JEE Mains 2 Apr 2026 Shift 1 (Previous Year Paper)

Given that:

( \alpha, \alpha + 2, \alpha \in \mathbf{Z} ), are the roots of the quadratic equation:

( x(x + 2) + (x + 1)(x + 3) + (x + 2) (x + 4) + \ldots + (x + n - 1) (x + n + 1) = 4n ), where ( n \in \mathbf{N} ).

Calculation:

( nx^2 + x(2 + 4 + 6 + \ldots + 2n) + (1 \cdot 3 + \ldots + (n - 1)(n + 1)) = 4n )

⇒ ( nx^2 + n(n + 1)x + \frac{n(n - 1)(2n + 5)}{6} = 4n )

⇒ ( x^2 + (n + 1)x + \frac{(n - 1)(2n + 5)}{6} = 4 )

Discriminant must be a perfect square:

( D = (n + 1)^2 - 4 \times \frac{(n - 1)(2n + 5)}{6} )

⇒ ( D = \frac{122 - 2n^2}{6} = 20 - \left( \frac{n^2 - 1}{3} \right) )

For a perfect square:

( \frac{n^2 - 1}{3} = 16 )

⇒ ( n = 7 )

Substitute ( n = 7 ) into the equation:

( x^2 + 8x + \frac{8 \times 15}{6} - 5 = 0 )

⇒ ( x^2 + 8x + 15 = 0 )

Roots: ( x = -3, -5 )

( \alpha = -5, \alpha + 2 = -3 )

( \alpha + n = 7 - 5 = 2 )

∴ The correct answer is option (3).

Given that:

(50\left(\frac{2x}{1+3i}-\frac{y}{1-2i}\right)=31+17i)

Calculation:

(50\left(\frac{2x}{1+3i}-\frac{y}{1-2i}\right)=31+17i)

⇒ (\frac{1}{1+3i}=\frac{1-3i}{10}) and (\frac{1}{1-2i}=\frac{1+2i}{5})

⇒ (10x-30xi-10y-20yi=31+17i)

⇒ Real part: (10(x-y)=31), Imaginary part: (-30x-20y=17)

⇒ (x=0.9,;y=-2.2)

⇒ (10(x-3y)=75)

∴ The correct answer is option (4).

Given that:

( x + 2y + z = 5 )

( 2x + y + \alpha z = 5 )

( 8x + 4y + \beta z = 18 )

Calculation:

( \Delta = \begin{vmatrix} 1 & 2 & 1 \ 2 & 1 & \alpha \ 8 & 4 & \beta \end{vmatrix} = 0 )

( \Delta = 1(\beta - 4\alpha) - 2(2\beta - 8\alpha) + 1(0) = 0 )

( \beta - 4\alpha - 4\beta + 16\alpha = 0 )

( -3\beta + 12\alpha = 0 )

( 4\alpha = \beta )

( \frac{\beta}{\alpha} = 4 )

∴ The correct answer is option (2).

Given that:

( A = \begin{bmatrix} 1 & 2 \ 1 & \alpha \end{bmatrix}, ; A^2 - 4A + I = O )

( B = \begin{bmatrix} 3 & 3 \ \beta & 2 \end{bmatrix}, ; B^2 - 5B - 6I = O )

Calculation:

\text{Trace of } A: ; \alpha + 1 = 4 \Rightarrow \alpha = 3

\text{Determinant of } B: ; \text{det(B)} = 6 - 3\beta = -6 \Rightarrow 3\beta = 12 \Rightarrow \beta = 4

B - A = \begin{bmatrix} 2 & 1 \ 3 & -1 \end{bmatrix}, ; B + A = \begin{bmatrix} 4 & 5 \ 5 & 5 \end{bmatrix}

(B - A)(B + A) = \begin{bmatrix} 2 & 1 \ 3 & -1 \end{bmatrix} \begin{bmatrix} 4 & 5 \ 5 & 5 \end{bmatrix} \Rightarrow \begin{bmatrix} 13 & 15 \ 7 & 10 \end{bmatrix}

\text{adj(A + B)} = \begin{bmatrix} 5 & -5 \ -5 & 4 \end{bmatrix}, ; \text{det(adj(A + B))} = (5)(4) - (-5)(-5) = 20 - 25 = -5

∴ The correct answer is option (3).

Given that:

( \text{AP}_1 \text{ (Set A)} = { 1, 6, 11, 16, \dots, 101 \text{ terms} } )

( \text{AP}_2 \text{ (Set B)} = { 9, 16, \dots, 71 \text{ terms} } )

( D = \text{L.C.M. } { d_1, d_2 } = 35 )

( 1^{\text{st}} \text{ Common term} = 16 )

Calculation:

( 16 + (n - 1) \cdot 35 \leq 499 )

⇒ ( n \leq 14.8 )

⇒ ( n = 14 )

( A \cap B = { 16, 51, 86, 121, 156, 191, 226, 261, 296, 331, 366, 401, 436, 471 } )

Divisible by 3:

( = { 51, 156, 261, 366, 471 } )

∴ The correct answer is option (2).

Given that:

( {}^5C_1 \times \frac{7!}{3!} + {}^5C_2 \times \frac{7!}{2!2!} )

Calculation:

( {}^5C_1 \times \frac{7!}{3!} = 5 \times \frac{5040}{6} = 5 \times 840 = 4200 )

( {}^5C_2 \times \frac{7!}{2!2!} = 10 \times \frac{5040}{4} = 10 \times 1260 = 12600 )

⇒ ( 4200 + 12600 = 16800 )

∴ The correct answer is option (3).

Given that:

( {}^{36}C_{r+1} = \frac{6({}^{35}C_r)}{k^2 - 3} )

Calculation:

( \frac{36}{r+1} \cdot {}^{35}C_r = 6 \cdot \frac{{}^{35}C_r}{k^2 - 3} )

⇒ ( k^2 - 3 = \frac{r+1}{6} )

⇒ ( k^2 = \frac{r+19}{6} )

( 0 \leq r \leq 35 )

⇒ ( 3.1 \leq \frac{r+19}{6} \leq 9 )

⇒ ( k^2 = 4, 9 )

⇒ ( k = \pm 2, \pm 3 )

⇒ ( (r, k) = (5, \pm 2), (35, \pm 3) )

∴ The correct answer is option (2).

Given that:

( 21 = \frac{15 + 12.5k + 490 + 1215 + 27.5 + 27.5k + 162.5}{90 + 2k} )

Calculation:

( 21(90 + 2k) = 15 + 12.5k + 490 + 1215 + 27.5 + 27.5k + 162.5 )

⇒ ( 1890 + 42k = 1910 + 40k )

⇒ ( 42k - 40k = 1910 - 1890 )

⇒ ( 2k = 20 )

⇒ ( k = 10 )

( k ) satisfies the equation ( 2x^2 - 23x - 10 = 0 ).

∴ The correct answer is option (1).

Given that:

( \text{Midpoints: } \left( \frac{5}{2}, 7 \right), \left( \frac{5}{2}, 3 \right), (4, 5) )

( \text{Incentre: } (h, k) )

Calculation:

( I = \left( \frac{24}{9}, 5 \right) )

⇒ ( h = \frac{24}{9} = \frac{8}{3}, ; k = 5 )

⇒ ( 3h + k = 3 \times \frac{8}{3} + 5 = 8 + 5 = 13 )

∴ The correct answer is option (3).

Given that:

(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1,; a < b)

((4, 3)) lies on the ellipse

(e = \frac{\sqrt{5}}{3})

Calculation:

(e^2 = 1 - \frac{a^2}{b^2} \Rightarrow \frac{a^2}{b^2} = 1 - e^2 = 1 - \frac{5}{9} = \frac{4}{9})

(\Rightarrow \frac{a^2}{b^2} = \frac{4}{9} \dots (1))

( \frac{16}{a^2} + \frac{9}{b^2} = 1 \dots (2))

From (1) and (2):

(a^2 = 20,; b^2 = 45)

(LR = \frac{2a^2}{b} = \frac{2(20)}{3\sqrt{5}} = \frac{8\sqrt{5}}{3})

∴ The correct answer is option (4).

Given that:

( K = \sin 10^\circ \sin 50^\circ \sin 70^\circ )

Calculation:

( K = \sin 10^\circ \sin 50^\circ \sin 70^\circ )

⇒ ( K = \frac{1}{4} \sin 30^\circ )

⇒ ( K = \frac{1}{8} )

( \sin \frac{10K\pi}{3} = \sin\left( 10 \times \frac{1}{8} \times \frac{\pi}{3} \right) )

⇒ ( \sin \frac{10K\pi}{3} = \sin \frac{5\pi}{12} )

⇒ ( \sin \frac{5\pi}{12} = \frac{\sqrt{3} + 1}{2\sqrt{2}} )

∴ The correct answer is option (1).

Given that:

( S = { x \in [-\pi, \pi] : \sin x (\sin x + \cos x) = a, a \in \mathbf{Z} } )

Calculation:

( \sin x (\sin x + \cos x) \in \left[ \frac{1-\sqrt{2}}{2}, \frac{1+\sqrt{2}}{2} \right] )

⇒ ( a = 0, 1 ) (since ( a \in \mathbf{Z} ))

Case 1: ( a = 0 )

( \sin x (\sin x + \cos x) = 0 )

⇒ ( \sin x = 0 ) or ( \sin x + \cos x = 0 )

( \sin x = 0 \Rightarrow x = -\pi, 0, \pi ) (3 solutions)

( \sin x + \cos x = 0 \Rightarrow \tan x = -1 \Rightarrow x = -\frac{\pi}{4}, \frac{3\pi}{4} ) (2 solutions)

Total solutions for ( a = 0 ): 3 + 2 = 5

Case 2: ( a = 1 )

( \sin x (\sin x + \cos x) = 1 )

⇒ ( \sin^2 x + \sin x \cos x = 1 )

⇒ ( 2\sin^2 x + 2\sin x \cos x = 2 )

⇒ ( 1 - \cos 2x + \sin 2x = 2 )

⇒ ( \sin 2x - \cos 2x = 1 )

Squaring both sides:

( \sin^2 2x - 2\sin 2x \cos 2x + \cos^2 2x = 1 )

⇒ ( 1 - \sin 4x = 1 )

⇒ ( \sin 4x = 0 \Rightarrow 4x = -4\pi, -3\pi, \dots, 3\pi, 4\pi )

⇒ ( x = -\pi, -\frac{3\pi}{4}, -\frac{\pi}{2}, -\frac{\pi}{4}, 0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \pi )

Total solutions for ( a = 1 ): 4

Total solutions: ( 5 + 4 = 9 )

∴ The correct answer is option (4).

Given that:

( \frac{x+1}{3} = \frac{y+a}{5} = \frac{z+b+1}{7} )

( \frac{x-2}{1} = \frac{y-b}{4} = \frac{z-2a}{7} )

Calculation:

( 3r_1 - 1 = r_2 + 2 \Rightarrow r_2 = 3r_1 - 3 )

( 7r_1 - b - 1 = 0 \Rightarrow b = 7r_1 - 1 )

( 7r_2 + 2a = 0 \Rightarrow 2a = -7r_2 \Rightarrow a = -\frac{7}{2}r_2 )

( r_2 = 3r_1 - 3 \Rightarrow a = -\frac{7}{2}(3r_1 - 3) \Rightarrow a = -\frac{21}{2}r_1 + \frac{21}{2} )

( a + b = -\frac{21}{2}r_1 + \frac{21}{2} + 7r_1 - 1 )

( a + b = \left(-\frac{21}{2} + 7\right)r_1 + \left(\frac{21}{2} - 1\right) )

( a + b = \frac{14}{2}r_1 + \frac{19}{2} \Rightarrow r_1 = 1 )

( a + b = 7 + 2 = 9 )

∴ The correct answer is option (4).

Given that:

( |\vec{a}| = 2, ; |\vec{b}| = 3 )

( E = 3\left| \left( 3\vec{a} + 2\vec{b} \right) \right| + 4\left| \left( 3\vec{a} - 2\vec{b} \right) \right| )

Calculation:

( E = 3\sqrt{9a^2 + 4b^2 + 12\vec{a}.\vec{b}} + 4\sqrt{9a^2 + 4b^2 - 12\vec{a}.\vec{b}} )

⇒ ( a = 2, ; b = 3, ; \vec{a}.\vec{b} = 6 \cos \theta )

⇒ ( E = 3\sqrt{36 + 36 + 12 \times 6 \cos \theta} + 4\sqrt{36 + 36 - 72 \cos \theta} )

⇒ ( E = 3\sqrt{72 + 72 \cos \theta} + 4\sqrt{72 - 72 \cos \theta} )

⇒ ( E = 18\sqrt{2} \sqrt{1 + \cos \theta} + 24\sqrt{2} \sqrt{1 - \cos \theta} )

⇒ ( E = 36 \cos \frac{\theta}{2} + 48 \sin \frac{\theta}{2} )

⇒ ( E_{\text{max}} = \sqrt{(36)^2 + (48)^2} = \sqrt{1296 + 2304} = \sqrt{3600} = 60 )

∴ The correct answer is option (3).

Given that:

( \vec{L} ) passes through point (1, 1, 1) and is perpendicular to ( 2\hat{i} + 2\hat{j} + \hat{k} ) and ( \hat{i} + 2\hat{j} + \hat{k} ).

Calculation:

( \text{Direction vector of } \vec{L} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 2 \ 2 & 2 & 1 \end{vmatrix} = -2\hat{i} + 3\hat{j} - 2\hat{k} )

Equation of line: ( \frac{x-1}{-2} = \frac{y-1}{3} = \frac{z-1}{-2} = \lambda )

Let ( P(a, b, c) \equiv (2\lambda + 1, -3\lambda + 1, 2\lambda + 1) ) be the foot of perpendicular from (0, 0, 0).

( \vec{OM} \cdot (-2\hat{i} + 3\hat{j} - 2\hat{k}) = 0 )

⇒ ( -2(2\lambda + 1) + 3(-3\lambda + 1) - 2(2\lambda + 1) = 0 )

⇒ ( \lambda = -\frac{1}{17} )

Substitute ( \lambda = -\frac{1}{17} ):

( a = \frac{-2}{17} + 1, ; b = \frac{3}{17} + 1, ; c = \frac{-2}{17} + 1 )

⇒ ( 34(a+b+c) = 100 )

∴ The correct answer is option (3).

Given that:

( \lim_{x \to 2} \frac{\sin(x^3 - 5x^2 + ax + b)}{(\sqrt{x-1} - 1)\log_e(x-1)} = m )

Calculation:

( \because ) Denominator = 0 at (x = 2)

( \Rightarrow ) Numerator = 0 at (x = 2)

( \Rightarrow 2^3 - 5(2)^2 + a(2) + b = 0 )

( \Rightarrow 2a + b = 12 ) ... (1)

( m = \lim_{x \to 2} \frac{\frac{\sin(x^3 - 5x^2 + ax + b)}{(x^3 - 5x^2 + ax + b)} \cdot (x^3 - 5x^2 + ax + b)}{\frac{\sqrt{x-1} - 1}{\sqrt{x-1} + 1} \cdot \frac{\log_e(1 + (x-2))}{(x-2)} \cdot (x-2)} )

( \Rightarrow m = \lim_{x \to 2} 2 \frac{x^3 - 5x^2 + ax + b}{(x-2)^2} )

( \Rightarrow m = \lim_{x \to 2} 2 \frac{3x^2 - 10x + a}{2(x-2)} )

( \because ) Denominator = 0 at (x = 2)

( \Rightarrow ) Numerator = 0 at (x = 2)

( \Rightarrow 3(2)^2 - 10(2) + a = 0 )

( \Rightarrow a = 8 )

Substituting (a = 8) in (1):

( 2(8) + b = 12 )

( \Rightarrow b = -4 )

( \therefore m = \lim_{x \to 2} 2 \left( \frac{6x - 10}{2} \right) )

( \Rightarrow m = 2 )

( \therefore a + b + m = 8 - 4 + 2 = 6 )

∴ The correct answer is option (2).

Given that:

( \frac{dy}{dx} = y(2 + \ln x) )

Calculation:

( \int \frac{1}{y} , dy = \int (2 + \ln x) , dx )

⇒ ( \ln y = 2x + x \ln x - x + C )

⇒ ( \ln y = x + x \ln x + C )

Given point (1, e): ( \ln e = 1 + 1 \ln 1 + C )

⇒ ( 1 = 1 + 0 + C \Rightarrow C = 0 )

⇒ ( \ln y = x + x \ln x )

⇒ ( f(x) = y = e^{x + x \ln x} )

⇒ ( f(e) = e^{e + e} = e^{2e} )

∴ The correct answer is option (3).

Given that:

( f(x) = \begin{cases} \left| \frac{\sin x}{x} \right| & , \quad x \neq 0 \ 1 & , \quad x = 0 \end{cases} ), ( x \in (-2\pi, 2\pi) )

Calculation:

( \lim_{x \to 0} \left| \frac{\sin x}{x} \right| = 1 = f(0) ), hence ( f(x) ) is continuous.

( f'(x) = \frac{d}{dx} \left| \frac{\sin x}{x} \right| = \frac{x \cos x - \sin x}{x^2} )

( f'(x) = 0 \Rightarrow \tan x = x )

From the graph, there are 3 solutions to ( \tan x = x ) in ( (-2\pi, 2\pi) ).

Additionally, ( f'(x) ) does not exist at ( x = -\pi, \pi ).

Total critical points = 3 (from ( \tan x = x )) + 2 (where ( f'(x) ) does not exist) = 5.

∴ The correct answer is option (3).

Given that:

(\int_{0}^{3}\dfrac{e^x+e^{-x}}{[x]!},dx)

Calculation:

(\int_{0}^{3}\dfrac{e^x+e^{-x}}{[x]!},dx)

⇒ (\int_{0}^{1}(e^x+e^{-x}),dx+\int_{1}^{2}(e^x+e^{-x}),dx+\dfrac{1}{2}\int_{2}^{3}(e^x+e^{-x}),dx)

⇒ (\left[e^x-e^{-x}\right]{0}^{1}+\left[e^x-e^{-x}\right]{1}^{2}+\dfrac{1}{2}\left[e^x-e^{-x}\right]_{2}^{3})

⇒ (\left(e-\dfrac{1}{e}\right)+\left(e^2-e-\dfrac{1}{e^2}+\dfrac{1}{e}\right)+\dfrac{1}{2}\left(e^3-e^2-\dfrac{1}{e^3}+\dfrac{1}{e^2}\right))

⇒ (\dfrac{1}{2}\left(e^2+e^3-\dfrac{1}{e^2}-\dfrac{1}{e^3}\right))

∴ The correct answer is option (2).

Given that:

( (1 + \sin x) \frac{dy}{dx} + (y + 1) \cos x = 0, ; y(0) = 0 )

Calculation:

( \int \frac{dy}{y+1} = -\int \frac{\cos x}{1+\sin x} dx )

⇒ ( \ln(y+1) = -\ln(1+\sin x) + C )

⇒ ( y+1 = \frac{e^{C}}{1+\sin x} )

Using ( y(0) = 0 ):

⇒ ( 1+1 = \frac{e^{C}}{1+\sin 0} )

⇒ ( C = 0 )

⇒ ( y+1 = \frac{1}{1+\sin x} )

Given ( y = -\frac{1}{2}, ; x = \alpha ):

⇒ ( -\frac{1}{2} + 1 = \frac{1}{1+\sin \alpha} )

⇒ ( \frac{1}{2} = \frac{1}{1+\sin \alpha} )

⇒ ( 1+\sin \alpha = 2 )

⇒ ( \sin \alpha = 1 )

⇒ ( \alpha = \frac{\pi}{2} )

∴ The correct answer is option (4).

Given that:

( \log_{0.6} \left| \frac{2x - 5}{x^2 - 4} \right| \geq 0 )

( \left| \frac{2x - 5}{x^2 - 4} \right| \leq 1 )

( x \neq \frac{5}{2} )

( -1 \leq \frac{2x - 5}{x^2 - 4} \leq 1 )

( \frac{(x+1)^2 - 10}{(x-2)(x+2)} \geq 0 )

( x \in (-\infty, -1-\sqrt{10}] \cup (-2, 2) \cup [-1+\sqrt{10}, \infty) )

( \frac{(x-1)^2}{(x-2)(x+2)} \geq 0 )

( x \in (-\infty, -2) \cup (2, \infty) \cup {1} )

Calculation:

( (1) \cap (2) \cap (3) )

( x \in (-\infty, -1-\sqrt{10}] \cup {1} \cup \left[ -1+\sqrt{10}, \frac{5}{2} \right) \cup \left( \frac{5}{2}, \infty \right) )

( a + b + c + d + e = -2 + 1 + 5 = 4 )

Given that:

( a_1 + a_2 + \ldots + a_n = 6n^3 )

( a_1 + a_2 + \ldots + a_n + a_{n+1} = 6(n+1)^3 )

Calculation:

( a_{n+1} = 6(n+1)^3 - 6n^3 )

⇒ ( a_{n+1} = 6((n+1)-n)((n+1)^2 + n^2 + n(n+1)) )

⇒ ( a_{n+1} = 6(1)(3n^2 + 3n + 1) )

⇒ ( a_n = 6(3(n-1)^2 + 3(n-1) + 1) )

⇒ ( a_n = 6(3n^2 - 3n + 1) )

⇒ ( \sum_{k=1}^6 \left( \frac{6(3k^2 + 3k + 1) - 6(3k^2 - 3k + 1)}{36} \right)^2 )

⇒ ( \sum_{k=1}^6 \left( \frac{36k}{36} \right)^2 = \sum_{k=1}^6 k^2 = \frac{6 × 7 × 13}{6} = 91 )

Given that:

(8b^2 - 4ac < 0 \Rightarrow 2b^2 < ac)

Calculation:

Total combinations of (b, a, c):

(1 \times 4 + 1 \times 3 + 1 \times 4 + 1 \times 4 + 1 \times 2 + 1 \times 2 = 19)

Valid combinations satisfying (2b^2 < ac):

(2 + 3 + 4 + 4 + 2 + 2 = 17)

Required probability:

(\frac{17}{64} = \frac{m}{n}), where (m = 17, n = 64)

(m + n = 17 + 64 = 81)

Given that:

( r^2 = \left( \frac{\sqrt{14}}{2} \right)^2 + \left( r - \frac{1}{\sqrt{2}} \right)^2 )

Calculation:

( r^2 = \frac{7}{2} + \left( r - \frac{1}{\sqrt{2}} \right)^2 )

⇒ ( r^2 = \frac{7}{2} + r^2 - \sqrt{2}r + \frac{1}{2} )

⇒ ( \sqrt{2}r = 4 )

⇒ ( r = 2\sqrt{2} )

⇒ ( r^2 = 8 )

Given that:

( \int_a^b f(x) dx + \int_{f(a)}^{f(b)} f^{-1}(x) dx = bf(b) - af(a) )

( \alpha = (2\sqrt{3})(4) - (0)(2) )

Calculation:

( \alpha = 8\sqrt{3} )

( \alpha^2 = 192 )

CONCEPT:

Stoichiometry of the Reaction

• Stoichiometry refers to the quantitative relationship between reactants and products in a chemical reaction.
• In the given reaction:

3Fe(s) + 4H₂O(g) → Fe₃O₄(s) + 4H₂(g)

• From the balanced chemical equation:
• 3 moles of Fe react with 4 moles of H₂O.
• The molar ratio of Fe to H₂O is 3:4.
• Molar masses:
• Molar mass of Fe = 56 g/mol
• Molar mass of H₂O = 18 g/mol

EXPLANATION:

• Given mass of steam (H₂O) = 18 g
• Number of moles of steam:

n_steam = ⅛18 = 1 mol

• From the stoichiometric ratio, moles of Fe required:

n_Fe = ¾

• Mass of Fe required:

Mass of Fe = n_Fe × Molar mass of Fe

= ¾ × 56

= 42 g

Therefore, the mass of iron converted into Fe₃O₄ by the action of 18 g of steam is 42 g.

CONCEPT:

Ionization Energy and Its Dependence on Nuclear Charge (Z)

• The ionization energy is the energy required to remove an electron from an atom or ion in its gaseous state.
• For hydrogen-like ions (single electron systems like ( \text{Li}^{2+} )), the ionization energy can be calculated using the formula:

( \text{E} = \text{E}_{\text{H}} \times \text{Z}^2 )

• Here:
• ( \text{E}_{\text{H}} ) is the ionization energy of a hydrogen atom in its ground state.
• ( \text{Z} ) is the atomic number of the element.

EXPLANATION:

• The given reaction is:

( \text{Li}^{2+}\text{(g)} \rightarrow \text{Li}^{3+}\text{(g)} + \text{e}^- )

• Here, ( \text{Li}^{2+} ) is a hydrogen-like ion with a single electron in its orbit. Its ionization energy can be calculated using the formula:

( \text{E} = \text{E}_{\text{H}} \times \text{Z}^2 )

• The ionization energy of hydrogen, ( \text{E}_{\text{H}} = 2.18 \times 10^{-18} \text{ J atom}^{-1} ).
• The atomic number of lithium, ( \text{Z} = 3 ).
• Substitute the values into the formula:

( \text{E}_{\text{Li}^{2+}} = 2.18 \times 10^{-18} \times (3)^2 )

( \text{E}_{\text{Li}^{2+}} = 2.18 \times 10^{-18} \times 9 )

( \text{E}_{\text{Li}^{2+}} = 1.962 \times 10^{-17} \text{ J atom}^{-1} )

Therefore, the energy required for the process is ( 1.962 \times 10^{-17} \text{ J atom}^{-1} ).

CONCEPT:

Bond Order and Unpaired Electrons in Molecular Orbital Theory

• Bond Order:

Bond Order = ^{(Number of bonding electrons - Number of antibonding electrons)}⁄₂

• The bond order is a measure of the number of chemical bonds between a pair of atoms. It is given by the formula:
• A higher bond order corresponds to a shorter and stronger bond, while a lower bond order corresponds to a longer and weaker bond.
• Unpaired Electrons:
• The number of unpaired electrons in a molecule is determined based on the electronic configuration of its molecular orbitals.
• Molecules with unpaired electrons are paramagnetic, while those without unpaired electrons are diamagnetic.

EXPLANATION:

• The electronic configurations of the molecular orbitals for the given oxygen species are as follows:
• O+2O2+

: Bond Order = 2.5, Unpaired Electrons = 1

• O2O2

: Bond Order = 2, Unpaired Electrons = 2

• O−2O2−

: Bond Order = 1.5, Unpaired Electrons = 1

• O2−2O22−

: Bond Order = 1, Unpaired Electrons = 0

• Statement I: The bond length is inversely proportional to the bond order. Therefore, the correct sequence of bond lengths is:

O+2<O2<O−2<O2−2O2+<O2<O2−<O22−

This makes Statement I true.

• Statement II: The number of unpaired electrons decreases in the following order:

O2>O+2>O−2>O2−2O2>O2+>O2−>O22−

This makes Statement II true.

Conclusion: Both Statement I and Statement II are true.

CONCEPT:

Enthalpy of Formation (( \Delta_{\text{f}}H ))

• The enthalpy of formation (( \Delta_{\text{f}}H )) of a compound is the change in enthalpy when 1 mole of a compound is formed from its constituent elements in their standard states under standard conditions.
• To calculate the enthalpy of formation for a reaction, the enthalpy changes (( \Delta_{\text{r}}H )) from individual steps involved in the reaction can be summed up using Hess's Law.
• Hess's Law states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps that lead to the reaction.

EXPLANATION:

• Given data:
• Step (i): ( 2\text{Al(s)} + 6\text{HCl(aq)} \rightarrow \text{Al}_2\text{Cl}_6\text{(aq)} + 3\text{H}2\text{(g)} ), ( \Delta{\text{r}}H_1 = -1200 , \text{kJ/mol} )
• Step (ii): ( \text{H}_2\text{(g)} + \text{Cl}2\text{(g)} \rightarrow 2\text{HCl(g)} ), ( \Delta{\text{r}}H_2 = -164 , \text{kJ/mol} ) (multiplied by 3: ( -164 \times 3 , \text{kJ/mol} ))
• Step (iii): ( \text{HCl(g)} + \text{aq} \rightarrow \text{HCl(aq)} ), ( \Delta_{\text{r}}H_3 = -83 , \text{kJ/mol} ) (multiplied by 6: ( -83 \times 6 , \text{kJ/mol} ))
• Step (iv): ( \text{Al}_2\text{Cl}_6\text{(aq)} \rightarrow \text{Al}_2\text{Cl}6\text{(s)} + \text{aq} ), ( \Delta{\text{r}}H_4 = +663 , \text{kJ/mol} )
• Target reaction: ( 2\text{Al(s)} + 3\text{Cl}_2 \rightarrow \text{Al}_2\text{Cl}_6\text{(s)} )
• Using Hess's Law:
• ( \Delta_{\text{f}}H = \Delta_{\text{r}}H_1 + \Delta_{\text{r}}H_2 + \Delta_{\text{r}}H_3 + \Delta_{\text{r}}H_4 )
• ( \Delta_{\text{f}}H = -1200 - (164 \times 3) - (83 \times 6) + 663 )
• ( \Delta_{\text{f}}H = -1200 - 492 - 498 + 663 )
• ( \Delta_{\text{f}}H = -1527 , \text{kJ/mol} )

Therefore, the enthalpy of formation (( \Delta_{\text{f}}H )) of anhydrous solid ( \text{Al}_2\text{Cl}_6 ) is ( -1527 , \text{kJ/mol} ).

CONCEPT:

Freezing Point Depression and Degree of Dissociation

• The depression in the freezing point of a solution is given by the equation:

ΔT_f = iK_fm

where:

• ΔT_f = depression in freezing point
• i = van't Hoff factor
• K_f = molal freezing point depression constant
• m = molality of the solution
• The van't Hoff factor, i, is related to the degree of dissociation (α) by the formula:

i = 1 + (n - 1)α

where:

• n = number of ions produced per molecule upon dissociation
• α = degree of dissociation
• The dissociation constant (K_a) is given by the formula:

K_a = (Cα²) / (1 - α)

where:

• C = concentration of the solute
• α = degree of dissociation

EXPLANATION:

• Given data:
• Mass of fluoroacetic acid = 19.5 g
• Molar mass of fluoroacetic acid = 78 g mol^-1
• Mass of water = 500 g = 0.5 kg
• ΔT_f = 1°C
• K_f = 1.86 K kg mol^-1
• Calculate the number of moles of fluoroacetic acid:

= ^19.5 / ₇₈ = 0.25 mol

• Calculate the molality (m):

m = ^{Number of moles of solute} / _{Mass of solvent in kg}

= ^0.25 / _0.5 = 0.5 mol kg^-1

• ΔT_f = iK_fm

1 = i × 1.86 × 0.5

i = ¹ / _{1.86 × 0.5} = ² / _1.86

• i = 1 + α (since fluoroacetic acid dissociates into 2 particles: H⁺ and CH₂FCOO^-)

² / _1.86 = 1 + α

α = ² / _1.86 - 1 = 0.075

• K_a = ^Cα² / _{1 - α}

Here, C = molality = 0.5 mol kg^-1

K_a = ^{0.5 × (0.075)²} / _{1 - 0.075}

K_a = ^{0.5 × 0.005625} / _0.925 = ^0.0028125 / _0.925

K_a ≈ 3 × 10^-3

Therefore, the dissociation constant (K_a) of fluoroacetic acid is approximately 3 × 10^-3.

CONCEPT:

Solubility Product and Molarity Relation

• The solubility product constant (( K_{\text{sp}} )) represents the equilibrium constant for the dissolution of a sparingly soluble salt.
• The molarity of ions in solution can be derived from the solubility product constant using the stoichiometric relationship between the ions and the salt.

EXPLANATION:

• For ( \text{Ag}_2\text{CrO}_4 ):
• The dissociation equation is:

( \text{Ag}_2\text{CrO}_4 \rightleftharpoons 2\text{Ag}^+ + \text{CrO}_4^{2-} )

• The solubility product is given by:

( K_{\text{sp}} = [\text{Ag}^+]^2[\text{CrO}_4^{2-}] )

• If the molarity of ( \text{Ag}_2\text{CrO}_4 ) is ( S_1 ), then:
• ( [\text{Ag}^+] = 2S_1 )
• ( [\text{CrO}_4^{2-}] = S_1 )
• Substitute these values into the solubility product equation:

( K_{\text{sp}} = (2S_1)^2(S_1) )

( K_{\text{sp}} = 4S_1^3 )

• Given ( K_{\text{sp}} = 32x ), solve for ( S_1 ):

( 4S_1^3 = 32x )

( S_1 = 2x^{1/3} )

• For ( \text{AgBr} ):
• The dissociation equation is:

( \text{AgBr} \rightleftharpoons \text{Ag}^+ + \text{Br}^- )

• The solubility product is given by:

( K_{\text{sp}} = [\text{Ag}^+][\text{Br}^-] )

• If the molarity of ( \text{AgBr} ) is ( S_2 ), then:
• ( [\text{Ag}^+] = S_2 )
• ( [\text{Br}^-] = S_2 )
• Substitute these values into the solubility product equation:

( K_{\text{sp}} = S_2^2 )

• Given ( K_{\text{sp}} = 4y ), solve for ( S_2 ):

( S_2^2 = 4y )

( S_2 = 2y^{1/2} )

• Calculate the ratio ( \left( \frac{\text{S}_1}{\text{S}_2} \right) ):

( \frac{\text{S}_1}{\text{S}_2} = \frac{2x^{1/3}}{2y^{1/2}} )

( \frac{\text{S}_1}{\text{S}_2} = \frac{\sqrt[3]{x}}{\sqrt{y}} )

Therefore, the ratio ( \left( \frac{\text{molarity of Ag}_2\text{CrO}_4}{\text{molarity of AgBr}} \right) ) is ( \frac{\sqrt[3]{x}}{\sqrt{y}} ).

CONCEPT:

Standard Cell Potential (E⁰_cell)

• The standard cell potential (E⁰_cell) is the difference in the standard reduction potentials of the cathode and the anode in an electrochemical cell.
• The standard cell potential can be calculated using the equation:

E⁰_cell = E⁰_cathode - E⁰_anode

EXPLANATION:

• In the given reaction:

Fe(s) + 2OH^-(aq) + 2AgBr(s) → Fe(OH)₂(s) + 2Ag(s) + 2Br^-(aq)

• Iron (Fe) is oxidized to Fe(OH)₂, so Fe is the anode.
• Silver bromide (AgBr) is reduced to Ag, so AgBr is the cathode.
• Given standard reduction potentials:
• E⁰_AgBr = +0.07 V (for AgBr(s) + e^- → Ag(s) + Br^-(aq))
• E⁰_Fe = -0.88 V (for Fe(OH)₂(s) + 2e^- → Fe(s) + 2OH^-(aq))
• Using the standard cell potential equation:
• E⁰_cell = E⁰_cathode - E⁰_anode
• = 0.07 V - (-0.88 V)
• = 0.07 V + 0.88 V
• = 0.95 V
• Key points to note:
• Fe is oxidized in the reaction, not reduced.
• E⁰_cell is not an extensive property; it is an intensive property, as it does not depend on the amount of material.

Therefore, the correct overall reaction is:

Fe(s) + 2OH^-(aq) + 2AgBr(s) → Fe(OH)₂(s) + 2Ag(s) + 2Br^-(aq)

The standard cell potential (E⁰_cell) is 0.95 V.

CONCEPT:

Zero-order and First-order Reaction Time Relationships

• For a zero-order reaction, the rate of reaction is constant and independent of the concentration of the reactant.
• For a first-order reaction, the rate of reaction is proportional to the concentration of the reactant.
• The times required for 50% completion (( \text{t}{1/2} )) and 100% completion (( \text{t}{100%} )) of the reaction differ for zero-order and first-order reactions.

EXPLANATION:

• Zero-order Reaction:
• For a zero-order reaction, the time required for the reaction to complete 100% is given by:

( \text{t}_{100%} = \frac{\text{a}}{\text{k}} )

• The time required for 50% completion is:

( \text{t}_{1/2} = \frac{\text{a}}{2\text{k}} )

• By dividing ( \text{t}{100%} ) by ( \text{t}{1/2} ), we get:

( \text{t}{100%} = 2\text{t}{1/2} )

• First-order Reaction:
• For a first-order reaction, the time required for 100% completion is theoretically infinite, because the reactant concentration never truly reaches zero.
• The time required for 50% completion is:

( \text{t}_{1/2} = \frac{\ell\text{n } 2}{\text{k}} )

• Since ( \text{t}_{100%} ) is infinite, we represent the relationship as:

( \text{t}{100%} = (\text{t}{1/2})^\infty )

So, For a zero-order reaction, ( \text{t}{100%} = 2\text{t}{1/2} ).

• For a first-order reaction, ( \text{t}{100%} = (\text{t}{1/2})^\infty ).

Therefore, the correct relationship is:

( \text{t}{100%} = 2\text{t}{1/2} ) for zero-order reactions and ( \text{t}{100%} = (\text{t}{1/2})^\infty ) for first-order reactions.

CONCEPT:

Ionization Enthalpy

• The ionization enthalpy is the energy required to remove an electron from the outermost shell of an isolated gaseous atom.
• The first ionization enthalpy corresponds to the removal of the first electron, the second ionization enthalpy corresponds to the removal of the second electron, and so on.
• Ionization enthalpy is influenced by factors such as atomic size, nuclear charge, shielding effect, and electron configuration.

EXPLANATION:

• Statement (I): The first ionization enthalpy of the elements Na, Mg, Cl, and Ar follows the order Na > Mg > Cl > Ar.
• This statement is false because the correct order of first ionization enthalpy is Na < Mg < Cl < Ar.
• The trend in ionization energy across a period increases from left to right due to increasing nuclear charge and decreasing atomic size.
• Na has the lowest ionization enthalpy, followed by Mg, Cl, and Ar, which has the highest due to its stable noble gas configuration.
• Statement (II): Among Ca, Al, Fe, and B, the third ionization enthalpy is very high for Ca.
• This statement is true.
• Calcium (Ca) has a very high third ionization enthalpy because after losing two electrons, it attains a stable noble gas configuration (similar to Argon: [Ar]).
• Removing a third electron from this stable configuration requires a significantly higher amount of energy.

So, Statement I is false but Statement II is true.

CONCEPT:

Oxidizing Power of Halogens

• Oxidizing power refers to the ability of a substance to oxidize another substance, i.e., to accept electrons.
• Among the halogens, the oxidizing power decreases down the group in the periodic table in the following order: F₂ > Cl₂ > Br₂ > I₂.
• This trend is due to the decreasing standard reduction potentials of the halogens as we move down the group, making them less effective oxidizing agents.

Layer Test

• The "Layer Test" is a chemical test used to identify halogens in aqueous solutions based on their relative oxidizing powers.
• In this test, a more powerful oxidizing agent (e.g., Cl₂) is used to oxidize halide ions (e.g., Br^- or I^-) to their molecular forms (Br₂ or I₂), which can then be detected by their characteristic colors.
• For example:
• Cl₂ oxidizes bromide ions (Br^-) to bromine (Br₂), which appears as an orange-brown color in aqueous solution:

NaBr + Cl₂-water → Br₂ + NaCl

• Cl₂ also oxidizes iodide ions (I^-) to iodine (I₂), which appears as a violet color in organic solvents or a brown color in aqueous solution:

NaI + Cl₂-water → I₂ + NaCl

EXPLANATION:

• Statement (I): "Oxidizing power of halogens decreases in the order F₂ > Cl₂ > Br₂ > I₂, which is the basis of 'Layer test'" is true. This is because fluorine has the highest oxidizing power, followed by chlorine, bromine, and iodine, consistent with their standard reduction potentials.
• Statement (II): "'Layer test' to identify Br₂ and I₂ in aqueous solution involves the oxidation of bromide or iodide into Br₂ or I₂ respectively with Cl₂, which is a type of displacement redox reaction" is also true. This statement correctly describes the chemical basis of the layer test, where Cl₂, being a stronger oxidizing agent, displaces Br₂ and I₂ from their respective halide ions.
• Since both statements are correct and Statement (II) explains the process mentioned in Statement (I), the correct interpretation is that both Statement I and Statement II are true.

Therefore, the answer is that both Statement I and Statement II are true.

CONCEPT:

Oxidation-Reduction Reactions and Changes in Color

• In an acidic medium, ( \text{K}_2\text{Cr}_2\text{O}_7 ) (potassium dichromate) undergoes a reduction reaction. The dichromate ion ( \text{Cr}_2\text{O}_7^{2-} ) is reduced to ( \text{Cr}^{3+} ), causing a color change from orange to green.
• This color change occurs when the dichromate ion reacts with a reducing agent, which donates electrons to reduce ( \text{Cr}_2\text{O}_7^{2-} ).
• Common reducing agents include species that can themselves undergo oxidation, such as ( \text{Fe}^{2+}, \text{Sn}^{2+}, \text{I}^-, \text{S}^{2-} ), etc.

EXPLANATION:

• To determine which species will cause the color change of ( \text{K}_2\text{Cr}_2\text{O}_7 ) in acidic medium, we need to identify species that act as reducing agents and undergo oxidation.
• For the species in the given options:
• ( \text{Fe}^{2+} ): Oxidized to ( \text{Fe}^{3+} ), donating electrons and reducing ( \text{Cr}_2\text{O}_7^{2-} ). Thus, it causes the color change.
• ( \text{Sn}^{2+} ): Oxidized to ( \text{Sn}^{4+} ), donating electrons and reducing ( \text{Cr}_2\text{O}_7^{2-} ). Thus, it causes the color change.
• ( \text{I}^- ): Oxidized to ( \text{I}_2 ), donating electrons and reducing ( \text{Cr}_2\text{O}_7^{2-} ). Thus, it causes the color change.
• ( \text{S}^{2-} ): Oxidized to sulfur (( \text{S} )), donating electrons and reducing ( \text{Cr}_2\text{O}_7^{2-} ). Thus, it causes the color change.
• Other species, such as ( \text{Fe}^{3+} ), ( \text{Sn}^{4+} ), or ( \text{SO}_4^{2-} ), cannot act as reducing agents in this context, as they are already in their oxidized states.

Conclusion: The species ( \text{Fe}^{2+}, \text{Sn}^{2+}, \text{I}^-, \text{S}^{2-} ) will change the orange color of ( \text{K}_2\text{Cr}_2\text{O}_7 ) to green in acidic medium, as they are all capable of acting as reducing agents.

CONCEPT:

Crystal Field Splitting Energy (CFSE) and Ligand Strength

• The crystal field splitting energy (( \Delta_0 )) is directly proportional to the strength of the ligand in the spectrochemical series.
• Stronger ligands cause greater splitting of the d-orbitals, resulting in higher ( \Delta_0 ) values.
• The spectrochemical series for the given ligands is as follows:

CN^- > en (ethylene diamine) > H₂O > F^-

Thus, CN^- is the strongest ligand, and F^- is the weakest ligand.

• Additionally, chelation (e.g., by ethylene diamine) increases the ligand strength compared to monodentate ligands of similar type.

EXPLANATION:

• The given complexes are:
• A. ( [\text{Cr(CN)}_6]^{3-} )
• B. ( [\text{CrF}_6]^{3-} )
• C. ( [\text{Cr(H}_2\text{O)}_6]^{3+} )
• D. ( [\text{Cr(en)}_3]^{3+} )
• The crystal field splitting energy (( \Delta_0 )) values are:
• I. 15,060 cm^-1
• II. 17,400 cm^-1
• III. 22,300 cm^-1
• IV. 26,600 cm^-1
• Based on the ligand strength:
• ( [\text{Cr(CN)}_6]^{3-} ) (CN^-, strongest ligand) should have the highest splitting energy, i.e., ( \Delta_0 ) = 26,600 , \text{cm}^{-1} ) (IV).
• ( [\text{Cr(en)}_3]^{3+} ) (chelating ligand, moderate strength) should have the second-highest splitting energy, i.e., ( \Delta_0 ) = 22,300 , \text{cm}^{-1} ) (III).
• ( [\text{Cr(H}_2\text{O)}_6]^{3+} ) (H₂O, weaker ligand) should have the second-lowest splitting energy, i.e., ( \Delta_0 ) = 17,400 , \text{cm}^{-1} ) (II).
• ( [\text{CrF}_6]^{3-} ) (F^-, weakest ligand) should have the lowest splitting energy, i.e., ( \Delta_0 ) = 15,060 , \text{cm}^{-1} ) (I).

MATCHING:

FINAL ANSWER:

A-IV, B-I, C-II, D-III

CONCEPT:

Distillation and Separation of Mixtures

• Distillation is a separation technique based on the difference in boiling points of substances.
• Simple distillation is effective when the boiling points of the components differ significantly, while fractional distillation is used for closer boiling points.
• An azeotropic mixture is a mixture of two or more liquids that boils at a constant temperature and retains its composition in the vapor phase, making it impossible to separate by conventional fractional distillation.

EXPLANATION:

• Statement (I): 1,2,3-Trihydroxypropane (commonly known as glycerol) cannot be separated from water by simple distillation because glycerol has a very high boiling point (approximately 290°C) compared to water (100°C). Moreover, glycerol decomposes at high temperatures, so it is typically separated by distillation under reduced pressure to avoid decomposition.
• Statement (II): An azeotropic mixture cannot be separated by fractional distillation because the composition of the liquid and vapor phases remains constant during boiling, preventing further separation. This statement is true.
• Based on the above analysis:
• Statement (I) is false because glycerol cannot be separated from water by simple distillation.
• Statement (II) is true because azeotropic mixtures cannot be separated by fractional distillation.

So, Statement I is false but Statement II is true.

CONCEPT:

Rate of S_N1 Reaction

• The rate of an S_N1 reaction is directly proportional to the stability of the intermediate carbocation.
• Carbocation stability is influenced by factors like resonance, hyperconjugation, and inductive effects.
• Order of carbocation stability:

Ph-⁺C_H2 > CH₃-⁺C_H2

• Resonance stabilization is more effective than hyperconjugation in stabilizing carbocations.

EXPLANATION:

• Statement (I): Benzyl chloride reacts faster in an S_N1 mechanism than ethyl chloride.
• This is true because the benzyl carbocation (Ph-⁺C_H2) formed during the reaction is stabilized by resonance with the aromatic ring.
• In contrast, the ethyl carbocation (CH₃-⁺C_H2) is stabilized only by hyperconjugation, which is less effective compared to resonance.
• Statement (II): Ethyl carbocation intermediate is less stabilized by hyperconjugation than benzyl carbocation by resonance.
• This is also true because the benzyl carbocation has multiple resonance structures, allowing delocalization of the positive charge over the aromatic ring, which provides significant stabilization.
• On the other hand, the ethyl carbocation relies only on hyperconjugation, which provides limited stabilization compared to resonance.
• Since both statements are true and align with the principles of carbocation stability, the correct answer is that both Statement (I) and Statement (II) are true.

Therefore, the stability of the benzyl carbocation due to resonance explains the faster reaction rate of benzyl chloride in the S_N1 mechanism compared to ethyl chloride.

CONCEPT:

IUPAC Nomenclature Priority of Functional Groups

• In IUPAC nomenclature, the priority of functional groups is based on the specific rules established by IUPAC. The priority order is used to determine the principal functional group, which dictates the suffix of the compound name.
• The priority order is determined based on the oxidation state of the carbon atom to which the functional group is attached. Functional groups with higher oxidation states have higher priority.
• In general, the priority order from highest to lowest is:
• Carboxylic acids (-COOH)
• Esters (-COOR)
• Amides (-CONH₂)
• Aldehydes (-CHO)
• Ketones (>C=O)
• Nitriles (-CN)
• Alcohols (-OH)
• Amines (-NH₂)
• Alkenes (C=C)
• Alkynes (C≡C)

EXPLANATION:

• Memonics: "Crazy Elephants Always Ask Kind Neighbors About Amazing Airplanes Always."
• Functional groups provided in the question:
• -CONH₂ (Amide)
• -CHO (Aldehyde)
• C=O (Ketone)

• -NH₂ (Amine)
• -C≡C- (Alkyne)
• Based on the priority order of functional groups:
• Amides (-CONH₂) have the highest priority among the groups provided.
• Aldehydes (-CHO) have the next highest priority.
• Ketones (>C=O) follow aldehydes in priority.
• Amines (-NH₂) come after ketones.
• Alkynes (-C≡C-) have the lowest priority among the given groups.
• Thus, the correct decreasing order of priority is:

-CONH₂, -CHO, >C=O, -NH₂, -C≡C-

Therefore, the correct answer is Option 3.

CONCEPT:

Electrophilic Attack on Aromatic Compounds

• Electrophilic substitution reactions occur when an electrophile replaces a substituent (usually a hydrogen atom) in an aromatic ring.
• The position at which the electrophile attacks depends on the electron density of the aromatic ring and the nature of the substituents already present.
• Electron-donating groups (+M or +I effect) increase electron density at certain positions, making them more favorable for electrophilic attack.
• Electron-withdrawing groups (-M or -I effect) decrease electron density at specific positions, making them less favorable for electrophilic attack.

EXPLANATION:

• In the given molecule, the nitrogen atom is attached to the aromatic ring and exhibits a +M (resonance donating) effect.
• The +M effect of nitrogen increases the electron density at the ortho and para positions of the aromatic ring.
• Among the labeled positions ("r", "u", "p", and "s"), position "u" is predominantly favored for electrophilic attack because it is para to the nitrogen atom, and the +M effect enhances the electron density at this position.
• Position "r" also experiences an increase in electron density due to the +M effect, but it is less favorable than "u" because steric hindrance near the nitrogen may play a minor role in reducing reactivity at the ortho position.

Therefore, the preferred site for the attack of the electrophile is predominantly at "u" because of the +M nature of nitrogen.

CONCEPT:

The question involves distinguishing between different chemical compounds in mixtures using specific reagents. The concept relies on the unique chemical reactions of these compounds with the reagents provided.

• Phenol vs Ethanol: Phenol reacts with neutral ( FeCl_3 ) to give a violet color, whereas ethanol does not react with neutral ( FeCl_3 ).
• Diethyl amine vs Ethyl amine: Ethyl amine (a primary amine) reacts with ( CHCl_3/KOH ) to give a carbylamine (foul-smelling test), whereas diethyl amine (a secondary amine) does not give this test.
• Acetaldehyde vs Acetone: Acetaldehyde (an aldehyde) gives a positive test with ammonical silver nitrate (Tollens' test), forming a silver mirror, whereas acetone (a ketone) does not.
• Benzoic acid vs Cinnamic acid: Cinnamic acid reacts with bromine water due to the presence of a double bond in its structure, whereas benzoic acid does not react with bromine water.

EXPLANATION:

1. A. Diethyl amine + Ethyl amine: Ethyl amine (primary amine) reacts with ( CHCl_3/KOH ), producing a foul smell (carbylamine test). Diethyl amine (secondary amine) does not react. Therefore, the reagent used is II. ( CHCl_3 + KOH, \Delta ).
2. B. Acetaldehyde + Acetone: Acetaldehyde gives a positive test with ammonical silver nitrate (Tollens' test), forming a silver mirror, whereas acetone does not. Therefore, the reagent used is IV. Ammonical silver nitrate.
3. C. Ethanol + Phenol: Phenol reacts with neutral ( FeCl_3 ) to produce a violet color, whereas ethanol does not. Therefore, the reagent used is III. Neutral ( FeCl_3 ).
4. D. Benzoic acid + Cinnamic acid: Cinnamic acid reacts with bromine water due to the presence of a double bond, whereas benzoic acid does not. Therefore, the reagent used is I. Bromine water.

So the correct matching is A-II, B-IV, C-III, D-I

CONCEPT:

Electrophilic Substitution Reaction (ESR) Reactivity

• The reactivity of an aromatic compound towards electrophilic substitution reactions (ESR) depends on the electron density of the benzene ring.
• Electron-donating groups increase the nucleophilicity of the benzene ring, enhancing its reactivity with electrophiles.
• Electron-withdrawing groups decrease the nucleophilicity of the benzene ring, reducing its reactivity with electrophiles.

EXPLANATION:

• Rate of ESR ( \propto ) nucleophilicity of benzene ring.
• The nucleophilicity of the benzene rings follows the order is P > R > Q
• Thus, the reactivity of these compounds towards the diazonium salt in a slightly acidic medium is P > R > Q

Therefore, the correct order regarding the reactivity of these compounds with ( Ph-N \equiv N \ Cl^{(-)} ) is P > R > Q.

CONCEPT:

Vitamins and Their Chemical Names

• Vitamins are organic compounds essential for various metabolic processes in the human body.
• Each vitamin has a specific chemical name that represents its structure and function.
• Matching vitamins with their chemical names helps in understanding their roles in human nutrition.

EXPLANATION:

• The matching of vitamins with their chemical names is as follows:
• Vitamin B₁ (A): Its chemical name is Thiamine. Thiamine plays a crucial role in energy metabolism and nerve function.
• Vitamin B₂ (B): Its chemical name is Riboflavin. Riboflavin is involved in energy production and cellular function.
• Vitamin B₆ (C): Its chemical name is Pyridoxine. Pyridoxine is important for amino acid metabolism and neurotransmitter synthesis.
• Vitamin C (D): Its chemical name is Ascorbic acid. Ascorbic acid is an antioxidant and is essential for collagen synthesis and immune function.

Therefore, the correct matching is: A - III, B - IV, C - I, D - II.

CONCEPT:

Flame Test and Identification of Ions

• The flame test is used to identify certain metal ions based on the characteristic color they produce when heated in a flame.

• Flame Colors:

• Ba+2Ba+2

  produces an apple green color.

• Ca+2Ca+2

  produces a brick red color.

• Sr+2Sr+2

  produces a crimson red color.

• Reactions for Identification:

• Ba+2+CrO2−4→BaCrO4Ba+2+CrO42−→BaCrO4

: Yellow precipitate indicates the presence of Ba+2Ba+2

.

• The sodium carbonate extract test with ammonium molybdate and concentrated nitric acid produces a canary yellow precipitate if PO3−4PO43−

  is present.

EXPLANATION:

• From the flame test:
• The apple green color in the flame test confirms the presence of Ba+2Ba+2

.

• From the reaction with K2CrO4K2CrO4

:

• The formation of a yellow precipitate confirms the presence of Ba+2Ba+2

  as BaCrO4BaCrO4

.

• From the sodium carbonate extract test:

• Heating the sodium carbonate extract with concentrated HNO3HNO3

  and ammonium molybdate produces a canary yellow precipitate, confirming the presence of PO3−4PO43−

.

• Based on the observations:
• The cation present in the salt is Ba+2Ba+2

.

• The anion present in the salt is PO3−4PO43−

.

Therefore, the cation and anion present in the salt are Ba2+Ba2+

and PO3−4PO43−

respectively.

CONCEPT:

Coordination Chemistry and Precipitation Reaction

• The given reaction involves a coordination complex ( [Cr(H_2O)_6]Cl_3 ), which is a hexaaquachromium(III) chloride complex.
• When this complex reacts with an excess of AgNO₃ (silver nitrate), the chloride ions present outside the coordination sphere precipitate as AgCl (silver chloride).
• The number of moles of AgCl precipitated depends on the number of chloride ions outside the coordination sphere of the complex.

EXPLANATION:

• The reaction can be written as:

( [Cr(H_2O)_6]Cl_3 + 3AgNO_3 \xrightarrow{excess} 3AgCl(s) + Cr(H_2O)_6_3 )

• In the given coordination complex ( [Cr(H_2O)_6]Cl_3 ), all six water molecules are within the coordination sphere of the chromium ion, while the three chloride ions are outside the coordination sphere.
• When reacted with excess AgNO₃, each chloride ion forms a precipitate of AgCl:

( Cl^- + Ag^+ \rightarrow AgCl(s) )

• Thus, one mole of ( [Cr(H_2O)_6]Cl_3 ) reacts to produce 3 moles of AgCl precipitate.
• The ratio of the moles of the complex reacted to the moles of AgCl precipitated is given as:

( \frac{\text{no. of moles of complex reacted}}{\text{no. of moles of AgCl precipitated}} \times 100 )

= ( \frac{1}{3} \times 100 )

= 33.3%

Therefore, the percentage of the moles of the complex reacted with respect to the moles of AgCl precipitated is 33.3%.

CONCEPT:

Mathematical Addition

• The addition of numbers is a fundamental arithmetic operation.
• It involves combining two or more numbers to find their total.
• The general form for addition is:

a + b = c, where c is the sum of a and b.

EXPLANATION:

Therefore, the sum of 1 + 4 + 4 + 2 + 3 is 14.

CONCEPT:

Combustion of Alkanes

• Alkanes are hydrocarbons with the general formula ( C_nH_{2n+2} ).
• When alkanes undergo combustion in the presence of oxygen (( O_2 )), they produce carbon dioxide (( CO_2 )) and water (( H_2O )) as products.
• The balanced chemical equation for the combustion of an alkane is:

( C_nH_{2n+2} + \left( \frac{3n+1}{2} \right) O_2 \rightarrow nCO_2 + (n+1) H_2O )

EXPLANATION:

• In the given question, we are tasked with identifying the alkane based on the combustion equation.
• The given combustion equation includes the term ( \left( \frac{3n+1}{2} \right) ) for the oxygen (( O_2 )) coefficient.
• We are also provided with the condition ( \frac{3n+1}{2} = 8 ).
• Solving this equation:
• ( \frac{3n+1}{2} = 8 )
• ( 3n+1 = 16 ) (multiplying both sides by 2)
• ( 3n = 15 ) (subtracting 1 from both sides)
• ( n = 5 ) (dividing both sides by 3)
• Since ( n = 5 ), the alkane is ( C_5H_{12} ) (pentane).
• Substituting ( n = 5 ) into the combustion equation:

( C_5H_{12} + 8O_2 \rightarrow 5CO_2 + 6H_2O )

Therefore, the alkane is pentane (( C_5H_{12} )), and the number of oxygen molecules required for complete combustion is 8.

CONCEPT:

Arrhenius Equation and Activation Energy

• The Arrhenius equation relates the rate constants of a reaction at two different temperatures with the activation energy of the reaction.
• The equation is given as:

ℓn(K2K1)=EaR[T2−T1T1T2]ℓn(K2K1)=EaR[T2−T1T1T2]

• Here:
• K1K1

and K2

are the rate constants at temperatures T1

and T2

respectively.

• Ea

is the activation energy of the reaction.

• R

is the universal gas constant (8.3 J mol−1 K−1

).

• T1

and T2

are the absolute temperatures in Kelvin.

EXPLANATION:

• In the given question:
• K1=1.5×103

, K2=4.5×103

• Ea=60×103 J mol−1

, T1=300 K

• Substitute these values into the Arrhenius equation:

ℓn(4.5×1031.5×103)=60×1038.3[T2−300300.T2]

• Simplify the natural logarithm:
• ℓn(4.5×1031.5×103)=ℓn 3
• ℓn 3=1.0986
• Rearrange the equation:

1.0986=60×1038.3[T2−300300.T2]

• Calculate:
• 60×1038.3=7228.92
• 1.0986=7228.92[T2−300300.T2]
• T2−300300.T2=1.09867228.92
• T2−300300.T2=0.000152
• T2−300=0.000152⋅300.T2
• T2−300=0.0456⋅T2
• T2−0.0456⋅T2=300
• 0.9544⋅T2=300
• T2=3000.9544
• T2=314.4 K
• Convert T2

to Celsius:

• t=T2−273
• t=314.4−273
• t=41.4∘C

Therefore, the temperature T2

is 314.4 K, which corresponds to 41.4∘C

.

CONCEPT:

Equilibrium and Gibbs Free Energy

• At equilibrium under standard pressure conditions, the Gibbs free energy change (( \Delta G^\circ )) is zero.
• The relationship between Gibbs free energy and equilibrium temperature can be expressed using thermodynamic principles.

EXPLANATION:

• The given equation is:

105 – 35 log T = 0

• We solve this equation to find the value of temperature (( T )):
• Rearranging the equation:

35 log T = 105

• Dividing both sides by 35:

log T = 105 / 35

log T = 3

• Converting from logarithmic to exponential form:

( T = 10^{3} )

( T = 1000 , \text{K} )

• Converting temperature from Kelvin to Celsius:

( T (\text{in } ^\circ \text{C}) = T (\text{in K}) - 273 )

( T (\text{in } ^\circ \text{C}) = 1000 - 273 )

( T (\text{in } ^\circ \text{C}) = 727^\circ \text{C} )

Therefore, the temperature at equilibrium is ( T = 1000 , \text{K} ) or ( T = 727^\circ \text{C} ).

( \theta = \frac{5t^4}{40} - \frac{t^3}{3} )

( \frac{d\theta}{dt} = \frac{5}{40}(4t^3) - \frac{3t^2}{3} = \frac{t^3}{2} - t^2 )

( \alpha = \frac{d^2\theta}{dt^2} = \frac{3t^2}{2} - 2t )

at t = 10

( \alpha = \frac{3}{2}(10)^2 - 2 \times 10 = 130 rad/sec^2 )

Initial capacitance

( C_i = \frac{A \epsilon_0}{d} = C )

Final capacitance

( C_1 = \frac{A \epsilon_0 \times 5}{\frac{d}{2}} = 10C )

( C_2 = \frac{A \epsilon_0}{\frac{d}{2}} = 2C )

( C_f = \frac{C_1 C_2}{C_1 + C_2} = \frac{10 \times 2}{10 + 2} C = \frac{5}{3} C )

% increase = ( \frac{\frac{5}{3} C - C}{C} \times 100 = 66.67 )

( \Delta Q_p = nC_p \Delta T = \frac{7}{2} nR \Delta T )

( \Delta U = nC_v \Delta T = \frac{5}{2} nR \Delta T )

W = ( nR \Delta T )

( \Delta Q : \Delta U : W :: 7 : 5 : 2 )

( V_1 = V_2 )

( \frac{kQ_1}{8} = \frac{kQ_2}{18} \quad \left{ V_{\text{surface}} = \frac{kQ}{R} \right} )

( \frac{Q_1}{Q_2} = \frac{4}{9} )

( \frac{E_{S_1}}{E_{S_2}} = \frac{kQ_1}{r_1^2} \frac{r_2^2}{kQ_2} \quad \left{ E_{\text{surface}} = \frac{kQ}{R^2} \right} )

( = \frac{Q_1}{Q_2} \left( \frac{r_2}{r_1} \right)^2 = \frac{4}{9} \left( \frac{9}{4} \right)^2 = \frac{9}{4} )

( v = \frac{\text{coeff of t}}{\text{coeff of x}} = \frac{200}{\frac{1}{150} \times 100} = 300\text{m/s} )

( \vec{E}_A = \frac{2kp_1}{x^3} (-\hat{i}) ) {Axial point}

( \vec{E}_B = \frac{kp_2}{x^3} (-\hat{j}) ) {Equatorial point}

Now : ( \tan \theta = \left| \frac{\vec{E}_B}{\vec{E}_A} \right| )

( \Rightarrow \tan 60^\circ = \frac{p_2}{2p_1} )

( \Rightarrow \frac{p_2}{p_1} = 2\sqrt{3} )

will be updated soon.

Sol. Power dissipated : ( P = \frac{\varepsilon^2}{R} )

Induced emf ( \varepsilon = -NA \frac{dB}{dt} )

Resistance of coil : ( R = \rho \frac{\ell}{\pi r^2} )

Now : ( \ell = N(2\pi R_{coil}) )

Also : ( A = \pi R_{coil}^2 \Rightarrow R_{coil} = \sqrt{\frac{A}{\pi}} )

Thus ; ( R = \frac{\rho N(2\pi) \sqrt{\frac{A}{\pi}}}{\pi r^2} )

( P = \frac{N^2 A^2 \left( \frac{dB}{dt} \right)^2}{\frac{2\rho N}{r^2} \sqrt{\frac{A}{\pi}}} \Rightarrow P \propto NA^{3/2}r^2 ......(1) )

( P' \propto (2N)(2A)^{3/2}(3r)^2 ......(2) )

( \frac{(2)}{(1)} : \frac{P'}{P} = 36\sqrt{2} \Rightarrow P' = (\sqrt{2})(36)P )

Thus : ( \alpha = 36 )

Sol. ( B_1 = \frac{\mu_0 I}{4\pi r} \times 2 + \frac{\mu_0 I}{4r} = \frac{\mu_0 I}{4r} \left( \frac{2}{\pi} + 1 \right) )

( B_2 = \frac{\mu_0 I}{4\pi r} + \frac{\mu_0 I}{4r} = \frac{\mu_0 I}{4r} \left( \frac{1}{\pi} + 1 \right) )

( \frac{B_1}{B_2} = \frac{2 + \pi}{1 + \pi} )

Sol. We have for thin prism minimum deviation :

( \delta_m = (\mu - 1)A )

( \Rightarrow \delta_m = (1.5 - 1)A \Rightarrow \delta_m = \frac{A}{2} )

Also, ( i = \frac{A + \delta_m}{2} \Rightarrow i = \frac{A + \frac{A}{2}}{2} \Rightarrow i = \frac{3A}{4} )

Thus ( \frac{i}{\delta_m} = \frac{3}{2} )

Sol. ( \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R} )

( \frac{1.54}{v} - \frac{1}{(-40)} = \frac{1.54 - 1}{-20} )

( \frac{1.54}{v} = -\frac{1}{40} - \frac{0.54}{20} )

( \frac{1.54}{v} = \frac{-2.08}{40} )

( v = -29.61 \text{ cm} )

( \frac{h_i}{h_o} = \frac{\mu_1 v}{\mu_2 u} = \frac{1(-29.61)}{1.54(-40)} = 0.48 )

( h_i = 0.48 \times 2 = 0.96 \approx 1 \text{cm} )

Sol. ( \frac{hc}{\lambda} - \phi = 3eV_0 \dots (i) )

( \frac{hc}{2\lambda} - \phi = eV_0 \dots (ii) )

( \frac{hc}{\lambda} - \phi = 3 \left[ \frac{hc}{2\lambda} - \phi \right] )

( \phi = \frac{hc}{4\lambda} )

( \frac{hc}{\lambda_0} = \frac{hc}{4\lambda} )

( \lambda_0 = 4\lambda )

( \alpha = 4 )

Sol. ( E_y = 300 \sin \omega \left( t - \frac{x}{v} \right) )

( F_M = qvB \dots (i) )

( F_e = qE \dots (ii) )

( \frac{F_e}{F_m} = \frac{qE}{qvB} = \frac{E}{vB} = \frac{c}{v} = \frac{3 \times 10^8}{1.5 \times 10^6} = 200 )

Sol.  L = ( \frac{nh}{2\pi} )

( \frac{3h}{\pi} = \frac{nh}{2\pi} )

n = 6

E = ( \frac{-13.6}{n^2} ) eV

= ( \frac{-13.6}{6^2} \approx -0.38 ) eV

Sol.  ( V_1 = V_2 )

( \frac{4}{3} \pi R^3 = 512 \left( \frac{4}{3} \pi r^3 \right) )

( R = (512)^{1/3} r )

1 = 8r

r = ( \frac{1}{8} ) mm

( \Delta SE = SE_2 - SE_1 )

= ( 512 \left[ T \times 4\pi r^2 \right] - \left[ T \times 4\pi R^2 \right] )

= ( 4\pi T [512r^2 - R^2] )

= ( 4\pi (0.08) \left[ 512 \left( \frac{1}{8} \right)^2 - 1^2 \right] \times 10^{-6} \approx 7 \times 10^{-6} )

( \alpha = 7 )

( \frac{1}{2} \epsilon_0 E^2 ) = Energy density

( \therefore \left[ \frac{1}{2} \epsilon_0 E^2 \right] = \frac{ML^2T^{-2}}{L^3} = ML^{-1}T^{-2} )

( \therefore ) a = 1, b = -1, c = -2

( \therefore ) 2a - b + c = 2 + 1 - 2 = 1

Y = ( \frac{F\ell}{A\Delta\ell} )

( \therefore \frac{\Delta Y}{Y} = \frac{\Delta F}{F} + \frac{\Delta \ell}{\ell} + 2\frac{\Delta d}{d} + \frac{\Delta(\Delta\ell)}{\Delta\ell} )

( \left[ A = \pi \frac{d^2}{4} \Rightarrow \frac{\Delta A}{A} = \frac{2\Delta d}{d} \right] )

Since F is known exactly ( \Rightarrow \Delta F = 0 )

( \frac{\Delta Y}{Y} = \frac{0.1}{150} + \frac{2 \times 0.001}{0.08} + \frac{0.001}{0.5} = 0.0277 )

( \therefore \Delta Y = 0.0277 Y )

Y = ( \frac{100 \times 1.5}{\frac{\pi}{4} \left( \frac{0.08}{100} \right)^2 \times 0.5 \times 10^{-2}} = 5.97 \times 10^{10} N/m^2 )

( \therefore \Delta Y = 0.0277 \times 5.97 \times 10^{10} N/m^2 = 1.65 \times 10^9 N/m^2 )

( \vec{v} = -x\hat{i} + 2y\hat{j} - z\hat{k} )

( \vec{a} = \frac{d\vec{v}}{dt} = -\frac{dx}{dt}\hat{i} + 2\frac{dy}{dt}\hat{j} - \frac{dz}{dt}\hat{k} )

= ( -(-x)\hat{i} + 2(2y)\hat{j} - (-z)\hat{k} = x\hat{i} + 4y\hat{j} + z\hat{k} )

( \therefore \vec{a} ) at (1,2,4) = ( \hat{i} + 8\hat{j} + 4\hat{k} )

( \therefore |\vec{a}| = 9 m/s^2 )

Given ( \vec{r} = 10t^2\hat{i} + 5t^3\hat{j} \Rightarrow \vec{r}{t=1} = 10\hat{i} + 5\hat{j} )

( \vec{v} = \frac{d\vec{r}}{dt} = 20t\hat{i} + 15t^2\hat{j} \Rightarrow \vec{v}{t=1sec} = 20\hat{i} + 15\hat{j} )

( \vec{a} = \frac{d^2\vec{r}}{dt^2} = 20\hat{i} + 30t\hat{j} \Rightarrow \vec{a}_{t=1sec} = 20\hat{i} + 30\hat{j} )

( \therefore \vec{P} = m\vec{v} = 2\hat{i} + 1.5\hat{j} )

( \vec{F} = m\vec{a} = 2\hat{i} + 3\hat{j} )

( \vec{\tau} = \vec{r} \times \vec{F} = 20\hat{k} )

( \vec{L} = \vec{r} \times \vec{P} = 5\hat{k} )

From Kepler's law,

( T^2 = \frac{4\pi^2}{GM_s}r^3 )

( \therefore T_{P_1} = \frac{2\pi}{\sqrt{G(2M)}} R^{3/2} )

( T_{P_2} = \frac{2\pi}{\sqrt{G(4M)}} (2R)^{3/2} )

( \therefore \frac{T_{P_2}}{T_{P_1}} = \frac{(2)^{3/2}}{\sqrt{4}} = \frac{\sqrt{2}}{1} \times \frac{2\sqrt{2} \times \sqrt{2}}{2} = 2 )