Class 12th

49.  (B) KMnO₄ oxidises HCl into Cl₂  which is also an oxidizing agent.

HCl is not used in titrations involving KMnO₄  as it produces unsatisfactory results. When HCl  is added to KMnO₄, it itself acts as reducing agent and coverts HCl to Cl₂. 

48.  (C)  both have similar atomic radius

As an effect of lanthanide contraction, zirconium and hafnium have similar radii of 160 pm and 159 pm respectively. Due to this similarity in their size, they show similar physical and chemical properties. 

Lanthanide contraction: Because the elements in Row 3 have 4f electrons. These electrons do not shield well, causing a greater nuclear charge. This greater nuclear charge has a greater pull on the electrons and results in the decrease in their size and atomic radii. 

47.  (D) in covalent compounds fluorine can form single bond only while oxygen forms double bond

Oxygen has the capacity to form multiple bonds which enables it to form a variety of covalent compounds. 

In (Mn₂O₇) also, 6 oxygen are doubly bonded to two manganese atoms and one oxygen is forming bridge between two. 

While in (MnF₄), four fluorine atoms are singly bonded to manganese atom giving it a +4 oxidation state.

Therefore, due to capability of oxygen to have multiple bonds in covalent compounds, manganese is having higher oxidation state of +7 in (Mn₂O₇) . 

 46.  (C) Sn⁴⁺

Cr₂O₇^2-  +  14H⁺  +  3Sn²⁺  → 2Cr₃ +  +  3Sn⁴⁺  +  7H₂O

In the above redox reaction, oxidation of Sn²⁺  is taking place. It gets converted to Sn⁴⁺  whereas, reduction of chromium is occurring from +6 oxidation state to +3. 

45.  (A) is incorrect.

Copper lies below hydrogen in electrochemical series and hence cannot liberate hydrogen from acids. 

Cu  +  2H₂SO₄ → CuSO₄ +  SO₂ +  2H₂O

44.  (C) IO₃^- 

Iodide is oxidised to iodate.  

2KMnO₄ +  KI  + H₂O → 2KOH  +  2MnO₂ +  KIO₃

               iodide                                               iodate 

43.  (B) 3.87 B.M. 

Cr →  (Z = 24)

Cr3 +  →  (Z = 21)

1s² 2s² 2p⁶ 3s² 3p⁶ 3d³ 4s⁰

n  =  3 ( 3 unpaired electrons )

μ = $\sqrt[]{n(n+2)}$

μ = $\sqrt[]{3(3+2)}$

μ = $\sqrt[]{3\left(5\right)}$

μ = $\sqrt[]{15}$

=  3.87 B.M. 

42.  (D) They are chemically very reactive.

Interstitial compounds/alloys are substances that are formed when a small atom like carbon, hydrogen, boron, nitrogen can occupy space in their lattices. All the properties mentioned above are true for interstitial compounds except (D), they are chemically inert.

41.  (A) [Xe] 4f⁷ 5d¹ 6s²

Gadolinium belongs to the 4f series; it has atomic no.= 64. It has extra stability due to a half-filled 4f subshell.

Gd  (Z=64) ?   [Xe] 4f⁶ 5d² 6s² 

40.  (A) V₂O₅, Cr₂O₃

Oxides in the lower oxidation state are basic and in the higher oxidation state they are acidic. They are amphoteric in the intermediate oxidation state. Therefore, V₂O₅, Cr₂O₃ are amphoteric in nature.