Which one of the following reactions will occur on heating ${\rm{AgN}}{{\rm{O}}_3}$ above its melting point?

$2{\rm{AgN}}{{\rm{O}}_3} \to 2{\rm{AgN}}{{\rm{O}}_2} + {{\rm{O}}_2}$

$2{\rm{AgN}}{{\rm{O}}_3} \to 2{\rm{Ag}} + 2{\rm{N}}{{\rm{O}}_2} + {{\rm{O}}_2}$

$2{\rm{AgN}}{{\rm{O}}_3} \to 2{\rm{Ag}} + {{\rm{N}}_2} + 3{{\rm{O}}_2}$

$2{\rm{AgN}}{{\rm{O}}_3} \to 2{\rm{Ag}} + 2{\rm{NO}} + 2{{\rm{O}}_2}{\rm{\;}}$

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NEET • CHEMISTRY • d & f- BLOCK ELEMENTS

D & f- BLOCK ELEMENTS practice · Hard

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1 Hard

1 of 46

Application +4 | -1

A compound in which a metal ion

{\mathrm{M}^{x + }}

(Z=25) has a spin only magnetic moment of

\sqrt {24}

BM. The number of unpaired electrons in the compound and the oxidation state of the metal ion are respectively.

View answer and explanation

Correct option: D

4 and 3

Option D (4 and 3) is correct because a magnetic moment of

\sqrt{24}

BM corresponds to 4 unpaired electrons, and with Z=25, an oxidation state of +3 leaves 4 d-electrons unpaired.

2 Hard

2 of 46

Analyzing +4 | -1

Assertion :

{\text{C}}{{\text{e}}^{ + 4}}

acts as oxidising agent in aqueous medium

Reason :

+4

is common oxidation state of lanthanides

View answer and explanation

Correct option: C

If Assertion is True but the Reason is False.

Option C (If Assertion is True but the Reason is False) is correct because

{\text{C}}{{\text{e}}^{ + 4}}

acts as an oxidizing agent, but +4 is not the common oxidation state of lanthanides.

3 Hard

3 of 46

Analyzing +4 | -1

Assertion :

{\text{E}}{{\text{u}}^{2 + }}\,\& \,{\text{Y}}{{\text{b}}^{{\text{2 + }}}}

are reducing agents for their ions.

Reason: Both ions have stable half - filled configuration.

View answer and explanation

Correct option: C

If Assertion is True but the Reason is False.

Option C is correct because while Eu2+ and Yb2+ are indeed reducing agents, the reason given about stable half-filled configuration is not entirely accurate or complete.

4 Hard

4 of 46

Understanding +4 | -1

{NaCl}

reacts with conc.

{H}_2 {SO}_4

and

{K}_2 {Cr}_2 {O}_7

to give reddish fumes (B), which react with

{NaOH}

to give yellow solution (C). (B) and (C) respectively are:

View answer and explanation

Correct option: A

{CrO}_2 {Cl}_2, {Na}_2 {CrO}_4

The correct answer is

{CrO}_2 {Cl}_2, {Na}_2 {CrO}_4

because the reddish fumes (B) are chromyl chloride, and it reacts with NaOH to form sodium chromate (yellow solution).

5 Hard

5 of 46

Remembering +4 | -1

Oxygen is absorbed by molten Ag, which is evolved on cooling and the silver particles are scattered; the phenomenon is known as:

View answer and explanation

Correct option: B

Spitting of silver

Option B (Spitting of silver) is correct because the phenomenon of oxygen being absorbed by molten silver and then released upon cooling, causing the silver to scatter, is known as spitting of silver.

6 Hard

6 of 46

Understanding +4 | -1

The reaction of

\mathrm{H}_2 \mathrm{O}_2

with potassium permanganate in acidic medium leads to the formation of mainly

View answer and explanation

Correct option: A

\mathrm{Mn}^{2+}

Option A (

\mathrm{Mn}^{2+}

) is correct because in acidic medium,

\mathrm{H}_2\mathrm{O}_2

reduces

\mathrm{KMnO}_4

\mathrm{Mn}^{2+}

7 Hard

7 of 46

Understanding +4 | -1

Assertion :

{\text{KMn}}{{\text{O}}_{\text{4}}}

is dark pink coloured compound.

Reason : In the

{\text{KMn}}{{\text{O}}_{\text{4}}}

charge transfer spectrum occurs.

View answer and explanation

Correct option: A

If both Assertion & Reason are True & the Reason is correct explanation of the Assertion.

Option A (If both Assertion & Reason are True & the Reason is correct explanation of the Assertion) is correct because

KMnO_4

is dark pink due to ligand to metal charge transfer.

8 Hard

8 of 46

Understanding +4 | -1

Gold is extracted by hydrometallurgical process, based on its property

View answer and explanation

Correct option: B

Of being less reactive

Option C (To form complexes which are water soluble) is correct because gold is extracted by forming a water-soluble complex with cyanide ions in the hydrometallurgical process.

9 Hard

9 of 46

Application +4 | -1

\mathrm{~A} ~0.081 \mathrm{~g}

sample of pyrolusite ore (impure

\mathrm{MnO}_2

) is treated with

1.651 \mathrm{~g}

of oxalic acid

\left(\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4 \cdot \mathrm{H}_2 \mathrm{O}\right)

in an acidic medium. Following this reaction the excess oxalic acid is titrated with

30.6 \mathrm{~mL}

0.1 \mathrm{~M} \mathrm{~KMnO}_4

, the percentage of

\mathrm{~MnO}_2

in the ore will be

(\mathrm{Mn}=55)

\mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4+\mathrm{MnO}_2+2 \mathrm{H}^{+} \longrightarrow \mathrm{Mn}^{2+}+2 \mathrm{H}_2 \mathrm{O}+2 \mathrm{CO}_2

5 \mathrm{H}_2 \mathrm{C}_2 \mathrm{O}_4+2 \mathrm{MnO}_4^{-}+6 \mathrm{H}^{+} \longrightarrow 2 \mathrm{Mn}^{2+}+8 \mathrm{H}_2 \mathrm{O}+10 \mathrm{CO}_2

View answer and explanation

Correct option: D

82 \%

Option D (82%) is correct because it is the calculated percentage of MnO2 in the pyrolusite ore based on the given reactions and titration data.

10 Hard

10 of 46

Remembering +4 | -1

Railway wagon axles are made by heating iron rods embedded in charcoal powder. This process is known as

View answer and explanation

Correct option: A

Tempering

Option B (Case hardening) is correct because case hardening involves heating steel in a carbon-rich environment to create a hard outer layer.

11 Hard

11 of 46

Remembering +4 | -1

When steel is heated red hot and then slowly cooled, the process is known as:

View answer and explanation

Correct option: A

Annealing

Option A (Annealing) is correct because annealing involves heating steel to a high temperature and then slowly cooling it to relieve internal stresses and soften the metal.

12 Hard

12 of 46

Understanding +4 | -1

{\mathrm{F}_2}

is formed by reacting

{\mathrm{K}_2}\mathrm{Mn}{\mathrm{F}_6}

with

View answer and explanation

Correct option: B

\mathrm{Sb}{\mathrm{F}_5}

Option B is correct because SbF5 acts as a fluoride ion acceptor, oxidizing Mn(IV) in K2MnF6 to Mn(V) and releasing F2.

13 Hard

13 of 46

Analyzing +4 | -1

A transition metal

\mathrm{M}

can exist in two oxidation states +1 and +2 . It forms a sulphide whose experimental formula is given by

\mathrm{M}_{\mathrm{x}} \mathrm{S}

where

2 > x > 1

. Then the ratio of metal ion in +1 state to those in +2 state in sulphide is given by

View answer and explanation

Correct option: B

\frac{2(x-1)}{2-x}

Option B (

\frac{{2(x-1)}}{{2-x}}

) is correct because it represents the ratio of M+1 to M+2 ions based on the given formula MxS and charge balance.

14 Hard

14 of 46

Understanding +4 | -1

Which of the following is an acidic oxide?

View answer and explanation

Correct option: C

\mathrm{M}{\mathrm{n}_2}{\mathrm{O}_7}

Option C (

\mathrm{M}{\mathrm{n}_2}{\mathrm{O}_7}

) is correct because higher oxidation states of manganese tend to form acidic oxides, and

Mn_2O_7

has Mn in the +7 oxidation state.

15 Hard

15 of 46

Understanding +4 | -1

Consider the following reaction:

\mathrm{NaCl}+\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathbf{A}+\mathrm{K}_2 \mathrm{SO}_4+\mathrm{Na}_2 \mathrm{SO}_4+\mathrm{H}_2 \mathrm{O}

The formula and colour of the product '

A

' respectively will be

View answer and explanation

Correct option: D

\mathrm{CrO}_2 \mathrm{Cl}_2

, red

The correct answer is

\mathrm{CrO}_2 \mathrm{Cl}_2

, red because chromyl chloride (

\mathrm{CrO}_2 \mathrm{Cl}_2

) is formed in the reaction, which has a reddish-brown color.

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16 Hard

16 of 46

Application +4 | -1

Magnetic moment of manganese in

{\left( {\mathrm{N}{\mathrm{H}_4}} \right)_2}\mathrm{MnB}{\mathrm{r}_2}

View answer and explanation

Correct option: B

5.91 BM

Option B (5.91 BM) is correct because in (NH4)2MnBr2, Mn is in the +2 oxidation state with a d5 configuration, leading to 5 unpaired electrons and a magnetic moment of approximately 5.92 BM.

17 Hard

17 of 46

Understanding +4 | -1

Which one of the following reactions will occur on heating

\mathrm{AgN}{\mathrm{O}_3}

above its melting point?

View answer and explanation

Correct option: C

2\mathrm{AgN}{\mathrm{O}_3} \to 2\mathrm{AgN}{\mathrm{O}_2} + {\mathrm{O}_2}

Option C (

2\mathrm{AgN}{\mathrm{O}_3} \to 2\mathrm{AgN}{\mathrm{O}_2} + {\mathrm{O}_2}

) is correct because silver nitrate decomposes to silver nitrite and oxygen upon heating.

18 Hard

18 of 46

Analyzing +4 | -1

Assertion :

{\text{U}}{{\text{F}}_6}

is more covalent than

{\text{U}}{{\text{F}}_4}

.

Reason : Fluorine is smaller in size.

View answer and explanation

Correct option: B

If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion.

Option A (If both Assertion & Reason are True & the Reason is correct explanation of the Assertion) is correct because smaller fluorine size leads to higher polarization and increased covalent character, making

{\text{U}}{{\text{F}}_6}

more covalent than

{\text{U}}{{\text{F}}_4}

19 Hard

19 of 46

Understanding +4 | -1

Copper exhibits only +2 oxidation state in its stable compounds. Why?

View answer and explanation

Correct option: B

+ 2

state compounds of copper are formed by exothermic reactions.

Option B is correct because the stability of Cu in the +2 oxidation state is attributed to the exothermic nature of the reactions forming its compounds.

20 Hard

20 of 46

Understanding +4 | -1

Transition metal complex with highest value of crystal field splitting

({\Delta _0})

will be

View answer and explanation

Correct option: D

{\left[ {Os{{({H_2}O)}_6}} \right]^{3 + }}

Option B (

\mathrm{A}{\mathrm{g}^ + }

) is correct because

Ag^+

forms a linear complex

[Ag(CN)_2]^-

with a coordination number of 2.

21 Hard

21 of 46

Understanding +4 | -1

Impurities of Cu and Ag from gold are removed by

View answer and explanation

Correct option: D

Both (2) and (3)

Impurities of Cu and Ag from gold are removed by boiling impure gold with concentrated

{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}

and electrolytically, as these methods effectively separate the metals based on their chemical properties.

22 Hard

22 of 46

Understanding +4 | -1

A clock spring is heated to a high temperature and then suddenly plunged into cold water. This treatment will cause the metal to become:

View answer and explanation

Correct option: C

Hard and brittle (case hardening)

Heating a clock spring to a high temperature and then suddenly plunging it into cold water causes the metal to become hard and brittle due to case hardening.

23 Hard

23 of 46

Application +4 | -1

From sodium agrentocyanide

\mathrm{Na}\left[ {\mathrm{Ag}{{\left( {\mathrm{CN}} \right)}_2}} \right]

, silver is precipitated by adding a powder of:

View answer and explanation

Correct option: B

Zinc

The correct answer is Zinc because Zinc is a stronger reducing agent than silver and can displace silver from the complex.

24 Hard

24 of 46

Analyzing +4 | -1

Assertion : Actinides show much higher range of oxidation states

Reason : Energy difference between 5f and 6d orbitals is large

View answer and explanation

Correct option: C

If Assertion is True but the Reason is False.

Option C (If Assertion is True but the Reason is False) is correct because actinides show a higher range of oxidation states, but the energy difference between 5f and 6d orbitals is *small*, not large.

25 Hard

25 of 46

Understanding +4 | -1

Which of the following statements are correct?

(i) Interstitial compounds contain non-metal atoms trapped inside the metal crystal whereas alloys are homogeneous blend of metals.

(ii) Steel and bronze are alloys of transition and nontransition metals.

(iii) Some boride containing interstitial compounds are very hard comparable to that of diamond.

(iv) Interstitial compounds are chemically more reactive than parent metal.

View answer and explanation

Correct option: A

(i) and (iii)

Option A ((i) and (iii)) is correct because interstitial compounds contain non-metal atoms trapped inside the metal crystal, and some boride interstitial compounds are very hard.

26 Hard

26 of 46

Remembering +4 | -1

Silver halides are used in photography because they are:

View answer and explanation

Correct option: A

Photosensitive

Option A (Photosensitive) is correct because silver halides are photosensitive, meaning they decompose upon exposure to light, which is the basis of photography.

27 Hard

27 of 46

Understanding +4 | -1

The equilibrium

\mathrm{C}{\mathrm{r}_2}\mathrm{O}_7^{2 - } + 2e\leftrightharpoons2\mathrm{CrO}_4^{2 - }

View answer and explanation

Correct option: B

Exists in basic medium

Option B (Exists in basic medium) is correct because dichromate (

\mathrm{C}{\mathrm{r}_2}\mathrm{O}_7^{2 - }

) exists in acidic medium, while chromate (

\mathrm{CrO}_4^{2 - }

) exists in basic medium.

28 Hard

28 of 46

Remembering +4 | -1

Which of the following is called white vitriol?

View answer and explanation

Correct option: D

\mathrm{ZnS}{\mathrm{O}_4} \cdot 7{\mathrm{H}_2}\mathrm{O}

Option D (

\mathrm{ZnS}{\mathrm{O}_4} \cdot 7{\mathrm{H}_2}\mathrm{O}

) is correct because Zinc sulfate heptahydrate (

\mathrm{ZnS}{\mathrm{O}_4} \cdot 7{\mathrm{H}_2}\mathrm{O}

) is commonly known as white vitriol.

29 Hard

29 of 46

Analyzing +4 | -1

{I}_2+{Na}_2 {CO}_3

sol

^{\mathrm{n}} . \stackrel{\Delta}{\longrightarrow} {X}+{Y}

If '

{X}

' gives coloured ppt. with

{Pb}\left({CH}_3 {COO}\right)_2

solution, then '

{Y}

' will not be respond to which of the following

View answer and explanation

Correct option: B

{Y}+{Cr}_2 {O}_7{ }^{2-}\text{(aq)} +{OH}^{-}(\mathrm{aq})

Option B (

{Y}+{Cr}_2 {O}_7{ }^{2-}\text{(aq)} +{OH}^{-}(\mathrm{aq})

) is correct because the reaction of iodine with sodium carbonate produces iodide and iodate. Iodide can be oxidized by dichromate in acidic medium, but not in basic medium.

30 Hard

30 of 46

Analyzing +4 | -1

Consider the following statements

I.

Sc

has a low value of

E_{{M^{ + 3}}/{M^{ + 2}}}^0

II.

Zn

has the highest value of

E_{{M^{ + 3}}/{M^{ + 2}}}^0

III.

Mn

and

Fe

have comparatively high and low values of

E_{{M^{ + 3}}/{M^{ + 2}}}^0

respectively

The correct set of statements is (Choose the low values of

E_{{M^{ + 3}}/{M^{ + 2}}}^0

respectively).

The correct set of statements is (Choose the appropriate option)

View answer and explanation

Correct option: B

I, II and III

Option B (I, II and III) is correct because all three statements accurately describe the trends in reduction potentials of Sc, Zn, Mn, and Fe.

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Key Takeaways

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