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GOVERNMENT OF KARNATAKA
DEPARTMENT OF PRE UNIVERSITY EDUCATION, BANGALORE
II PU 2014 VIKASANA TELE PROGRAME
SUBJECT : CHEMISTRY
TOPIC : The d- and f- block Elements
MANJUNATH BHAT
LECTURER IN CHEMISTRY
GOVT. PRE UNIVERSITY COLLEGE,
BRAHMAVAR, UDUPI DIST.
MOBILE NO. 9448416987
EMAIL :[email protected]
1
The transition elements (d block) and inner transition
elements (f block) in the periodic table
The d-block of the periodic table contains the elements of the
groups 3-12 in which the (n-1)d - orbital's are progressively
filled with electrons in each of the four long periods.
There are mainly three series of the transition metals,
3d series - Sc to Zn
4d series - Y
to Cd
5d series - La to Hg (omitting Ce to Lu).
6d series - starting from Ac is still incomplete
The d–block elements are also known as Transition metals ,
they possess properties that are in between the s– and p–
.
block elements
All transition elements are metals, whereas main-group
elements in each period change from metal to nonmetal
Characteristic properties of transition elements are;
Variable oxidation states - due to the availability of both ns & (n – 1 )
d electrons for bond formation.
Paramagnetic behavior – due to the presence of one or more
unpaired electrons.
Colored compound formation - due to presence of unpaired
electron & d-d transition of these electrons.
Catalytic activity - due to their ability to adopt multiple
oxidation states and to form complexes and provide a
suitable surface for the reaction to take place.
Alloy formation – due to similar radii .
Complex formation - due to Small size and high nuclear
charge of cations
The properties of transition metal compounds are related to the
electron configuration of the metal ion.
The electrons in (n-1) d orbitals are very singificant , because they
influence their properties , such as variable o.s, coloured compounds ,
magnetic character, catalytic property , complex formation etc.
Element
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Z
21
22
23
24
25
26
27
28
29
30
Electronic Arrangement
3d
[Ar]
[Ar]
[Ar]
[Ar]
[Ar]
[Ar]
[Ar]
[Ar]
[Ar]
[Ar]
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
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
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4s
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Which one of the following pairs of ions have the same electronic
configuration;
a)
Ni2+,
Co3+
8 4s
0 , Co3+ [Ar] 3d64s0
2+
2+
Ni2+
[Ar]
3d
b) Fe , Mn
Fe2+ - [Ar] 3d64s0 , Mn2+ [Ar]3d54s0
c) Fe 3+, Mn 2+
Ans: c
Fe3+ - [Ar] 3d54s0 ,Mn2+[Ar]3d54s0
d) Sc+3 , Ti+3
Sc+3 - [Ar] 3d04s0 , Ti+3 - [Ar]3d24s0
Electronic configuration of a transition element X in +3
oxidation state is [Ar]3d 5 4s 0 .What is its atomic number?
a.25
b. 26
X +3 - [ Ar]3d 5  X - [Ar]3d 6 4s 2
i.e atomic no. is 26
c. 27
Ans. b
d. 24
Which of the following do not have valence electron in
3d sub shell?
a) Fe (III)
Fe3+ - [Ar] 3d54s0
Cr+ - [Ar] 3d54s0
c) Cr (I)
Ans. d
2+ - [Ar] 3d54s0
MnMn
b)
(II)
Sc3+ - [Ar] 3d04s0
d)valence
Sc (III)
Fe (III), Mn (II) , Cr (I) all have
electron in 3d
sub shell, but Sc (III) has no electron in 3d sub shell
The transitional elements
elements show paramagnetism,
form
alloyspair effect
a.Transition
Show diamagnetism
b.Undergo
inert
and do not undergo inert pair effect . But they Show variable oxidation
states. In addition to the electrons of ns electrons of incompletely filled (n1)d involve in bonding.
c. Do not form alloys
Ans. d
d. Show variable oxidation states
Which one of the following statement is not true regarding variable
oxidation state of transition metals.
dblock elements contain partially filled penultimate orbitals i.e d
a.Theorbital.
variable oxidation state is due to incomplete filling of (n-1) d
orbitals
Oxidation state of a transition metal varies , in which they
differ from
other
by unity
. E.g Vanadium
b. Oxidation
stateeach
differ
from
each other
by unity shows o.s. of +2,
+3, +4 and +5
Generallystate
non differ
transition
o.s differ
by of
a unit
c. Oxidation
fromelements
each other
by unit
twoof two.
E.g S shows o.s of -2 , +2, +4 and +6.
d. It isIt due
to to
involvement
inbonding
bonding
is due
involvementofof ns
nsand
and(n-1)
(n-1) d electrons
electrons in
Ans. c
Transitional elements exhibit variable valences because they
release electrons from the following orbital's.
a.ns orbitals
b. (n - 1)d and ns orbitals
In transition elements , ns and (n-1)d orbital's have
very little difference in energy , hence they release
electrons from both these orbital's and exhibit
variable valences
c. ns and np orbital's
Ans. b
d. (n - 1)d orbital's
Out of Cr2+ and Cr3+,which one is stable in aqueous
solution?
a.Cr2+
b.Cr3+
Cr3+ is more stable in aqueous solution due to higher hydration
energy which is due to smaller size and higher charge.
c. Both Cr2+ & Cr3+
Ans. b.
d. None of the two
In transition elements , the incoming electron occupies
(n-1)d orbital in preference to.
a.ns
b. np
In transition elements , (n-1)d orbital's are filled with electrons
after filling ns orbital. (n-1)d orbital has lower energy than np
orbital , hence (n-1)d is filled in preference to np orbital.
c. (n-1)p
Ans.b.
d. (n-1)s
Zn and Cd do not show variable valence , because :
a. They have only two electrons in outermost sub shells
b. Their d – sub shells are incomplete
c. Their d – sub shells are complete
d. They are relatively soft metals
Zn - [Ar] 4s2 3d10
Cd - [Kr] 5s2 4d10
Since their d – sub shells are complete , they show fixed valence of 2.
Ans. c
Which of the following has the maximum number of
unpaired ‘d’ electrons?
a.Co3+
b. Zn2+
c. Cu+
d. Ni3+
Cu+ - [Ar] 3d104s0 – 0
Co3+ - [Ar] 3d64s0 - 4
Zn2+ - [Ar] 3d104s0 - 0
Ans. a
Ni3+ - [Ar] 3d7 4s0 – 3
In which of the following compounds the radius of
chromium is smallest?
a. CrO2
c. K2Cr2O7
Ans. c
The ionic radii decreases as charge on cation
b. Cr
2 O3
increases ( i.e higher oxidation
state)
In CrO2 o.s of Cr is + 4, Cr2O3 and CrCl3 - + 3 and in K2Cr2O7 -- +6
Hence radius of chromium is smallest in
d.CrCl
3
K2Cr2O7
8.
The outer most electronic configuration of an ion in +3
oxidation state is [Ar] 3d54s0, the element belong to
The electronic configuration of an ion in +3
54s0
oxidation
state
is
[Ar]
3d
a. III group
b. 18th group
The electronic configuration of the element
is
[Ar] 3d64s2.
c. 5th group
d. 8th group
It belong to 8th group
Ans. d
The minimum oxidation state shown by transition element
is +2 which of the following is an exception?
a. Fe
Fe- [Ar] 3d64s2
c. Ni
Ni- [Ar] 3d84s2
b. Co
Co- [Ar] 3d74s2
d. Cu
Cu- [Ar] 3d104s1
Oxidation States of the 1st Row Transition Metals
Iron , cobalt and nickel shows common minimum
oxidation state of +2, copper shows +2 as well as +1
oxidation states.
Ans. d
22.1
Within each transition series the oxidation state,
a.Decreases regularly in moving from left to right across a
period .
b. First increases till the middle of the period & then
decreases
c. . First decreases till the middle of the period & then
increases
d. Increases regularly in moving from left to right across a
period
Oxidation States of the 1st Row Transition Metals
(most stable oxidation numbers are shown in red)
O.S first increases till the middle
decreases .
Ans. b
of
the period & then
22.1
Which of the following transition metal does not show
the variable oxidation state ?
a.Cu
b. Zn
Cu- [Ar] 3d104s1
c. Mn
Mn- [Ar]
Zn- [Ar] 3d104s2
d. V
3d54s2
V- [Ar] 3d34s2
Mn, V and Cu shows variable oxidation state
Zn is the last element in the 3d series which shows +2 o.s.
Ans. b
First ionisation enthalpy of Cr is lower than that of Zn,
because,
a. Cr is smaller atom than Zn
b. Removal of an electron from 4s level needs less energy.
c. Removal of one electron from 4s orbital of Cr does not
change the d configuration.
d. Removal of one electron from 4s orbital of Cr change the
d configuration.
IE1 of Cr is lower, because removal of one electron
from 4s orbital Cr does not change the d
configuration .
Cr (z= 24 , 3d5 4s1)  Cr+ (3d5 4s0)
------ IE1
IE1 value for Zn is higher, because removal of
electron from 4s level needs more energy.
Zn (3d10 4s2)  Zn+ (3d10 4s1)

Ans. c
IE1 (Zn) > IE1 (Cr)
------ IE1
Species
having same number
Zn+2 ion is isoelectronic
with
of electrons are
known as iso electronic species .
Zn+2 - [Ar] 3d10 4s0
a. Cu+
Cu+
b. Cu+2
- [Ar]3d10 4s0
Cu+2
c. Ni+2
Ni+2
- [Ar]3d8
Ans. a
-[Ar] 3d9 4s0
d. Fe+2
4s0
Fe+2
- [Ar]3d6
4s0
Which of the following shows positive E0 value?
a.Mn/Mn2+
c.Cu/Cu2+
Ans. c
b. Ni/Ni2+
Mn2+, Zn2+ shows negative E0 value due
stability of half filled and completely filled
configurations, but for Ni2+ it is due to highest
–ve enthalphy of hydration
Cu2+ shows +ve E0 value.
d. Zn/Zn2+
Standard reduction potential of transition metals is
a.Negative
c. Zero
Ans .d
b. positive
T.M ions like Mn2+, Zn2+ shows negative E0 value
due stability of half filled and completely filled
2+ it is due to highest –ve
configurations. Ind.Nimay
be positive
or
negative
enthalphy of hydration. Cu2+ shows +ve E0 value
due to higher + hydration ethalpy.
Metallic radii of some transition elements are given
below. Which of these elements will have highest
density? Element
Fe
Co
Ni
Cu
Metallic radii/pm
126
125
a.Fe
b.Ni
c. Co
d. Cu
124
128
As atomic radius decreases, density increases along the
period, hence Cu has highest density
Ans. d
Which one of the following transition metal has lowest
density?
a. Copper
b. Nickel
As atomic radius decreases density increases along the
period, Scandium has larger atomic radius , hence it has
lowest density (d = 3.0g/cm3)
c. Scandium
Ans. c
d. Zinc
The melting points and boiling points up to middle of
3d- series
a.Increases
b. decreases
As the no. of unpaired electrons increases, metallic bond becomes
strong due involvement of ns & (n-1)d electrons, increases up to
middle , they become max. in the element having d5 confn (Cr).
Then they show down word trends.
c. Remains same
d. no regular trend
The m.p. of Mn is lower because it has stable half filled
conf. so electrons are held tightly, delocalization is less &
metallic bond is weak.
Ans . a
Which one of the following first row transition metal is
expected to have the highest 3rd IE?
V - [Ar]
3d34s2
Mn - [Ar] 3d54s2
a. V (z =Removal
23) of
b.Mn
3rd electron from Mn+2 requires large
amount(z
of =
energy due its stable d5 configuration.
Cr - [Ar] 3d54s1
c.Cr (z = 24)
Ans. b
25)
Fe - [Ar] 3d64s2
d. Fe (z = 26)
Which of the following oxide of vanadium is most basic ?
a. VO
2
When metal in oxide has higher o.s itb.VO
is acidic
and when metal in oxide has lower o.s it is basic.
V2O5 - acidic oxide.
VO, V2O3 - basic oxides.
c. V2O5, VO2 - is amphoteric
Ans. a
d.V2O3
Which of the following is not an acidic oxide?
a. Mn2O7
b.CrO3
In Mn2O7 ,CrO3 , V2O5, are acidic oxides.
Fe2O3 is amphoteric
c. V2O5,
Ans. d
d.Fe2O3
Which of the following compound of manganese is
amphoteric?
a. MnO
b. Mn2O3
Soln. When metal in oxide has higher o.s it is acidic, metal in
oxide has lower o.s it is basic and metal in oxide has intermidiate
o.s , it is amphoteric.
MnO and Mn2O3 are in lower o.s are basic oxides Mn2O7 is an
acidic oxide . Mn2O5 is amphoteric
c. Mn2O7
Ans. d
d. Mn2O5
The “spin-only” magnetic moment [µ] of Ni 2+ in aqueous
solution would be
(Ni ,z = 28)
. a. 2.84
b. 4.90
Ni +2 - [Ar]3d8 4s0 it contains 2 unpaired electrons
If n = 2
c. 0
Ans. a
µ = n(n+2) = 2.84
d. 1.73
The spin only magnetic moment of an ion is 3.87 B.M the
unpaired electrons present in it are,
a. 2
b. 3
µ = n(n+2)
(3.87) 2 = n(n+2)
15 = n 2 + 2n
By approx n = 3
c. 4
Ans = b
d. 5
Titanium shows magnetic moment of 1.73 BM in its
compound. What is the oxidation state of Ti in the
compound?
a. +1
b.+4
µ = n(n+2)
(1.73) 2 = n(n+2)
3.0 = n 2 + 2n
By apporx n = 1
No. of unpaired electron = 1
Ti (z = 22) is in +3 o.s Ti +3 - [Ar]3d1 4s0
c.+3
Ans. c
d. +2
The magnetic moment of Manganese is 5.92 BM in its
compound. What is the symbol of the ion?
a. Mn+
b. Mn 2+
µ = n(n+2)
(5.92) 2 = n(n+2)
35 = n 2 + 2n
By apporx n = 5
No. of unpaired electron = 5
Mn (z = 25) in +2 o.s has 5 unpaired electrons. Mn+2 - [Ar]3d5 4s0
c. Mn3+
Ans. b
d. Mn 7+
Which one of the following cyano complexes would exhibit
the lowest value of paramagnetic behavior?
a. [Cr(CN) 6] 3Cr+3 -[Ar]d3- 3
c.[Fe(CN) 6] 3-
b. [Mn(CN) 6] 3Mn+3 -[Ar]d4 - 4
d.[Co(CN) 6] 3Co+3 -[Ar]d4- 4
5- 5
Fe+3 -[Ar]d
(At. No. Cr = 24, Mn = 25, Fe = 26, Co = 27)
Paramagnetic behavior depends on the no. of unpaired electrons
Ans. a
Which one of the following shows highest magnetic moment ?
(a) Fe2+
(b) Co2+
Fe+2 -[Ar]d6 - 4
Co+2 -[Ar]d7 - 3
(c) Cr3+
Cr+3 -[Ar]d3 -
Ans.a
(d) Ni2+
3
Ni+2 -[Ar]d2 - 2
Which pair of compounds is expected to show similar
colour in aqueous medium?
a. FeCl2 and CuCl2
b. FeCl2 and MnCl2
Fe+2 (FeCl2 ) - 4
unpaired electrons
Cu+2 (CuCl2)
electron.
- 1 unpaired
 VOCl2 and CuC l2 expected to show similar colour in
aqueous medium .
c. VOCl2 and FeCl2
Mn+2 (MnCl2 ) - 5
unpairedAns:
electrons
d
d. VOCl2 and CuCl2
V+4 (VOCl2 ) - 1 unpaired
electron.
Generally transition elements form coloured salts due to
the presence of unpaired electrons. Which of the
following compounds are coloured ?
a.TiCl4
b. CuF2
c. ZnF2
d. Cu2Cl2
Ti4+ has 3d0 Configuration
Cu+ and Zn2+ both have 3d10 Configuration
Cu2+ also has 3d9 Configuration.
Ans. b CuF2
Ti(H2O)6]3+ is purple in
complimentary colour of
colour
because
a.Greenish Yellow
b.Red
c.Green
d.Blue
is
Each color has a complementary
color; the one opposite it on the
artist’s wheel.
The color an object exhibits
depends on the wavelengths of
light that it absorbs.
Purple colour is complimentary colour of Greenish Yellow.
The color of
+
[Ti(H2O)6]3
Green and yellow light are
absorbed
while
other
wavelengths are transmitted.
This gives a purple color.
The hydrated Ti3+ ion is purple.
Ans. a
When a compound of transition element is dissolved in a solution
of salt then it produces
a.Simple ions
b. complex ions
Eg .4KCN + Fe(CN)2 ----- [Fe(CN)6 ] 4 - + 4 K +
c. double salts
Ans. b
d.strong anions
d-block element form complexes due to
a.Large size and high nuclear charge of cations
b. Small size and high nuclear charge of cations
d-block elements and their ions show strong tendency to form
complexes due to Small size and high nuclear charge of cations
c. Small size and low nuclear charge
d. Availabilty of 4s electrons
Ans.b
The catalytic activity of the transition metals and their compounds
is due to
a.The chemical reactivity
b. Presence of unpaired electrons
Their ability
adopt multiple oxidation states (form
c.Their
unfilledtod-orbitals
unstable intermediates) and availability of adsorption sites
d.Their ability to adopt multiple oxidation states and
availability of adsorption sites
Ans d.
V2O5 acts as catalyst , because
a.it has large surface area.
b.It can form unstable intermediates .
c.It can form unstable intermediates .which readily change into
products.
d. All the above statement are correct
It has large surface area.
It can form unstable intermediates .
It can form unstable intermediates .which readily change into
products.
Ans. d.
When KMnO4 solution is added to oxalic acid solution, the
decolourisation is slow in the beginning, but becomes instantaneous
after some time because
a. CO 2 5C
is formed
as the -product.
O 2− + 2MnO
+ 16H+ —> 2Mn2+ + 8H O + 10CO
2
4
4
2
b. Reaction is exothermic.
c. MnO4– catalyses the reaction.
d. Mn2+ acts as autocatalyst.
Ans. d
2
In the reaction,
2 MnO4- + 6H+ + 5NO2-  2Mn 2+ + 3H2O + 5NO3 a.MnO4- is reduced
b.Oxidation number of Mn decreases from + 7 to + 2
c. Oxidation number of N increases from + 3 to + 5
- + 6H+ + 5NO -  2Mn 2+ + 3H O + 5NO –
2
MnO
d. All the4 above statements
are correct
2
2
3
Ans. d
KMnO4 acts as an oxidising agent in acidic medium. The
number of moles of KMnO4 that will be needed to react
with one mole of sulphide ions in acidic solution is
a 2/5
b 3/5
S2− + 2/5MnO4− + 16/5H+ —> 2/5Mn2+ + 8/5H2O + S
c 4/5
Ans. a
d 1/5
Which one of the following does not decolourise an acidified
KMnO4 solution?
a.SO2
b. FeCl3
2KMnO4 + 5SO2 + 2H2 O  2MnSO4 + K2SO4 +2H2 SO4
2KMnO4 + 8H2 SO4 + 10FeSO4  2MnSO4+ K2SO4+8H2O+5Fe 2(SO4 )3
c. H2 O2
2KMnO4 + 3H2 SO4 +5 H2 O2  2MnSO4 + K2SO4 +8H2 O+ 5O 2
Ans. b
d.FeSO4
Why is HCl not used to make the medium acidic in oxidation
reactions of KMnO4?
a.Both HCl and KMnO4 act as oxidising agents.
b.KMnO4oxidises HCl into Cl2 which is also an oxidizing
c. KMnO4 is a weaker oxidising agent than HCl.
d.KMnO4 acts as a reducing agent in the presence of HCl
Ans. b
2KMnO4 + 16HCl 2 MnCl2 + 2KCl + 8H2O + 5Cl2
The oxidation state of chromium in the reaction between KI
and acidified potassium dichromate solution changes from
a. +4 to +2
Cr 2 O 7 2− + 14H + + 6I −  3 I2 + 2Cr
c. +2 to +4
Ans. b
b. +6 to +3
3+ +
7H 2 O
d. +3 to +6
When acidified K2Cr2O7 solution is added to Sn2+ salts,
then Sn2+ changes to
a. Sn
b. Sn 3+
Cr2O72- + 14H+ + 3Sn2+  3Sn 4+ + 7H2O + 2Cr3+
c. Sn 4+
Ans. c
d. Sn +
Potassium chromate is
dichromate by adding;
a. Potassium
hydroxide
2−
2Cr O
+ 2 H +  Cr
4
2O7
converted
2−
into
potassium
b.
Sulphuric
acid
+H O
2
chromates and dichromate ions exist in equilibrium
and are inter convertible by altering the pH of the
solution.
c.acetic acid
Ans b
d. ammonium hydroxide
When hydrogen peroxide is
treated with cold acidified
K2Cr2O7 solution containing ether a blue colour is obtained
this is due to,
a. Chromium sulphate
b.Potassium chromate
K2Cr 2O7 + H2 SO4 +2H2O2 2K2SO4 +CrO5 + 5H2O
c. Perchromic acid
Ans d
d.Chromium pentoxide
Which of the following statements is not correct?
a. K2Cr2O7 solution in acidic medium is orange color
K2Cr2O7 solution in acidic medium is orange colour
b.K2Cr2−2O7 solution
becomes
yellow on increasing the pH of the
+
2−
2Cr O 4
+2H
 Cr 2 O 7
+ H2O
solution beyond 7 .
K2Cr2O7 solution becomes yellow on increasing the pH of the solution beyond 7
2−passing
On
H 2S
acidified
Crc.2 O
+ 2 OH - 
2Crthrough
O 4 2− + Hthe
7
2O .
K2Cr2O7 solution a milky
is observed .
K2Crcolour
2O7 + 4H2 SO4 +3H2SK2SO4 + Cr 2 (SO4) 3 +7H2O + 3S
gives milky
colour .
d. Na Cr O is prefered over K Cr OSulphur
in volumetric
analysis
2
2
7
2
Ans. d
2
7
KMnO4 acts as an oxidising agent in alkaline medium. When alkaline
KMnO4 is treated with KI, iodide ion is oxidised to ____________
b. IO–
a. I2
2 MnO4- + 16H+ +10 I -  2Mn 2+ + 8H2O + 5I2
c. IO3 –
Ans. a
d. IO4 –
Non-stoichiometric compounds of transition elements are
called-----------a.Hydrates
b.Hydrides
Hydrates : In which water molecules are
Hydrides : A binary compound of hydrogen
chemically bound to a compound. E.g
with metal. E.g NaH
CoCl2.6H2O.
c.Interstitial compounds
d.Binary compounds
Interstitial compounds are non stoichiometric compds
formed when small atoms (like H,C or N ) are
trapped inside the crystal lattice of T. metals. E.g
VH0.56 ,TiH1.7
Ans. c
Which one of the following is not the characteristic property of
interstitial compounds?
Interstitial
compounds
are
formed
when
small
atoms
(like
H,C
or
a.They have high melting points in comparison to pure metals.
N ) are trapped inside the crystal lattice of T. metals
b.They are very hard.
They have
highmetallic
melting points
in comparison to pure metals. They are very hard.
c.They
retain
conductivity.
.
They retain metallic conductivity.
They have
chemical properties
as the parent metals
d.They
aresame
chemically
very reactive
Ans.d
Lanthanides and Actinides
The lanthanides are also called the rare earth elements.
The atomic properties of the lanthanides vary little across the
period, and their chemical properties are also very similar.
General electron configuation of Ln - [Xe] 4f 1-14 5d 0-16s2
The +3 oxidation state is common for both lanthanides and
actinides.
General electron configuation of Ac - [Rn] 5f 1-14 6d 0-17s2
All actinides are radioactive, and have very similar physical and
chemical properties.
The element with electronic configuration [Xe] 4f 75d16s2 is
a. Representative element
b.Transition element
[Xe] 4f 75d16s2 z = 64 i. e lanthanoid , Gd
c. Lanthanoids
Ans c.
d. Actinoids
Which of the following oxidation state is common for all
lanthanoids?
a. +2
b.+3
[Xe] 4f 1-14 5d 0-16s2 common oxidation
state of lanthanides is +3
c. +4
Ans. b
d. +5
Which one of the following ion is not a good reducing
agent in solution ?
a.Sm2+
b.Eu2+
c. Ce4+
d.Yb 2+
The most stable oxidation state of lanthanides is +3.
Hence the ions in +2 oxidation state tend to change +3
state by loss of electron acting as reducing agents
Ln2+  Ln3+ + e (e.g Sm2+, Eu2+, Yb2+ )
Those in +4 oxidation state tend to change to +3 oxidation
state by gain of electron acting as a good oxidising agent
in aqueous solution. Ce4+ + e  Ce3+
Ans c
Identify the incorrect statement among the following
a. d-Block elements show irregular and erratic chemical
properties among themselves
b. La and Lu have partially filled d orbitals and no other
partially filled orbitals
c. The chemistry of various lanthanoids is very similar
d. 4f and 5f orbitals are equally shielded
Sol. 4f and 5f belongs to different energy levels, hence the
shielding effect is on them is not the same.
Shielding of 4f is more than 5f.
Ans. d
Number of elements belonging to 4f series are --------a. 6
b. 10
4f series from Ce to Lu contains
14 elements.
c. 28
Ans. d
d. 14
Lanthanoid contraction is caused due to,
a. The appropriate shielding on outer electrons by 4f electrons
from the nuclear charge
b.The aprpriate shielding on outer electrons by 5d electrons from the
nuclear charge
c. The same effective nuclear charge from Ce to Lu .
d.The imperfect shielding on outer electrons by 4f electrons from the
nuclear charge.
Ans. c
Although Zirconium belongs to 4d transition series and
Hafnium to 5d transition series even then they show similar
physical and chemical properties because___________.
Both
similar atomic radius Zr 160pm
a. both belong
to have
d-block
and Hf 159pm
b.both have same number of electrons.
c.both have similar atomic radius.
d.both belong to the same group of the periodic table
Ans. d
Lanthanoids and actinoids resemble in
a.Electronic configuration
b. Ionisation energy
Ln - [Xe] 4f 1-14 5d 0-16s2 ,
Ac- [Rn] 5f 1-14 6d 0-17s2
c. Oxidation state
Both show a dominant o.s of +3
Ans. c
d. Formation of complexes
Ln – Show lesser tendency for
complex formation
Ac- Show greater tendency for
complex formation
Which one of the following set correctly represents the
increase in the paramagnetic property of the ion?
a.Mn2+
<
+2 -[Ar]d5 - 5
Mn
2+
V < Cr2+ < Cu2+
+2 -[Ar]d
Crb.
- 4V2+
Cu2+4 <
< Cr2+ < Mn2+
Paramagnetic behavior depends on the no. of unpaired electrons
c. Cu2+ < Cr2+ < V2+ < Mn2+
Cu+2-[Ar]d9 -
Ans. b
1
d.Mn2+ < Cu2+ < Cr2+ < V2+
V+2 -[Ar]d3 - 4
The magnetic moment is associated with its spin angular
momentum and orbital angular momentum. Spin only
magnetic moment value of Cr3+ ion is
a. 2.84 B.M.
b. 3.87 B.M.
Cr +3 - [Ar]3d3 4s0 it contains 3 unpaired electrons
If n = 3
µ = n(n+2) = 3.87
c. 3.47 B.M.
Ans. b
d. 3.57 B.M.
KMnO 4 can be prepared from K2MnO4as per the reaction ;
3MnO4 2− + 2H2 O  2MnO4− + MnO2 + 4OH −
The reaction goes to completion by removing OH − ion by
adding
4KMnO + 4KOH 2 K MnO + O + 2H O
4
a. KOH
2
4
2
2
b. CO 2
3K2MnO4 + 2CO2 2 KMnO4 +MnO2 + 2K2CO3
HCl c.SO
react with OH − ion and gives water hence reaction
goes
d.HCl
2
to completion
Ans. d
Which of the following reactions are disproportionation
reactions?
i. Cu+ → Cu2+ + Cu
ii. 3MnO42- + 4H+ →2MnO4– + MnO2 + 2H2O
iii. 2KMnO4 →K2MnO4 + MnO2 + O2
iv. 2MnO4– + 3Mn2+ + 2H2O →5MnO2 + 4H+
a. i, ii
b. i, ii, iii
A particular oxidation state, which is relatively less stable compared to
other oxidation states, undergoes disproportionation.
Manganese(VI) which is relatively less stable changes over to
manganese(VII) and manganese(IV) in acid solution.
3MnO42- + 4H+  2MnO4- + MnO2 + 2H2O
c. ii, iii,
d. i, iv
Similarly
Cu+ iv
→ Cu2+ + Cu
Ans. a
Which of the following statements is not correct?
a. Copper liberates hydrogen from acids.
b.In its higher oxidation states, manganese forms stable
compounds with oxygen and fluorine.
c.Mn3+ and Co3+ are oxidising agents in aqueous solution.
d. Ti2+ and Cr2+ are reducing agents in aqueous solution
Ans.a