d and f Block Elements — Page 16 | JEE Chemistry

Q151

Most common oxidation states of Ce (cerium) are :

A +3, +4

B +2, +3

C +2, +4

D +3, +5

Correct Answer

Option A

Solution

Common oxidation states of

Ce

(Cerium) are

+3

and

+4

Q152

Which one of the following statements is correct?

A From a mixed precipitate of AgCl and AgI, ammonia solution dissolves only AgCl

B Ferric ions give a deep green precipitate on adding potassium ferrocyanide solution

C On boiling a solution having K+, Ca2+ and HCO3 ions we get a precipitate of K2Ca(CO3)2.

D Manganese salts give a violet borax bead test in the reducing flame

Correct Answer

Option A

Solution

Between AgCl and AgI, AgI is less soluble, hence ammonia can dissolve ppt. of AgCl only due to formation of complex as given below: AgCl + 2NH3 $\to$ [Ag(NH3)2]Cl

Q153

Which one of the following nitrates will leave behind a metal on strong heating?

A Copper nitrate

B Manganese nitrate

C Silver nitrate

D Ferric nitrate

Correct Answer

Option C

Solution

AgN{O_3}

on heating till red hot decomposes as follows:

AgN{O_3} \to Ag + N{O_2} + {1 \over 2}{O_2}

Q154

Cerium (Z = 58) is an important member of the lanthanoids. Which of the following statements about cerium is incorrect?

A The common oxidation states of cerium are +3 and +4

B Cerium (IV) acts as an oxidizing agent

C The +4 oxidation state of cerium is not known in solutions

D The +3 oxidation state of cerium is more stable than the +4 oxidation state

Correct Answer

Option C

Solution

The

+4

oxidation state of cerium is also known in solution.

Q155

Excess of KI reacts with CuSO4 solution and then Na2S2O3 solution is added to it. Which of the statements is incorrect for this reaction?

A Cu2I2 is reduced

B Evolved I2 is reduced

C Na2S2O3 is oxidized

D CuI2 is formed

Correct Answer

Option D

Solution

4\mathop {KI}\limits^{ - 1} + 2CuS{O_4} \to \mathop {{I_2}}\limits^0 + C{u_2}{I_2} + 2{K_2}S{O_4}

\mathop {{I_2}}\limits^0 + 2N{a_2}\mathop {{S_2}}\limits^{2 + } {O_3} \to N{a_2}\mathop {{S_4}}\limits^{ + 2.5} {O_6} + \mathop {2NaI}\limits^{ - 1}

In this

Cu{I_2}

is not formed

Q156

Of the following outer electronic configurations of atoms, the highest oxidation state is achieved by which one of them?

A

\left( {n - 1} \right){d^3}n{s^2}

B

\left( {n - 1} \right){d^5}n{s^1}

C

\left( {n - 1} \right){d^8}n{s^2}

D

\left( {n - 1} \right){d^5}n{s^2}

Correct Answer

Option D

Solution

\left( {n - 1} \right){d^5}n{s^2}\,\,

attains the maximum

O.S.

of

+7

Q157

The lanthanide contraction is responsible for the fact that :

A Zr and Y have about the same radius

B Zr and Nb have similar oxidation state

C Zr and Hf have about the same radius

D Zr and Zn have the same oxidation

Correct Answer

Option C

Solution

In the entire d-block, Only in group 3, the size increase from top to bottom due to increase of no of shells in an atom.

That is why size of Sc < Y < La.

And shell size in group 3 is 3d < 4d < 5d.

But from group 4 to 12, the size of 3d < 4d but 4d $\cong$ 5d due to lanthanide contraction.

That is why Zr and Hf have about the same radius.

Q158

Which of the following factors may be regarded as the main cause of lanthanide contraction?

A Poor shielding of one of 4f electron by another in the subshell

B Effective shielding of one of 4f electrons by another in the subshell

C Poorer shielding of 5d electrons by 4f electrons

D Greater shielding of 5d electrons by 4f electrons

Correct Answer

Option C

Solution

The main reason of lanthanide contraction is, poor shielding of 4f electrons on the outer most shell electrons.

In lanthanides, there is poorer shielding of

5d

electrons by

4f

electrons resulting in greater attraction of the nucleus over

5d

electrons and reduction of the atomic radii.

Q159

Calomel

\left( {H{g_2}C{l_2}} \right)

on reaction with ammonium hydroxide gives :

A

HgO

B

H{g_2}O

C

N{H_2} - Hg - Hg - Cl

D

HgN{H_2}Cl

Correct Answer

Option D

Solution

H{g_2}C{l_2} + 2N{H_4}OH\overset{\,}\longrightarrow HgN{H_2}Cl + N{H_4}Cl + 2{H_2}O

Q160

On heating mixture of

C{u_2}O

and

C{u_2}S

will give :

A

C{u_2}S{O_3}

B

CuO + CuS

C

Cu + S{O_3}

D

Cu + S{O_2}

Correct Answer

Option D

Solution

2C{u_2}O + C{u_2}S\overset{\,}\longrightarrow 6Cu + S{O_2}

self reduction