Class 12th

 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.

39.  (A) 25

2MnO₄^-  +  5S^2-  + 16H⁺  → 2Mn²⁺  +  5S  + H₂O

From the reaction,

For 5 moles of S, two moles of KMnO₄ were required.

Therefore, for one  mole of S, the number of moles of KMnO₄ required will be = 25.

Given curve is $y=cosx$

$y=sinx$ for $0\le x\le \frac{\pi }{2}$

And $y-axis$

We know that $sinx=cosx$ at $x=\frac{\pi }{4}and<\frac{\pi }{4}<\frac{\pi }{2}$ i.e,  $cos\frac{\pi }{4}=sin\frac{\pi }{4}=1/\surd 2$

So the point of intersection is at $x=\frac{\pi }{4}$

The given area of the circle is $x^2+y^2=16-------(1)$ is a circle with centre (0,0) and radius, $\pi =4$ and the parabola is $y^2=6x$ -------------(2)

Solving (1) and (2) for x and y.

$\begin{array}{l}{x}^2+6x=16\\ =x^2+6x-16=0\\ =x^2+8x-2x-16=0\\ =x\left(x+8\right)-2\left(x+8\right)=0\\ =\left(x+8\right)\left(x-2\right)=0\\ =x=-8\&x=2\end{array}$

For, $x=-8,y^2=6\left(-8\right)=-48$

Which is not possible.

For, $x=2,y^2=6\left(2\right)=12$

$y=\pm \mathrm{2\surd 3}$

$areaOACBO=2*\left\{area\left(OADO\right)+area\left(ACDA\right)\right\}$