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d and f Block Elements

NEET Chemistry · 90 questions · Page 6 of 9 · Click an option or "Show Solution" to reveal answer

Q51
More number of oxidation states are exhibited by the actinoids than by the lanthanoids. The main reason for this is
A more active nature of the actinoids
B more energy difference between 5f and 6d orbitals than that between 4f and 5d orbitals
C lesser energy difference between 5f and 6d orbitals than that between 4f and 5d orbitals
D greater metallic character of the lanthanoids than that of the corresponding actinoids.
Correct Answer
Option C
Solution

The 5f-orbitals extend into space beyond the 6s and 6p-orbitals and participate in bonding.

This is in direct contrast to the lanthanides where the 4f-orbitals are buried deep inside in the atom, totally shielded by outer orbitals and thus unable to take part in bonding.

Q52
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)
A Ni 2+ , Cu +
B Ni 2+ , Ti 3+
C Sc 3+ , Ti 3+
D Sc 3+ , Co 2+ .
Correct Answer
Option B
Solution

For Ni +2 = [Ar] 3d 8 4s 0 (2 Unpaired electrons) Ti +3 = [Ar] 3d 8 4s 0 (1 unpaired electron) Sc +3 = [Ar] (0 unpaired electron) Cu + = [Ar] 3d 10 (0 unpaired electron) Hence, Ni +2 and Ti 3+ are coloured in aqueous solution due to the presence of unpaired electrons in d-subshell.

Q53
Four successive members of the first row transition elements are listed below with their atomic numbers. Which one of them is expected to have the highest third ionisation enthalpy?
A Vanadium (Z = 23)
B Chromium (Z = 24)
C Manganese (Z = 25)
D Iron (Z = 26)
Correct Answer
Option C
Solution

For third ionization enthalpy electronic configuration of V 2+ (21) : [Ar] 18 3d 3 4s 0 Cr 2+ (22) : [Ar] 18 3d 4 4s 0 Mn 2+ (23) : [Ar] 18 3d 5 4s 0 Fe 2+ (24) : [Ar] 18 3d 5 4s 1 Mn has most stable configuration due to half filled d-orbital.

Hence 3rd ionization energy will be highest for Mn.

Q54
The aqueous solution containing which one of the following ions will be colourless? (Atomic number : Sc = 21, Fe = 26, Ti = 22, Mn = 25)
A Sc 3+
B Fe 2+
C Ti 3+
D Mn 2+
Correct Answer
Option A
Solution

If the transition metal ion has unpaired electron then it shows colour.

Sc 3+ : [Ar] 18 3d 0 4s 0 Fe 2+ : [Ar] 18 3d 5 4s 1 Ti 3+ : [Ar] 18 3d 1 4s 0 Mn 2+ : [Ar] 18 3d 5 4s 0 Hence, Sc 3+ do not contain unpaired electron and hence it will not undergo d – d transition and do not show colour.

Q55
The number of moles of KMnO 4 reduced by one mole of KI in alkaline medium is
A one
B two
C five
D one fifth
Correct Answer
Option B
Solution

The chemical reactions involved in this case are as follows : KI + H 2 O →\to KOH + HI 2KMnO 4 + 2KOH → 2K 2 MnO 4 + H 2 O + [O] Thus, for overall reaction, two moles of KMnO 4 are reduced by two moles of KOH or KI.

Hence, one mole of KI is required to reduce one mole of KMnO 4 .

Q56
More number of oxidation states are exhibited by the actinoids than by the lanthanoids. The main reason for this is
A more active nature of the actinoids
B more energy difference between 5f and 6d orbitals than that between 4f and 5d orbitals
C lesser energy difference between 5f and 6d orbitals than that between 4f and 5d orbitals
D greater metallic character of the lanthanoids than that of the corresponding actinoids.
Correct Answer
Option C
Solution

The 5f-orbitals extend into space beyond the 6s and 6p-orbitals and participate in bonding.

This is in direct contrast to the lanthanides where the 4f-orbitals are buried deep inside in the atom, totally shielded by outer orbitals and thus unable to take part in bonding.

Q57
lanthanoids are
A 14 elements in the sixth period (atomic no. 90 to 103) that are filling 4 f sublevel
B 14 elements in the seventh period (atomic number = 90 to 103) that are filling 5 f sublevel
C 14 elements in the sixth period (atomic number = 58 to 71) that are filling the 4 f sublevel
D 14 elements in the seventh period (atomic number = 50 to 71) that are filling 4 f sublevel.
Correct Answer
Option C
Solution

As sixth period can accommodate only 18 elements in the table, 14 members of 4f series (atomic number 58 to 71) are separately accommodated in a horizontal row below the periodic table.

These are called as lanthanides.

Q58
Among the following series of transition metal ions, the one where all metal ions have 3d 2 electronic configuration is [At. Nos. Ti = 22, V = 23, Cr = 24, Mn = 25]
A Ti 3+ , V 2+ , Cr 3+ , Mn 4+
B Ti + , V 4+ , Cr 6+ , Mn 7+
C Ti 4+ , V 3+ , Cr 2+ , Mn 3+
D Ti 2+ , V 3+ , Cr 4+ , Mn 5+
Correct Answer
Option D
Solution

Ti +2 = [Ar] 18 3d 2 4s 0 V 3+ = [Ar] 18 3d 2 4s 0 Cr +4 = [Ar] 18 3d 2 4s 0 Mn +5 = [Ar] 18 3d 2 4s 0

Q59
The correct order of ionic radii of Y 3+ , La 3+ , Eu 3+ and Lu 3+ is (Atomic nos. Y = 39, La = 57, Eu = 63, Lu = 71)
A Y 3+ < La 3+ < Eu 3+ < Lu 3+
B Y 3+ < Lu 3+ < Eu 3+ < La 3+
C Lu 3+ < Eu 3+ < La 3+ < Y 3+
D La 3+ < Eu 3+ < Lu 3+ < Y 3+
Correct Answer
Option B
Solution

In lanthanide series there is a regular decrease in the atomic as well as ionic radii of trivalent ions (M 3+ ) as the atomic number increases.

Although the atomic radii do show some irregularities but ionic radii decreases from La(103 pm) to Lu (86pm).

Q60
The basic character of the transition metal monoxides follows the order (Atomic no's. Ti = 22, V = 23, Cr = 24, Fe = 26)
A VO > CrO > TiO > FeO
B CrO > VO > FeO > TiO
C TiO > FeO > VO > CrO
D TiO > VO > CrO > FeO
Correct Answer
Option D
Solution

The basic character of the transition metal monoxide is TiO > VO > CrO > FeO because basic character of oxides decrease with increase in atomic number.

Oxides of transitional metals in low oxidation state i.e., + 2 and + 3 are generally basic except Cr 2 O 3 .

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