an alloy of lathanide,iron and carbon

an alloy of lathanide and copper

an alloy of lanthanide and nickel

MCQ Questions for Class 12 Engineering Chemistry The D-And F-Block Elements Quiz 5

The aqueous solution of the following salts will be colored in the case of _________.

0%
$Zn(NO_{3})_{2}$
0%
$LiNO_{3}$
0%
$Co(NO_{3})_{2}$
0%
$CrCl_{3}$

Explanation

As only

$Co(NO_{3})_{2}$ (

$d^7$ configuration of

$Co^{2+}$ ) and

$CrCl_{3}$ (

$d^3$ configuration of

$Cr^{3+}$ ) have d electrons for d-d transition, so they will show colour and other compounds will not show any colour.

The correct order of unpaired electrons present in d orbitals for elements

$Fe,\ Ti,\ Cr$ and

$Co$ is :

0%
$Cr=Fe>Co>Ti$
0%
$Fe>Co>Cr>Ti$
0%
$Cr>Fe>Co>Ti$
0%
$Cr=Fe>Ti>Co$

Explanation

$Cr= [Ar]\ 4s^1\ 3d^{5};$ 6 unpaired electrons

$Fe= [Ar]\ 4s^2\ 3d^{6};$ 4 unpaired electrons

$Co= [Ar]\ 4s^2\ 3d^{7};$ 3 unpaired electrons

$Ti= [Ar]\ 4s^2\ 3d^{2};$ 2 unpaired electrons

$Cr > Fe > Co > Ti$

Hence, option

$C$ is correct.

Statement 1: Lanthanides have much less tendency to form complexes than actinides.
Statement 2: Compared to actinides, the lanthanides have relatively larger size of atoms and less nuclear charge.

0%
Statement 1 is True, statement 2 is True, statement 2 is a correct explanation of statement 1
0%
Statement 1 is True, statement 2 is True, statement 2 is not a correct explanation of statement 1
0%
Statement 1 is true, Statement 2 is False
0%
Statement 1 is False, Statement 2 is True

The magnetic moment of an ion is close to $36\times 10^{-24}$ Joule /Tesla. The number of unpaired electrons of the ion are ____________.

0%
4
0%
2
0%
1
0%
3

Explanation

As magnetic moment is

$\sqrt{n(n+2)}$ Bohr magneton, where n is no. of unpair electron (Bohr Magneton is

$9.273\times 10^{-24}J.tesla^{-1}$ ) , magnetic moment is

$36\times 10^{-24}J.tesla^{-1} = 3.89 B.M.$ so unpair electron is 3.

The radius of $La^{3+}$ (At. No. of La = 57) is 1.06 A $^{0}$ . Which one of the following given values will be closest to the radius of $Lu^{3+}$ ?( Atomic number of Lu=71)

0%
$1.40 A^0$
0%
$1.06 A^0$
0%
$0.85 A^0$
0%
$1.60 A^0$

Explanation

Because of lanthanoid contraction as we move from left to right in a row, radii decreases.
So radii of $Lu^{+3}$ should be lesser than $1.06A^0$ .

Which of the following ions will exhibit colour in aqueous solutions?

0%
La $^{3+}$ (z = 57)
0%
Ti $^{3+}$ (z = 22)
0%
Lu $^{3+}$ (z = 71)
0%
Sc $^{3+}$ (z = 21)

Explanation

The ion which have unpaired electron shows colour

$(A)$

$La=[Xe]5d^16s^2$

$La^{+3}=[Xe]\quad 0$ unpaired electron.

$(B)$

$Ti=[Ar]3d^24s^2$

$Ti^{+3}=[Ar]3d^1\quad 1$ unpaired electron.

$(C)$

$Lu=[Xe]4f^{14}5d^16s^2$

$Lu^{+3}=[Xe]4f^{14}\quad 0$ unpaired electron.

$(D)$

$Sc=[Ar]3d^14s^2$

$Sc^{+3}=[Ar]\quad 0$ unpaired electron.

The main reason for larger number of oxidation states exhibited by the actinoids than the corresponding lanthanoids, is:

0%
the lesser energy difference between 5f and 6d orbitals than between 4f and 5d orbitals
0%
more energy difference between 5f and 6d orbitals than between 4f and 5d orbitals
0%
greater reactive nature of the actinoids than the lanthanoids
0%
larger atomic size of actinoids than the lanthanoids

Explanation

Actinoids shows larger oxidation states due to poor shielding of both

$4f$ and

$5f$ orbital electrons as a result these orbital have almost similar energy and hence take part in bond formation. Also, there is very small energy between

$5f, 6d$ and

$7s$ subshells.

In which of the following pairs are both the ions coloured in aqueous solution?
[At. No. :

$Sc = 21, Ti = 22, Ni = 28, Cu = 29, Co = 27$ ]

0%
$Ni^{2+}, Ti^{3+}$
0%
$Sc^{3+}, Ti^{3+}$
0%
$Sc^{3+}, Co^{2+}$
0%
$Ni^{2+}, Cu^{+}$

Explanation

$(I)$

$Ni\quad [Ar]3d^84s^2$

$Ni^{+2}\quad [Ar]3d^8$

$2$ unpaired

$e^-s\longrightarrow$ paramagnetic.

$(II)$

$Cu\quad [Ar]3d^{10}4s^1$

$Cu^{+}\quad [Ar]3d^10$

$0$ unpaired

$e^-s\longrightarrow$ dimagnetic.

$(III)$

$Ti\quad [Ar]3d^24s^2$

$Ti^{+3}\quad [Ar]3d^1$

$1$ unpaired

$e^-s\longrightarrow$ paramagnetic.

$(IV)$

$Sc\quad [Ar]3d^14s^2$

$Sc^{+3}\quad [Ar]$

$0$ unpaired

$e^-s\longrightarrow$ dimagnetic.

$(V)$

$Co\quad [Ar]3d^74s^2$

$Co^{+2}\quad [Ar]3d^7$

$3$ unpaired

$e^-s\longrightarrow$ paramagnetic.

Ions which are paramagnetic show colour in aqueous solution.

Out of

$TiF^{2-}_6$ and

$CoF^{3-}_6$ ,

$Cu_2Cl_2$ and

$NiCl^{2-}_4$ (Z of Ti = 22, Co = 27, Cu = 29, Ni = 28) the colourless species are:

0%
$CoF^{3-}_6\; and \;NiCl^{2-}_4$
0%
$TiF^{2-}_6 \; and \;CoF^{3-}_6$
0%
$Cu_2Cl_2 \; and \;NiCl^{2-}_4$
0%
$TiF^{2-}_6 \; and \;Cu_2Cl_2$

Explanation

$TiF_6^{2-}\quad Ti^{+4}\rightarrow [Ar]\quad 0$ unpair

$e^-\Rightarrow$ Dimagnetic/ colourless

$CoF_6^{3-}\quad Co^{+3}\rightarrow [Ar]3d^6\quad 4$ unpair

$e^-\Rightarrow$ Paramagnetic/ coloured

$Cu_2Cl_2\quad Cu^{+}\rightarrow [Ar]3d^{10}\quad 0$ unpair

$e^-\Rightarrow$ Dimagnetic/ colourless

$NiCl_4^{2-}\quad Ni^{+2}\rightarrow [Ar]3d^8\quad 2$ unpair

$e^-\Rightarrow$ Paramagnetic/ coloured

Which one of the following ions is the most stable in aqueous solution?
(At. No. are:

$Ti = 22, V = 23, Cr = 24, Mn = 25$ )

0%
$Mn^{3+}$
0%
$Cr^{3+}$
0%
$V^{3+}$
0%
$Ti^{3+}$

Explanation

In aqueous solution the configuration of the ions are following:

$Cr^{3+}(21)= 3d^34s^0 \Rightarrow {t_{2g}}^3 e_g^0$

$V^{3+}(20)= 3d^24s^0 \Rightarrow {t_{2g}}^2 e_g^0$

$Ti^{3+}(19)= 3d^14s^0\Rightarrow {t_{2g}}^1 e_g^0$

$Mn^{3+}(22)= 3d^44s^0\Rightarrow {t_{2g}}^4 e_g^0$

We know that half-filled and fully filled orbitals are more stable than another half nor fully filled orbitals. Hence among the given ions the only ion which occupies half-filled

$t_{2g}$ orbitals is Chromium (III) ion.

$Note:$ Although

$Mn^{3+}$ ion has half-filledr

$t_{2g}$ orbital but there is one more electron filled in

$e_{g}$ orbital which reduces the stability.

Hence option A is the correct answer.

Which one of the following ions will be colourless in an aqueous solution?

0%
$Fe^{2+}$
0%
$Mn^{2+}$
0%
$Ti^{3+}$
0%
$Sc^{3+}$

Explanation

$(A)$

$Fe= [Ar]3d^64s^2$

$Fe^{2+}= [Ar]3d^6$

$4$ unpaired

$e^-s\longrightarrow$ paramagnetic.

$(B)$

$Mn= [Ar]3d^54s^2$

$Mn^{2+}= [Ar]3d^5$

$5$ unpaired

$e^-s\longrightarrow$ paramagnetic.

$(C)$

$Ti= [Ar]3d^24s^2$

$Ti^{3+}= [Ar]3d^1$

$1$ unpaired

$e^-s\longrightarrow$ paramagnetic.

$(D)$

$Sc= [Ar]3d^14s^2$

$Sc^{+3}= [Ar]$

$0$ unpaired

$e^-s\longrightarrow$ diamagnetic.

Note: Ions which are paramagnetic show colour in an aqueous solution.

The catalytic activity of transition metals and their compounds is ascribed mainly to:

0%
their unfilled d-orbitals
0%
their ability of adopt variable oxidation states
0%
their chemical reactivity
0%
their magnetic behaviour

Explanation

Transition metals show variable oxidation states, due to which they can perform as a catalyst.

For example: Contact process, i.e., manufacturing of sulphuric acid.

$SO_3+\overset {+5}{V_2}O_5\longrightarrow SO_3+\overset {+4}{V_2}O_4$

$V_2O_4+O_2\longrightarrow SO_3+\overset {+4}{V_2}O_4$

$V_2O_4+O_2\longrightarrow \overset {+5}{V_2}O_5$

The number of oxidation states is exhibited by the actinoids more than by the lanthanide. The main reason for this is:

0%
more energy difference between 5f and 6d orbitals than that between 4f and 5d orbitals.
0%
the lesser energy difference between 5f and 6d orbitals than between 4f and 5d orbitals.
0%
the greater metallic character of the lanthanoids than that of the corresponding actinoids.
0%
more active nature of the actinoids.

Explanation

Actinides show larger oxidation states due to poor shielding of both

$4f$ and

$5f$ orbital electrons, as a result, their orbitals have almost similar energy and hence take part in bond formation. Thus the energy gap between

$5f, 6d$ and

$7s$ become very small.

Which of the following statement is not correct?

0%
$La(OH)_3$ is less basic than $Lu(OH)_3$
0%
In Lanthanide series ionic radius of $Ln^{+3}$ ions decreases
0%
La is actually an element of transition series rather Lanthanide
0%
Aomic radius of Zr and Hf are same because of Lanthanide contraction

Explanation

$La(OH)_3$ is less basic than

$Lu(OH)_3$ is an incorrect statement.

Basic strength

$La(OH)_3>Lu(OH)_3$

Due to Lanthanide contraction as the size of the lanthanide ions decreases from

$La^{+3}$ to

$Lu^{+3}$ , the covalent character of hydroxides increases and hence basic strength decreases.

Which shows maximum magnetic moment among the bivalent ions of the first transition series ?

0%
$Fe^{2+}$
0%
$Co^{2+}$
0%
$Ni^{2+}$
0%
$Mn^{2+}$

Explanation

As we know,
Magnetic moment

$\mu= \sqrt {n(n+2)}$ , n is no of unpaired electron.

So higher the number of unpaired electrons, the higher is the magnetic moment.

$Fe^{2+} \rightarrow d^6\rightarrow$ 4 unpaired electrons

$Co^{2+} \rightarrow d^7\rightarrow$ 3 unpaired electrons

$Ni^{2+} \rightarrow d^8\rightarrow$ 2 unpaired electrons

$Mn^{2+} \rightarrow d^5 \rightarrow$ 5 unpaired electrons so it will have the highest magnetic moment.

The increasing order of paramagnetism of the following compound is:

(I) $MnSO_{4}.4H_{2}O$

(II) $FeSO_{4}.7H_{2}O$

(III) $NiSO_{4}.6H_{2}O$

(IV) $CuSO_{4}.5H_{2}O$

0%
I < II < III < IV
0%
IV < III < II < I
0%
III < IV < II < I
0%
III < IV < I < II

Explanation

${ Mn }^{ 2+ },{ Fe }^{ 2+ },{ Ni }^{ 2+ },{ Cu }^{ 2+ } -$ central metal atoms

${ 3d }^{ 5 }\quad { 3d }^{ 6 }\quad { 3d }^{ 8 }\quad { 3d }^{ 9 }$

$n=5\quad n=4\quad n=2\quad n=1$

$\Rightarrow I>II>III>IV$

decreasing order of paramagnetism.

The transition metals are mostly :

0%
diamagnetic
0%
paramagnetic
0%
neither diamagnetic nor paramagnetic
0%
both diamagnetic and paramagnetic

Salt of which of the elements are mostly blue in colour ?

0%
Cu
0%
Zn
0%
Fe
0%
Cr

Explanation

$Cu\longrightarrow$ Blue colour

$Zn\longrightarrow$ White colour

$Fe\longrightarrow$ Green colour

$Cr\longrightarrow$ Orange colour

Which of the following shows highest magnetic moment ?

0%
$Co^{+3}$
0%
$Cr^{+3}$
0%
$Fe^{+3}$
0%
$Ni^{+2}$

Explanation

$Cr^{3+}$

$\rightarrow$

$3d^0$

$Co^{3+}$

$\rightarrow$

$3d^6$

$Fe^{3+}$

$\rightarrow$

$3d^5$

$Ni^{2+}$

$\rightarrow$

$3d^8$
therefore

$Fe^{3+}$ has the highest number of unpaired electrons and is always paramagnetic and shows highest magnetic moment.

Variable valency is characteristic of:

0%
halogen
0%
transition elements
0%
alkali metals
0%
noble gas

Explanation

In transition metals,

$(n-1)$ d and ns electrons have nearly same energy level. Hence, most of these electrons take part in chemical bonding. Hence, transition metals show variable valency.

Colour of

$Fe^{2+}$ ions is:

0%
blue
0%
light green
0%
very dark green
0%
yellow

Explanation

Colour of

$Fe^{2+}$ is light green.

Due to this many salts having

$Fe^{2+}$ ion are also green in colour.

For ex:

$FeSO_4$

Of the following metals, the most reactive metal is:

0%
$Fe$
0%
$Ni$
0%
$Pt$
0%
$Au$

Explanation

According to reactive series,

Most reactive metal is Iron

$(Fe)$

Reactivity order is:

$Fe>Ni>Au>Pt$

Ammonium dichromate on heating gives:

0%
chromic oxide and nitrogen
0%
chromic acid and ammonia
0%
chromic acid and nitrogen
0%
chromic oxide and ammonia

Explanation

$\underbrace {(NH_4)_2Cr_2O_7}_{Ammonium\quad Dichromate\quad(orange\quad colour)}\longrightarrow \underbrace {Cr_2O_3}_{chromic\quad oxide\quad(green\quad colour)}+4H_2O+\underbrace {N_2}_{Nitrogen\quad(colourless\quad gas)}$

A compound of vanadium possesses a magnetic moment of 1.73 BM. The oxidation state of vanadium in this compounds is :

0%
1
0%
2
0%
4
0%
can not be predicted.

Explanation

$1.73=\sqrt {n(n+2})$

$n=1$ , where n is the number of unpaired electrons.
Thus, vanadium in oxidation state +4 has one unpaired electron.

The ionic radius of

$_{57}La^{3+}$ is 1.06 A. Which one of the following given values will be closest to the ionic radius of

$_{71}Lu^{3+}?$

0%
1.06 A
0%
1.56 A
0%
0.85 A
0%
1.35 A

Explanation

As we move along a row radii decreases so ionic radius of Lu

$^{3+}$ should be smaller than that of La

$^{3+}$ because of lanthanide contraction across the period.

Which of the following does not corrode when exposed to the atmosphere?

0%
iron
0%
copper
0%
gold
0%
silver

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Assertion is incorrect but Reason is correct

Explanation

$Cu{ SO }_{ 4 }\cdot 5{ H }_{ 2 }O$ contains

$Cu^{2+}$ ion with

$3d^9$ electronic configuration and contains one unpaired electron.hence, it is blue in colour.

$Fe{ SO }_{ 4 }\cdot 7{ H }_{ 2 }O$ contains

$Fe^{2+}$ ion and with

$3d^6$ electronic configuration and contains 4 unpaired electrons. Hence, it is green in colour.

Which of the following ion involved in the above process will show paramagnetism?

0%
${ V }^{ 5+ }$
0%
${ V }^{ 4+ }$
0%
${ O }^{ 2- }$
0%
${ VO }_{ 3 }^{ - }$

Explanation

${ V }^{ 4+ }$ has

$3d^1$ electronic configuration. It contains on unpaired electron. Hence, it is paramagnetic in nature.

Knowing that the chemistry of lanthanids (Ln) is dominated by its +3 oxidation state, which of the following statements are correct?
(I) The ionic sized of Ln (III) decreases in general with increasing atomic number.
(II) Ln (III) compounds are generally colourless.
(III) Ln (III) hydroxides are mainly basic in character.
(IV) Because of the large size of the Ln (III) ions, the bonding in its compounds is predominently ionic in character.

0%
I, III and III
0%
II, III and IV
0%
I, II and IV
0%
I, III and IV

Explanation

The ionic radii of trivalent lanthanides steadily decreases with increase in atomic number due to lanthanide contraction.
In lanthanides (At no. of elements 58 to 71) the electronic configuration of three outermost shelts are

$(n-2)^{f-14}, (n-1)s^2 p^6d^{10 - 1}ns^2$ .
So, most of the trivalent lanthanoid compounds except that of

$La^{+3}$ and

$Lu^{+3}$ are coloured, both in the solid state and in the aqueous solution. These ions are coloured due to the presence of unpaired f-electrons.
Due to small electronegativity and large size of the Ln (III) ions the bonding in its compounds is predominently ionic in character.

Lanthanide contraction implies:

0%
decrease in density
0%
decrease in mass
0%
decrease in ionic radii
0%
decrease in radioactivity

Which of the following elements is placed just after actinide series?

0%
Rutherfordium
0%
Unnilpentium
0%
Unnilquadium
0%
Ununbium

All

$Zn(+II)$ compounds are white because:

0%
${ Zn }^{ 2+ }$ has a ${d}^{10}$ configuration and the $d$ -subshell is full
0%
${ Zn }^{ 2+ }$ shows $d-d$ transition
0%
${ Zn }^{ 2+ }$ has no electron in the $4s-$ subshell
0%
$Zn$ is not a transition element

Explanation

The electronic configuration of

$Zn^{2+}$ is

$3d^{10}4s^0$ . All the electrons are paired.
Hence, all

$Zn^{2+}$ compounds are colorless or white.

Which of the following factors favour the formation of complexes by d-block elements?

0%
Small size.
0%
High nuclear charge.
0%
Presence of low energy vacant orbitals to accept lone pair of electrons donated by ligands.
0%
All of the above

Explanation

The d-block elements are small in size and have high electropositive density. They consist of

$(n-1)d$ free orbitals to accept the free electrons from the ligand and hence, form complexes easily.

To protect iron against corrosion, the most durable metal plating on it, is :

0%
nickel plating
0%
tin plating
0%
copper plating
0%
zinc plating

Explanation

Hint: A thin layer of metal has been put to the outside of a substance and is known as plating.

Explanation:

Metal plating creates a sacrificial coating that shields the iron from corrosion. Instead of oxidizing iron, this metal layer oxidizes itself, protecting the iron.

For plating, a metal's oxidation should be preferred over iron, and it should therefore have a lower reduction potential than iron.

Zinc has a lower reduction potential than iron and thus it can be utilized for plating among the available possibilities. Galvanization is the name of the procedure.

Final Answer:

The correct answer is option $(D)$ .

The main reason for larger number of oxidation state exhibited by the actinides than that corresponding lanthanide, is:

0%
lesser energy difference between 5f and 6d orbitals than between 4f and 5d-orbitals
0%
larger atomic size of actinides than the lanthanides
0%
more energy differenc between 5f and 6d orbitals than between 4f and 5d-orbitals
0%
greater reactive nature of the actinides than the lanthanides

Misch metal is:

0%
an alloy of lathanide and copper
0%
an alloy of lanthanide and nickel
0%
an alloy of lathanide,iron and carbon
0%
an alloy of calcium and copper

The atomic radii of cerium and promethium are quite similar because

0%
they belong to the same period in periodic table
0%
they have similar electronic configuration
0%
f-orbitals have poor shielding effect
0%
nuclear charge is higher on cerium than promethium

Explanation

The most significant characteristic of lanthanides is the contraction of the atomic size with increase in atomic number called as lanthanide contraction. While moving along the lanthanides series from

$^{58}Ce$ to

$^{71}Lu$ , a regular decrease in the size of the atom/ion and an increase in atomic number is observed. This decrease in size is called the lanthanide contraction. The decrease in size, though continuous, is not regular. Lanthanide contraction takes place due to the imperfect shielding of a

$4f$ - electron which increases by one more in the same sub-shell.
So, due to this poor shielding effect,

$Ce$ and

$Pm$ have similar radii.

The transition metal used as a catalyst is :

0%
nickel
0%
platinum
0%
cobalt
0%
all of these

The magnetic moment order is correctly given in:

0%
$Fe^{2+} < Fe^{3+} < Mn^{4+} < Cr^{3+}$
0%
$Cr^{3+} = Mn^{4+} < Fe^{2+} < Fe^{3+}$
0%
$Cr^{3+} = Mn^{4+} < Fe^{3+} < Fe^{2+}$
0%
$Fe^{2+} < Cr^{3+} < Fe^{3+} < Mn^{4+}$

Explanation

As we know,
magnetic moment

$=\sqrt {n(n+2)}$ where n is no of unpaired electron
here,

$Cr^{3+}, Mn^{4+}, Fe^{2+}$ and

$Fe^{3+}$ have

$3,3,4,5$ unpaired electrons.
so order is

$Cr^{3+} = Mn^{4+} < Fe^{2+} < Fe^{3+}$

$Fe$ is made passive by:

0%
$dil.\ H_2SO_4$
0%
$dil.\ HCl$
0%
$conc. \ HNO_3$
0%
$conc.\ H_2SO_4$

The highest magnetic moment is shown by

0%
$Ni$
0%
$Co$
0%
$Fe$
0%
$Sc$

Explanation

The expression for magnetic moment is

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

$n$ is the number of unpaired electrons. Greater is the number of unpaired electrons, greater is the magnetic moment.

$Sc,\ Ni,\ Fe$ and

$Co$ have

$1,\ 2,\ 4$ and

$3$ electrons respectively. Therefore,

$Fe$ has the highest magnetic moment.

Which of the following is not an actinide ?

0%
Curium
0%
Californium
0%
Uranium
0%
Terbium

Explanation

Hint: In actinides, the valence electrons enter into $5f$ orbital

Explanation:

The electronic configuration of Curium is as the last electron is entering in $5f$ orbital, so this is an actinide.
The electronic configuration of Californium is and as the last electron is entering in $5f$ orbital so, it is an actinide.
The electronic configuration of Uranium is $[Rn]5{f^3}$ and as the last electron is entering in $5f$ orbital so it is an actinide.
The electronic configuration of terbium is $[Xe]4{f^9}6{s^2}$ as here last electron is entering in the $4f$ orbital so, it is not an actinide. it is lanthanide.
The actinide series is shown as below;

Final answer: The correct answer is $D$ .

Which plays a major role in the formation of complex compound ?

0%
Transition metal
0%
Lanthanides and actinides
0%
Representative elements
0%
p-block element

Explanation

Because of the presence of unpaired electrons in the inner d-orbital, transition element forms coordination complexes with appropriate ligand.

$K_3Fe(CN)_6$

$Ni(CO)_4$

Which of the following lanthanide is commonly used ?

0%
Lanthanium
0%
Nobelium
0%
Thorium
0%
Cerium

Transition elements does not show :

0%
paramagnetism
0%
colour
0%
fixed valency
0%
all of these

Which of the following electronic configuration will be associated with the lowest magnetic moment?

0%
${ d }^{ 2 }$
0%
${ d }^{ 3 }$
0%
${ d }^{ 4 }$
0%
${ d }^{ 9 }$

Explanation

The magnetic moment,

$\mu_s = \sqrt {n(n+2)} \mu_B$ where n is the number of unpaired electrons.

$d^2$ has 2 unpaired electrons,

$d^3$ has 3 unpaired electrons,

$d^4$ has 4 unpaired electrons and

$d^9$ has 1 unpaired electron.

Hence

$d^9$ will have the lowest magnetic moment.

The magnetic moment of a transition metal of

$3d$ -series is

$6.92BM$ . Its electronic configuration will be:

0%
${ \left( 3d \right) }^{ 4 }{ \left( 4s \right) }^{ 2 }$
0%
${ \left( 3d \right) }^{ 5 }{ \left( 4s \right) }^{ 1 }$
0%
${ \left( 3d \right) }^{ 10 }$
0%
${ \left( 3d \right) }^{ 5 }{ \left( 4s \right) }^{ 0 }$

Explanation

The magnetic moment,

$\mu_s = \sqrt {n(n+2)} B M$ where n is the number of unpaired electrons.

Option B has 6 unpaired electrons and the corresponding magnetic moment is

$6.92\mu_B$ .

$\displaystyle [{ Fe\left( { H }_{ 2 }{ O } \right) }_{ 5 }{ No{ ] }^{ 2+ } }$ is formed as brown ring in

$\displaystyle { NO }_{ 3 }^{ + }$ test.

$\displaystyle Fe$ in this complex has _________ unpaired electrons.

0%
one
0%
two
0%
three
0%
four

Explanation

The magnetic moment of nitroso ferrous sulphate is 3.87 BM. This indicates presence of 3 unpaired electrons. This is due to

$3d^7$ electronic configuration of

$Fe^+$ ion.

Arrange the following ions in order of their magnetic moment:

$(i) {V}^{4+}$

$(ii) {Mn}^{4+}$

$(iii) {Fe}^{3+}$

$(iv){Ni}^{2+}$

[Atomic number of

$V=23, Mn=25,Fe=26,Ni=28$ ]

0%
$(ii)> (iii)> (i)> (iv)$
0%
$(iii)> (ii)> (iv)> (i)$
0%
$(i)> (iv)> (iii)> (ii)$
0%
$(iv)> (iii)> (i)> (ii)$

Explanation

The transition metal with more unpaired electrons corresponds to the higher magnetic moment.

$V^{4+}\rightarrow4s^0 3d^1$ , 1 unpaired electron;

$Mn^{4+}\rightarrow4s^0 3d^3$ , 3 unpaired electron;

$Fe^{3+}\rightarrow4s^0 3d^5$ , 5 unpaired electron and

$Ni^{2+}\rightarrow4s^0 3d^8$ , 2 unpaired electron.

Hence, option B is correct.

The position in the periodic table to which the

$Al$ metal belongs :

0%
Alkali metals
0%
Alkaline earth metals
0%
$d$ - block elements
0%
$p$ - block elements

Explanation

Electronic configuration of aluminium is $[Ne] 3s^{2}3p^{1}$ .

In the above configuration, the valence electron enters into p-orbital. So $Al$ is a p-block element.

Hence option D is correct.

CBSE Questions for Class 12 Engineering Chemistry The D-And F-Block Elements Quiz 5 - 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 5 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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