Column-I (Alloys) Column-II (Constituents) (A) TiCl$$_4$$ (p) Adams catalyst in reduction (B) PdCl$$_2$$ (q) In preparation of (CH$$_3$$)$$_2$$SiCl$$_2$$ (C) Pt/PtO (r) Used as the Natta catalyst in polythene production (D) Cu (s) Wake process for converting C$$_2$$H$$_4$$ to CH$$_3$$CHO

(A) $$\rightarrow$$ r ; (B) $$\rightarrow$$ s ; (C) $$\rightarrow$$ p ; (D) $$\rightarrow$$ q

(A) $$\rightarrow$$ s ; (B) $$\rightarrow$$ r ; (C) $$\rightarrow$$ p ; (D) $$\rightarrow$$ q

(A) $$\rightarrow$$ q ; (B) $$\rightarrow$$ s ; (C) $$\rightarrow$$ p ; (D) $$\rightarrow$$ r

(A) $$\rightarrow$$ p ; (B) $$\rightarrow$$ s ; (C) $$\rightarrow$$ r ; (D) $$\rightarrow$$ q

Match the catalysts to the correct processes: Catalyst Process A. $$TiCl_3$$ i. Wacker process B. $$PdCl_2$$ ii. Ziegler - Natta polymerization C. $$CuCl_2$$ iii. Contact process D. $$V_2O_5$$ iv. Deacon's process

(A) - (ii), (B) - (i), (C) - (iv), (D) - (iii)

(A) - (iii), (B) - (ii), (C) - (iv), (D) - (i)

(A) - (ii), (B) - (iii), (C) - (iv), (D) - (i)

(A) - (iii), (B) - (i), (C) - (ii), (D) - (iv)

MCQ Questions for Class 12 Engineering Chemistry The D-And F-Block Elements Quiz 11

$CuSO_{4}$ is paramagnetic while

$ZnSO_{4}$ is diamagnetic because

0%
$Cu^{2+}$ ion has $3d^{9}$ configuration while $Zn^{2+}$ ion has $3d^{10}$ configuration
0%
$Cu^{2+}$ ion has $3d^{5}$ configuration while $Zn^{2+}$ ion has $3d^{6}$ configuration
0%
$Cu^{2+}$ has half filled orbitals while $Zn^{2+}$ has fully filled orbitals
0%
$CuSO_{4}$ is blue in colour while $ZnSO_{4}$ is white.

Chrome yellow is chemically known as

0%
lead chromate
0%
lead sulphate
0%
lead iodide
0%
basic lead acetate

Which of the following complex ion has a magnetic moment same as

$[Cr(H_2O)_6]^{3+}$ ?

0%
$[Mn(H_2O)_6]^{4+}$
0%
$[Mn(H_2O)_6]^{3+}$
0%
$[Fe(H_2O)_6]^{3+}$
0%
$[Cu(H_3H)_4]^{2+}$

Explanation

compound

$[Mn(H_2O)_6]^{4+}$ have same properties to like magnetic moments like

$[Cr(H_2O)_6]^3+$

A blood red colour is obtained when ferric chloride solution reacts eith:

0%
$KCN$
0%
$KSCN$
0%
q $K_4[Fe(CN)_6]$
0%
$K_3[Fe(CN)_6]$

Explanation

$FeCl_3$ +3KSCN → blood red colour

$Fe(SCN)_3+3KCl$

These questions consists of two statements each, printed as assertion and reason, while answering these questions you are required to choose any one of the following responses.

Assertion :

$CrO_3$ reacts with

$HCl$ to form thornyl chloride gas.
Reason : Chromyl chloride

$(CrO_2Cl_2)$ has tetrahedral shape.

0%
If both assertion and reason are true and the reason is a correct explanation of assertion
0%
If both assertion and reason are true but reason is not a correct explanation of assertion
0%
If assertion is true but the reason is false
0%
If assertion is false but the reason is true

Explanation

Assertion :

$CrO_3$ reacts with

$HCl$ to form thornyl chloride gas.

Reason : Chromyl chloride

$(CrO_2Cl_2)$ has tetrahedral shape.

both are true but not show correct explanation

Transition metal and their compounds are used as catalysts in industry and in biological system. For example, in the contact process, vanadium compounds in the

$+5$ state

$(V_2O_5$ or

$VO_3^$ ) are used to oxidise

$SO_2$ to

$SO_3$ :

$SO_2 + \frac{1}{2}O_2 \overset{V_2O_5}{\rightarrow}SO_3$
It is thought that the actual oxidation process takes place in two stages. In the first step,

$V^{5+}$ in the presence of oxide ions converts

$SO_2$ to

$SO_3$ . At the same time,

$V^{5+}$ is reduced to

$V^{4+}$ .

$2V^{5+}+O^{2-} + SO_2 2V^{4+}+ SO_3$ . In the second step,

$V^{5+}$ is regenerated from

$V^{4+}$ by oxygen :

$2V^{4+}+\frac{1}{2}O_2 \rightarrow 2V^{5+} + O^{2-}.$
The overall process is, of course, the sum of these two steps:

$SO_2 + \frac{1}{2}O_2 \rightarrow SO_3$

During the course of the reaction

0%
catalyst undergoes changes in oxidation state
0%
catalyst increases the rate constant
0%
catalyst is regenerated in its original form when the reactants form the products
0%
all are correct

Amongst

$Tif_{6}^{2-}, CoF_{6}^{3-}, Cu_2Cl_2 and NiCl_{4}^{2-}$ the colourless species are:

0%
$CoF_{6}^{3-}$ and $NiCl_{4}^{2-}$
0%
$TiF_{6}^{2-}$ and $CoF_{6}^{3-}$
0%
$Cu_2Cl_2$ and $NiCl_{4}^{2-}$
0%
$TiF_{6}^{2-}$ and $Cu_2Cl_2$

Explanation

$TiF_{6}^{2-}$ and

$Cu_2Cl_2$ these are two molecule show coloueless species

When

$CO_2$ is passes into aqueous

$K_2CrO_4$ solution, this changes to another colour. What is the another colour?

0%
Black
0%
Yellow
0%
Pink
0%
Orange

Explanation

$CO_2+ H_2O \rightleftharpoons H_2 CO_3$ ,

$H_2CO_3 \rightleftharpoons H^+ + HCO^-_3$ ,

$H^+$ change

$K_2CRO_4$ into

$K_2Cr_2O_7$ ,

$\underset {yellow}{2CrO^{2-}_4} + 2H^+ \longrightarrow \underset {orange}{Cr_2O^{2-}_7} + H_2O$

Which of the following pairs has almost same colour?

0%
$_{60}Nd^{3+}.\ _{69}Tm^{3+}$
0%
$_{70}Yb^{3+},\ _{59}Pr^{3+}$
0%
$_{63}Eu^{2+},\ _{65}Tb^{3+}$
0%
$_{64}Gd^{3+},\ _{67}Ho^{3+}$

Explanation

The color of ions depends upon no of unpair electrons for transition.

Electronic configuration of

$_{63}Eu^{2+}, is [Xe] 4f^7$ and

$_{65}Tb^{3+} is [Xe]4f^8$ .

They both have almost same number of electrons for the transition, so they show the almost same color.

Hence, option C is correct.

What would happen when a solution of potassium chromate is treated with an excess of dilute nitric acid?

0%
$Cr_2O_{7}^{2-}$ and $H_{2}O$ are formed.
0%
$CrO_{7}^{2-}$ is reduced to $+3$ state of $Cr$ .
0%
$CrO_{7}^{2-}$ is oxidized to $+7$ state of $Cr$ .
0%
$Cr^{3+}$ and $Cr_{2}O_{7}^{2-}$ are formed.

Explanation

When potassium chromate is treated with an excess of dilute nitric acid, the following reaction takes place:

$2K_2CrO_4+2HNO_3\rightarrow K_2Cr_2O_7+2KNO_3+H_2O$
Oxidation state of

$Cr$ in

$K_2CrO_4$ and

$K_2Cr_2O_7$ is +6.

$N$ in

$HNO_3$ can not be oxidised as it is already present in its highest oxidation state.

What would happen when a solution of potassium chromate is treated with an excess of dilute nitric acid?

0%
$CrO_{4}^{2-}$ is oxidized to +7 state of Cr
0%
$Cr^{3+}$ and $Cr_{2}O_{7}^{2-}$ are formed
0%
$Cr_{2}O_{7}^{2-}$ and $H_{2}O$ are formed
0%
$CrO_{4}^{2-}$ is reduced to +3 state of Cr

Explanation

Correct Answer: Option C

The oxidation state of chromium in $K_2Cr_2O_7$ is $+6$ , whereas the oxidation state of nitrogen in $HNO_3$ is $+5$ .

Since the maximum oxidation state of nitrogen is $+5$ , so it won’t get oxidize further. Rather it will react to form dichromate ion and a water molecule ( $Cr_2O_{7}^{2-}$ and $H_2O$ ).

The equation involved is:

$2K_2CrO_4+2HNO_3\rightarrow K_2Cr_2O_7+2KNO_3+H_2O$

Therefore, when potassium chromate solution is treated with an excess of dilute nitric acid, $Cr_2O_{7}^{2-}$ and $H_2O$ are formed.

The actinoids exhibit more oxidation states in general than the lanthanoids. This is because:

0%
the 5f orbitals are more buried than the 4f orbitals.
0%
there is a similarity between 4f and 5f orbitals in their angular part of the wave function.
0%
the actinoids are more reactive than the lanthanoids.
0%
the 5f orbitals extend further from the nucleus than the 4f orbitals.

Explanation

Hint: Variability in oxidation states corresponds to degeneracy in the outer orbitals.

Correct Answer: Option D

Explanation:

In actinoids, the outermost

$5f$ orbitals are away from the nucleus. This makes them similar in energies to

$6d$ and

$7s$ orbitals. Thus, electrons can easily transition between these orbitals, and once they reach the valence orbital they can be ionized, This leads to the variability in oxidation states. In the case of lanthanoids, the outermost

$4f$ orbitals are too close to the nucleus.

A larger number of oxidation states are exhibited by the actinides than those by the lanthanides. The main reason being:

0%
more reactive nature of the actinides than the lanthanides
0%
$4f$ orbitals are more diffused than the $5f$ orbitals
0%
lesser energy difference between $5f$ and $6d$ orbitals than between $4f$ and $5d$ orbitals
0%
more energy difference between $5f$ and $6d$ orbitals than between $4f$ and $5d$ orbitals

Explanation

A larger number of oxidation states are exhibited by the actinides than those by the lanthanides because energy difference between

$5f$ and

$6d$ orbitals is less than that between

$4f$ and

$5d$ orbitals.

Hence, excitation of electrons can take place easily in actinides and thus, they exhibit a large number of oxidation states.

So, option C is correct.

The actinoids exhibit more number of oxidation states in general than lanthanoids because:

0%
the 5f orbitals are more buried than the 4f orbitals
0%
there is a similarity between 4f and 5f orbitals in their angular part of the wave function
0%
the actinoids are more reactive than the lanthanoids
0%
the 5f orbitals extend further from the nucleus than the 4f orbitals

Explanation

Correct Answer: Option D

Explanation:

The common oxidation number of lanthanoid is $+3$ but $+2$ and $+4$ ions are also found in the solution.

Actinoids show $+3$ as oxidation state but some Actinoids are stable till $+7$ oxidation states.

Actinoids have more oxidation states than the Lanthanoids as $5f$ orbitals are at a greater distance from the nucleus than $4f$ orbitals.

Therefore lanthanoids are larger in size than in comparison to actinoids.

Statements that are true of d-and f-block elements are:

1. The colour of

$K_{2}Cr_{2}O_{7}$ is due to d-d transition.

Actinides have larger number of oxidation states than lanthanides.

Tripositive ions of

$Pr(At. No. =59)$ and

$Tm(At. No. =69)$ have the same (green) colour.

Actinides have paramagnetic moments less than theoretically predicted values.

0%
2,3,4
0%
1,2,3
0%
2,4
0%
1,2,4

Explanation

$K_{2}Cr_{2}O_{7}$ the color is due to charge transfer and not d-d transition.

Actinides exhibit a larger number of oxidation states than lanthanides because, there is a very small energy gap between 5f, 6d, and 7s orbitals.

$Pr^{3+}$ has two 4f electrons (both unpaired).

$Tm^{3+}$ has twelve 4f electrons (5 pairs and 2 unpaired). Both

$Pr^{3+}$ and

$Tm^{3+}$ , with 2 unpaired electrons each, have the same colour.

5f electrons in actinides are less effectively shielded resulting in quenching of orbital contribution of magnetic moments. So their magnetic moments are less than theoretically predicted values.

Regarding the magnetic properties of lanthanides and actinides the correct statement is:

0%
Lanthanides are weakly paramagnetic while actinides are strongly paramagnetic.
0%
Quenching of orbital contribution is greater in lanthanides than in actinides.
0%
5f electrons in actinides are too diffuse and are less effectively shielded which results in considerable quenching of orbital contribution, and hence the magnetic moments of actinides are significantly less than theoretically predicted values.
0%
Because of diffused f-electrons there is no quenching of orbital contribution towards magnetic moment.

Transition metals are less reactive than alkali metals because they have:

0%
high ionisation potential and low melting point
0%
high ionisation potential and high melting point
0%
low ionisation potential and low melting point
0%
low ionisation potential and high melting point

The larger number of oxidation states are exhibited by the actinides than those by lanthanoids, the main reason being:

0%
more energy difference between 5f and 6d than between 4f and 5d orbitals
0%
more reactive nature of the actinoids than the lanthanoids
0%
4f orbitals more diffused than the 5f orbitals
0%
lesser energy difference between 5f and 6d than between 4f and 5d orbitals

Dipositive and tripositive ions of A and B possess same number of electron pairs, but different number of unpaired electrons. Both are paramagnetic. Identify the atomic numbers of A and B.

0%
24, 25
0%
29, 30
0%
21, 22
0%
22, 24

Explanation

Let us assume the atomic number of A and B be 22 and 24 respectively.

${ A }^{ +2 } (20): { 1s }^{ 2 }{ 2s }^{ 2 }2p^{ 6 }{ 3s }^{ 2 }{ 3p }^{ 6 }{ 4s }^{ 0 }{ 3d }^{ 2 }$ (2 unpaired electron, hence paramagnetic)

${ B }^{ +3 } (21): { 1s }^{ 2 }{ 2s }^{ 2 }2p^{ 6 }{ 3s }^{ 2 }{ 3p }^{ 6 }{ 4s }^{ 0 }{ 3d }^{ 3 }$ (3 unpaired electron, hence paramagnetic)

Hence, D is the answer.

Atoms of the transition element are smaller than those of the s-block elements. This is because of :

0%
usual contraction in size across a horizontal period
0%
orbital electrons are added to the penultimate d-shell rather than to the outermost shell of the atom
0%
both A and B
0%
none of the above

The value of the magnetic moment of a particular ion is 2.83 Bohr magneton. The ion is :

0%
$Fe^{2+}$
0%
$Ni^{2+}$
0%
$Mn^{2+}$
0%
$Co^{3+}$

Explanation

The value of

$\mu=$ 2.83 BM (

$=\sqrt{n(n+2)}$ ) corresponds to the presence of two unpaired electrons.

So, the ion is

$Ni^{2+}(3d^8)$ since it has two unpaired electrons in d-subshell.

Increasing order of magnetic moment among the following species is:

$\displaystyle Na^{+},Fe^{+3},Co^{+2},Cr^{+2}$

0%
$Na^{+} < Co^{2+} < Cr^{2+} < Fe^{3+}$
0%
$Na^{+} < Fe^{3+} < Cr^{2+} < Co^{2+}$
0%
$Cr^{2+} < Co^{2+} < Na^{+} < Fe^{3+}$
0%
$Co^{2+}< Na^{+} < Cr^{2+} < Fe^{3+}$

Explanation

Magnetic moment is

$\mu = \sqrt { n(n+2) }$ , where n is no. of unpaired electron so more no. of unpaired electron means more magnetic moment.

$Na^{+}$ has zero unpaired electron

$Co^{2+}$ has 3 unpaired electrons

$Cr^{2+}$ has 4 unpaired electrons

$Fe^{3+}$ has 5 unpaired electron

The order is

$Na^{+} < Co^{2+} < Cr^{2+} < Fe^{3+}$ .

Transition elements form complexes very readily because of:

0%
small cation size
0%
vacant d-orbitals
0%
large nuclear charge
0%
all of the above

Explanation

As transition metals have vacant

$d$ -orbitals, they exhibt variable valencies and form complexes. Besides, they have smaller size and larger nuclear charge. Hence, they have large charge density which can stablize charged complexes. Therefore, transition metals form complexes very readily.

Select the correct order of ions with respect to their ionic radii.

0%
$Y^{3+} < Lu^{3+} < Eu^{3+} < La^{3+}$
0%
$Lu^{3+} < Y^{3+} < Eu^{3+} < La^{3+}$
0%
$Lu^{3+} < Eu^{3+} < La^{3+} < Y^{3+}$
0%
$La^{3+} < Eu^{3+} < Lu^{3+} < Y^{3+}$

Explanation

We know that as we move down the group, radii increases and as we move across a period, radii decreases. The correct order of ionic radii is as follows:

$Y^{3+}< Lu^{3+}< Eu^{3+} < La^{3+}$

This is because

$Eu$ and

$Lu$ are the members of the lanthanide series which have smaller radii than the transition metals.

$Y^{3+}$ ion has a lower radius as comparison to

$La^{3+}$ because it lies immediately above it in the periodic table and hence, it has a lower effective nuclear charge and a larger size.

Magnetic moments of

$M^{2+}$ ions (divalent cations ) of 3d-series in the decreasing order is :

$Zn^{2+} > Sc^{2+} = Cu^{2+} > Ti^{2+} = Ni^{2+} > V^{2+}=Co^{2+} > Cr^{2+} = Fe^{2+} > Mn^{2+}$

Is the above statement true?

0%
True
0%
False

What would happen when as solution of potassium chromate is treated with an excess of dilute nitric acid?

0%
$Cr^{3+}$ and $Cr_2O^{2-}_7$ are formed
0%
$Cr_2O^{2-}_7$ ang $H_2O$ are formed
0%
$CrO^{2-}_4$ is reduced to $Cr^{3+}$
0%
None of the above

Spin only magnetic moment of the compound

$Hg[Co{(SCN)}_{4}]$ is :

0%
$\sqrt 3$
0%
$\sqrt 15$
0%
$\sqrt 24$
0%
$\sqrt 8$

Explanation

Cobalt is present as

$Co^{+2}$ which has [Ar]

$4s^03d^7$ configuration, which means it has 3 unpaired electrons. So the spin only magnetic moment of the compound is

$\sqrt{3(3+2)} = \sqrt{15}$

Least paramagnetic property is shown by:

0%
$Fe$
0%
$Mn$
0%
$Ni$
0%
$Cu$

Which of the following statements are correct with reference to the ferrous and ferric ions?

0%
${Fe}^{3+}$ gives brown color with potassium ferricyanide
0%
${Fe}^{2+}$ gives blue precipitate with potassium ferricyanide
0%
${Fe}^{3+}$ gives red color with potassium thiocyanate
0%
${Fe}^{2+}$ gives brown color with ammonium thiocyanate

Explanation

$Fe^{2+}$ on reaction with potassium ferricyanide will give a dark blue precipitate. This precipitate is known as Turnbull's blue and it is identical to Prussian blue.

$K^{+}(aq) + Fe^{+}_{2}(aq) + [Fe(CN)_{6}]^{3-} (aq) \rightleftharpoons KFe[Fe(CN)_{6}](s)$

$Fe^{3+}$ on reaction with potassium thiocyanate (KSCN) gives a blood red solution.

$FeCl_{3} + 3KSCN \rightarrow Fe(SCN)_{3} + 3KCl$

blood red color

Hence,option B and C is correct.

An inorganic salt is lemon yellow in colour. It becomes orange in colour like methyl orange when it is acidic and again becomes yellow when it is alkaline. The inorganic salt will be :

0%
copper nitrate
0%
ferric chloride
0%
potassium chromate
0%
potassium ferricyanide

Explanation

2CrO₄^2-(aq) (yellow) + 2H⁺(aq)

$\rightarrow$ Cr₂O₇^2-(aq) (orange) + H₂O(l)

The addition of acid encourages the equilibrium towards the right, producing more orange-coloured dichromate(VI) ions. The addition of hydroxide ions causes the concentration of hydrogen ions to decrease, and this brings the equilibrium back to the left-hand side, regenerating yellow chromate(VI) ions.

Copper becomes green when exposed to moist air for a long period of time due to the

0%
formation of a layer of cupric oxide on the surface of copper.
0%
formation of a layer of basic carbonate of copper on the surface of copper.
0%
formation of a layer of cupric hydroxide on the surface of copper.
0%
none of the above.

Explanation

Copper metal when exposed to air turns green in colour due to corrosion. When copper vessel is exposed to air in rainy season, the metal reacts with gases, moisture and atmospheric gases to form a mixture of copper carbonate and copper hydroxide. This gives a green colour to the surface of copper metal.

The reaction is as follows:

$2Cu + H_2O + CO_2 + O_2 \rightarrow Cu(OH)_2 + CuCO_3$

Hence, copper becomes green when exposed to moist air for a long period of time due to the formation of a layer of basic carbonate of copper on the surface of copper.

$FeCl_3.6H_2O + C(CH_3)_2 (CH_3O)_2 \rightarrow$ Products.

Reaction products are:

0%
$FeCl_3,\ CH_3OH$ and $CH_3COCH_3$
0%
$CH_3O)_3Fe,\ HCl$ and $H_2O$
0%
$FeCl_2,\ HCl$ and $CH_3COCH_3$
0%
$Fe(OH)_3,\ FeCl_3$ and $CH_3COCH_3$

Explanation

Reaction:

$FeCl_3.6H_2O + C(CH_3)_2 (CH_3O)_2 \rightarrow FeCl_3+ CH_3OH + CH_3COCH_3$

In an experiment when placed in the weak magnetic field, calomel was slightly repelled by the magnetic field. This experimental observation suggests that:

0%
$Hg^+$ ion has no unpaired electron.
0%
mercurous ion has a formula $Hg_2^{2+}$ instead of $Hg^+$
0%
this experimental observation is not correct and actually mercurous salts are paramagnetic due to $6s$ unpaired electron
0%
sometimes mercurous ion may exist as $Hg_2^{2+}$

Explanation

Mercurous ion has formula

$Hg_2^{2+}$ instead of

$Hg^+$ .
It has been proved on the basis of magnetic moment determination and equilibrium studies on

$Hg(I)$ and

$Hg(II)$ salts. It is interesting to note that

$Hg(I)$ salts are diamagnetic, yet

$Hg^+$ contains unpaired electron. In solutions

$Hg_2^{2+}$ disproportionates as:

$Hg_2^{2+}$

$\rightleftharpoons$

$Hg_{(l)}$

$+ Hg^{2+}$ , but equilibrium constant is very low.

Column-I (Alloys)	Column-II (Constituents)
(A) TiCl $_4$	(p) Adams catalyst in reduction
(B) PdCl $_2$	(q) In preparation of (CH $_3$ ) $_2$ SiCl $_2$
(C) Pt/PtO	(r) Used as the Natta catalyst in polythene production
(D) Cu	(s) Wake process for converting C $_2$ H $_4$ to CH $_3$ CHO

0%
(A) $\rightarrow$ s ; (B) $\rightarrow$ r ; (C) $\rightarrow$ p ; (D) $\rightarrow$ q
0%
(A) $\rightarrow$ q ; (B) $\rightarrow$ s ; (C) $\rightarrow$ p ; (D) $\rightarrow$ r
0%
(A) $\rightarrow$ r ; (B) $\rightarrow$ s ; (C) $\rightarrow$ p ; (D) $\rightarrow$ q
0%
(A) $\rightarrow$ p ; (B) $\rightarrow$ s ; (C) $\rightarrow$ r ; (D) $\rightarrow$ q

Explanation

$TiCl_4-$ Ziegler-Natta catalyst

This catalyst is used in polythene production.

$PdCl_2$

This is used as an oxidizing agent in the Wake process to oxidize alkene to aldehyde.

$PtO_2-$ Adams catalyst

It is used in reduction reactions to convert nitro compounds to amines.

Cu is used in the preparation of

$(CH_3)_2SiCl_2$ .

How many of the following ions have the same magnetic moments?
Fe

$^{2+}$ Mn

$^{2+}$ Cr

$^{2+}$ Ni

$^{2+}$

0%
1
0%
2
0%
3
0%
4

Explanation

As we know,

$\mu = \sqrt {n(n+2)}$
n = no of unpaired electron
Fe

$^{2+}$ = 4 unpaired electron
Mn

$^{2+}$ = 5 unpaired electron
Cr

$^{2+}$ = 4 unpaired electron
Ni

$^{2+}$ = 2 unpaired electron.

A metal

$M$ which is not affected by strong acids like conc.

$HNO_3, conc. H_2SO_4$ and conc. solution of alkalies like

$NaOH, KOH$ forms

$MCl_3$ which finds use for toning in photography. The metal

$M$ is :

0%
$Ag$
0%
$Hg$
0%
$Au$
0%
$Cu$

Explanation

Gold and silver both find use for toning in photography. However

$Au$ is not affected by strong acids and strong alkalies.

$Au$ forms

$AuCl_3$ , whereas

$Ag$ forms

$AgCl$ . Hence, metal M is

$Au$ .

Transition metals and their compounds catalyse reactions because:

0%
they have completely filled $s$ - subshell.
0%
they have a comparable size due to poor shielding of d-subshell.
0%
they introduce an entirely new reaction mechanism with lower activation energy.
0%
they have variable oxidation states.

Catalytic activity of transition metals depends on:

0%
their ability to exist in different oxidation states
0%
the size of the metal atoms
0%
the number of empty atomic orbitals available
0%
none of these

Match the catalysts to the correct processes:

Catalyst	Process
A. $TiCl_3$	i. Wacker process
B. $PdCl_2$	ii. Ziegler - Natta polymerization
C. $CuCl_2$	iii. Contact process
D. $V_2O_5$	iv. Deacon's process

0%
(A) - (iii), (B) - (ii), (C) - (iv), (D) - (i)
0%
(A) - (ii), (B) - (i), (C) - (iv), (D) - (iii)
0%
(A) - (ii), (B) - (iii), (C) - (iv), (D) - (i)
0%
(A) - (iii), (B) - (i), (C) - (ii), (D) - (iv)

Explanation

A Ziegler–Natta catalyst, named after Karl Ziegler and Giulio Natta, is a catalyst used in the synthesis of polymers of 1-alkenes (alpha-olefins). The catalyst

$TiCl_3$ is used for Ziegler - Natta polymerization.

The Wacker Oxidation is an industrial process, which allows the synthesis of ethanal from ethene by palladium-catalyzed oxidation with oxygen. The catalyst

$PdCl_2$ is used for the Wacker process.

The Deacon process is a process used during the manufacture of alkalis (the initial end product was sodium carbonate) by the Leblanc process. The catalyst

$CuCl_2$ is used for Deacon's process.

The contact process is the current method of producing sulfuric acid in the high concentrations needed for industrial processes. The catalyst

$V_2O_5$ is used for the Contact process.

Hence, option

$B$ is correct.

Which of the following statement(s) is (are) not correct with reference to ferrous and ferric ions?

0%
$Fe^{3+}$ gives brown colour with potassium ferricyanide
0%
$Fe^{2+}$ gives blue ppt with potassium ferricyanide
0%
$Fe^{3+}$ gives red colour with potassium sulphocyanide
0%
$Fe^{2+}$ gives brown colour with potassium sulphocyanide

Explanation

$Fe^{3+} + 2KCNS \rightarrow [Fe(SCN)]^{2+}$ (Red coloured ppt.)

$Fe^{3+}$ gives red ppt with KCNS and not

$Fe^{2+}$ .

Which of the following ion involved in the above process will show paramagnetism?

0%
$V^{5+}$
0%
$V^{4+}$
0%
$O^{2-}$
0%
$VO_3^-$

Explanation

As here in this process, the

$V^{5+}$ first changes to

$V^{4+}$ and then again,

$V^{4+}$ changes to

$V^{5+}$ .

$V^{4+}$ has

$3d^1$ configuration, so it has 1 unpaired electron, due to which this can show paramagnetism.
Whereas

$V^{5+}$ , has

$3d^0$ configuration, hence, due to no unpaired electrons, it does not show paramagnetism.
In the same way, others also do not show paramagnetism.

The bivalent metal ion having maximum paramagnetic behaviour among the first transition series elements is :

0%
$Mn^{2+}$
0%
$Cu^{2+}$
0%
$Sc^{2+}$
0%
$Cu^+$

Explanation

$Mn^{2+}$ will have 5 unpaired electrons in its electronic configuration so it will have maximum paramagnetic behaviour.

Statement 1 : A piece of zinc placed in a blue copper nitrate solution will displace the copper from the solution, producing copper metal and a colorless

$\displaystyle { Zn }^{ 2+ }$ solution.
Statement 2 : Copper is a much more active metal than zinc.

0%

Both Statement 1 and Statement 2 are correct and Statement 2 is the correct explanation of Statement 1.
0%

Both Statement 1 and Statement 2 are correct and Statement 2 is not the correct explanation of Statement 1.
0%

Statement 1 is correct but Statement 2 is not correct.
0%

Statement 1 is not correct but Statement 2 is correct.
0%

Both the Statement 1 and Statement 2 are not correct.

Explanation

Any oxidizing agent (the species on the left of the half reaction) can oxidize any reducing agent (the species on the right of half reaction) that appears below it but cannot oxidize the specie located above it in the electrochemical series.
Cu is above Zn in the electrochemical series. Hence, Cu(II) can oxidize Zn.
Statement 1 : A piece of zinc placed in a blue copper nitrate solution will displace the copper from the solution, producing copper metal and a colorless Zn

$^{2+}$ solution.
Statement 2 : Copper is a less active metal than zinc.
Statement 1 is correct but Statement 2 is not correct.
Hence, the option (C) is the correct answer.

The electrical conductivity of copper is due to which of the following?

0%
Hydrogen bonding
0%
Ionic bonding
0%
Network bonding
0%
London dispersion force
0%
Metallic bonding

Potassium maganate

$(K_2MnO_4)$ is formed when:

0%
Chlorine is passed into aqueous $KMnO_4$ solution
0%
Manganese dioxide is fused with potassium hydroxide in air
0%
Formaldehyde reacts with potassium permanganate in the presence of a strong alkali
0%
Potassium permanganate reacts with concentrated sulphuric acid.

Explanation

Potassium maganate

$(K_2MnO_4)$ is formed when magnese dioxide fused with potassium hydroxide in air or formaldehyde reacts with potassium permanganate in the presence of a strong alkali.

$4KOH + 2MnO_2 +O_2 \rightarrow 2K_2MnO_4 + 2H_2O$

$HCHO+2KMnO_4 + 2KOH \rightarrow 2K_2MnO_4 + H_2O+HCOOH$

Identify the order in which the spin only magnetic moment (in BM) increases for the following four ions:
(I)

$Fe^{2+}$ (II)

$Ti^{2+}$ (III)

$Cu^{2+}$ (IV)

$V^{2+}$

0%
$I, II, IV, III$
0%
$IV, I, II, III$
0%
$III, IV, I, II$
0%
$III, II, IV, I$

Explanation

$(I) Fe^{2+}$ has

$3d^6$ configuration.
Hence, it has 4 unpaired electrons.

$(II) Ti^{2+}$ has

$3d^2$ configuration.
So, it has 2 unpaired electrons.

$(III) Cu^{2+}$ has

$3d^9$ configuration.
So, it has 1 unpaired electron.

$(IV) V^{2+}$ has

$3d^3$ configuration.
So, it has 3 unpaired electrons.

More the unpaired electrons more will be the spin only magnetic moment. So the order of increasing unpaired electrons and spin only magnetic moment of species is as follows:

$(III)<(II)<(IV)<(I)$

Statement 1: Transition metal compounds generally exhibit bright colors.
Statement 2: The electrons in the partially filled d orbitals are easily promoted to excited states.

0%

Both Statement 1 and Statement 2 are correct and Statement 2 is the correct explanation of Statement 1.
0%

Both Statement 1 and Statement 2 are correct and Statement 2 is not the correct explanation of Statement 1.
0%

Statement 1 is correct but Statement 2 is not correct.
0%

Statement 1 is not correct but Statement 2 is correct.
0%
Both the Statement1 and Statement 2 are not correct.

The compound pictured below undergoes decomposition when heated.
Which of the following CORRECTLY represents a possible set of products of this decomposition?

Explanation

The given compound is (NH

$_4)_2 Cr_2O_7$

Now,

(NH

$_4)_2 Cr_2O_7$ (s)

$^\Delta\rightarrow$

$Cr_2O_3$ (s)

$+N_2(g) +4H_2O$ (g)

When ammonium dichromate is heated , it gives out orange sparks and green

$Cr_2O_3$ crystals and thrown out into the alr producing volcanic eruption like effect.

When

$MnO_2$ is fused with

$KOH$ and

$KNO_3$ , a coloured compound is formed, the product and its colour is:

0%
$K_2MnO_4,$ Green
0%
$KMnO_4,$ Purple
0%
$Mn_2O_3,$ Blue
0%
$K_3MnO_2$ , Red

Explanation

When

$MnO_2$ is fused with

$KOH$ and

$KNO_3$ potassium magnate

$K_2MnO_4$ is formed .

$K_2MnO_4$ is green coloured

In the oxidation of oxalic acid by potassium permanganate, in the presence of dil.

$H_2SO_4$ one of the products _______ acts as an autocatalyst.

0%
$K_2SO_4$
0%
$MnSO_4$
0%
$MnO_2$
0%
$Mn_2O_3$

Explanation

In the oxidation of oxalic acid by potassium permanganate, in the presence of dil.

$\displaystyle H_2SO_4$ one of the products

$\displaystyle MnSO_4$ acts as an autocatalyst.

$\displaystyle 2KMnO_4 +5H_2C_2O_4+3H_2SO_4\rightarrow 2MnSO_4+K_2SO_4+10CO_2 + 8H_2O$

CBSE Questions for Class 12 Engineering Chemistry The D-And F-Block Elements Quiz 11 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry The D-And F-Block Elements Quiz 11 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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