The d- and f- Block Elements Test

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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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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.

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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

The d- and f- B...

Question 1

copper nitrate

ferric chloride

potassium chromate

potassium ferricyanide

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.

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 $$CH_3COCH_3$$

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

Question 5

1

2

3

4

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

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 $$Hg_2^{2+}$$

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

Question 10

$$Ag$$

$$Hg$$

$$Au$$

$$Cu$$

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