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8 months agoContributor-Level 10
61. Although Zr and Hf lie in same group but As an effect of lanthanide contraction, zirconium and hafnium have similar radius of 160pm and 159 pm respectively. Due to this similarity in their size, they show similar physical and chemical properties. As a result, Zr and Hf are difficult to separate. Both the statements assertion and reason are correct but reason is not the correct explanation of assertion.
New answer posted
8 months agoContributor-Level 10
60. All halogens combine with copper to form copper halides except iodine. The reason behind this is that Cu2+ oxidises iodide (-1) to iodine (0). Therefore, Cu2+ iodide does not exist .Both the statements assertion and reason are correct and reason is the correct explanation of assertion.
New answer posted
8 months agoContributor-Level 10
59. (B) and (C)
Electronic configuration of cerium:
Ce - 4f1 5d1 6s2
Ce4+? 4f0 5d0 6s0
Lead has the tendency to lose its four outer electrons to attain Noble gas configuration and this removal of electrons corresponds to the f? configuration and cause it to be stable in +4 oxidation state
New answer posted
8 months agoContributor-Level 10
58. (B) and (C)
A compound act as oxidising agent when it's central atom is reduced to its lower oxidation state. This occurs only when the lower oxidation state of metal is more stable than the higher oxidation states.
In metal WO3 and CrO42- , both W and Cr are stable in their higher oxidation state and won't get reduced to their lower oxidation state. Therefore, will not act as oxidising agents.
New answer posted
8 months agoContributor-Level 10
56. (B) and (C)
Magnetic moment of an electron/dipole moment is caused by its intrinsic properties of spin and electric charge. It depends upon the number of unpaired electrons in its valence shell. The more the number of unpaired electrons, the greater will be the value of magnetic moment.
Electronic configuration:
Ti3+ = 3d1 = 1 unpaired electron
Mn2+ = 3d5 = 5 unpaired electrons
Fe2+ = 3d6 = 4 unpaired electrons
Co2+ = 3d7 = 3 unpaired electrons
Out of all given options, Mn2+ and Fe2+ have unpaired el
New answer posted
8 months agoContributor-Level 10
55. (B) and (C)
Ce (Z = 57) = [Xe] 4f5 5d0 6s2 , O.S. = + 3, + 4
Eu (Z = 63) = [Xe] 4f7 5d0 6s2, O.S. = + 2, + 3
Yb (Z = 70) = [Xe] 4f14 5d0 6s2, O.S. = + 2, + 3
Ho (Z = 67) = [Xe] 4f11 5d0 6s2, O.S. = + 3
New answer posted
8 months agoContributor-Level 10
54. (A) and (B)
Electronic configuration:
92U - 5f3 6d1 7s2
93Np - 5f4 6d1 7s2
New answer posted
8 months agoContributor-Level 10
53. (B) & (D)
Actinides show a variety of oxidation states like +3, +5, +6 and +7.
From the given option, plutonium and neptunium show all oxidation states up to +7
New answer posted
8 months agoContributor-Level 10
52. (B) and (C)
For the heavier metals of d-block, higher oxidation state is more favourably stable than the lower oxidation state. For example: Mo (VI) and W (VI) are more stable than Cr (VI). For lighter d-block elements like chromium, +3 oxidation state is more stable. So, Cr (VI) in dichromate act as a strong oxidising agent by reducing itself from +6 to +3 oxidation states While Mo (VI) and W (VI) does not.
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