Class 12th

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, $x=2,y^2=6\left(2\right)=12$

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

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

38. (C) Tm

In the periodic table, actinides lie from atomic number 90 to 103. 

Thulium is not an actinide; it is a lanthanide. 

The given curve is $y=x\left|x\right|$

$=\left\{\left(x.xifx\ge 0\right)\left(x.\left(-x\right)ifx\le 0\right)\right\}=\left\{\left(x^{2}if,x\ge 0\right)\left(-x^{2}if,x\le 0\right)\right\}$

Which is in the form of a parabola nad the lines are $x=-1\&x-axis$

At $x=1>0,y=1^2=1$

At $$$x=-1<0,y=-1^2=-1$

Shaded area of the Ist quadrant

$\begin{array}{l}={\displaystyle \underset{0}{\overset{1}{\int }}ydx}={\displaystyle \underset{0}{\overset{1}{\int }}{x}^{2}dx}\\ ={\left[\frac{x^3}{3}\right]}_0^1\\ =\frac{1}{3}\end{array}$

Shaded area of the IInd quadrant

$\begin{array}{l}={\displaystyle \underset{-1}{\overset{0}{\int }}ydx}={\displaystyle \underset{-1}{\overset{0}{\int }}{x}^{2}dx}\\ ={\left[\frac{-x^3}{-3}\right]}_{-1}^0\\ =\frac{1}{3}\end{array}$

$\therefore$ Total area of the enclosed region $=\frac{1}{3}+\frac{1}{3}$

$=\frac{2}{3}uni{t}^2$

$\therefore$ Option (c) is correct.

37. (D) Mn²⁺ acts as auto catalyst

Mn²⁺  formed in the reaction acts as an autocatalyst. Auto catalyst is a compound that is one of the products that are formed in the reaction itself catalyses the reaction.

Reaction: 2MnO₄^-  +  16H⁺  +  5C₂O₄^2- → 2Mn₂ +  +  10CO₂ +  8H₂O

55.  (ii) Both assertion and reason are true but reason is not the correct explanation of assertion.

Explanation: Hydrometallurgy is used to extract copper from low-grade ore. Hydrometallurgy entails dissolving the ore in a suitable reagent and then precipitating it. In this method, more electropositive metal is used, allowing pure metal to be displaced.

Given is $y=x^3$ and the ines $x=-2\&x=1$

For $y=x^3$

$\begin{array}{l}area\left(OAB\right)={\displaystyle \underset{0}{\overset{1}{\int }}ydx}\\ ={\displaystyle \underset{0}{\overset{1}{\int }}{x}^{3}dx}\\ ={\left[\frac{x^4}{4}\right]}_0^1=\frac{1}{4}\\ area\left(ODC\right)={\displaystyle \underset{-2}{\overset{0}{\int }}ydx}\\ ={\displaystyle \underset{-2}{\overset{0}{\int }}{x}^{3}dx}\\ =\left|{\left[\frac{x^4}{4}\right]}_{-2}^0\right|=\left|\left[\frac{0^4}{4}-\frac{{\left(-2\right)}^4}{4}\right]\right|=4\end{array}$

$\therefore$ Total area of the bounded region $=\frac{1}{4}+4$

$=\frac{17}{4}uni{t}^2$