d and f Block Elements — Page 7 | NEET Chemistry

Q61

Which one of the following characteristics of the transition metals is associated with their catalytic activity?

A High enthalpy of atomization

B Paramagnetic behaviour

C Colour of hydrated ions

D Variable oxidation states

Correct Answer

Option D

Solution

The catalytic activity of transitional metals is due to their variable oxidation states.

Q62

Zn gives H 2 gas with H 2 SO 4 and HCl but not with HNO 3 because

A Zn act as oxidising agent when react with HNO 3

B HNO 3 is weaker acid than H 2 SO 4 and HCl

C In electrochemical series Zn is above hydrogen

D NO

_3^ -

is reduced in preference to hydronium ion.

Correct Answer

Option D

Solution

Zinc is on the top position of hydrogen in electrochemical series.

So Zn displaces H 2 from dilute H 2 SO 4 and HCl with liberation of H 2 .

Zn + H 2 SO 4 $\to$ ZnSO 4 + H 2 On the other hand HNO 3 is an oxidising agent.

Hydrogen obtained in this reaction is converted into H 2 O.

Zn + 2HNO 3 $\to$ Zn(NO 3 ) 2 + 2H 2HNO 3 $\to$ H 2 O + 2NO 2 + O 2H + O $\to$ H 2 O.

Q63

An atom has electronic configuration 1s 2 2s 2 2p 6 3s 2 3p 6 3d 3 4s 2 , you will place it in

A fifth

B fifteenth

C second

D third.

Correct Answer

Option A

Solution

Given electronic configuration of atom 1s 2 2s 2 2p 6 3s 2 3p 6 3d 3 4s 2 .

In the configuration, the last electron of the atom is filled in d sub-shell as 3d 3 .

Thus this element belongs to d-block of the periodic table with group no.

V.

Q64

Which of the following shows maximum number of oxidation states?

A Cr

B Fe

C Mn

D V

Correct Answer

Option C

Solution

Mn : [Ar] 3d 5 4s 2 Shows +2, +3, +4, +5, +6 & +7 oxidation states

Q65

General electronic configuration of lanthanides is

A

\left( {n - 2} \right){f^{1 - 14}}\,\left( {n - 1} \right){s^2}{p^6}{d^{0 - 1}}\,n{s^2}

B

\left( {n - 2} \right){f^{0 - 14}}\,\left( {n - 1} \right){d^{0 - 1}}\,n{s^2}

C

\left( {n - 2} \right){f^{0 - 14}}\,\left( {n - 1} \right){d^{10}}\,n{s^2}

D

\left( {n - 2} \right){d^{0 - 1}}\left( {n - 1} \right)\,{f^{1 - 14}}\,n{s^2}

Correct Answer

Option A

Solution

General electronic configuration of lanthanides is

\left( {n - 2} \right){f^{1 - 14}}\,\left( {n - 1} \right){s^2}{p^6}{d^{0 - 1}}\,n{s^2}

.

Q66

The most convenient method to protect the botton of ship made of iron is

A coating it with red lead oxide

B white tin plating

C connecting it with Mg block

D connecting it with Pb block.

Correct Answer

Option B

Solution

The most convenient method to protect the bottom of the ship made of iron is white tin plating preventing the build up of barnacles.

Q67

Which of the following statement is not correct?

A La(OH) 3 is less basic than Lu(OH) 3 .

B In lanthanide series ionic radius of Lu +3 ion decreases.

C La is actually an element of transition series rather lanthanides.

D Atomic radius of Zn and Hf are same because of lanthanide contraction.

Correct Answer

Option A

Solution

In lanthanides the basic character of hydroxides decreases as the ionic radius decreases.

La 3+ ions are larger than Li 3+ so it easily gives OH – ion thus La(OH) 3 is more basic than Li(OH) 3 .

Q68

Which of the following shows maximum number of oxidation states?

A Cr

B Fe

C Mn

D V

Correct Answer

Option C

Solution

Mn : [Ar] 3d 5 4s 2 Shows +2, +3, +4, +5, +6 & +7 oxidation states.

Q69

Which ion is colourless?

A Cr 4+

B Se 3+

C Ti 3+

D V 3+

Correct Answer

Option B

Solution

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

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

Q70

Given below are two statements: one is labelled as "Assertion A" and the other is labelled as "Reason R" Assertion A : In the complex

\mathrm{Ni}(\mathrm{CO})_{4}

and

\mathrm{Fe}(\mathrm{CO})_{5}

, the metals have zero oxidation state. Reason R : Low oxidation states are found when a complex has ligands capable of

\pi

-donor character in addition to the

\sigma

-bonding. In the light of the above statements, choose the most appropriate answer from the options given below

A A is not correct but R is correct

B A is correct but R is not correct

C Both A and R are correct but R is NOT the correct explanation of A

D Both A and R are correct and R is the correct explanation of A

Correct Answer

Option B

Solution

Option B: "A is correct but R is incorrect" is the right answer.

Here is the reasoning: Assertion A is correct.

In the complex

\mathrm{Ni(CO)_4}

, if we assume the oxidation state of Ni to be x, we can set up the equation: $x + (0 \times 4) = 0$ From which we find that (x = 0).

Similarly, in

\mathrm{Fe(CO)_5}

, if we let the oxidation state of Fe be y, the equation becomes: $y + (0 \times 5) = 0$ Solving this gives (y = 0).

Therefore, the oxidation states of both Ni and Fe are zero in their respective complexes, verifying that Assertion A is true.

Regarding Reason R, it is incorrect.

The low oxidation states in these complexes are stabilized through synergic bonding, which involves the ligand being a $\pi$ -acceptor, not a $\pi$ -donor.

The CO ligand in these complexes accepts electron density from the metal into its antibonding molecular orbitals, characterizing a $\pi$ -acid ligand, and not a $\pi$ -donor ligand as stated.

Consequently, Reason R does not hold true.