d and f Block Elements — Page 17 | JEE Chemistry

Q161

A metal,

M

forms chlorides in its

+2

and

+4

oxidation states. Which of the following statements about these chlorides is correct?

A

MC{l_2}

is more ionic than

MC{l_4}

B

MC{l_2}

is more easily hydrolysed than

MC{l_4}

C

MC{l_2}

is more volatile than

MC{l_4}

D

MC{l_2}

is more soluble in anhydrous ethanol than

MC{l_4}

.

Correct Answer

Option A

Solution

Metal atom in the lower oxidation state forms the ionic bond and in the higher oxidation state the covalent bond.

Because higher oxidation state means small size and great polarizing power and hence greater the covalent character.

Hence

MC{l_2}

is more ionic than

MC{l_4}.

Q162

The actinoids exhibits more number of oxidation states in general than the lanthanoids. This is because :

A the 5f orbitals are more buried than the 4f orbitals

B there is a similarity between 4f and 5f orbitals in their angular part of the wave function

C the actinoids are more reactive than the lanthanoids

D the 5f orbitals extend further from the nucleus than the 4f orbitals

Correct Answer

Option D

Solution

NOTE : More the distance between nucleus and outer orbitals, lesser will be force of attraction on them.

Distance between nucleus and

5f

orbitals is more as compared to distance between

4f

orbital and nucleus. So actinoids exhibit more number of oxidation states in general than the lanthanoids.

Q163

Amount of oxalic acid present in a solution can be determined by its titration with

KMn{O_4}

solution in the presence of

{H_2}S{O_4}.

The titration gives unsatisfactory result when carried out in the presence of

HCl,

because

HCl

:

A gets oxidised by oxalic acid to chlorine

B furnishes

{H^ + }

ions in addition to those from oxalic acid

C reduces permanganate to

M{n^{2 + }}

D oxaidises oxalic acid to carbon doxide and water

Correct Answer

Option C

Solution

The titration of oxalic acid with

KMnO{}_4

in presence of

HCl

gives unsatisfactory result because of the fact that

KMn{O_4}

can also oxidise

HCl

along with oxalic acid.

HCl

on oxidation gives

C{l_2}

and

HCl

reduces

KMnO{}_4

to

M{n^{2 + }}

thus the correct answer is

(c).

Q164

The INCORRECT statement is :

A the color of [CoCl(NH3)5]2+ is violet as it absorbs the yellow light.

B the gemstone, ruby, has Cr3+ ions occupying the octahedral sites of beryl.

C the spin-only magnetic moment of [Ni(NH3)4(H2O)2]2+ is 2.83 BM.

D the spin-only magnetic moments of [Fe(H2O)6]2+ and [Cr(H2O)6]2+ are nearly similar.

Correct Answer

Option B

Solution

(A) Complementarty color of yellow is violet.

As [CoCl(NH3)5]2+ absorbs yellow so it will show complementary color violet.

(B)The building block of ruby is aluminium oxide (Al2O3) containing about 0.5 – 1% Cr3+ ions which are randomly distributed in the position normally occupied by Al3+ ions.

(C) Total number of unpaired electons in [Fe(H2O)6]2+ = 4 and total number of unpaired electons in [Cr(H2O)6]2+ = 4 So, the spin-only magnetic moments are same in both cases.

(D) Total number of unpaired electons in [Ni(NH3)4(H2O)2]2+ = 2 So, $\mu$ =

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

BM = 2.83 BM

Q165

Consider the hydrated ions of Ti2+, V2+, Ti3+, and Sc3+. The correct order of their spin-only magnetic moments is :

A Sc3+ < Ti3+ < V2+ < Ti2+

B Sc3+ < Ti3+ < Ti2+ < V2+

C Ti3+ < Ti2+ < Sc3+ < V2+

D V2+ < Ti2+ < Ti3+ < Sc3+

Correct Answer

Option B

Solution

As we know that

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

where n = no. of unpaired electrons i.e. greater the no. of unpaired electron more will be the spin-only magnetic moments. .

Electronic configuration of the given transition metal ions are Sc3+ (Z = 21) = 1s22s22p63s23p6 $\therefore$ n = 0 Ti2+ (Z = 22) = 1s22s22p63s23p63d2 $\therefore$ n = 2 Ti3+ (Z = 22) = 1s22s22p63s23p63d1 $\therefore$ n = 1 V2+ (Z = 23) = 1s22s22p63s23p63d3 $\therefore$ n = 3 So the correct increasing order of magnetic moment is Sc3+ < Ti3+ < Ti2+ < V2+

Q166

The maximum number of possible oxidation states of actinoides are shown by :

A nobelium (No) and lawrencium (Lr)

B actinium (Ac) and thorium (Th)

C neptunium (Np) and plutonium (Pu)

D berkelium (Bk) and californium (Cf)

Correct Answer

Option C

Solution

Actinoid 'Th' shows oxidation state = +4 Actinoid 'Ac' shows oxidation state = +3 Actinoid 'Pu' shows oxidation states = +3, +4, +5, +6, +7 Actinoid 'Np' shows oxidation states = +3, +4, +5, +6, +7 Actinoid 'Bk' shows oxidation states = +3, +4 Actinoid 'Cm' shows oxidation states = +3, +4 Actinoid 'Lr' shows oxidation states = +3 $\therefore$ Maximum oxidation state is shown by Np and Pu.

Q167

The statement that is INCORRECT about the interstitial compounds is :

A They are chemically reactive

B They have metallic conductivity

C They have high melting points

D They are very hard

Correct Answer

Option A

Solution

Interstitial compound are – (i) very hard (ii) chemically inert (iv) high melting point.

As interstitial compounds are chamically inert.

Q168

5 g of zinc is treated separately with an excess of (a) dilute hydrochloric acid and (b) aqueous sodium hydroxide. The ratio of the volumes of H2 evolved in these two reactions is :

A 1 : 2

B 1 : 1

C 1 : 4

D 2 : 1

Correct Answer

Option B

Solution

Zn + 2dil.

HCl $\to$ ZnCl2 + H2 Zn + 2NaOH $\to$ Na2ZnO2 + H2 From one mole of Zn, 1 mol of H2 is produced by both NaOH and HCl.

The ratio of the volume of H2 is 1 : 1

Q169

\underline A \,\,\overset{{4KOH,{O_2}}}\longrightarrow \,\,\mathop {2\underline B }\limits_{\left( {Green} \right)} \,\, + \,\,2{H_2}O

3\underline B \,\,\overset{{4HCl}}\longrightarrow \,\,\mathop {2\underline C }\limits_{\left( {Purple} \right)} \,\, + \,\,Mn{O_2} + 2{H_2}O

2\underline C \,\,\overset{{{H_2}O.KI}}\longrightarrow \,\,\mathop {2\underline A }\limits_{\left( {Purple} \right)} \,\, + \,\,2KOH\,\, + \,\,\underline D

In the above sequence of reactions,

{\underline A }

and

{\underline D }

, respectively, are :

A Kl and KMnO4

B Kl and K2MnO4

C KlO3 and MnO2

D MnO2 and KlO3

Correct Answer

Option D

Solution

Mn{O_2}(A) \,\,\overset{{4KOH,{O_2}}}\longrightarrow \,\,\mathop {2K_2MnO_4(B) }\limits_{\left( {Green} \right)} \,\, + \,\,2{H_2}O

3K_2MnO_4(B) \,\,\overset{{4HCl}}\longrightarrow \,\,\mathop {2K_2MnO_4(C) }\limits_{\left( {Purple} \right)} \,\, + \,\,Mn{O_2} + 2{H_2}O

2K_2MnO_4(C) \,\,\overset{{{H_2}O.KI}}\longrightarrow \,\,\mathop {2 Mn{O_2}(A) }\limits_{\left( {Purple} \right)} \,\, + \,\,2KOH\,\, + \,\,KIO_3(D)

$\therefore$ A $\to$ MnO2 D $\to$ KIO3

Q170

The correct order of atomic radii is :

A Ho > N > Eu > Ce

B Ce > Eu > Ho > N

C N > Ce > Eu > Ho

D Eu > Ce > Ho > N

Correct Answer

Option D

Solution

In lanthanide series from left to right atomic radius decreases but the exception is Europiun (Eu) and Ytterbium (Yb).

So the correct order is : Eu > Ce > Ho > N