d and f Block Elements — Page 3 | NEET Chemistry

Q21

The electronic configuration of Eu (Atomic No. 63), Gd (Atomic No. 64) and Tb (Atomic No.65) are

A [Xe]4f 6 5d 1 6s 2 , [Xe]4f 7 5d 1 6s 2 and [Xe]4f 8 5d 1 6s 2

B [Xe]4f 7 6s 2 , [Xe]4f 7 5d 1 6s 2 and [Xe]4f 9 6s 2

C [Xe]4f 7 6s 2 , [Xe]4f 8 6s 2 and [Xe]4f 8 5d 1 6s 2

D [Xe]4f 6 5d 1 6s 2 , [Xe]4f 7 5d 1 6s 2 and [Xe]4f 9 6s 2

Correct Answer

Option B

Solution

Eu (63) = [Xe] 4f 7 6s 2 Gd (64) = [Xe] 4f 7 5d 1 6s 2 Tb (65) = [Xe] 4f 9 6s 2

Q22

Assuming complete ionisation, same moles of which of the following compounds will require the least amount of acidified KMnO 4 for complete oxidation?

A FeSO 3

B FeC 2 O 4

C Fe(NO 2 ) 2

D FeSO 4

Correct Answer

Option D

Solution

FeSO 4 will require the least amount of acidified KMnO 4 for complete oxidation.

Q23

Gadolinium belongs to 4f series. Its atomic number is 64. Which of the following is the correct electronic configuration of gadolinium?

A [Xe]4f 9 5s 1

B [Xe]4f 7 5d 1 6s 2

C [Xe]4f 6 5d 2 6s 2

D [Xe]4f 8 6d 2

Correct Answer

Option B

Solution

Gd(64) = [Xe]4f 7 5d 1 6s 2

Q24

Which of the following processes does not involve oxidation of iron?

A Formation of Fe(CO) 5 from Fe.

B Liberation of H 2 from steam by iron at high temperature.

C Rusting of iron sheets.

D Decolourisation of blue CuSO 4 solution by iron.

Correct Answer

Option A

Solution

Fe + 5CO $\to$ Fe(CO) 5 Thus, formation of Fe(CO) 5 from Fe does not involve oxidation of iron because there is no change in oxidation state.

Rest in all options there is a change in oxidation state of Fe.

Q25

Because of lanthanoid contraction, which of the following pairs of elements have nearly same atomic radii? (Numbers in the parenthesis are atomic numbers)

A Zr(40) and Hf(72)

B Zr(40) and Ta(73)

C Ti(22) and Zr(40)

D Zr(40) and Nb(41)

Correct Answer

Option A

Solution

Because of the lanthanoid contraction Zr (atomic radii 160 pm) and Hf (atomic radii 158 pm) have nearly same atomic radii.

Q26

Magnetic moment 2.84 B.M. is given by (At. nos. Ni = 28, Ti = 22, Cr = 24, Co = 27)

A Cr 2+

B Co 2+

C Ni 2+

D Ti 3+

Correct Answer

Option C

Solution

Magnetic moment ( $\mu$ ) =

\sqrt {n\left( {n + 2} \right)}

For $\mu$ = 2.84 2.84 =

\sqrt {n\left( {n + 2} \right)}

$\Rightarrow$ n = 2 Cr 2+ – [Ar]3d 4 4s 0 , 4 unpaired electrons Co 2+ – [Ar]3d 7 4s 0 , 3 unpaired electrons Ni 2+ – [Ar]3d 8 4s 0 , 2 unpaired electrons Ti 3+ – [Ar]3d 1 4s 0 , 1 unpaired electron

Q27

Reason of lanthanoid contraction is

A negligible screening effect of 'f'- orbitals

B increasing nuclear charge

C decreasing nuclear charge

D decreasing screening effect.

Correct Answer

Option A

Solution

The shape of f-orbitals is very much diffused and they have poor shielding effect.

The effective nuclear charge increases which causes the contraction in the size of electron charge cloud.

This contraction in size is quite regular and known as lanthanoid contraction.

Q28

The reaction of aqueous KMnO 4 with H 2 O 2 in acidic conditions gives

A Mn 4+ and O 2

B Mn 2+ and O 2

C Mn 2+ and O 3

D Mn 4+ and MnO 2

Correct Answer

Option B

Solution

Hydrogen peroxide is oxidised to H 2 O and O 2 .

2KMnO 4 + 3H 2 SO 4 + 5H 2 O 2 $\to$ K 2 SO 4 + 2MnSO 4 + 8H 2 O + 5O 2 Thus, Mn 2+ and O 2 are produced.

Q29

Magnetic moment 2.83 BM is given by which of the following ions? (At. nos. Ti = 22, Cr = 24, Mn = 25, Ni = 28)

A Ti 3+

B Ni 2+

C Cr 3+

D Mn 2+

Correct Answer

Option B

Solution

Magnetic moment ( $\mu$ ) =

\sqrt {n\left( {n + 2} \right)}

For $\mu$ = 2.84 2.84 =

\sqrt {n\left( {n + 2} \right)}

$\Rightarrow$ n = 2 Cr 2+ – [Ar]3d 4 4s 0 , 4 unpaired electrons Co 2+ – [Ar]3d 7 4s 0 , 3 unpaired electrons Ni 2+ – [Ar]3d 8 4s 0 , 2 unpaired electrons Ti 3+ – [Ar]3d 1 4s 0 , 1 unpaired electron

Q30

Sc (Z = 21) is a transition element but Zn (Z = 30) is not because

A both Sc 3+ and Zn 2+ ions are colourless and form white compounds.

B in case of Sc, 3d orbitals are partially filled but in Zn these are filled.

C last electron is assumed to be added to 4s level in case of Zn.

D both Sc and Zn do not exhibit variable oxidation states.

Correct Answer

Option B

Solution

A transition element must have incomplete d-subshell.

Zinc have completely filled d subshell having 3d 10 configuration.

Hence do not show properties of transition elements to any appreciable extent except for their ability to form complexes.