d and f Block Elements

JEE Chemistry · 278 questions · Page 28 of 28 · Click an option or "Show Solution" to reveal answer

Q271
When Cu2+\mathrm{Cu}^{2+} ion is treated with KI\mathrm{KI}, a white precipitate, X\mathrm{X} appears in solution. The solution is titrated with sodium thiosulphate, the compound Y\mathrm{Y} is formed. X\mathrm{X} and Y\mathrm{Y} respectively are :
A .tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal;} .tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;} .tg .tg-0lax{text-align:left;vertical-align:top} X=CuI2\mathrm{X=CuI_2} Y=Na2S4O6\mathrm{Y=Na_2S_4O_6}
B .tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal;} .tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;} .tg .tg-0lax{text-align:left;vertical-align:top} X=Cu2I2\mathrm{X=Cu_2I_2} Y=Na2S4O6\mathrm{Y=Na_2S_4O_6}
C .tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal;} .tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;} .tg .tg-0lax{text-align:left;vertical-align:top} X=CuI2\mathrm{X=CuI_2} Y=Na2S2O3\mathrm{Y=Na_2S_2O_3}
D .tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal;} .tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;} .tg .tg-0lax{text-align:left;vertical-align:top} X=Cu2I2\mathrm{X=Cu_2I_2} Y=Na2S4O5\mathrm{Y=Na_2S_4O_5}
Correct Answer
Option B
Solution

2Cu2++4KICu2I2 White ppt. +I22 \mathrm{Cu}^{2+}+4 \mathrm{KI} \longrightarrow \underset{\text{ White ppt. }}{\mathrm{Cu}_{2} \mathrm{I}_{2}}+\mathrm{I}_{2} I2+Na2 S2O32Nal+Na2 S4O6\mathrm{I}_{2}+\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} \longrightarrow 2 \mathrm{Nal}+\mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6} X=Cu2I2\mathrm{X}=\mathrm{Cu}_{2} \mathrm{I}_{2} Y=Na2 S4O6\mathrm{Y}=\mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}

Q272

Match the catalysts to the correct processes : .tg .tg Catalyst Process

List - IList - II
(A) TiCl3 (i) Wacker process
(B) PdCl2 (ii) Ziegler - Natta polymerization
(C) CuCl2 (iii) Contact process
(D) V2O5 (iv) Deacon's process
A (A)(ii),(B)(iii),(C)(iv),(D)(i)\left( A \right) - \left( {ii} \right),\,\,\left( B \right) - \left( {iii} \right),\,\,\left( C \right) - \left( {iv} \right),\,\,\left( D \right) - \left( i \right)
B (A)(iii),(B)(i),(C)(ii),(D)(iv)\left( A \right) - \left( {iii} \right),\,\,\left( B \right) - \left( i \right),\,\,\left( C \right) - \left( {ii} \right),\,\,\left( D \right) - \left( {iv} \right)
C (A)(iii),(B)(ii),(C)(iv),(D)(i)\left( A \right) - \left( {iii} \right),\,\,\left( B \right) - \left( {ii} \right),\,\,\left( C \right) - \left( {iv} \right),\,\,\left( D \right) - \left( i \right)
D (A)(ii),(B)(i),(C)(iv),(D)(iii)\left( A \right) - \left( {ii} \right),\,\,\left( B \right) - \left( i \right),\,\,\left( C \right) - \left( {iv} \right),\,\,\left( D \right) - \left( {iii} \right)
Correct Answer
Option D
Solution

To match the catalysts to the correct processes : TiCl3TiCl_3 is well-known as a catalyst in Ziegler-Natta polymerization which is used for the production of polymers of olefins.

So, (A)(ii)(A) - (ii). PdCl2PdCl_2 is used in the Wacker process, which is used to convert ethene to acetaldehyde.

Hence, (B)(i)(B) - (i). V2O5V_2O_5 is used in the Contact process to manufacture sulfuric acid by catalyzing the oxidation of sulfur dioxide to sulfur trioxide.

Therefore, (D)(iii)(D) - (iii). CuCl2CuCl_2 is used in Deacon's process for the oxidation of hydrochloric acid to chlorine.

Hence, (C)(iv)(C) - (iv).

Comparing these pairings with the options given, we find that the correct option is :

(A)(ii),(B)(i),(C)(iv),(D)(iii)\left( A \right) - \left( {ii} \right),\,\,\left( B \right) - \left( {i} \right),\,\,\left( C \right) - \left( {iv} \right),\,\,\left( D \right) - \left( {iii} \right)
Q273

Match with , match the gas evolved during each reaction. _{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \stackrel{\Delta}{\longrightarrow}$$

List - IList - II
(B) KMnO4+HCl\mathrm{KMnO}_{4}+\mathrm{HCl} \rightarrow (I) H2\mathrm{H}_{2}
(C) Al+NaOH+H2O\mathrm{Al}+\mathrm{NaOH}+\mathrm{H}_{2} \mathrm{O} \rightarrow (II) N2\mathrm{N}_{2}
(D) NaNO3Δ\mathrm{NaNO}_{3} \stackrel{\Delta}{\longrightarrow} (III) O2\mathrm{O}_{2}
() (IV) Cl2\mathrm{Cl}_{2}
A (A)(II),(B)(III),(C)(I),(D)(IV)(\mathrm{A})-(\mathrm{II}),(\mathrm{B})-(\mathrm{III}),(\mathrm{C})-(\mathrm{I}),(\mathrm{D})-(\mathrm{IV})
B (A)(III),(B)(I),(C)(IV),(D)(II)(\mathrm{A})-(\mathrm{III}),(\mathrm{B})-(\mathrm{I}),(\mathrm{C})-(\mathrm{IV}),(\mathrm{D})-(\mathrm{II})
C (A)(II),(B)(IV),(C)(I),(D)(III)(\mathrm{A})-(\mathrm{II}),(\mathrm{B})-(\mathrm{IV}),(\mathrm{C})-(\mathrm{I}),(\mathrm{D})-(\mathrm{III})
D (A)(III),(B)(IV),(C)(I),(D)(II)(\mathrm{A})-(\mathrm{III}),(\mathrm{B})-(\mathrm{IV}),(\mathrm{C})-(\mathrm{I}),(\mathrm{D})-(\mathrm{II})
Correct Answer
Option C
Solution

(NH4)2Cr2O7ΔN2+4H2O+Cr2O3\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \stackrel{\Delta}{\longrightarrow} \mathrm{N}_{2}+4 \mathrm{H}_{2} \mathrm{O}+\mathrm{Cr}_{2} \mathrm{O}_{3} KMnO4+HClKCl+MnCl2+Cl2+H2O\mathrm{KMnO}_{4}+\mathrm{HCl} \longrightarrow \mathrm{KCl}+\mathrm{MnCl}_{2}+\mathrm{Cl}_{2}+\mathrm{H}_{2} \mathrm{O} Al+NaOH+H2ONa(Al(OH)4)+H2\mathrm{Al}+\mathrm{NaOH}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{Na}\left(\mathrm{Al}(\mathrm{OH})_{4}\right)+\mathrm{H}_{2} 2NaNO3( s)Δ2NaNO2( s)+O22 \mathrm{NaNO}_{3}(\mathrm{~s}) \stackrel{\Delta}{\longrightarrow} 2 \mathrm{NaNO}_{2}(\mathrm{~s})+\mathrm{O}_{2}

Q274

Match with .tg .tg

List - IList - II
(A) Bronze (I) Cu, Ni
(B) Brass (II) Fe, Cr, Ni, C
(C) UK silver coin (III) Cu, Zn
(D) Stainless steel (IV) Cu, Sn
A (A)-(IV), (B)-(II), (C)-(III), (D)-(I)
B (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
C (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
D (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
Correct Answer
Option B
Solution

To match the items from List I to List II, we need to understand the composition of each material: Bronze is an alloy composed of copper (Cu) and tin (Sn).

Brass is an alloy made of copper (Cu) and zinc (Zn).

The UK silver coin primarily consists of copper (Cu) and nickel (Ni).

Stainless steel is an alloy that includes iron (Fe), chromium (Cr), nickel (Ni), and carbon (C).

Based on these definitions, the correct matches are: (A) Bronze corresponds to (IV) Cu, Sn (B) Brass corresponds to (III) Cu, Zn (C) UK silver coin corresponds to (I) Cu, Ni (D) Stainless steel corresponds to (II) Fe, Cr, Ni, C These pairings illustrate the composition of each material and ensure accurate alignment between List I and List II.

Q275
Which of the following arrangements does not represent the correct order of the property stated against it?
A Ni2+ < Co2+ < Fe2+ < Mn2+ : ionic size
B Co3+ < Fe3+ < Cr3+ < Sc3+ : stability in aqueous solution
C Sc < Ti < Cr < Mn : number of oxidation states
D V2+ < Cr2+ < Mn2+ < Fe2+ : paramagnetic behaviour
Correct Answer
Option D
Solution

(1) .tg .tg V = 3d34s2 V2+ = 3d3 = 3 unpaired electron Cr = 3d54s1 Cr2+ = 3d4 = 4 unpaired electron Mn = 3d54s2 Mn2+ = 3d5 = 5 unpaired electron Fe = 3d64s2 Fe2+ = 3d6 = 4 unpaired electron hence the correct order of paramagnetic behaviour

V2+<Cr2+=Fe2+<Mn2+{V^{2 + }} < Cr{}^{2 + } = F{e^{2 + }} < M{n^{2 + }}

(2)

\,\,\,\,

For the same oxidation state, the-ionic radii generally

dcdc

-creases as the atomic number increases in a particular transition series. Hence the order is

Mn++>Fe++>Co++>Ni++M{n^{ + + }} > F{e^{ + + }} > C{o^{ + + }} > N{i^{ + + }}

(3) In solution, the stability of the compounds depends upon electrode potentials,

SEPSEP

of the transitions metal ions are given as

Co3+/Co=+1.97,Fe3+/Fe=+0.77;C{o^{3 + }}/Co = + 1.97,\,\,\,F{e^{3 + }}/Fe = + 0.77;
Cr3+/Cr2+=0.41,Sc3+C{r^{3 + }}/C{r^{2 + }} = - 0.41,\,\,S{c^{3 + }}

is highly stable as it does not show

+2+2
O.O.
S.S.

(4)

Sc(+2),(+3)Sc - \left( { + 2} \right),\left( { + 3} \right)
Ti(+2),(+3),(+4)Ti - \left( { + 2} \right),\left( { + 3} \right),\left( { + 4} \right)
Cr(+1),(+2),(+3),(+4),(+5),(+6)Cr - \left( { + 1} \right),\left( { + 2} \right),\left( { + 3} \right),\left( { + 4} \right),\left( { + 5} \right),\left( { + 6} \right)
Mn(+2),(+3),(+4),(+5),(+6),(+7)Mn - \left( { + 2} \right),\left( { + 3} \right),\left( { + 4} \right),\left( { + 5} \right),\left( { + 6} \right),\left( { + 7} \right)
i.e.Sc<Ti<Cr=Mni.e.\,Sc < Ti < Cr = Mn
Ea=53598.6J/mol{E_a} = 53598.6J/mol
=53.6kJ/mol.= 53.6\,kJ/mol.
Q276
Pair of transition metal ions having the same number of unpaired electrons is
A Ti3+,Mn2+\mathrm{Ti}^{3+}, \mathrm{Mn}^{2+}
B Ti2+,Co2+\mathrm{Ti}^{2+}, \mathrm{Co}^{2+}
C Fe3+,Cr2+\mathrm{Fe}^{3+}, \mathrm{Cr}^{2+}
D V2+,Co2+\mathrm{V}^{2+}, \mathrm{Co}^{2+}
Correct Answer
Option D
Solution

.tg .tg Configuration No. of unpaired e^- (1) V3+\mathrm{V}^{3+} \Rightarrow [Ar]3 d34 s0[\mathrm{Ar}] 3 \mathrm{~d}^3 4 \mathrm{~s}^0 3 Co2+\mathrm{Co}^{2+} \Rightarrow [Ar]3 d74 s0[\mathrm{Ar}] 3 \mathrm{~d}^7 4 \mathrm{~s}^0 3 (2) Ti2+\mathrm{Ti}^{2+} \Rightarrow [Ar]3 d24 s0[\mathrm{Ar}] 3 \mathrm{~d}^2 4 \mathrm{~s}^0 2 Co2+\mathrm{Co}^{2+} \Rightarrow [Ar]3 d74 s0[\mathrm{Ar}] 3 \mathrm{~d}^7 4 \mathrm{~s}^0 3 (3) Fe3+\mathrm{Fe}^{3+} \Rightarrow [Ar]3 d54 s0[\mathrm{Ar}] 3 \mathrm{~d}^5 4 \mathrm{~s}^0 5 Cr2+\mathrm{Cr}^{2+} \Rightarrow [Ar]3 d44 s0[\mathrm{Ar}] 3 \mathrm{~d}^4 4 \mathrm{~s}^0 4 (4) Ti3+\mathrm{Ti}^{3+} \Rightarrow [Ar]3 d14 s0[\mathrm{Ar}] 3 \mathrm{~d}^1 4 \mathrm{~s}^0 1 Mn2+\mathrm{Mn}^{2+} \Rightarrow [Ar]3 d54 s0[\mathrm{Ar}] 3 \mathrm{~d}^5 4 \mathrm{~s}^0 5 So V2+&Co2+\mathrm{V}^{2+} \& \mathrm{Co}^{2+} same number of unpaired electron.

Q277
The incorrect statement is :
A bronze is an alloy of copper and tin
B cast iron is used to manufacture wrought iron
C german silver is an alloy of zinc, copper and nickel
D brass is an alloy of copper and nickel
Correct Answer
Option D
Solution

Bronze contains 75 – 90% Cu and 10 – 25% Sn Cast iron is used to make wraugut iron.

German silver contains 56% Cu, 24% Zn and 20% Ni.

Brass contains 60% Cu and 40% Zn.

Note : .tg .tg Alloys Composition Brass 60% Cu + 40% Zn Bronze 75-90% Cu + 10-25% Sn Gun-metal 90% Cu + 8% Sn + 2% Zn Aluminium bronze 90% Cu + 10% Al Monel metal 30% Cu + 67% Ni + 3% (Fe + Mn) German Silver 56% Cu + 24% Zn + 20% Ni Manganin 78-86% Cu + 13-18% Mn + 1-4% Ni Bell metal 80% Cu + 20% Sn Constantan 60% Cu + 40% Ni

Q278
The major components of German Silver are :
A Zn, Ni and Ag
B Cu, Zn and Ni
C Cu, Zn and Ag
D Ge, Cu and Ag
Correct Answer
Option B
Solution

Composition of German Silver \to 56% Cu + 24% Zn + 20% Ni Note : .tg .tg Alloys Composition Brass 60% Cu + 40% Zn Bronze 75-90% Cu + 10-25% Sn Gun-metal 90% Cu + 8% Sn + 2% Zn Aluminium bronze 90% Cu + 10% Al Monel metal 30% Cu + 67% Ni + 3% (Fe + Mn) German Silver 56% Cu + 24% Zn + 20% Ni Manganin 78-86% Cu + 13-18% Mn + 1-4% Ni Bell metal 80% Cu + 20% Sn Constantan 60% Cu + 40% Ni

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