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New answer posted
10 months agoContributor-Level 10
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
New answer posted
10 months agoContributor-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.
New answer posted
10 months agoContributor-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.
New answer posted
10 months agoContributor-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.
New answer posted
10 months agoContributor-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.
New answer posted
10 months agoContributor-Level 10
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.
New answer posted
10 months agoContributor-Level 10
45. Option (ii) and (iii)
Explanation: Using oxidation method for extraction of chlorine from brine. The reactions involved are:
2Cl− + 2H2O → 2OH− + H2 + Cl2
For this reaction, the value of ΔG°=+422 kJ, which is positive. Using the formula ΔG°=−nE°F, we get a negative value of E° =−2.2 V.
New answer posted
10 months agoContributor-Level 10
26. As the positive charge of the ion increases or we can say that oxidation state of a transition element increases, its size decreases and as per Fajan's rule, more the charge on the metal ion, more is its tendency to form covalent compounds because positively charged cation attracts the electron cloud strongly towards itself.
New answer posted
10 months agoContributor-Level 10
44. Option (i) and (ii)
Explanation: Depressants are materials that are added for the separation of ores that prevent certain types of particles from coming to froth and forming bubbles. For example, an ore containing ZnS and PbS, NaCN is used as a depressant.
When sulfur ores are blown in hot air along with silica, the solidified metal which is obtained has a blistered appearance due to SO2 evolution.
New answer posted
10 months agoContributor-Level 10
25. The sum of sublimation energy and ionisation enthalpy to oxidise cu (s) to Cu2+ is so highly that it is not compensated by the hydration enthalpy of Cu. Due to this, the Eof Cu is positive.
While in case if Zn, the E value is negative or more negative than the expected value because when the electrons are removed from the 4s-orbital. Zn acquires a stable 3d10 configuration state.
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