Boards Sprint - d and f-block elements (03.12.2020).pdf

2. Date Topic of the day
19 Nov Solid state
21 Nov Solutions
24 Nov Electrochemistry
26Nov Chemical Kinetics
28 Nov Surface Chemistry
1 Dec p-Block Elements
3 Dec d & f Block Elements
5 Dec Coordination Compounds
8 Dec Haloalkanes & Haloarenes
10 Dec Alcohols, Phenols & Ethers
12 Dec Aldehydes, Ketones & Carboxylic acid
15 Dec Amines
17 Dec Biomolecules
12th
Board Sprint - Schedule

3. 3
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13. 1M
2M
3M
5M
Topic wise analysis 10 Years (2019-2010)

14. Topic wise analysis 10 Years (2019-2010)
VSA
(1 mark)
SA-I
(2 Marks)
SA-II
(3 Marks)
LA
(5 Marks)
Max. Questions
“ Gen.
Characteristics
d & f ”
Max. Questions
“General
Properties of
d-block”
-

17. Sc Ti
Y Zr
La Hf
V Cr Mn Fe Co Ni Cu Zn
Nb Mo Tc Ru Rh Pd Ag Cd
T
a
W Re Os Ir Pt Au Hg
General E.C.
(n-1)d1–10
ns0–2

20. ●Due to stability of
half filled orbitals
Strong metallic
bonds due to d5
-d5
interaction
Cr
Maximum at d5

21. Almost same size
Filling of 4f before 5d,
decreases the overall size
(Poor shielding by of
f-orbital)
Due to

22. d5
shows high
values due to
strong metallic
bonds
0
100
200
300
400
500
600
700
800
900
Series 1
Series 2
Series 3
Atomic number
ΔH
o
/kJ
mol
-1
Cr
Mo
W

23. Poor shielding
from f-e-
s
increase I.E down
the group
1st
I.E. < 2nd
I.E. < 3rd
I.E.

24. Sc Ti V Cr Mn Fe Co Ni Cu Zn
+3
+2
+3
+4
+2
+3
+4
+5
+2
+3
+4
+5
+6
+2
+3
+4
+5
+6
+7
+2
+3
+4
+6
+2
+3
+4
+2
+3
+4
+1
+2
+2
Highest
oxidation
number

25. Due to unpaired electrons
(Weakly attracted to magnet)
Due to unpaired electrons
(Strongly attracted to magnet)
Due to paired electrons
( Repelled by magnet)
Due to unpaired electrons
(Moderately attracted to magnet)

26. Magnetic moment
n = no. of unpaired electrons
𝝻 = Magnetic moment
Units = BM (Bohr magneton)
One unpaired electron
1.73 BM

27. Vacant d-orbitals
readily accepts
electrons from donor
atoms (Ligands)

28. Due to variable
reactivity, they are
known for catalysis
Ex: Fe in Haber’s
process

29. V2
O5
in Contact process
. . . .
O
V V
O
O
O
O

30. Raney-Ni in hydrogenation of alkenes

31. Due to
electronic
transitions in
d-orbitals

33. [Xe] 4f1-14
5d0-1
6s2
Lanthanides
[Rn] 5f1-14
6d0-1
7s2
Actinides

35. La
Lu
Atomic
Radius
Atomic No.
Lanthanide contraction
Due to increased nuclear charge
& decreased screening effect.

36. Ce = +2, +3, +4
Pr, Nd, Tb & Dy = +3, +4

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42. Very Short Answers (VSA)
1.Account for the following : Zn is not
considered as a transition element.
000(AI 2014, Delhi 2010)000
Zn has completely filled d-orbitals
even in oxidized state (3d10
4s0
)

43. Very Short Answers (VSA)
2. What are the different oxidation
states exhibited by the lanthanides?
000(Delhi 2014)000
Lanthanide show majorly - +3
Rarely they show - +2 and +4

44. Very Short Answers (VSA)
3. Assign reason for the following :
Copper (I) ion is not known in
aqueous solution
000(Delhi 2011)000
Undergoes disproportionation in water
To form stable oxidation state.

45. Very Short Answers (VSA)
4. Explain giving reasons : Transition
metals and their compounds generally
exhibit a paramagnetic behaviour
000(Delhi 2014)000
Most of the compounds have unpaired
electrons (Paramagnetic) in (n-1)d orbital

46. Very Short Answers (VSA)
5. Account for the following :
Mn2+
is more stable than Fe2+
towards oxidation to +3 state
000(Delhi 2014)000
Mn2+
- 3d5
= Half filled and difficult to
oxidize to +3.
Fe2+
- 3d6
: Easy to oxidize to +3 as it
forms half filled orbital

47. Short Answers (SA-I)
6. Give reasons :
(i) Mn shows the highest oxidation state of
+7 with oxygen but with Fluorine it shows
the highest oxidation state of +4
(ii) Transition metals show variable
oxidation states.
000(Delhi 2014)000

48. (i) With O: forms p𝝿 - d𝝿 thus maximum
oxidation of +7
With F: it cannot form p𝝿 - d𝝿
Short Answers (SA-I)
(ii) Transition metals use both ns and (n-1)d electrons
for bond formation
Thus, shows variable oxidation states.

49. Short Answers (SA-I)
7. Assign a reason for each of the following
observations :
(i) The transition metals (with the exception of
Zn, Cd and Hg) are hard and have high
melting and boiling points.
(ii) The ionisation enthalpies (First and
second) in the first series of the transition
elements are found to vary irregularly
000(2018)000

50. Short Answers (SA-I)
(ii) Variation in I.E. is due to variation in stability
brought by electronic configuration.
Ex: d0
, d5
, d10
are highly stable
(i) From left to right, size decreases due
to increase in effective nuclear charge.
This increases the density of the metal.
Higher density = Higher M.P. and B.P.

51. Short Answers (SA-I)
8. Assign reasons for the following :
(i) Transition metals and many of their
compounds act as good catalysts.
(ii) Transition metals are much harder than
the alkali metals.
000(AI 2014)000

52. Short Answers (SA-I)
(ii) Usage of ns and (n-1)d electrons form stronger
metallic bonds. Thus, they show hard metallic
nature.
(i) Due to variable oxidation states and
formation of complex, transition
elements exhibit catalytic property.
Ex: V2
O 5
in contact process

53. Short Answers (SA-I)
9. (i) d-block elements exhibit more
oxidation states than f-block elements.
(ii) The enthalpies of atomization of the
transition metals are high.
(iii) The variation in oxidation states of
transition metals is of different type
from that of the non-transition metals.
000( AI 2013)000
Short Answers (SA-II)

54. Short Answers (SA-I)
Short Answers (SA-II)
(i) Variable oxidation states are more in d than f,
due to small energy difference b/w ns & (n-1)d
(ii) High enthalpy of atomization is due to high no. of unpaired
electrons in d- orbitals. This forms strong metallic bonds.
(iii) Transition metals have oxidation states which differ by unity
Non-Transition metals differ by two or more.
Ex: Fe3+
and Fe2+
, Cu2+
and Cu+
Vs Pb2+
and Pb4+
, Sn2+
and
Sn4+

55. Short Answers (SA-I)
10. How would you account for the following :
(i) Metal-metal bonding is more extensive in the
4d and 5d series of transition elements than the
3d series.
(ii) Mn (III) undergoes disproportionation reaction
easily.
(iii) Name a member of the lanthanoid series
which is well known to exhibit +4 oxidation state.
000(AI 2011)000
Short Answers (SA-II)

56. Short Answers (SA-II)
(i) Shielding is low in 4d and 5d as compared to
3d. Hence, forms stronger metallic bonds
(ii) Mn3+
is less stable, but Mn2+
is half-filled and more stable.
Thus, Mn3+
in H2
O shows disproportionation.
2Mn3+
+ 2H2
O → Mn2+
+ MnO2
+ 4H+
(iii) Lanthanoids showing +4 oxidation state are CeIV

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