The d- and f- Block Elements Test

Result

The d- and f- Block Elements Test - 39

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The correct order of ionic radii of Ce, La, Pm, and Yb in +3 oxidation state is:

A
$$La^{3+} \, < Pm^{3+} \, < Ce^{3+} \, < Yb^{3+}$$

B
$$Yb^{3+} \, < Pm^{3+} \, < Ce^{3+} \, < La^{3+}$$

C
$$La^{3+} \, < Ce^{3+} \, < Pm^{3+} \, < Yb^{3+}$$

D
$$Yb^{3+} \, < Ce^{3+} \, < Pm^{3+} \, < La^{3+}$$

Solution

$${ La }^{ 3+ }>{ Ce }^{ 3+ }>{ Pm }^{ 3+ }>{ Yb }^{ 3+ }$$

Due to lanthanoid contraction ionic radii decreases.
Ionic radii decreases from $$La^{3+} -Lu^{3+}$$ in +3 oxidation state.

Hence, the correct option is $$C$$.
Question 2
1 / -0

Identify the correct Structure of dichromate ion.

A

B

C

D

Solution

Structure of dichromate ion is as shown.
Question 3
1 / -0

Which of the following statements concerning lanthanide elements is false?

A
All lanthanides are highly dense metals.

B
The characteristic oxidation state of lanthanide elements is +3.

C
Lanthanides are separated from one another by ion exchange method.

D
Ionic radii of trivalent lanthanides steadily increase with an increase in the atomic number.

Solution

Ionic radii decreases with increasing atomic number in Lanthanide series. So, the ionic radii of trivalent Lanthanide will also follow the same property.
Question 4
1 / -0

Which ion has magnetic moment $$5.90$$ B.M?

A
$$Fe^{2+}$$

B
$$Fe^{3+}$$

C
$$Co^{2+}$$

D
$$Mn^{4+}$$

Solution

$$ \displaystyle \mu = \sqrt {n(n+2)}=5.9$$ B.M.
$$ \displaystyle {n(n+2)}=5.9 \times 5.9 = 34.8$$

$$ \displaystyle n^2+2n= 34.8$$

$$ \displaystyle n^2+2n- 34.8=0$$

$$ \displaystyle n= \dfrac {-b \pm \sqrt {b^2-4ac}}{2a}$$

$$ \displaystyle n= \dfrac {-2 \pm \sqrt {2^2-4(1)(- 34.8)}}{2(1)}$$

$$ \displaystyle n= \dfrac {-2 \pm \sqrt {4+139}}{2}$$

$$ \displaystyle n= \dfrac {-2 \pm \sqrt {143}}{2}$$

$$ \displaystyle n= \dfrac {-2 \pm 12}{2}$$

$$ \displaystyle n= \dfrac {10}{2}$$ or $$ \displaystyle n= \dfrac {-14}{2}$$

$$ \displaystyle n= 5$$ or $$ \displaystyle n= -7$$

The negative value is discarded as the number of electrons cannot be negative.
$$ \displaystyle \therefore n= 5$$

$$ \displaystyle Fe ^{3+}$$ ion (with valence shell electronic configuration $$ \displaystyle 3d^5 4s^0$$ has magnetic moment 5.90 B.M. It has 5 unpaired electrons.

Hence, option $$B$$ is correct.
Question 5
1 / -0

Which of the following ion is paramagnetic?

A
$$Cu^{+}$$

B
$$Zn^{2+}$$

C
$$O_{2}^{2-}$$

D
$$Cr^{3+}$$

Solution

Ions with unpaired electron is paramagnetic. Elelctronic configuration and unpaired electron in the given ions are:
$$Cu^+:3d^{10}$$: 0 unpaired electron
$$Zn^{2+}:3d^{10}$$: 0 unpaired electron
$$O_2^{2-}:\sigma2p^2\pi2p^4\pi^{*}2p^4$$: 0 unpaired electron
$$Cr^{3+}:3d^{3}$$: 3 unpaired electron
thus option D is correct.
Question 6
1 / -0

Which of the following ion has the maximum theoretical magnetic moment?

A
$$V^{3+}$$

B
$$Cr^{3+}$$

C
$$Ti^{3+}$$

D
$$Co^{3+}$$

Solution

Metal ion with maximum number of unpaired electrons will have highest magnetic momemt.
$$V^{3+}$$: $$3d^2$$: 2 unpaired electrons
$$Cr^{3+}$$: $$3d^3$$: 3 unpaired electrons
$$Ti^{3+}$$: $$3d^1$$: 1 unpaired electrons
$$Co^{3+}$$: $$3d^6$$: 4 unpaired electrons
Thus $$Co^{3+}$$ will have maximum magnetic moment.
option D is correct
Question 7
1 / -0

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

A
$$1.40\mathring{ A}$$

B
$$1.06\mathring{ A}$$

C
$$0.85\mathring{ A}$$

D
$$1.60\mathring{ A}$$

Solution

The ionic radii of $$Ln^{3+}$$ reduces from $$La^{3+}$$ to $$Lu^{3+}$$ due to Lanthanide Contraction. Therefore the lowest value here must be the one closest ionic radii of $$Lu^{3+}\implies 0.85 A.$$
Question 8
1 / -0

Which of the following has highest value of magnetic moment?

A
$$Fe^{2+}$$

B
$$Co^{3+}$$

C
$$V^{3+}$$

D
$$Ti^{3+}$$

Solution

$$ Magnetic\ moment = \sqrt {n({n+2})}$$
$$ n$$= no. of unpaired electrons,
$$ Fe^{+2}$$ has $$4$$ unpaired electrons
Other compound have less unpaired electrons.
More the number of electron, more the magnetic moment. Thus $$Fe^{2+}$$ has the highest value of the magnetic moment among the given elements.
Question 9
1 / -0

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

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

B
$$Ce^{3+} > Yb^{3+} < Pm^{3+} < La^{3+}$$

C
$$Yb^{3+} > Pm^{3+} < La^{3+} < Ce^{3+}$$

D
none of these

Solution

Atomic and ionic radii of Lanthanides decrease from $$La$$ to $$Lu$$.
Their order of ionic radii: $$Yb^{+3}<Pm^{3+}< Ce^{3+}< La^{+3}$$
Question 10
1 / -0

Among $$V(Z=23)$$, $$Cr(Z=24)$$, $$Mn(Z=25)$$ and $$Fe(Z=26)$$, which will have the highest magnetic moment?

A
$$V$$

B
$$Cr$$

C
$$Mn$$

D
$$Fe$$

Solution

$$V$$,$$Cr$$,$$Mn$$ and $$Fe$$ has three, six, five and four unpaired $$e^-$$ respectively.
So, $$M_V=\sqrt {3(3+2)}=\sqrt {15}=3.18$$ BM
$$M_{Cr}=\sqrt {6(6+2)}=\sqrt {48}=6.92$$ BM
$$M_{Mn}=\sqrt{5(5+2)}=\sqrt {35}=5.91$$ BM
$$M_{Fe}=\sqrt{4(4+2)}=\sqrt {24}=4.89$$ BM
So, $$Cr$$ is having highest magnetic moment. It can be concluded that more is the number of unpaired $$e^-s$$, higher is the magnetic moment.