0%
Pm $^{3+}$

Explanation

$Lu^{3+}$ is the last of the Lanthanides, so it's ions will have the smallest ionic radius due to Lanthanide contraction.

The atomic and ionic radii (

$M^{3+}$ ions) of Lanthanide elements decrease with increase in atomic number. This effect is called as:

0%
Lanthanoid contraction
0%
Lanthanoid expansion
0%
Actinoid contraction
0%
none of these

Explanation

The atomic and ionic radii (

$M^{3+}$ ions) of Lanthanide elements decrease with increase in atomic number.

This effect is called Lanthanoid contraction. This is due to poor shielding of f electrons.

The separation of Lanthanides by ion exchange method is based on:

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basicity of the hydroxides
0%
size of the ions
0%
the solubility of their nitrates
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oxidation state of the ion

Which element among the Lanthanides has the smallest atomic radius ?

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Cerium
0%
Lutetium
0%
Europium
0%
Gadolinium.

Which of the following belongs to the actinide series of elements?

0%
Y
0%
Ta
0%
U
0%
Lu

The radius of

$La^{3+}$ (At. no. of La = 57) is 1.06A. Which one of the following given values will be closest to the radius of

$Lu^{3+}$ ? (At. no. of Lu = 71)

0%
1.40A
0%
1.06A
0%
0.85A
0%
1.60A

Explanation

A reduction in atomic/ionic size with increase in atomic number is a characteristic of lanthanoids. Hence, the ionic radius of

$\displaystyle Lu^{3+}$ will be smaller than the ionic radius of

$\displaystyle La^{3+}$ . It will be less than 1.06 A. Hence, the option C, 0.85 A is the correct answer.

The Lanthanoid contraction refers to decrease in the:

0%
valence electrons of the Lanthanide series
0%
ionic radius of the series
0%
the density of the series
0%
nuclear mass of the series

Transuranic elements begin with :

0%
Np
0%
Cm
0%
Pu
0%
U

Differentiating electron in inner transition elements enters the ________ orbital.

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s
0%
p
0%
d
0%
f

Many Lanthanoid elements are used to prepare:

0%
ceramic materials
0%
water softener
0%
superconducting materials
0%
enzyme catalysts

Explanation

Many Lanthanoid elements are used to prepare superconducting materials.

Examples include Lanthanum sesquicarbide superconductor

$La_2C_3$ . Gadolinium at very low temperatures becomes highly magnetic and may function as a superconductor.

The correct order of ionic radii of

$Yb^{3+},La^{3+},Eu^{3+}$ and

$Lu^{3+}$ is:

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$Yb^{3+}> Lu^{3+}> Eu^{3+}> La^{3+}$
0%
$Lu^{3+}< Yb^{3+}< Eu^{3+}< La^{3+}$
0%
$Lu^{3+}< Eu^{3+}< La^{3+}< Yb^{3+}$
0%
$La^{3+}< Eu^{3+}< Lu^{3+}< Yb^{3+}$

Explanation

In periodic table, as we move from left to right in any period, the atomic and ionic radii decreases. This is due to the increase in a nuclear charge in a period.

Thus the correct order of ionic radii in Lanthanoid is:

Ionic radii order:

$Lu^{3+}< Y^{3+}< Eu^{3+}< La^{3+}$

Which of the following statement regarding transition elements is incorrect?

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Their atoms contain partially filled d orbitals.
0%
They are capable of showing variable valencies.
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All of their ions are colourless.
0%
They form complexes readily.

Catalytic activity of transition elements and their compounds is due to their _______.

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small size
0%
vacant d-orbitals
0%
higher densities
0%
colour

Which has the highest ionic radii among given ions?

$Ce^{3+}, La^{3+}, Pm^{3+}$ and

$Yb^{3+}$

0%
$Yb^{3+}$
0%
$Ce^{3+}$
0%
$La^{3+}$
0%
$Pm^{3+}$

Explanation

Electronic configuration of ions are:

$Yb^{ 3+ }:Xe4f^{ 13 }$

$Ce^{ 3+ }:Xe4f^{ 1 }$

$La^{ 3+ }:Xe4f^0$

$Pm^{ 3+ }:Xe4f^4$

Since, the least shielding tendency of f-orbital electron, effective nuclear charge increases and therefore size decreases along the period. Therefore

$La^{ 3+ }$ has highest ionic radii.

This phenomenon is called lanthanoid contraction.

Which of the following compounds is most widely used as an oxidizing agent in the volumetric analysis?

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$CeO_{2}$
0%
$(NH_{4})_{2}[Ce(NO_{3})_{6}]$
0%
$Ce(SO_{4})_{2}$
0%
$CeF_{3}$

Explanation

Ceric sulphate

$Ce(SO_{4})_{2}$ is most widely used as an oxidizing agent in the volumetric analysis.

Ce has +3 and +4 oxidation states.

$Ce^{+4}$ readily gains an electron to form

$Ce^{+3}$ and is reduced. Hence, it easily oxidizes other substances.

The following belongs to d-block but it is not transition element

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Mn
0%
Fe
0%
Zn
0%
Cr

Chemically Zinc group elements closely resemble ______.

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I A group
0%
II A group
0%
III A group
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IV A group

The elements which exhibit both vertical and horizontal similarities are _______.

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inert gas elements
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representative elements
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rare elements
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transition elements

Which of the following is used as Catalyst in the hydrogenation of oil?

0%
$V_{2}O_{5}$
0%
Pd
0%
Fe
0%
Ni

d -block elements form complexes because they have:

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vacant orbitals
0%
small sizes
0%
higher nuclear charge
0%
all of the above

Which of the following is not an interstitial compound?

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$TiC$
0%
$MoC$
0%
$Fe_{0.82}O$
0%
$Cr_{2}O_{3}$

Explanation

$TiC$ ,

$MoC$ and

$Fe_{0.82}O$ are interstitial compounds. However,

$Cr_{2}O_{3}$ is not an interstitial compound.
Interstitial compounds are those compounds which are formed when small atoms like

$H,\ C$ or

$N$ are trapped inside the interstitial spaces in the crystal lattice of metals. They are usually non-stoichiometric in nature and cannot be represented by a definite composition and are neither typically ionic nor covalent.

The property seen in f-block elements is :

0%
Lanthanoid contraction
0%
Actinide contraction
0%
both (A) and (B)
0%
Lanthanoid expansion

Many lanthanoid elements are used to prepare:

0%
ceramic materials
0%
water softener
0%
superconducting materials
0%
enzyme catalysts

Oil is converted into fat by using ____ as a catalyst.

0%
Ni
0%
Fe
0%
Mn
0%
$V_{2}O_{5}$

Magnetic moment of diamagnetic substance in Bohr Magnetons is:

0%
1.73
0%
2.83
0%
50
0%
0

The atomic and ionic radii (M $^{3+}$ ions) of Lanthanide elements decrease with increase in atomic number. This effect is called:

0%
Lanthanoid contraction
0%
f-f transition
0%
Actinoid contraction
0%
shielding effect

The magnetic moment of an ion in its +3 oxidation state is 3.85 BM. The number of unpaired electrons present in the ion is _____________.

0%
1
0%
4
0%
3
0%
5

Explanation

Given that magnetic moment is 3.85 BM

Magnetic moment =

$\sqrt{n(n+2)}$

where n is no. of unpaired electron.

$\therefore$ 3.85 =

$\sqrt{n(n+2)}$

No. of unpaired electron present in ion is 3.

Hence, the correct option is

$C$

In dichromate ion ________________________.

0%
4 Cr-O bonds are equivalent
0%
6 Cr-O bonds are equivalent
0%
8 Cr-O bonds are equivalent
0%
all Cr-O bonds are equivalent

Arrange $Ce^{3+}, La^{3+}, Pm^{3+}$ and $Yb^{3+}$ in increasing order of their ionic radii.

0%
$Yb^{3+} < Pm^{3+} < Ce^{3+} < La^{3+}$
0%
$Ce^{3+} < Yb^{3+} < Pm^{3+} < La^{3+}$
0%
$Yb^{3+} < Pm^{3+} < La^{3+} < Ce^{3+}$
0%
$Pm^{3+} < La^{3+} < Ce^{3+} < Yb^{3+}$

Explanation

As we move from left to right in a period, radii decrease due to an increase in nuclear charge and lanthanide contraction.

So the order is

$Yb^{3+} < Pm^{3+} < Ce^{3+} < La^{3+}.$

The radius of $La^{3+}$ (atomic number of La $=$ 57) is 1.06A. Which one of the following given values will be closest to the radius of $Lu^{3+}$ ? (atomic number of Lu=71)

0%
1.40A
0%
1.06A
0%
0.85A
0%
1.60A

Explanation

As we move from left to right in a raw, radii decreases.
So radii of

$Lu^{+3}$ should be lesser than 1.06A.

Transition elements show positive oxidation state, generally, due to:

0%
large atomic size
0%
low ionization energy
0%
low electronegativity
0%
high electronegativity

Which one of the following ion exhibits colour in aqueous solution ?

0%
Sc $^{3+}$
0%
Ni $^{2+}$
0%
Ti $^{4+}$
0%
Zn $^{+2}$

Explanation

Only

$Ni^{+2}$ has incomplete d-orbitals

$d^8$ so it is colourfull.

$Sc^{+3}, Ti^{+4}; d^0$ configuration and

$Zn^{+2}, d^{10}$ configuration so they are colourless.

Select the correct order with respect to paramagnetic property of the ions.

0%
$Mn^{2{+}} > Cr^{2+} > V^{2{+}} > Ti^{2+}$
0%
$Ti^{2{+}} > V^{2{+}} > Mn^{2{+}} > Cr^{{2+}}$
0%
$V^{2{+}} > Ti^{2{+}} > Mn^{2{+}} > Cr^{2{+}}$
0%
$Ti^{2{+}} > V^{2{+}} > Mn^{2{+}} > Cr^{2{+}}$

Explanation

The expression for magnetic moment is

$\sqrt{n(n+2)}$ , where

$n$ is number of unpaired electrons.

So, greater is the number of unpaired electrons, greater will be the magnetic moment of an ion.

$Mn^{2+}$ has

$5$ unpaired electrons.

$Cr^{2+}$ has

$4$ unpaired electrons.

$V^{2+}$ has

$3$ unpaired electrons.

$Ti^{2+}$ has

$2$ unpaired electrons.

Hence, the correct order is as follows:

$Mn^{2+}>Cr^{2+}>V^{2+}>Ti^{2+}$

Variable valency of transition elements is on account of :

0%
incomplete s - orbitals
0%
incomplete d - orbitals
0%
completely filled d - orbitals
0%
incomplete p-orbitals

Which of the following pair has the same size?

0%
Fe $^{2+}$ , Ni $^{2+}$
0%
Zr $^{4+}$ , Ti $^{4+}$
0%
Zr $^{4+}$ , Hf $^{4+}$
0%
Zn $^{2+}$ , Hf $^{4+}$

Explanation

$Zr^{+4}$ and

$Hf^{4+}$ have same size due to lanthanide contraction.

Lanthanide contraction is caused due to poor shielding effect of the

$4f$ electrons.

Colour in transition metal compounds is attributed to :

0%
small size of metal ions
0%
absorption of light in UV region
0%
complete $(ns)$ subshell
0%
incomplete $(n-1)d$ subshell

Transition metals are less reactive because of their :

0%
High ionization potential and low melting point
0%
High ionization potential and high melting point
0%
Low ionization potential and low melting point
0%
Low ionization potential and high melting point

Which of the following arrangements does not represent the correct order of the property stated against it?

0%
$Ni^{2+} < Co^{2+} < Fe^{2+} < Mn^{2+}$ : Ionic size
0%
$Co^{3+} < Fe^{3+} < Cr^{3+} < Sc^{3+}$ : Stability in aqueous solution
0%
$Sc < Ti < Cr < Mn$ : Number of oxidation states
0%
$V^{2+} < Cr^{2+} < Mn^{2+} < Fe^{2+}$ : Paramagnetic behaviour

Explanation

The exothermic hydration enthalpies of the given trivalent cations are:

$Sc^{+3}=3960kJ/mole$

$Fe^{+3}=4429kJ/mole$

$Co^{+3}=4653kJ/mole$

$Cr^{+3}=4563kJ/mole$
Hence,

$Sc^{+3}$ is least hydrated, i.e., least stable (not most stable).

$F$ e

$^{+2}$ contains 4 unpaired electrons whereas

$Mn^{+2}$ contains 5 unpaired electrons.

CBSE Questions for Class 12 Engineering Chemistry The D-And F-Block Elements Quiz 4 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry The D-And F-Block Elements Quiz 4 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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