d and f Block Elements

NEET Chemistry · 90 questions · Page 5 of 9 · Click an option or "Show Solution" to reveal answer

Q41

Which of the following oxidation states is the most common among the lanthanoids?

A 4

B 2

C 5

D 3

Correct Answer

Option D

Solution

The common stable oxidation state of all the lanthanoids is +3.

The oxidation state of +2 and +4 are also exhibited by some of the elements.

These oxidation states are only stable in those cases where stable 4f 0 , 4f 7 or 4f 14 configurations are achieved.

Q42

Match (substances) with (processes) employed in the manufacture of the substances and select the correct option. <br> <table class=tg> <tbody><tr> <th class=tg-0xlv colspan=2>Column ${\rm I}$ <br>(Substances)</th> <th class=tg-4qqe></th> <th class=tg-0xlv colspan=2>Column ${\rm I}$ ${\rm I}$ <br>(Processes)</th> </tr> <tr> <td class=tg-c4o0>

List - I		List - II
(A)	</td> <td class=tg-c4o0>Sulphuric acid</td> <td class=tg-x1hj></td> <td class=tg-c4o0>	(i)	</td> <td class=tg-c4o0>Haber's process</td> </tr> <tr> <td class=tg-c4o0>
(B)	</td> <td class=tg-c4o0>Steel</td> <td class=tg-x1hj></td> <td class=tg-c4o0>	(ii)	</td> <td class=tg-c4o0>Bessemer's process</td> </tr> <tr> <td class=tg-c4o0>
(C)	</td> <td class=tg-c4o0>Sodium hydroxide</td> <td class=tg-x1hj></td> <td class=tg-c4o0>	(iii)	</td> <td class=tg-c4o0>Leblanc process</td> </tr> <tr> <td class=tg-c4o0>
(D)	</td> <td class=tg-c4o0>Ammonia</td> <td class=tg-x1hj></td> <td class=tg-c4o0>	(iv)	</td> <td class=tg-c4o0>Contact process</td> </tr> </tbody></table>

A (A)

\to

(i); (B)

\to

(iv); (C)

\to

(ii); (D)

\to

(iii)

B (A)

\to

(i); (B)

\to

(ii); (C)

\to

(iii); (D)

\to

(iv)

C (A)

\to

(iv); (B)

\to

(iii); (C)

\to

(ii); (D)

\to

(i)

D (A)

\to

(iv); (B)

\to

(ii); (C)

\to

(iii); (D)

\to

(i)

Correct Answer

Option D

Solution

(1) Sulphuric acid (iv) Contact process (2) Steel (ii) Bessemer’s process (3) Sodium hydroxide (iii) Leblanc process (4) Ammonia (i) Haber’s process

Q43

Which of the following pairs has the same size?

A Fe 2+ , Ni 2+

B Zr 4+ , Ti 4+

C Zr 4+ , Hf 4+

D Zn 2+ , Hf 4+

Correct Answer

Option C

Solution

Due to lanthanide contraction, the size of Zr and Hf (atom and ions) become nearly similar.

Q44

Which of the following ions has electronic configuration [Ar]3d 6 ? (At. nos. Mn = 25, Fe = 26, Co = 27, Ni = 28)

A Ni 3+

B Mn 3+

C Fe 3+

D Co 3+

Correct Answer

Option D

Solution

The electronic configuration of the given ions is : Ni 3+ : [Ar]3d 7 4s 0 Mn 3+ : [Ar]3d 4 4s 0 Fe 3+ : [Ar]3d 5 4s 0 Co 3+ : [Ar]3d 6 4s 0 Thus, Co 3+ is the ion with the desired configuration.

Q45

Which of the following ions will exhibit colour in aqueous solutions?

A La 3+ (Z = 57)

B Ti 3+ (Z = 22)

C Lu 3+ (Z = 71)

D Se 3+ (Z = 21)

Correct Answer

Option B

Solution

La 3+ : 54 e – = [Xe] Ti 3+ : 19 e – = [Ar] 4s 0 3d 1 (Colour) Lu 3+ : 68 e – = [Xe] 4f 14 Sc 3+ : 18 e – = [Ar] As ions which have unpaired electrons exhibit colour in solution.

Ti 3+ has an outer electronic configuration of 4s 0 3d 1 , i.e., 1 unpaired electron.

Thus its solution will be coloured.

Q46

Which one of the elements with the following outer orbital configurations may exhibit the largest number of oxidation states?

A 3d 5 4s 1

B 3d 5 4s 2

C 3d 2 4s 2

D 3d 3 4s 2

Correct Answer

Option B

Solution

All the given elements are d-block elements.

For d-block element, it can lose its electrons from 3d and 4s subshell completely.

Thus, the element which has higher number of electrons in 3d and 4s subshells exhibits largest number of oxidation states.

Among the given options, 3d 5 , 4s 2 has maximum number of electrons that is, 5 + 2 = 7.

Hence, it has largest number of oxidation states.

Q47

The correct order of decreasing second ionisation enthalpy of Ti(22), V(23), Cr(24) and Mn(25) is

A Mn > Cr > Ti > V

B Ti > V > Cr > Mn

C Cr > Mn > V > Ti

D V > Mn > Cr > Ti

Correct Answer

Option C

Solution

Ti ; Z (22) is 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 2 V ; Z (23) is 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 3 Cr ; Z (24) is 1s 2 2s 2 2p 6 3s 2 3p 6 3d 5 4s 1 Mn ; Z (25) is 1s 2 2s 2 2p 6 3s 2 3d 5 4s 2 The second electron in all the cases (except Cr) is taken out from 4s-orbital and for Cr it is taken from completely half filled 3d-orbital.

The force required for removal of second electron will be more for Mn than others (except for Cr) due to having more positive charge.

Based on this we find the correct order Mn > V > Ti. i.e.

Cr > Mn > V > Ti.

Q48

Identify the incorrect statement among the following:

A Lanthanoid contraction is the accumulation of successive shrinkages.

B As a result of lanthanoid contraction, the properties of 4d series of the transition elements have no similarities with the 5d series of elements.

C Shielding power of 4

f

electrons is quite weak.

D There is a decrease in the radii of the atoms or ions as one proceeds from La to Lu.

Correct Answer

Option B

Solution

In each vertical column of transition elements, the elements of second and third transition series resemble each other more closely than the elements of first and second transition series on account of lanthanide contraction.

Hence the properties of elements of 4d series of the transition elements resemble with the properties of the elements of 5d series of the transition elements.

Q49

Which one of the following ions is the most stable in aqueous solution? (At. No. Ti = 22, V = 23, Cr = 24, Mn = 25)

A V 3+

B Ti 3+

C Mn 3+

D Cr 3+

Correct Answer

Option C

Solution

The exactly half-filled and completely filled d-orbital have extra stability and stability is directly proportional to number of unpaired electrons.

Mn +3 (22) = 3d 4 4s 0 (4 unpaired electrons) So, Mn 3+ ion is most stable in aqueous solution.

Rest all have less number of unpaired electrons.

Q50

Copper sulphate dissolves in excess of KCN to give

A Cu(CN) 2

B CuCN

C [Cu(CN) 4 ] 3

-

D [Cu(CN) 4 ] 2

-

Correct Answer

Option C

Solution

Copper sulphate on treatment with excess of KCN forms complex K 3 [Cu(CN) 4 ] or [Cu(CN) 4 ] 3– 2CuSO 4 + 10KCN $\to$ 2K 3 [Cu(CN) 4 ] + 2K 2 SO 4 + (CN) 2

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