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Payal Gupta

Contributor-Level 10

32.  (B) CuF2

(A) Ag2SO4  (Ag+ )→ 5d106s0

(B) CuF2  (Cu2+ )→ 3d94s0

(C) ZnF2  (Zn2+)→ 3d104s0

(D) Cu2Cl2  (Cu+)→ 3d106s0

Unpaired electrons present in any compound impart colour to the salt of transition metal. Only CuF2 has unpaired electrons in its 3d orbital that's why it is white coloured in its solid-state while rest of the salts are colourless.

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Payal Gupta

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31.  (D) Cu

Density is mass by volume. As we move from left to right for a long period, the atomic radii decrease. Hence, volume decreases. Also, an increase in atomic masses is observed. 

So overall the density increases out of the above option from iron to copper, copper will have the highest density.

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alok kumar singh

Contributor-Level 10

49. Option (i) A (3) B (4) C (2) D (1)

Sapphire is a gemstone containing Co.

Hence, option (A) from column I is matched with option (3) from column II. The Sphalerite single is ZnS.

Hence, option (B) from column I is matched with option (4) from column II. NaCN is also used as a depressant.

Hence, option (C) from column I is matched with option (2) from column II. Al2O3 is also called corundum.

Hence, option (D) from column I is matched with option (1) from column II.

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Payal Gupta

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30. (A) Cu (II) is more stable .

Electronic configuration of Cuis [Ar] 3d10 4s1

Cu (I) - [Ar] 3d10 4s0

Cu (II)- [Ar] 3d9 4s0

Despite the fact that Cu (I)  has fully filled 3d-orbital but Cu (II) is more stable than Cu (I) due to the greater effective nuclear charge of Cu (II) as nucleus has to hold 17 electrons rather than 18 electrons like in Cu (I).

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Payal Gupta

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29.  (B) 26

Positive oxidation states indicate the loss of electrons from the atom. If X is in +3 oxidation state, then three electrons have been removed from it. Find the atomic number of the parent atom, we will add three in the given electronic number. i.e.

X3+  (Z = 23)  =  [Ar] 3d5

X  =  23 + 3  = 26  

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alok kumar singh

Contributor-Level 10

48. Correct code (i) A (4) B (2) C (3) D (1)

The cyanide process is used in the extraction of Au.

Hence, option (A) from column I is matched with option (4) from column II. Froth floatation process is used in the dressing of ZnS.

Hence, option (B) from column I is matched with option (2) from column II. Electrolytic reduction is used in the extraction of AI.

Hence, option (C) from column I is matched with option (3) from column II. Zone refining is used to get ultrapure Ge.

Hence, option (D) from column I is matched with option (1) from column II.

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alok kumar singh

Contributor-Level 10

47. Option (ii) A (4) B (3) C (1) D (2)

Coloured bands are found in chromatography.

Hence, option (A) from column I is matched with option (4) from column II. Impure metals are converted to volatile complexes in Mond's process.

Hence, option (B) from column I is matched with option (3) from column II. Purification of Ge and silicon is done using zone refining.

Hence, option (C) from column I is matched with option (1) from column II. Purification of mercury is done using fractional distillation.

Hence, option (D) from column I is matched with option (2) from column II.

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alok kumar singh

Contributor-Level 10

46. Option (ii) A (2) B (4) C (5) D (3)

Pendulum is always made of nickel steel.

Hence, option (A) from column I is matched with option (2) from column II. Malachite is the ore of copper.

Hence, option (B) from column I is matched with option (4) from column II. Calamine is the ore of zinc.

Hence, option (C) from column I is matched with option (5) from column II. Cryolite is an ore of aluminum.

Hence, option (D) from column I is matched with option (3) from column II.

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Payal Gupta

Contributor-Level 10

28. Ionisation enthalpies are the main factor that influence the reactivity of transition elements. Higher the ionisation enthalpy, lesser is the reacting of the transition element. 

When we move along the period from Sc to Cu, a regular increase in the ionisation enthalpy is observed which results in the almost regular decrease in the reactivity of elements.

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Payal Gupta

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27. As per (n + l) rule, 4s has lower energy than 3d-orbital. 

3d−n+l = 3+2 = 5

4s  -  n  +  l  =  4 + 0  =  4 

So, 4s are filled first.

After filling of electrons, 4s-orbital moves beyond 3d-orbital and 4s electrons are loosely held by the nucleus. Hence, electrons are removed first during the process of ionisation.

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