The d- and f- Block Elements Test

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The d- and f- Block Elements Test - 59

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Weekly Quiz Competition

Question 1
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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 :

A
copper nitrate

B
ferric chloride

C
potassium chromate

D
potassium ferricyanide

Solution

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.
Question 2
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Copper becomes green when exposed to moist air for a long period of time due to the

A
formation of a layer of cupric oxide on the surface of copper.

B
formation of a layer of basic carbonate of copper on the surface of copper.

C
formation of a layer of cupric hydroxide on the surface of copper.

D
none of the above.

Solution

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.
Question 3
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$$FeCl_3.6H_2O + C(CH_3)_2 (CH_3O)_2 \rightarrow$$ Products.
Reaction products are:

A
$$FeCl_3,\ CH_3OH$$ and $$CH_3COCH_3$$

B
$$CH_3O)_3Fe,\ HCl$$ and $$H_2O$$

C
$$FeCl_2,\ HCl$$ and $$CH_3COCH_3$$

D
$$Fe(OH)_3,\ FeCl_3$$ and $$CH_3COCH_3$$

Solution

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

Question 4

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

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

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

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

Solution

$$TiCl_4-$$ 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 $$(CH_3)_2SiCl_2$$.

Question 5
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How many of the following ions have the same magnetic moments?
Fe$$^{2+}$$ Mn$$^{2+}$$ Cr$$^{2+}$$ Ni$$^{2+}$$

A
1

B
2

C
3

D
4

Solution

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.
Question 6
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Directions For Questions

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 $$SO_3$$ : $$SO_2$$ + $$\displaystyle{\frac{1}{2}}O_2 \overset{V_2O_5} {\rightarrow}$$ $$SO_3$$. 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 $$V^{4+}$$. $$2V^{5+} + O^{2-} + SO_2 \rightarrow 2V^{4+} + SO_3$$. In the second step, $$V^{5+}$$ is regenerated from $$V^{4+}$$ by oxygen : $$2V^{4+} + \displaystyle{\frac{1}{2}O_2 \rightarrow} 2V^{5+} + O^{2-}$$. The overall process is, of course, the sum of these two steps: $$SO_2 + \displaystyle{\frac{1}{2}}O_2 \rightarrow SO_3$$.

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Catalytic activity of transition metals depends on:

A
their ability to exist in different oxidation states

B
the size of the metal atoms

C
the number of empty atomic orbitals available

D
none of these

Solution

The transition metals and their compounds are known for their homogeneous and heterogeneous catalytic activity. This activity is ascribed to their ability to adopt multiple oxidation states and to form complexes.
Question 7
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Which one of the following metals has the highest density?

A
Gold

B
Iron

C
Platinum

D
Lead
Question 8
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Directions For Questions

$$Hg$$ shows two oxidation states $$(I)$$ and $$(II)$$ in its compounds. Chemistry of $$Hg(I)$$ in interesting as $$Hg^+$$ does not exist. 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 an unpaired electron. In solutions $$Hg_2^{2+}$$ disproportionates as : $$Hg_2^{2+}$$ $$\rightleftharpoons$$ $$Hg_{(l)}$$ $$+ Hg^{2+}$$, but equilibrium constant is very low. But when anions like $$S^{2-}$$ are added the formation of insoluble $$HgS$$ promotes the disproportionation. The same reason is used to explain the non-existence of some mercurous salts.

...view full instructions

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

A
$$Hg^+$$ ion has no unpaired electron.

B
mercurous ion has a formula $$Hg_2^{2+}$$ instead of $$Hg^+$$

C
this experimental observation is not correct and actually mercurous salts are paramagnetic due to $$6s$$ unpaired electron

D
sometimes mercurous ion may exist as $$Hg_2^{2+}$$

Solution

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.
Question 9
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Which of the following statement(s) is (are) not correct with reference to ferrous and ferric ions?

A
$$Fe^{3+}$$ gives brown colour with potassium ferricyanide

B
$$Fe^{2+}$$ gives blue ppt with potassium ferricyanide

C
$$Fe^{3+}$$ gives red colour with potassium sulphocyanide

D
$$Fe^{2+}$$ gives brown colour with potassium sulphocyanide

Solution

$$Fe^{3+} + 2KCNS \rightarrow [Fe(SCN)]^{2+}$$(Red coloured ppt.)
$$Fe^{3+}$$ gives red ppt with KCNS and not $$Fe^{2+}$$.
Question 10
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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 :

A
$$Ag$$

B
$$Hg$$

C
$$Au$$

D
$$Cu$$

Solution

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 $$Au$$.

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Re-Attempt Weekly Quiz Competition

The d- and f- Block Elements Test - 59

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The d- and f- Block Elements Test - 59

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Question 1

copper nitrate

ferric chloride

potassium chromate

potassium ferricyanide

Solution

Question 2

formation of a layer of cupric oxide on the surface of copper.

formation of a layer of basic carbonate of copper on the surface of copper.

formation of a layer of cupric hydroxide on the surface of copper.

none of the above.

Solution

Question 3

$$FeCl_3,\ CH_3OH$$ and $$CH_3COCH_3$$

$$CH_3O)_3Fe,\ HCl$$ and $$H_2O$$

$$FeCl_2,\ HCl$$ and $$CH_3COCH_3$$

$$Fe(OH)_3,\ FeCl_3$$ and $$CH_3COCH_3$$

Solution

Question 4

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

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

Solution

Question 5

1

2

3

4

Solution

Question 6

their ability to exist in different oxidation states

the size of the metal atoms

the number of empty atomic orbitals available

none of these

Solution

Question 7

Gold

Iron

Platinum

Lead

Question 8

$$Hg^+$$ ion has no unpaired electron.

mercurous ion has a formula $$Hg_2^{2+}$$ instead of $$Hg^+$$

this experimental observation is not correct and actually mercurous salts are paramagnetic due to $$6s$$ unpaired electron

sometimes mercurous ion may exist as $$Hg_2^{2+}$$

Solution

Question 9

$$Fe^{3+}$$ gives brown colour with potassium ferricyanide

$$Fe^{2+}$$ gives blue ppt with potassium ferricyanide

$$Fe^{3+}$$ gives red colour with potassium sulphocyanide

$$Fe^{2+}$$ gives brown colour with potassium sulphocyanide

Solution

Question 10

$$Ag$$

$$Hg$$

$$Au$$

$$Cu$$

Solution

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