Class 12th

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.

The given equation of the sides of triangle is

$y=2x+1$ --------------------(1)

$y=3x+1$ -------------------(2)

$x=4$ -------------------------(3)

Solving eqn (1) and (2) for x & y we get

$\begin{array}{l}3x+1=2x+1\\ =3x-2x=1-1\\ =x=0\\ \&y=2*0+1=1\end{array}$

$\therefore$ The point of inersection of line (1)and (2)is A (0,1)

Putting x=4 in eq (1) and (2)we get,

$\begin{array}{l}y=2*4+1=8+1=9\\ \&y=3*4+1=12+1=13\end{array}$

$\therefore$ The point of intersection of line (1)and (3) is B(4,9) and C (4,13)

Hence the required area enclosed ABC

$\begin{array}{l}={\displaystyle \underset{0}{\overset{4}{\int }}{y}_{line(2)}dx-}{\displaystyle \underset{0}{\overset{4}{\int }}{y}_{line(1)}dx}\\ ={\displaystyle \underset{0}{\overset{4}{\int }}\left[3x+1\right]}dx-{\displaystyle \underset{0}{\overset{4}{\int }}\left[2x+1\right]}dx\\ ={\left[\frac{3x^2}{2}+x\right]}_0^4-{\left[\frac{2x^2}{2}+x\right]}_0^4\\ =\left[\left(\frac{3}{2}{\left(4\right)}^2+4\right)-\left(\frac{3*0^2}{2}+0\right)\right]-\left[\left(4^2+4\right)-\left(0^2+\right)\right]\\ =24+4-20=8uni{t}^2\end{array}$

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.

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. 

Let A (-1,0),B(1,3) and C (3,2) be the vertices of a triangle ABC

So, equation of line AB is $y-0=\frac{3-0}{1-\left(-1\right)}\left(x-\left(-1\right)\right)$

$=y=\frac{3}{2}\left(x+1\right)$ -------------(1)

Equation of line BC is $y-3=\frac{2-3}{3-1}\left(x-1\right)$

$=y=\frac{-1}{2}\left(x-1\right)+3=\frac{-1}{2}x+\frac{1}{2}+3=\frac{-x}{2}+\frac{7}{2}$ ---------------(2)

Equation of line AC is $y-0=\frac{2-0}{3-\left(-1\right)}\left[x-\left(-1\right)\right]$

$=y=\frac{2}{4}\left(x+1\right)=\frac{1}{2}\left(x+1\right)$ ------------------------------(3)

$\therefore$ Area of $?$ ABC= area ( $?ABE$ ) $+area\left(BCDE\right)$ $-area\left(?ACD\right)$

$\begin{array}{l}={\displaystyle \underset{-1}{\overset{1}{\int }}{y}_{eq(1)}dx+{\displaystyle \underset{1}{\overset{3}{\int }}{y}_{eq(2)}dx-{\displaystyle \underset{-1}{\overset{3}{\int }}{y}_{eq3}dx}}}\\ ={\displaystyle \underset{-1}{\overset{1}{\int }}\frac{3}{2}\left(x+1\right)dx{\displaystyle \underset{1}{\overset{3}{\int }}\left(\frac{-x}{2}+\frac{7}{2}\right)dx-}}{\displaystyle \underset{-1}{\overset{1}{\int }}\frac{1}{2}\left(x+1\right)}\\ =\frac{3}{2}{\left[\frac{x^2}{2}+x\right]}_{-1}^1+\frac{1}{2}{\left[-\frac{x^2}{2}+7x\right]}_1^{3}dx-\frac{1}{2}{\left[\frac{x^2}{2}+x\right]}_{-1}^3\end{array}$

$\begin{array}{l}=\frac{3}{2}\left[\left(\frac{1^2}{2}+1^2\right)-\left(\frac{{\left(-1\right)}^2}{2}+\left(\pi \right)\right)\right]+\frac{1}{2}\left[\left(\frac{3^2}{2}+7*3\right)-\left(\frac{-1^2}{2}+7*1\right)\right]-\frac{1}{2}\left[\left(\frac{3^2}{2}+3\right)-\left(\frac{{\left(-1\right)}^2}{2}+\left(-1\right)\right)\right]\\ =\frac{3}{2}\left[\frac{1}{2}+1-\frac{1}{2}+1\right]+\frac{1}{2}\left[\frac{-9}{2}+21+\frac{1}{2}-7\right]-\frac{1}{2}\left[\frac{9}{2}+3-\frac{1}{2}+1\right]\\ =\frac{3}{2}\left[2\right]+\frac{1}{2}\left[10\right]-\frac{1}{2}\left[8\right]=3+5-4=8-4=4uni{t}^2\end{array}$

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.

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.

The equation of the curve is $y=x^2+2\Rightarrow x^2=y-2$ - (1) and

lines are

$y=x$ - (2)

$x=0$ - (3)

$x=3$ - (4)

Equation (1)is a parabola with vertex (0,2)

Equation (2)is a straight line passing origin with shape = $tan\theta =1=\theta =45^{?}$

$\therefore$ The required area enclosed OBCDO = area (ODCAO)-area (OBAO)

24. As an effect of lanthanide contraction, the second and third rows of transition elements resemble each other. For example, zirconium and hafnium have similar radius i.e. 160pm and 159pm respectively. Due to this similarity in their size, they show similar physical and chemical properties. 

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

43. Option (i) and (ii)

In the vapor phase refining method, the metal is converted into its volatile compound and then is collected elsewhere. This involves two techniques:

1. Mond Process for refining Nickel: in this process, a volatile complex, nickel tetracarbonyl is formed when nickel is heated with a stream of carbon monoxide.

2. Van Arkel Method for refining Zirconium or Titanium: this method is basically used for removal of oxygen and nitrogen present as impurities in the Zr or Ti metal. These are heated in an evacuated metal with iodine. As iodine is more covalent than these metals, it volatilizes.