JEE Main 2025 — 7 Apr Shift 2 (Previous Year Paper)

JEE Main · 2025 · with answers and explanations

questions
75
Total marks
300
Duration
180 minutes
Negative marking
−1

Subjects: Mathematics · Physics · Chemistry

Question 1

Mathematics · 4 marks · −1

The number of real roots of the equation x∣x−2∣+3∣x−3∣+1=0x|x-2|+3|x-3|+1=0 is :

  1. A.1
  2. B.2
  3. C.3
  4. D.4

Correct answer: A

Explanation: x ≥ 3: x2+x−8=0x^2+x-8=0 no root ≥ 3; 2 ≤ x < 3: x2−5x+10x^2-5x+10 no real root; x < 2: −x2−x+10=0-x^2-x+10=0 ⇒ x=−1−412x=\frac{-1-\sqrt{41}}{2} only: 1 root.

Question 2

Mathematics · 4 marks · −1

Let a⃗\vec{a} and b⃗\vec{b} be the vectors of the same magnitude such that ∣a⃗+b⃗∣+∣a⃗−b⃗∣∣a⃗+b⃗∣−∣a⃗−b⃗∣=2+1\frac{|\vec{a}+\vec{b}|+|\vec{a}-\vec{b}|}{|\vec{a}+\vec{b}|-|\vec{a}-\vec{b}|}=\sqrt{2}+1. Then ∣a⃗+b⃗∣2∣a⃗∣2\frac{|\vec{a}+\vec{b}|^2}{|\vec{a}|^2} is :

  1. A.2+22+\sqrt{2}
  2. B.2+422+4\sqrt{2}
  3. C.4+224+2\sqrt{2}
  4. D.1+21+\sqrt{2}

Correct answer: A

Explanation: ∣a+b∣=2acos⁡θ2|a+b|=2a\cos\frac\theta2, ∣a−b∣=2asin⁡θ2|a-b|=2a\sin\frac\theta2: 1+t1−t=2+1\frac{1+t}{1-t}=\sqrt2+1 with t=tan⁡θ2t=\tan\frac\theta2 ⇒ θ = 45°; ∣a+b∣2/a2=2+2cos⁡45∘=2+2|a+b|^2/a^2=2+2\cos45^\circ=2+\sqrt2.

Question 3

Mathematics · 4 marks · −1

Let the length of a latus rectum of an ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 be 10. If its eccentricity is the minimum value of the function f(t)=t2+t+1112f(t)=t^2+t+\frac{11}{12}, t ∈ R, then a2+b2a^2+b^2 is equal to :

  1. A.120
  2. B.125
  3. C.115
  4. D.126

Correct answer: D

Explanation: e = 11/12 − 1/4 = 2/3; 2b2a=10\frac{2b^2}{a}=10, b2=a2⋅59b^2=a^2\cdot\frac59 ⇒ a = 9, b2=45b^2=45: 126.

Question 4

Mathematics · 4 marks · −1

If the orthocenter of the triangle formed by the lines y = x + 1, y = 4x − 8 and y = mx + c is at (3, −1), then m − c is :

  1. A.0
  2. B.−2
  3. C.4
  4. D.2

Correct answer: A

Explanation: A = (3, 4); AH is vertical so the third side is horizontal (m = 0, y = c). Vertex B = (c−1, c) on y = x+1; BH ⟂ y = 4x−8 gives c = 0. So m − c = 0.

Question 5

Mathematics · 4 marks · −1

Let A = {(α, β) ∈ R × R : |α − 1| ≤ 4 and |β − 5| ≤ 6} and B = {(α, β) ∈ R × R : 16(α − 2)² + 9(β − 6)² ≤ 144}. Then

  1. A.A∪B={(x,y):−4≤x≤4,−1≤y≤11}A\cup B=\{(x, y): -4\le x\le4, -1\le y\le11\}
  2. B.A⊂BA\subset B
  3. C.B⊂AB\subset A
  4. D.neither A⊂BA\subset B nor B⊂AB\subset A

Correct answer: C

Explanation: A: −3≤α≤5-3\le\alpha\le5, −1≤β≤11-1\le\beta\le11; B: ellipse centred (2, 6) with semi-axes 3 (x) and 4 (y): x ∈ [−1, 5], y ∈ [2, 10] — inside A.

Question 6

Mathematics · 4 marks · −1

The number of solutions of the equation cos⁡2θcos⁡θ2+cos⁡5θ2=2cos⁡35θ2\cos2\theta\cos\frac{\theta}{2}+\cos\frac{5\theta}{2}=2\cos^3\frac{5\theta}{2} in [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] is :

  1. A.7
  2. B.5
  3. C.9
  4. D.6

Correct answer: A

Question 7

Mathematics · 4 marks · −1

If the locus of z ∈ C, such that Re(z−12z+i)+Re(zˉ−12zˉ−i)=2\mathrm{Re}\left(\frac{z-1}{2z+i}\right)+\mathrm{Re}\left(\frac{\bar{z}-1}{2\bar{z}-i}\right)=2, is a circle of radius r and center (a, b), then 15abr2\frac{15ab}{r^2} is equal to :

  1. A.16
  2. B.24
  3. C.12
  4. D.18

Correct answer: D

Question 8

Mathematics · 4 marks · −1

Let e1e_1 and e2e_2 be the eccentricities of the ellipse x2b2+y225=1\frac{x^2}{b^2}+\frac{y^2}{25}=1 and the hyperbola x216−y2b2=1\frac{x^2}{16}-\frac{y^2}{b^2}=1, respectively. If b < 5 and e1e2=1e_1e_2=1, then the eccentricity of the ellipse having its axes along the coordinate axes and passing through all four foci (two of the ellipse and two of the hyperbola) is :

  1. A.45\frac{4}{5}
  2. B.32\frac{\sqrt{3}}{2}
  3. C.35\frac{3}{5}
  4. D.74\frac{\sqrt{7}}{4}

Correct answer: C

Explanation: (1−b225)(1+b216)=1(1-\frac{b^2}{25})(1+\frac{b^2}{16})=1 ⇒ b2=9b^2=9. Foci (0, ±4) and (±5, 0): ellipse semi-axes 5 (x), 4 (y), e = 3/5.

Question 9

Mathematics · 4 marks · −1

If the equation of the line passing through the point (0,−12,0)\left(0, -\frac{1}{2}, 0\right) and perpendicular to the lines r⃗=λ(i^+aj^+bk^)\vec{r}=\lambda(\hat{i}+a\hat{j}+b\hat{k}) and r⃗=(i^−j^−6k^)+μ(−bi^+aj^+5k^)\vec{r}=(\hat{i}-\hat{j}-6\hat{k})+\mu(-b\hat{i}+a\hat{j}+5\hat{k}) is x−1−2=y+4d=z−c−4\frac{x-1}{-2}=\frac{y+4}{d}=\frac{z-c}{-4}, then a + b + c + d is equal to :

  1. A.14
  2. B.10
  3. C.13
  4. D.12

Correct answer: A

Question 10

Mathematics · 4 marks · −1

Consider the lines L1L_1 : x − 1 = y − 2 = z and L2L_2 : x − 2 = y = z − 1. Let the feet of the perpendiculars from the point P(5, 1, −3) on the lines L1L_1 and L2L_2 be Q and R respectively. If the area of the triangle PQR is A, then 4A24A^2 is equal to :

  1. A.147
  2. B.139
  3. C.143
  4. D.151

Correct answer: A

Explanation: Q = (1, 2, 0), R = (2, 0, 1); PQ⃗×PR⃗=(−4,1,3)×(−3,−1,4)=(7,7,7)\vec{PQ}\times\vec{PR}=(-4,1,3)\times(-3,-1,4)=(7,7,7); 4A2=1474A^2=147.

Question 11

Mathematics · 4 marks · −1

Let the system of equations x + 5y − z = 1, 4x + 3y − 3z = 7, 24x + y + λz = μ, λ, μ ∈ R, have infinitely many solutions. Then the number of the solutions of this system, if x, y, z are integers and satisfy 7 ≤ x + y + z ≤ 77, is :

  1. A.3
  2. B.6
  3. C.5
  4. D.4

Correct answer: A

Question 12

Mathematics · 4 marks · −1

If the range of the function f(x)=5−xx2−3x+2f(x)=\frac{5-x}{x^2-3x+2}, x ≠ 1, 2, is (−∞,α]∪[β,∞)(-\infty, \alpha]\cup[\beta, \infty), then α2+β2\alpha^2+\beta^2 is equal to :

  1. A.192
  2. B.188
  3. C.190
  4. D.194

Correct answer: D

Explanation: yx2+(1−3y)x+2y−5=0yx^2+(1-3y)x+2y-5=0 real: y2+14y+1≥0y^2+14y+1\ge0 ⇒ y≤−7−43y\le-7-4\sqrt3 or y≥−7+43y\ge-7+4\sqrt3; α2+β2=2(49+48)=194\alpha^2+\beta^2=2(49+48)=194.

Question 13

Mathematics · 4 marks · −1

Let y = y(x) be the solution of the differential equation (x2+1)y′−2xy=(x4+2x2+1)cos⁡x(x^2+1)y'-2xy=(x^4+2x^2+1)\cos x, y(0) = 1. Then ∫−33y(x)dx\int_{-3}^{3}y(x)dx is :

  1. A.18
  2. B.30
  3. C.36
  4. D.24

Correct answer: D

Explanation: (y1+x2)=ˊcos⁡x\left(\frac{y}{1+x^2}\right)\'=\cos x ⇒ y=(1+x2)(sin⁡x+1)y=(1+x^2)(\sin x+1); odd part integrates to 0: ∫−33(1+x2)dx=24\int_{-3}^3(1+x^2)dx=24.

Question 14

Mathematics · 4 marks · −1

Let p be the number of all triangles that can be formed by joining the vertices of a regular polygon P of n sides and q be the number of all quadrilaterals that can be formed by joining the vertices of P. If p + q = 126, then the eccentricity of the ellipse x216+y2n=1\frac{x^2}{16}+\frac{y^2}{n}=1 is :

  1. A.74\frac{\sqrt{7}}{4}
  2. B.12\frac{1}{2}
  3. C.12\frac{1}{\sqrt{2}}
  4. D.34\frac{3}{4}

Correct answer: C

Explanation: (n3)+(n4)=(n+14)=126\binom n3+\binom n4=\binom{n+1}{4}=126 ⇒ n = 8; e=1−816=12e=\sqrt{1-\frac{8}{16}}=\frac{1}{\sqrt2}.

Question 15

Mathematics · 4 marks · −1

A bag contains 19 unbiased coins and one coin with head on both sides. One coin drawn at random is tossed and head turns up. If the probability that the drawn coin was unbiased, is mn\frac{m}{n}, gcd (m, n) = 1, then n2−m2n^2-m^2 is equal to :

  1. A.64
  2. B.60
  3. C.72
  4. D.80

Correct answer: D

Explanation: 19/219/2+1=1921\frac{19/2}{19/2+1}=\frac{19}{21}: 441−361=80441-361=80.

Question 16

Mathematics · 4 marks · −1

If the sum of the second, fourth and sixth terms of a G.P. of positive terms is 21 and the sum of its eighth, tenth and twelfth terms is 15309, then the sum of its first nine terms is :

  1. A.757
  2. B.755
  3. C.750
  4. D.760

Correct answer: A

Explanation: r6=729r^6=729, r = 3; ar(1+9+81)=21ar(1+9+81)=21 ⇒ a=113a=\frac{1}{13}; S9=39−113⋅2=757S_9=\frac{3^9-1}{13\cdot2}=757.

Question 17

Mathematics · 4 marks · −1

Let a random variable X take values 0, 1, 2, 3 with P(X = 0) = P(X = 1) = p, P(X = 2) = P(X = 3) and E(X2)=2E(X)E(X^2)=2E(X). Then the value of 8p − 1 is :

  1. A.0
  2. B.2
  3. C.3
  4. D.1

Correct answer: B

Explanation: q = (1 − 2p)/2 each for 2 and 3: E(X)=p+5qE(X)=p+5q, E(X2)=p+13qE(X^2)=p+13q; p+13q=2p+10qp+13q=2p+10q ⇒ p = 3q ⇒ p = 3/8: 8p − 1 = 2.

Question 18

Mathematics · 4 marks · −1

Let f : R → R be a polynomial function of degree four having extreme values at x = 4 and x = 5. If lim⁡x→0f(x)x2=5\lim_{x\to0}\frac{f(x)}{x^2}=5, then f(2) is equal to :

  1. A.8
  2. B.12
  3. C.10
  4. D.14

Correct answer: C

Explanation: f=x2(ax2+bx+5)f=x^2(ax^2+bx+5); f=ˊx(4ax2+3bx+10)f\'=x(4ax^2+3bx+10) with roots 4, 5 ⇒ a=18a=\frac18, b=−32b=-\frac32; f(2)=4(12−3+5)=10f(2)=4(\frac12-3+5)=10.

Question 19

Mathematics · 4 marks · −1

Let ana_n be the nthn^{th} term of an A.P. If Sn=a1+a2+a3+…+an=700S_n=a_1+a_2+a_3+\ldots+a_n=700, a6=7a_6=7 and S7=7S_7=7, then ana_n is equal to :

  1. A.56
  2. B.70
  3. C.65
  4. D.64

Correct answer: D

Explanation: a4=1a_4=1, a6=7a_6=7 ⇒ d = 3, a = −8; n2(−16+3(n−1))=700\frac n2(-16+3(n-1))=700 ⇒ n = 25; a25=64a_{25}=64.

Question 20

Mathematics · 4 marks · −1

If the area of the region {(x, y) : 1 + x² ≤ y ≤ min {x + 7, 11 − 3x}} is A, then 3A is equal to

  1. A.46
  2. B.50
  3. C.47
  4. D.49

Correct answer: B

Explanation: Parabola meets y = x + 7 at x = −2 and y = 11 − 3x at x = 2; lines meet at x = 1. A=∫−21(x+6−x2)dx+∫12(10−3x−x2)dx=272+196=503A=\int_{-2}^1(x+6-x^2)dx+\int_1^2(10-3x-x^2)dx=\frac{27}{2}+\frac{19}{6}=\frac{50}{3}: 3A = 50.

Question 21

Mathematics · 4 marks · −1

For t > −1, let αt\alpha_t and βt\beta_t be the roots of the equation ((t+2)1/7−1)x2+((t+2)1/6−1)x+((t+2)1/21−1)=0\left((t+2)^{1/7}-1\right)x^2+\left((t+2)^{1/6}-1\right)x+\left((t+2)^{1/21}-1\right)=0. If lim⁡t→−1+αt=a\lim_{t\to-1^+}\alpha_t=a and lim⁡t→−1+βt=b\lim_{t\to-1^+}\beta_t=b, then 72(a+b)272(a+b)^2 is equal to ______.

Correct answer: 98

Explanation: Near t = −1 the coefficients ∝ 1/7, 1/6, 1/21: x27+x6+121=0\frac{x^2}{7}+\frac{x}{6}+\frac{1}{21}=0; a+b=−76a+b=-\frac76; 72⋅4936=9872\cdot\frac{49}{36}=98.

Question 22

Mathematics · 4 marks · −1

If the function f(x)=tan⁡(tan⁡x)−sin⁡(sin⁡x)tan⁡x−sin⁡xf(x)=\frac{\tan(\tan x)-\sin(\sin x)}{\tan x-\sin x} is continuous at x = 0, then f(0) is equal to ______.

Correct answer: 2

Explanation: Numerator ≈x3\approx x^3, denominator ≈x32\approx\frac{x^3}{2}: f(0) = 2.

Question 23

Mathematics · 4 marks · −1

Let the lengths of the transverse and conjugate axes of a hyperbola in standard form be 2a and 2b, respectively, and one focus and the corresponding directrix of this hyperbola be (−5, 0) and 5x + 9 = 0, respectively. If the product of the focal distances of a point (α,25)(\alpha, 2\sqrt{5}) on the hyperbola is p, then 4p is equal to ______.

Correct answer: 189

Explanation: ae = 5, a/e = 9/5 ⇒ a = 3, b = 4; α2=9(1+2016)=814\alpha^2=9(1+\frac{20}{16})=\frac{81}{4}; p=e2α2−a2=259⋅814−9=1894p=e^2\alpha^2-a^2=\frac{25}{9}\cdot\frac{81}{4}-9=\frac{189}{4}: 189.

Question 24

Mathematics · 4 marks · −1

If ∫(1x+1x3)(3x−24+x−2623)dx=−α3(α+1)(3xβ+xγ)α+1α+C\int\left(\frac{1}{x}+\frac{1}{x^3}\right)\left(\sqrt[23]{3x^{-24}+x^{-26}}\right)dx=-\frac{\alpha}{3(\alpha+1)}\left(3x^{\beta}+x^{\gamma}\right)^{\frac{\alpha+1}{\alpha}}+C, x > 0, (α, β, γ ∈ Z), where C is the constant of integration, then α + β + γ is equal to ______.

Correct answer: 19

Question 25

Mathematics · 4 marks · −1

The sum of the series 2×1×20C4−3×2×20C5+4×3×20C6−5×4×20C7+⋯+18×17×20C202\times1\times{}^{20}C_4-3\times2\times{}^{20}C_5+4\times3\times{}^{20}C_6-5\times4\times{}^{20}C_7+\cdots+18\times17\times{}^{20}C_{20}, is equal to ______.

Correct answer: 34

Question 26

Physics · 4 marks · −1

The unit of 2Iϵ0c\sqrt{\frac{2I}{\epsilon_0c}} is : (I = intensity of an electromagnetic wave, c : speed of light)

  1. A.Nm
  2. B.Vm
  3. C.NC−1C^{-1}
  4. D.NC

Correct answer: C

Explanation: I=12ϵ0cE02I=\frac12\epsilon_0cE_0^2 ⇒ the expression is E0E_0.

Question 27

Physics · 4 marks · −1

A helicopter flying horizontally with a speed of 360 km/h at an altitude of 2 km, drops an object at an instant. The object hits the ground at a point O, 20 s after it is dropped. Displacement of 'O' from the position of helicopter where the object was released is : (use acceleration due to gravity g = 10 m/s² and neglect air resistance)

  1. A.252\sqrt{5} km
  2. B.4 km
  3. C.222\sqrt{2} km
  4. D.7.2 km

Correct answer: C

Explanation: Horizontal 100 × 20 = 2 km, vertical 2 km: 222\sqrt2 km.

Question 28

Physics · 4 marks · −1

Match List - I with List - II. List - I: (A) Isothermal (B) Adiabatic (C) Isobaric (D) Isochoric. List - II: (I) ΔW (work done) = 0 (II) ΔQ (supplied heat) = 0 (III) ΔU (change in internal energy) ≠ 0 (IV) ΔU = 0. Choose the correct answer from the options given below :

  1. A.(A)-(III), (B)-(II), (C)-(I), (D)-(IV)
  2. B.(A)-(II), (B)-(IV), (C)-(I), (D)-(III)
  3. C.(A)-(IV), (B)-(I), (C)-(III), (D)-(II)
  4. D.(A)-(IV), (B)-(II), (C)-(III), (D)-(I)

Correct answer: D

Question 29

Physics · 4 marks · −1

A capillary tube of radius 0.1 mm is partly dipped in water (surface tension 70 dyn/cm and glass water contact angle ≈ 0°) with 30° inclined with the vertical. The length of water risen in the capillary is ______ cm. (Take g = 9.8 m/s²)

  1. A.715\frac{71}{5}
  2. B.685\frac{68}{5}
  3. C.572\frac{57}{2}
  4. D.825\frac{82}{5}

Correct answer: D

Explanation: h=2Trρg=2(0.07)10−4(1000)(9.8)=14.29h=\frac{2T}{r\rho g}=\frac{2(0.07)}{10^{-4}(1000)(9.8)}=14.29 cm; length =h/cos⁡30∘=16.5≈825=h/\cos30^\circ=16.5\approx\frac{82}{5} cm.

Question 30

Physics · 4 marks · −1

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : The radius vector from the Sun to a planet sweeps out equal areas in equal intervals of time and thus areal velocity of planet is constant. Reason (R) : For a central force field the angular momentum is a constant. In the light of the above statements, choose the most appropriate answer from the options given below :

  1. A.(A) is correct but (R) is not correct
  2. B.Both (A) and (R) are correct and (R) is the correct explanation of (A)
  3. C.(A) is not correct but (R) is correct
  4. D.Both (A) and (R) are correct but (R) is not the correct explanation of (A)

Correct answer: B

Question 31

Physics · 4 marks · −1

The equation of a wave travelling on a string is y = sin[20πx + 10πt], where x and t are distance and time in SI units. The minimum distance between two points having the same oscillating speed is :

  1. A.10 cm
  2. B.5.0 cm
  3. C.2.5 cm
  4. D.20 cm

Correct answer: B

Explanation: λ = 0.1 m; speed ∝ |cos| repeats every λ/2 = 5 cm.

Question 32

Physics · 4 marks · −1

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Refractive index of glass is higher than that of air. Reason (R) : Optical density of a medium is directly proportionate to its mass density which results in a proportionate refractive index. In the light of the above statements, choose the most appropriate answer from the options given below :

  1. A.(A) is not correct but (R) is correct
  2. B.(A) is correct but (R) is not correct
  3. C.Both (A) and (R) are correct but (R) is not the correct explanation of (A)
  4. D.Both (A) and (R) are correct and (R) is the correct explanation of (A)

Correct answer: B

Question 33

Physics · 4 marks · −1

Which one of the following forces cannot be expressed in terms of potential energy ?

  1. A.Gravitational force
  2. B.Restoring force
  3. C.Frictional force
  4. D.Coulomb's force

Correct answer: C

Question 34

Physics · 4 marks · −1

An object with mass 500 g moves along x-axis with speed v=4xv=4\sqrt{x} m/s. The force acting on the object is :

  1. A.6 N
  2. B.5 N
  3. C.8 N
  4. D.4 N

Correct answer: D

Explanation: a=vdvdx=8a=v\frac{dv}{dx}=8; F = 0.5 × 8 = 4 N.

Question 35

Physics · 4 marks · −1

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : The density of the copper (2964Cu{}^{64}_{29}Cu) nucleus is greater than that of the carbon (612C{}^{12}_{6}C) nucleus. Reason (R) : The nucleus of mass number A has a radius proportional to A1/3A^{1/3}. In the light of the above statements, choose the most appropriate answer from the options given below :

  1. A.Both (A) and (R) are correct and (R) is the correct explanation of (A)
  2. B.(A) is correct but (R) is not correct
  3. C.(A) is not correct but (R) is correct
  4. D.Both (A) and (R) are correct but (R) is not the correct explanation of (A)

Correct answer: C

Explanation: R ∝ A^{1/3} makes nuclear density independent of A, so A is false.

Question 36

Physics · 4 marks · −1

A transparent block A having refractive index μ = 1.25 is surrounded by another medium of refractive index μ = 1.0 as shown in figure. A light ray is incident on the flat face of the block with incident angle θ as shown in figure. What is the maximum value of θ for which light suffers total internal reflection at the top surface of the block ?

Figure for question 36
  1. A.cos⁡−1(3/4)\cos^{-1}(3/4)
  2. B.sin⁡−1(3/4)\sin^{-1}(3/4)
  3. C.tan⁡−1(3/4)\tan^{-1}(3/4)
  4. D.tan⁡−1(4/3)\tan^{-1}(4/3)

Correct answer: B

Explanation: sin⁡θC=45\sin\theta_C=\frac45 ⇒ refraction angle at entry r=90∘−θCr=90^\circ-\theta_C with sin⁡r=35\sin r=\frac35; sin⁡θ=1.25×35=34\sin\theta=1.25\times\frac35=\frac34.

Question 37

Physics · 4 marks · −1

The helium and argon are put in the flask at the same room temperature (300 K). The ratio of average kinetic energies (per molecule) of helium and argon is : (Give : Molar mass of helium = 4 g/mol, Molar mass of argon = 40 g/mol)

  1. A.10 : 1
  2. B.1:10\sqrt{10}
  3. C.1 : 1
  4. D.1 : 10

Correct answer: C

Question 38

Physics · 4 marks · −1

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Magnetic monopoles do not exist. Reason (R) : Magnetic field lines are continuous and form closed loops. In the light of the above statements, choose the most appropriate answer from the options given below :

  1. A.Both (A) and (R) are correct but (R) is not the correct explanation of (A)
  2. B.(A) is correct but (R) is not correct
  3. C.Both (A) and (R) are correct and (R) is the correct explanation of (A)
  4. D.(A) is not correct but (R) is correct

Correct answer: C

Question 39

Physics · 4 marks · −1

The dimension of μ0ϵ0\sqrt{\frac{\mu_0}{\epsilon_0}} is equal to that of : (μ0\mu_0 = Vacuum permeability and ϵ0\epsilon_0 = Vacuum permittivity)

  1. A.Inductance
  2. B.Resistance
  3. C.Voltage
  4. D.Capacitance

Correct answer: B

Question 40

Physics · 4 marks · −1

Match List - I with List - II. List - I: (A) Mass density (B) Impulse (C) Power (D) Moment of inertia. List - II: (I) [ML2T−3][ML^2T^{-3}] (II) [MLT−1][MLT^{-1}] (III) [ML2T0][ML^2T^0] (IV) [ML−3T0][ML^{-3}T^0]. Choose the correct answer from the options given below :

  1. A.(A)-(I), (B)-(III), (C)-(IV), (D)-(II)
  2. B.(A)-(IV), (B)-(II), (C)-(III), (D)-(I)
  3. C.(A)-(II), (B)-(III), (C)-(IV), (D)-(I)
  4. D.(A)-(IV), (B)-(II), (C)-(I), (D)-(III)

Correct answer: D

Question 41

Physics · 4 marks · −1

A dipole with two electric charges of 2 μC magnitude each, with separation distance 0.5 μm, is placed between the plates of a capacitor such that its axis is parallel to an electric field established between the plates when a potential difference of 5 V is applied. Separation between the plates is 0.5 mm. If the dipole is rotated by 30° from the axis, it tends to realign in the direction due to a torque. The value of torque is :

  1. A.2.5×10−92.5\times10^{-9} Nm
  2. B.5×10−95\times10^{-9} Nm
  3. C.2.5×10−122.5\times10^{-12} Nm
  4. D.5×10−35\times10^{-3} Nm

Correct answer: B

Explanation: p=10−12p=10^{-12}, E=104E=10^4: τ=pEsin⁡30∘=5×10−9\tau=pE\sin30^\circ=5\times10^{-9} N m.

Question 42

Physics · 4 marks · −1

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : The outer body of an air craft is made of metal which protects persons sitting inside from lightning-strikes. Reason (R) : The electric field inside the cavity enclosed by a conductor is zero. In the light of the above statements, choose the most appropriate answer from the options given below :

  1. A.(A) is correct but (R) is not correct
  2. B.Both (A) and (R) are correct but (R) is not the correct explanation of (A)
  3. C.Both (A) and (R) are correct and (R) is the correct explanation of (A)
  4. D.(A) is not correct but (R) is correct

Correct answer: C

Question 43

Physics · 4 marks · −1

A mirror is used to produce an image with magnification of 14\frac{1}{4}. If the distance between object and its image is 40 cm, then the focal length of the mirror is ______.

  1. A.10 cm
  2. B.12.7 cm
  3. C.15 cm
  4. D.10.7 cm

Correct answer: D

Explanation: Convex mirror, v = −u/4 behind: |u| + |u|/4 = 40 ⇒ u = −32, v = 8; 1f=18−132\frac1f=\frac18-\frac1{32} ⇒ f = 10.7 cm.

Question 44

Physics · 4 marks · −1

A photoemissive substance is illuminated with a radiation of wavelength λi\lambda_i so that it releases electrons with de-Broglie wavelength λe\lambda_e. The longest wavelength of radiation that can emit photoelectron is λo\lambda_o. Expression for de-Broglie wavelength is given by : (m : mass of the electron, h : Planck's constant and c : speed of light)

  1. A.λe=hλo2mc\lambda_e=\sqrt{\frac{h\lambda_o}{2mc}}
  2. B.λe=h2mc(1λi−1λo)\lambda_e=\sqrt{\frac{h}{2mc\left(\frac{1}{\lambda_i}-\frac{1}{\lambda_o}\right)}}
  3. C.λe=h2mc(1λi−1λo)\lambda_e=\frac{h}{\sqrt{2mc\left(\frac{1}{\lambda_i}-\frac{1}{\lambda_o}\right)}}
  4. D.λe=hλi2mc\lambda_e=\sqrt{\frac{h\lambda_i}{2mc}}

Correct answer: B

Explanation: K=hc(1λi−1λo)K=hc\left(\frac1{\lambda_i}-\frac1{\lambda_o}\right), λe=h2mK=h2mc(…)\lambda_e=\frac{h}{\sqrt{2mK}}=\sqrt{\frac{h}{2mc(\ldots)}}.

Question 45

Physics · 4 marks · −1

Consider the following logic circuit. [A and B feed an AND gate; B and NOT(A)... feed an OR gate; both outputs feed a NOR gate giving Y.] The output is Y = 0 when :

Figure for question 45
  1. A.A = 1 and B = 1
  2. B.A = 0 and B = 1
  3. C.A = 1 and B = 0
  4. D.A = 0 and B = 0

Correct answer: A

Question 46

Physics · 4 marks · −1

The electric field in a region is given by E⃗=(2i^+4j^+6k^)×103\vec{E}=\left(2\hat{i}+4\hat{j}+6\hat{k}\right)\times10^3 N/C. The flux of the field through a rectangular surface parallel to x-z plane is 6.0 Nm2C−1Nm^2C^{-1}. The area of the surface is ______ cm2cm^2.

Correct answer: 15

Explanation: ϕ=EyA\phi=E_yA ⇒ A=64000=1.5×10−3A=\frac{6}{4000}=1.5\times10^{-3} m² = 15 cm².

Question 47

Physics · 4 marks · −1

A parallel plate capacitor has charge 5×10−65\times10^{-6} C. A dielectric slab is inserted between the plates and almost fills the space between the plates. If the induced charge on one face of the slab is 4×10−64\times10^{-6} C then the dielectric constant of the slab is ______.

Correct answer: 5

Explanation: qi=q(1−1K)q_i=q(1-\frac1K) ⇒ K = 5.

Question 48

Physics · 4 marks · −1

Two cylindrical rods A and B made of different materials, are joined in a straight line. The ratios of lengths, radii and thermal conductivites of these rods are: LALB=12\frac{L_A}{L_B}=\frac{1}{2}, rArB=2\frac{r_A}{r_B}=2 and KAKB=12\frac{K_A}{K_B}=\frac{1}{2}. The free ends of rods A and B are maintained at 400 K, 200 K, respectively. The temperature of rods interface is ______ K, when equilibrium is established.

Correct answer: 360

Explanation: R∝LKr2R\propto\frac{L}{Kr^2}: RA/RB=1/2(1/2)(4)=14R_A/R_B=\frac{1/2}{(1/2)(4)}=\frac14; interface =400−2005=360=400-\frac{200}{5}=360 K.

Question 49

Physics · 4 marks · −1

M and R be the mass and radius of a disc. A small disc of radius R/3 is removed from the bigger disc as shown in figure. The moment of inertia of remaining part of bigger disc about an axis AB passing through the centre O and perpendicular to the plane of disc is 4xMR2\frac{4}{x}MR^2. The value of x is ______.

Figure for question 49

Correct answer: 9

Explanation: Removed mass M/9: I=MR22−[M9R218+M94R29]=MR22−MR218=49MR2I=\frac{MR^2}{2}-\left[\frac{M}{9}\frac{R^2}{18}+\frac M9\frac{4R^2}{9}\right]=\frac{MR^2}{2}-\frac{MR^2}{18}=\frac49MR^2: x = 9.

Question 50

Physics · 4 marks · −1

An inductor of reactance 100 Ω, a capacitor of reactance 50 Ω, and a resistor of resistance 50 Ω are connected in series with an AC source of 10 V, 50 Hz. Average power dissipated by the circuit is ______ W.

Correct answer: 1

Explanation: Z=502Z=50\sqrt2; P=V2RZ2=100×505000=1P=\frac{V^2R}{Z^2}=\frac{100\times50}{5000}=1 W.

Question 51

Chemistry · 4 marks · −1

The hydration energies of K+K^+ and Cl−Cl^- are −x and −y kJ/mol respectively. If lattice energy of KCl is −z kJ/mol, then the heat of solution of KCl is :

  1. A.−z − (x + y)
  2. B.z − (x + y)
  3. C.x + y + z
  4. D.+x − y − z

Correct answer: B

Explanation: ΔHsol=−ΔHlattice(formation)+ΔHhyd=z−(x+y)\Delta H_{sol}=-\Delta H_{lattice(formation)}+\Delta H_{hyd}=z-(x+y).

Question 52

Chemistry · 4 marks · −1

The descending order of basicity of following amines is : (A) aniline (B) p-methoxyaniline (C) p-nitroaniline (D) CH3NH2CH_3NH_2 (E) (CH3)2NH(CH_3)_2NH. Choose the correct answer from the options given below :

  1. A.E > D > A > B > C
  2. B.E > D > B > A > C
  3. C.B > E > D > A > C
  4. D.E > A > D > C > B

Correct answer: B

Question 53

Chemistry · 4 marks · −1

In SO2SO_2, NO2−NO_2^- and N3−N_3^- the hybridizations at the central atom are respectively :

  1. A.sp2sp^2, sp2sp^2 and sp2sp^2
  2. B.sp2sp^2, sp and sp
  3. C.sp, sp2sp^2 and sp
  4. D.sp2sp^2, sp2sp^2 and sp

Correct answer: D

Question 54

Chemistry · 4 marks · −1

The extra stability of half-filled subshell is due to : (A) Symmetrical distribution of electrons (B) Smaller coulombic repulsion energy (C) The presence of electrons with the same spin in non-degenerate orbitals (D) Larger exchange energy (E) Relatively smaller shielding of electrons by one another. Identify the correct statements :

  1. A.(B), (C) and (D) only
  2. B.(B), (D) and (E) only
  3. C.(A), (B), (D) and (E) only
  4. D.(A), (B) and (D) only

Correct answer: C

Question 55

Chemistry · 4 marks · −1

Match List - I with List - II. List - I (Conversion): (A) chlorobenzene → phenol (B) 4-nitrochlorobenzene → 4-nitrophenol (C) 2,4-dinitrochlorobenzene → 2,4-dinitrophenol (D) 2,4,6-trinitrochlorobenzene → picric acid. List - II (Reagents, Conditions used): (I) Warm, H2OH_2O (II) (a) NaOH, 368K; (b) H3O+H_3O^+ (III) (a) NaOH, 443K; (b) H3O+H_3O^+ (IV) (a) NaOH, 623K, 300 atm; (b) H3O+H_3O^+. Choose the correct answer from the options given below :

Figure for question 55
  1. A.(A)-(III), (B)-(IV), (C)-(II), (D)-(I)
  2. B.(A)-(IV), (B)-(III), (C)-(II), (D)-(I)
  3. C.(A)-(II), (B)-(III), (C)-(I), (D)-(IV)
  4. D.(A)-(IV), (B)-(III), (C)-(I), (D)-(II)

Correct answer: B

Question 56

Chemistry · 4 marks · −1

The correct statement amongst the following is :

  1. A.ΔfH298θ\Delta_fH^\theta_{298} is zero for O(g)
  2. B.The standard state of a pure gas is the pure gas at a pressure of 1 bar and temperature 273 K.
  3. C.The term 'standard state' implies that the temperature is 0°C.
  4. D.ΔfH500θ\Delta_fH^\theta_{500} is zero for O2O_2(g)

Correct answer: D

Explanation: Standard enthalpy of formation of an element in its reference state is zero at any specified temperature.

Question 57

Chemistry · 4 marks · −1

Choose the incorrect trend in the atomic radii (r) of the elements.

  1. A.rAt<rCsr_{At}<r_{Cs}
  2. B.rMg<rAlr_{Mg}<r_{Al}
  3. C.rBr<rKr_{Br}<r_K
  4. D.rRb<rCsr_{Rb}<r_{Cs}

Correct answer: B

Question 58

Chemistry · 4 marks · −1

Match List - I with List - II. List - I: (A) Solution of chloroform and acetone (B) Solution of ethanol and water (C) Solution of benzene and toluene (D) Solution of acetic acid in benzene. List - II: (I) Minimum boiling azeotrope (II) Dimerizes (III) Maximum boiling azeotrope (IV) ΔVmix=0\Delta V_{mix}=0. Choose the correct answer from the options given below :

  1. A.(A)-(II), (B)-(IV), (C)-(I), (D)-(III)
  2. B.(A)-(III), (B)-(I), (C)-(IV), (D)-(II)
  3. C.(A)-(III), (B)-(IV), (C)-(I), (D)-(II)
  4. D.(A)-(II), (B)-(I), (C)-(IV), (D)-(III)

Correct answer: B

Question 59

Chemistry · 4 marks · −1

Match List - I with List - II. List - I (Complex): (A) [Co(en)2Cl2]Cl[Co(en)_2Cl_2]Cl (B) [Pt(NH3)2Cl(NO2)][Pt(NH_3)_2Cl(NO_2)] (C) Hg[Co(SCN)4]Hg[Co(SCN)_4] (D) [Mg(EDTA)]2−[Mg(EDTA)]^{2-}. List - II (Primary valency and Secondary valency): (I) 3, 6 (II) 3, 4 (III) 2, 6 (IV) 2, 4. Choose the correct answer from the options given below :

  1. A.(A)-(III), (B)-(I), (C)-(II), (D)-(IV)
  2. B.(A)-(II), (B)-(III), (C)-(IV), (D)-(I)
  3. C.(A)-(I), (B)-(IV), (C)-(II), (D)-(III)
  4. D.(A)-(I), (B)-(III), (C)-(II), (D)-(IV)

Correct answer: C

Question 60

Chemistry · 4 marks · −1

The number of unpaired electrons responsible for the paramagnetic nature of the following complex species are respectively : [Fe(CN)6]3−[Fe(CN)_6]^{3-}, [FeF6]3−[FeF_6]^{3-}, [CoF6]3−[CoF_6]^{3-}, [Mn(CN)6]3−[Mn(CN)_6]^{3-}

  1. A.1, 1, 4, 2
  2. B.1, 5, 4, 2
  3. C.1, 5, 5, 2
  4. D.1, 4, 4, 2

Correct answer: B

Question 61

Chemistry · 4 marks · −1

A(g) → B(g) + C(g) is a first order reaction. Time: t, ∞; PsystemP_{system}: ptp_t, p∞p_\infty. The reaction was started with reactant A only. Which of the following expression is correct for rate constant k ?

  1. A.k=1tln⁡p∞ptk=\frac{1}{t}\ln\frac{p_\infty}{p_t}
  2. B.k=1tln⁡p∞(p∞−pt)k=\frac{1}{t}\ln\frac{p_\infty}{(p_\infty-p_t)}
  3. C.k=1tln⁡2(p∞−pt)ptk=\frac{1}{t}\ln\frac{2(p_\infty-p_t)}{p_t}
  4. D.k=1tln⁡p∞2(p∞−pt)k=\frac{1}{t}\ln\frac{p_\infty}{2(p_\infty-p_t)}

Correct answer: D

Explanation: p0=p∞/2p_0=p_\infty/2, pA=2p0−pt=p∞−ptp_A=2p_0-p_t=p_\infty-p_t: k=1tln⁡p∞/2p∞−ptk=\frac1t\ln\frac{p_\infty/2}{p_\infty-p_t}.

Question 62

Chemistry · 4 marks · −1

"P" is an optically active compound with molecular formula C6H12OC_6H_{12}O. When \'P\' is treated with 2, 4-dinitrophenylhydrazine, it gives a positive test. However, in presence of Tollens reagent, "P" gives a negative test. Predict the structure of "P".

  1. A.CH3−CO−CH2−CH2−CH2−CH3CH_3-CO-CH_2-CH_2-CH_2-CH_3
  2. B.CH3−CO−CH2−CH(CH3)−CH3CH_3-CO-CH_2-CH(CH_3)-CH_3
  3. C.H−CO−CH2−CH(CH3)−CH2−CH3H-CO-CH_2-CH(CH_3)-CH_2-CH_3
  4. D.CH3−CO−CH(CH3)−CH2−CH3CH_3-CO-CH(CH_3)-CH_2-CH_3

Correct answer: D

Question 63

Chemistry · 4 marks · −1

The number of optically active products obtained from the complete ozonolysis of the given compound is : H3C−CH=CH−CH(CH3)−CH=CH−CH(CH3)−CH=CH−CH3H_3C-CH=CH-CH(CH_3)-CH=CH-CH(CH_3)-CH=CH-CH_3 (the two stereocentres drawn with opposite wedge configurations)

  1. A.4
  2. B.0
  3. C.1
  4. D.2

Correct answer: B

Explanation: Products: 2 CH₃CHO and two OHC–CH(CH₃)–CHO fragments, each achiral (two identical CHO groups): 0.

Question 64

Chemistry · 4 marks · −1

Given below are two statements : Statement (I) : On hydrolysis, oligo peptides give rise to fewer number of α-amino acids while proteins give rise to a large number of β-amino acids. Statement (II) : Natural proteins are denatured by acids which convert the water soluble form of fibrous proteins to their water insoluble form. In the light of the above statements, choose the most appropriate answer from the options given below :

  1. A.Both Statement I and Statement II are incorrect
  2. B.Statement I is correct but Statement II is incorrect
  3. C.Statement I is incorrect but Statement II is correct
  4. D.Both Statement I and Statement II are correct

Correct answer: A

Question 65

Chemistry · 4 marks · −1

Given below are two statements : Statement (I) : cis-1,4-dichlorobut-2-ene (drawn with Cl on each end) is more polar than cis-1,4-dibromobut-2-ene. Statement (II) : Boiling point of 2,3-dibromobut-2-ene... [trans-1,2-dibromo isomer drawn] is lower than the cis dibromo isomer but it is more polar than the cis isomer. In the light of the above statements, choose the most appropriate answer from the options given below :

Figure for question 65
  1. A.Statement I is incorrect but Statement II is correct
  2. B.Statement I is correct but Statement II is incorrect
  3. C.Both Statement I and Statement II are correct
  4. D.Both Statement I and Statement II are incorrect

Correct answer: B

Question 66

Chemistry · 4 marks · −1

Liquid A and B form an ideal solution. The vapour pressures of pure liquids A and B are 350 and 750 mm Hg respectively at the same temperature. If xAx_A and xBx_B are the mole fraction of A and B in solution while yAy_A and yBy_B are the mole fraction of A and B in vapour phase then,

  1. A.xAxB>yAyB\frac{x_A}{x_B}>\frac{y_A}{y_B}
  2. B.xAxB=yAyB\frac{x_A}{x_B}=\frac{y_A}{y_B}
  3. C.(xA−yA)<(xB−yB)(x_A-y_A)<(x_B-y_B)
  4. D.xAxB<yAyB\frac{x_A}{x_B}<\frac{y_A}{y_B}

Correct answer: A

Explanation: yAyB=350xA750xB<xAxB\frac{y_A}{y_B}=\frac{350x_A}{750x_B}<\frac{x_A}{x_B}.

Question 67

Chemistry · 4 marks · −1

'X' is the number of acidic oxides among VO2VO_2, V2O3V_2O_3, CrO3CrO_3, V2O5V_2O_5 and Mn2O7Mn_2O_7. The primary valency of cobalt in [Co(H2NCH2CH2NH2)3]2(SO4)3[Co(H_2NCH_2CH_2NH_2)_3]_2(SO_4)_3 is Y. The value of X + Y is ______.

  1. A.2
  2. B.3
  3. C.5
  4. D.4

Correct answer: C

Explanation: Acidic: CrO3CrO_3, Mn2O7Mn_2O_7 (X = 2; V2O5V_2O_5 amphoteric); primary valency of Co(III) = 3: 5.

Question 68

Chemistry · 4 marks · −1

Mixture of 1 g each of chlorobenzene, aniline and benzoic acid is dissolved in 50 mL ethyl acetate and placed in a separating funnel. 5 M NaOH (30 mL) was added in the same funnel. The funnel was shaken vigorously and then kept aside. The ethyl acetate layer in the funnel contains :

  1. A.benzoic acid
  2. B.benzoic acid and aniline
  3. C.benzoic acid and chlorobenzene
  4. D.chlorobenzene and aniline

Correct answer: D

Question 69

Chemistry · 4 marks · −1

Given below are two statements : 1 M aqueous solutions of each of Cu(NO3)2Cu(NO_3)_2, AgNO3AgNO_3, Hg2(NO3)2Hg_2(NO_3)_2, Mg(NO3)2Mg(NO_3)_2 are electrolysed using inert electrodes. Given : EAg+/AgθE^\theta_{Ag^+/Ag} = 0.80 V, EHg22+/HgθE^\theta_{Hg_2^{2+}/Hg} = 0.79 V, ECu2+/CuθE^\theta_{Cu^{2+}/Cu} = 0.24 V and EMg2+/MgθE^\theta_{Mg^{2+}/Mg} = −2.37 V. Statement (I) : With increasing voltage, the sequence of deposition of metals on the cathode will be Ag, Hg and Cu. Statement (II) : Magnesium will not be deposited at cathode instead oxygen gas will be evolved at the cathode. In the light of the above statements, choose the most appropriate answer from the options given below :

  1. A.Statement I is correct but Statement II is incorrect
  2. B.Statement I is incorrect but Statement II is correct
  3. C.Both Statement I and Statement II are correct
  4. D.Both Statement I and Statement II are incorrect

Correct answer: A

Explanation: Hydrogen, not oxygen, would be evolved at the cathode.

Question 70

Chemistry · 4 marks · −1

The correct statements from the following are : (A) Tl3+Tl^{3+} is a powerful oxidising agent (B) Al3+Al^{3+} does not get reduced easily (C) Both Al3+Al^{3+} and Tl3+Tl^{3+} are very stable in solution (D) Tl+Tl^+ is more stable than Tl3+Tl^{3+} (E) Al3+Al^{3+} and Tl+Tl^+ are highly stable. Choose the correct answer from the options given below :

  1. A.(B), (D) and (E) only
  2. B.(A), (B), (C), (D) and (E)
  3. C.(A), (B), (D) and (E) only
  4. D.(A), (C) and (D) only

Correct answer: C

Question 71

Chemistry · 4 marks · −1

Butane reacts with oxygen to produce carbon dioxide and water following the equation given below. C4H10(g)+132O2(g)→4CO2(g)+5H2O(l)C_4H_{10}(g)+\frac{13}{2}O_2(g)\to4CO_2(g)+5H_2O(l). If 174.0 kg of butane is mixed with 320.0 kg of O2O_2, the volume of water formed in liters is ______. (Nearest integer) [Given : (a) Molar mass of C, H, O are 12, 1, 16 g mol−1mol^{-1} respectively, (b) Density of water = 1g mL−1mL^{-1}]

Correct answer: 138

Explanation: O2O_2 10 kmol limiting → water 56.5×10=7.69\frac{5}{6.5}\times10=7.69 kmol × 18 = 138.5 kg ≈ 138 L.

Question 72

Chemistry · 4 marks · −1

In Dumas' method 292 mg of an organic compound released 50 mL of nitrogen gas (N2N_2) at 300 K temperature and 715 mm Hg pressure. The percentage composition of 'N' in the organic compound is ______% (Nearest integer) (Aqueous tension at 300 K = 15 mm Hg)

Correct answer: 18

Explanation: N2N_2 mass =(700/760)(0.05)0.0821×300×28=0.0524=\frac{(700/760)(0.05)}{0.0821\times300}\times28=0.0524 g → 17.9% ≈ 18.

Question 73

Chemistry · 4 marks · −1

One litre buffer solution was prepared by adding 0.10 mol each of NH3NH_3 and NH4ClNH_4Cl in deionised water. The change in pH on addition of 0.05 mol of HCl to the above solution is ______ ×10−2\times10^{-2}. (Nearest integer) Given : pKbpK_b of NH3NH_3 = 4.745 and log⁡103\log_{10}3 = 0.477

Correct answer: 48

Explanation: pOH change =log⁡3=0.477=\log3=0.477 → 48 × 10⁻².

Question 74

Chemistry · 4 marks · −1

Identify the structure of the final product (D) in the following sequence of the reactions : Ph−CO−CH3→ΔPCl5Ph-CO-CH_3\xrightarrow[\Delta]{PCl_5} A →3 eq. NaNH2/NH3\xrightarrow{3\ eq.\ NaNH_2/NH_3} B →Acidify\xrightarrow{\text{Acidify}} C →1. B2H6; 2. H2O2/OH−\xrightarrow{\text{1. } B_2H_6;\ \text{2. } H_2O_2/OH^-} D. Total number of sp2sp^2 hybridised carbon atoms in product D is ______.

Correct answer: 7

Explanation: D = phenylacetaldehyde PhCH2CHOPhCH_2CHO: 6 aromatic + 1 carbonyl = 7.

Question 75

Chemistry · 4 marks · −1

The number of paramagnetic metal complex species among [Co(NH3)6]3+[Co(NH_3)_6]^{3+}, [Co(C2O4)3]3−[Co(C_2O_4)_3]^{3-}, [MnCl6]3−[MnCl_6]^{3-}, [Mn(CN)6]3−[Mn(CN)_6]^{3-}, [CoF6]3−[CoF_6]^{3-}, [Fe(CN)6]3−[Fe(CN)_6]^{3-} and [FeF6]3−[FeF_6]^{3-} with same number of unpaired electrons is ______.

Correct answer: 2

Explanation: [MnCl6]3−[MnCl_6]^{3-} and [CoF6]3−[CoF_6]^{3-} both have 4 unpaired electrons: 2.

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