the d- and f- Block Elements Practice Questions - DPP for JEE

 Page 1


1. (d) (n – 1)d
5
ns
2
 attains the maximum O.S. of + 7.
2. (b) Mn
2+
 (3d
5
) is more stable them Mn
3+
 (3d
4
).
3. (b) Mn - 3d
5
 4s
2 
The no. of various oxidation states possible are +  2, + 3, + 4, + 5, + 6
and + 7.
4. (c) The transition metals and their compounds are used as catalysts.
Because of the variable oxidation states they may form
intermediate compound with one of the reactants. These
intermediate provides a new path with lower activation energy.
5. (a) According to their positions in the periods, these values are in the
order:
Yb
3+
 < Pm
3+
 < Ce
3+
 < La
3+
At. Nos. 70 61         58        57
This is due to lanthanide contraction.
6. (a) When a solution of potassium chromate is treated with an excess
of dilute nitric acid. Potassium dichromate and H
2
O are formed.
2K
2
CrO
4 
+ 2HNO
3 
–?  K
2
Cr
2
O
7 
+ 2KNO
3 
+ H
2
O
Hence Cr
2
O
7
? – 
and H
2
O are formed.
7. (b) Mn
2+
 (d
5
)is more stable than Mn
3+
 (d
4
), thus
8. (c)
9. (d) O.N. of Cr in all the species is + 6. Configuration of Cr (VI) is (Ar)
. Hence no d-d electron transition.
Page 2


1. (d) (n – 1)d
5
ns
2
 attains the maximum O.S. of + 7.
2. (b) Mn
2+
 (3d
5
) is more stable them Mn
3+
 (3d
4
).
3. (b) Mn - 3d
5
 4s
2 
The no. of various oxidation states possible are +  2, + 3, + 4, + 5, + 6
and + 7.
4. (c) The transition metals and their compounds are used as catalysts.
Because of the variable oxidation states they may form
intermediate compound with one of the reactants. These
intermediate provides a new path with lower activation energy.
5. (a) According to their positions in the periods, these values are in the
order:
Yb
3+
 < Pm
3+
 < Ce
3+
 < La
3+
At. Nos. 70 61         58        57
This is due to lanthanide contraction.
6. (a) When a solution of potassium chromate is treated with an excess
of dilute nitric acid. Potassium dichromate and H
2
O are formed.
2K
2
CrO
4 
+ 2HNO
3 
–?  K
2
Cr
2
O
7 
+ 2KNO
3 
+ H
2
O
Hence Cr
2
O
7
? – 
and H
2
O are formed.
7. (b) Mn
2+
 (d
5
)is more stable than Mn
3+
 (d
4
), thus
8. (c)
9. (d) O.N. of Cr in all the species is + 6. Configuration of Cr (VI) is (Ar)
. Hence no d-d electron transition.
10. (c) The number of unpaired electrons in Ni
2+
(aq) = 2
Water is weak ligand hence no pairing will take place spin magnetic
moment
11. (b) has electronic configuration hence stable due to
half filled atomic orbitals.
12. (c)
13. (a)
14. (a)
(a) V = 3d
3
 4s
2 
; V
2+
 = 3d
3
 = 3 unpaired electrons
Cr = 3d
5
 4s
1 
; Cr
2+
 = 3d
4
 = 4 unpaired electrons
Mn = 3d
5
 4s
2 
; Mn
2+
 = 3d
5
 = 5 unpaired electrons
Fe = 3d
6
 4s
2 
; Fe
2+
 = 3d
6
 = 4 unpaired electrons
Hence the correct order of paramagnetic behaviour
V
2+
 < Cr
2+
 = Fe
2+
 < Mn
2+
(b) For the same oxidation state, the ionic radii generally decreases as the
atomic number increases in a particular transition series. hence the 
order is
Mn
++
 > Fe
++
 > Co
++
 > Ni
++
(c) In solution, the stability of the compound depends upon electrode
potentials, SEP of the transitions metal ions are given as
Co
3+ 
/ Co = + 1.97, Fe
3+
 / Fe = + 0.77 ;
Cr
3+
 / Cr
2+
 = – 0.41, Sc 
3+
 is highly stable as it does not show + 2 O. S.
(d) Sc – (+ 2), (+ 3)
Ti – (+ 2), (+ 3), (+ 4)
Page 3


1. (d) (n – 1)d
5
ns
2
 attains the maximum O.S. of + 7.
2. (b) Mn
2+
 (3d
5
) is more stable them Mn
3+
 (3d
4
).
3. (b) Mn - 3d
5
 4s
2 
The no. of various oxidation states possible are +  2, + 3, + 4, + 5, + 6
and + 7.
4. (c) The transition metals and their compounds are used as catalysts.
Because of the variable oxidation states they may form
intermediate compound with one of the reactants. These
intermediate provides a new path with lower activation energy.
5. (a) According to their positions in the periods, these values are in the
order:
Yb
3+
 < Pm
3+
 < Ce
3+
 < La
3+
At. Nos. 70 61         58        57
This is due to lanthanide contraction.
6. (a) When a solution of potassium chromate is treated with an excess
of dilute nitric acid. Potassium dichromate and H
2
O are formed.
2K
2
CrO
4 
+ 2HNO
3 
–?  K
2
Cr
2
O
7 
+ 2KNO
3 
+ H
2
O
Hence Cr
2
O
7
? – 
and H
2
O are formed.
7. (b) Mn
2+
 (d
5
)is more stable than Mn
3+
 (d
4
), thus
8. (c)
9. (d) O.N. of Cr in all the species is + 6. Configuration of Cr (VI) is (Ar)
. Hence no d-d electron transition.
10. (c) The number of unpaired electrons in Ni
2+
(aq) = 2
Water is weak ligand hence no pairing will take place spin magnetic
moment
11. (b) has electronic configuration hence stable due to
half filled atomic orbitals.
12. (c)
13. (a)
14. (a)
(a) V = 3d
3
 4s
2 
; V
2+
 = 3d
3
 = 3 unpaired electrons
Cr = 3d
5
 4s
1 
; Cr
2+
 = 3d
4
 = 4 unpaired electrons
Mn = 3d
5
 4s
2 
; Mn
2+
 = 3d
5
 = 5 unpaired electrons
Fe = 3d
6
 4s
2 
; Fe
2+
 = 3d
6
 = 4 unpaired electrons
Hence the correct order of paramagnetic behaviour
V
2+
 < Cr
2+
 = Fe
2+
 < Mn
2+
(b) For the same oxidation state, the ionic radii generally decreases as the
atomic number increases in a particular transition series. hence the 
order is
Mn
++
 > Fe
++
 > Co
++
 > Ni
++
(c) In solution, the stability of the compound depends upon electrode
potentials, SEP of the transitions metal ions are given as
Co
3+ 
/ Co = + 1.97, Fe
3+
 / Fe = + 0.77 ;
Cr
3+
 / Cr
2+
 = – 0.41, Sc 
3+
 is highly stable as it does not show + 2 O. S.
(d) Sc – (+ 2), (+ 3)
Ti – (+ 2), (+ 3), (+ 4)
Cr – (+ 1), (+ 2), (+ 3), (+ 4), (+ 5), (+ 6)
Mn – (+ 2), (+ 3), (+ 4), (+ 5), (+ 6), (+ 7)
i.e. Sc < Ti < Cr = Mn
15. (b) In lanthanides, there is poorer shielding of 5d electrons by 4f
electrons resulting in greater attraction of the nucleus over 5d
electrons and contraction of the atomic radii.
16. (d)
Eu      La Gd Am
O.S +2      +3 +3 +3,+4,+5,+6
17. (a)   
18. (c)
              
19. (a) Paramagnetic moment is directly proportional to number of
unpaired electrons present in the complex.
20. (d) d 
5
 –––– strong ligand field
t
2g                             
e
g
µ = =  = 1.73BM
d
3
–– in weak as well as in  strong field
t
2g 
                   e
g
µ = =  = 3.87 B.M.
d
4
– in weak ligand field
t
2g 
                    e
g
Page 4


1. (d) (n – 1)d
5
ns
2
 attains the maximum O.S. of + 7.
2. (b) Mn
2+
 (3d
5
) is more stable them Mn
3+
 (3d
4
).
3. (b) Mn - 3d
5
 4s
2 
The no. of various oxidation states possible are +  2, + 3, + 4, + 5, + 6
and + 7.
4. (c) The transition metals and their compounds are used as catalysts.
Because of the variable oxidation states they may form
intermediate compound with one of the reactants. These
intermediate provides a new path with lower activation energy.
5. (a) According to their positions in the periods, these values are in the
order:
Yb
3+
 < Pm
3+
 < Ce
3+
 < La
3+
At. Nos. 70 61         58        57
This is due to lanthanide contraction.
6. (a) When a solution of potassium chromate is treated with an excess
of dilute nitric acid. Potassium dichromate and H
2
O are formed.
2K
2
CrO
4 
+ 2HNO
3 
–?  K
2
Cr
2
O
7 
+ 2KNO
3 
+ H
2
O
Hence Cr
2
O
7
? – 
and H
2
O are formed.
7. (b) Mn
2+
 (d
5
)is more stable than Mn
3+
 (d
4
), thus
8. (c)
9. (d) O.N. of Cr in all the species is + 6. Configuration of Cr (VI) is (Ar)
. Hence no d-d electron transition.
10. (c) The number of unpaired electrons in Ni
2+
(aq) = 2
Water is weak ligand hence no pairing will take place spin magnetic
moment
11. (b) has electronic configuration hence stable due to
half filled atomic orbitals.
12. (c)
13. (a)
14. (a)
(a) V = 3d
3
 4s
2 
; V
2+
 = 3d
3
 = 3 unpaired electrons
Cr = 3d
5
 4s
1 
; Cr
2+
 = 3d
4
 = 4 unpaired electrons
Mn = 3d
5
 4s
2 
; Mn
2+
 = 3d
5
 = 5 unpaired electrons
Fe = 3d
6
 4s
2 
; Fe
2+
 = 3d
6
 = 4 unpaired electrons
Hence the correct order of paramagnetic behaviour
V
2+
 < Cr
2+
 = Fe
2+
 < Mn
2+
(b) For the same oxidation state, the ionic radii generally decreases as the
atomic number increases in a particular transition series. hence the 
order is
Mn
++
 > Fe
++
 > Co
++
 > Ni
++
(c) In solution, the stability of the compound depends upon electrode
potentials, SEP of the transitions metal ions are given as
Co
3+ 
/ Co = + 1.97, Fe
3+
 / Fe = + 0.77 ;
Cr
3+
 / Cr
2+
 = – 0.41, Sc 
3+
 is highly stable as it does not show + 2 O. S.
(d) Sc – (+ 2), (+ 3)
Ti – (+ 2), (+ 3), (+ 4)
Cr – (+ 1), (+ 2), (+ 3), (+ 4), (+ 5), (+ 6)
Mn – (+ 2), (+ 3), (+ 4), (+ 5), (+ 6), (+ 7)
i.e. Sc < Ti < Cr = Mn
15. (b) In lanthanides, there is poorer shielding of 5d electrons by 4f
electrons resulting in greater attraction of the nucleus over 5d
electrons and contraction of the atomic radii.
16. (d)
Eu      La Gd Am
O.S +2      +3 +3 +3,+4,+5,+6
17. (a)   
18. (c)
              
19. (a) Paramagnetic moment is directly proportional to number of
unpaired electrons present in the complex.
20. (d) d 
5
 –––– strong ligand field
t
2g                             
e
g
µ = =  = 1.73BM
d
3
–– in weak as well as in  strong field
t
2g 
                   e
g
µ = =  = 3.87 B.M.
d
4
– in weak ligand field
t
2g 
                    e
g
µ = =  = 4.89
d 
4
– in strong ligand field
t
2g 
                e
g
      
µ = =  = 2.82.
21. (b) In neutral or faintly alkaline medium thiosulphate is quantitatively
oxidized by KMnO
4
 to SO
4
? 2–
8KMnO
4
 + 3Na
2
S
2
O
3
 + H
2
O  3K
2
SO
4
 + 8MnO
2
 + 3Na
2
SO
4
 +
2KOH
22. (a) Cu, Hg and Ag are attacked by conc acids but gold is not attacked.
 forms a complex with HCl which is used for
tonning in photography.
23. (a) (acid medium) 
24. (a) When KI is added to mercuric iodide it disssolve in it and form
complex.
On heating HgI
2
 decomposes as
25. (c)
= 0.34 V
other has – ve 
 =  – 0.28 V
= – 0.25V
 = – 0.44V