JEE (Main) · 2025
JEE (Main) 2025 — 22 January, Shift 2
Which chapter each question came from, and how to solve each one. Keep the official paper open beside this page.
01
Summary
The heaviest chapters in each subject, and how the questions were spread by level.
Mathematics
24 Q
- Sets, Relations and Functions2
- Matrices and Determinants2
- Binomial Theorem2
5 easy · 19 medium · 0 hard
Physics
13 Q
- Physical World and Measurement2
- Properties of Bulk Matter2
- Electromagnetic Induction and Alternating Currents2
13 easy · 0 medium · 0 hard
Chemistry
21 Q
- Coordination Compounds3
- Organic Chemistry: Some Basic Principles and Techniques3
- Hydrocarbons3
9 easy · 11 medium · 1 hard
02
Questions
Tap Explain on any question for a short summary and our full working.
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 1 | Sets, Relations and FunctionsOne-one, into and onto functions | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | Single correctMedium | |
A medium single correct question on one-one, into and onto functions, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Functions $f : A \to B$ whose range contains $1$: all functions minus those that never take the value $1$: $4^4 - 3^4 = 256 - 81 = 175$.
A many-one function is one that is not one-one. The one-one functions from $A$ to $B$ are the $4! = 24$ bijections, and each of them takes the value $1$.
Many-one functions with $1 \in f(A)$: $175 - 24 = 151$.
Answer: (C) $151$. | ||||
| 2 | Quadratic EquationsRelations between roots and coefficients | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on relations between roots and coefficients, from Quadratic Equations. NCERT: Class 11: Complex Numbers and Quadratic Equations The discriminant $\cos^2\theta + 8 > 0$, so the roots are always real and distinct.
$\alpha + \beta = -\frac{\cos\theta}{2}$, $\alpha\beta = -\frac{1}{2}$.
$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \frac{\cos^2\theta}{4} + 1$.
$\alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2\alpha^2\beta^2 = \left(1 + \frac{\cos^2\theta}{4}\right)^2 - \frac{1}{2}$.
On $(0, 2\pi)$, $\cos^2\theta$ takes every value in $[0, 1]$ ($0$ at $\theta = \frac{\pi}{2}$, $1$ at $\theta = \pi$), and the expression increases with $\cos^2\theta$.
$m = 1 - \frac{1}{2} = \frac{1}{2}$, $M = \frac{25}{16} - \frac{1}{2} = \frac{17}{16}$.
$16(M + m) = 17 + 8 = 25$.
Answer: (D) $25$. | ||||
| 3 | Complex NumbersArgand diagram | Class 11: Complex Numbers and Quadratic Equations | Single correctMedium | |
A medium single correct question on argand diagram, from Complex Numbers. NCERT: Class 11: Complex Numbers and Quadratic Equations Put $z = x + iy$: $z(1 + i) = (x - y) + i(x + y)$ and $\bar{z}(1 - i) = (x - y) - i(x + y)$. Their sum is $2(x - y) = 4$, so the curve is the line $x - y = 2$.
The region is the disc with centre $(3, 0)$ and radius $1$. Distance of the centre from the line: $\frac{|3 - 0 - 2|}{\sqrt{2}} = \frac{1}{\sqrt{2}}$.
The chord subtends a central angle $2\theta$ with $\cos\theta = \frac{1}{\sqrt{2}}$, so $2\theta = \frac{\pi}{2}$.
Minor segment: $\frac{1}{2}r^2\left(\frac{\pi}{2} - \sin\frac{\pi}{2}\right) = \frac{\pi}{4} - \frac{1}{2}$.
Major segment: $\pi - \left(\frac{\pi}{4} - \frac{1}{2}\right) = \frac{3\pi}{4} + \frac{1}{2}$.
$|\alpha - \beta| = \frac{\pi}{2} + 1$.
Answer: (A) $1 + \frac{\pi}{2}$. | ||||
| 4 | Matrices and DeterminantsConsistency of linear systems | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on consistency of linear systems, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants Eliminate $x$. $R_2 - 2R_1$: $y + (a - 4)z = a - 11$. $R_3 + R_1$: $-2y + (b + 2)z = 2b + 6$.
Add twice the first of these to the second: $(2a + b - 6)z = 2a + 2b - 16$.
Infinitely many solutions need $0 \cdot z = 0$: $2a + b = 6$ and $a + b = 8$.
So $a = -2$, $b = 10$, and $7a + 3b = -14 + 30 = 16$.
Answer: (C) $16$. | ||||
| 5 | Matrices and DeterminantsAdjoint and inverse | Class 12: Matrices; Class 12: Determinants | Single correctMedium | |
A medium single correct question on adjoint and inverse, from Matrices and Determinants. NCERT: Class 12: Matrices; Class 12: Determinants For an $n \times n$ matrix $M$, $\mathrm{adj}(\mathrm{adj}\,M) = |M|^{n-2}M$; for $n = 3$ it is $|M|\,M$.
With $M = 2A$: $|M| = 2^3 \cdot \frac{1}{2} = 4$, so $B = 4 \cdot 2A = 8A$.
$|B| = 8^3 \cdot \frac{1}{2} = 256$ and trace $(B) = 8 \times 3 = 24$.
$|B| + \text{trace}(B) = 280$.
Answer: (B) $280$. | ||||
| 6 | Sequences and SeriesArithmetic progressions | Class 11: Sequences and Series | Single correctEasy | |
A easy single correct question on arithmetic progressions, from Sequences and Series. NCERT: Class 11: Sequences and Series There are $k$ odd-placed and $k$ even-placed terms, and each even-placed term exceeds the odd-placed term before it by $d$.
So (sum of even terms) $-$ (sum of odd terms) $= kd = 55 - 40 = 15$.
Last term $-$ first term $= (2k - 1)d = 2kd - d = 30 - d = 27$, so $d = 3$ and $k = 5$.
Answer: (B) $5$. | ||||
| 7 | Permutations and CombinationsSimple applications | Class 11: Permutations and Combinations | Single correctEasy | |
A easy single correct question on simple applications, from Permutations and Combinations. NCERT: Class 11: Permutations and Combinations The girls' block and the boys' block can be in either order: $2$ ways.
Girls within their block: $3! = 6$.
Boys within their block with $B_1$, $B_2$ not adjacent: $4! - 2 \cdot 3! = 24 - 12 = 12$.
Total $= 2 \times 6 \times 12 = 144$.
Answer: (D) $144$. | ||||
| 8 | Binomial TheoremBinomial theorem for a positive integral index | Class 11: Binomial Theorem | Single correctMedium | |
A medium single correct question on binomial theorem for a positive integral index, from Binomial Theorem. NCERT: Class 11: Binomial Theorem With $a = \sqrt{x^3 - 1}$, the odd powers of $a$ cancel:
$(x + a)^5 + (x - a)^5 = 2\left[x^5 + 10x^3a^2 + 5xa^4\right]$.
Put $a^2 = x^3 - 1$: $2\left[x^5 + 10x^3(x^3 - 1) + 5x(x^3 - 1)^2\right] = 2\left[5x^7 + 10x^6 + x^5 - 10x^4 - 10x^3 + 5x\right]$.
So $\alpha = 10$, $\beta = 2$, $\gamma = -20$, $\delta = 10$.
$10u + 2v = 18 \Rightarrow 5u + v = 9$; $-20u + 10v = 20 \Rightarrow -2u + v = 2$.
Subtracting: $7u = 7$, $u = 1$, $v = 4$, and $u + v = 5$.
Answer: (C) $5$. | ||||
| 9 | ProbabilityAddition and multiplication theorems | Class 11: Probability; Class 12: Probability | Single correctMedium | |
A medium single correct question on addition and multiplication theorems, from Probability. NCERT: Class 11: Probability; Class 12: Probability $12x^2 - 7x + 1 = (3x - 1)(4x - 1)$, roots $\frac{1}{3}$ and $\frac{1}{4}$.
$P(A|B) = \frac{0.1}{P(B)}$ and $P(B|A) = \frac{0.1}{P(A)}$, so $\{P(A), P(B)\} = \{0.3, 0.4\}$.
$P(\bar{A} \cup \bar{B}) = 1 - P(A \cap B) = 0.9$.
$P(\bar{A} \cap \bar{B}) = 1 - P(A \cup B) = 1 - (0.3 + 0.4 - 0.1) = 0.4$.
Ratio $= \frac{0.9}{0.4} = \frac{9}{4}$.
Answer: (D) $\frac{9}{4}$. | ||||
| 10 | Conic SectionsParabola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on parabola, from Conic Sections. NCERT: Class 11: Conic Sections $48 = 16a$, so $a = 3$: the parabola is $y^2 = 12x$, focus $(3, 0)$, directrix $x = -3$.
In parametric form $(3t^2, 6t)$, P has $t_1 = \frac{2}{\sqrt{3}}$. For a focal chord $t_1t_2 = -1$, so $t_2 = -\frac{\sqrt{3}}{2}$ and $Q = \left(\frac{9}{4}, -3\sqrt{3}\right)$.
$M = (-3, 4\sqrt{3})$, $N = (-3, -3\sqrt{3})$.
PQMN is a trapezium with parallel horizontal sides $PM = 4 + 3 = 7$ and $QN = \frac{9}{4} + 3 = \frac{21}{4}$, and height $4\sqrt{3} + 3\sqrt{3} = 7\sqrt{3}$.
Area $= \frac{1}{2}\left(7 + \frac{21}{4}\right) \cdot 7\sqrt{3} = \frac{49}{8} \cdot 7\sqrt{3} = \frac{343\sqrt{3}}{8}$.
Answer: (B) $\frac{343\sqrt{3}}{8}$. | ||||
| 11 | Conic SectionsHyperbola | Class 11: Conic Sections | Single correctMedium | |
A medium single correct question on hyperbola, from Conic Sections. NCERT: Class 11: Conic Sections Read as: the distance between the foci of E is $2\sqrt{3}$, and so is the distance between the foci of H. With eccentricities $e_1$, $e_2$: $ae_1 = Ae_2 = \sqrt{3}$.
$\frac{e_1}{e_2} = \frac{1}{3}$ gives $\frac{A}{a} = \frac{e_1}{e_2} = \frac{1}{3}$, so $a = 3A$; with $a - A = 2$: $A = 1$, $a = 3$.
Ellipse: $b^2 = a^2 - a^2e_1^2 = 9 - 3 = 6$. Latus rectum $\frac{2b^2}{a} = \frac{12}{3} = 4$.
Hyperbola: $B^2 = A^2e_2^2 - A^2 = 3 - 1 = 2$. Latus rectum $\frac{2B^2}{A} = 4$.
Sum $= 8$.
Answer: (B) $8$. | ||||
| 12 | TrigonometryTrigonometric identities | Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions | Single correctEasy | |
A easy single correct question on trigonometric identities, from Trigonometry. NCERT: Class 11: Trigonometric Functions; Class 12: Inverse Trigonometric Functions First equation: $2\sin^2\theta = 1 - 2\sin^2\theta$, so $\sin^2\theta = \frac{1}{4}$.
Second: $2 - 2\sin^2\theta = 3\sin\theta$, i.e. $2\sin^2\theta + 3\sin\theta - 2 = 0$, $(2\sin\theta - 1)(\sin\theta + 2) = 0$, so $\sin\theta = \frac{1}{2}$ (which satisfies the first as well).
In $[0, 2\pi]$: $\theta = \frac{\pi}{6}, \frac{5\pi}{6}$. Sum $= \pi$.
Answer: (C) $\pi$. | ||||
| 13 | Vector AlgebraScalar and vector products | Class 12: Vector Algebra | Single correctEasy | |
A easy single correct question on scalar and vector products, from Vector Algebra. NCERT: Class 12: Vector Algebra $\vec{a}\cdot\vec{a} = \vec{b}\cdot\vec{b} = 1$, $\vec{a}\cdot\vec{b} = \frac{1}{2}$.
$(\lambda\vec{a} + 2\vec{b})\cdot(3\vec{a} - \lambda\vec{b}) = 3\lambda - \frac{\lambda^2}{2} + 3 - 2\lambda = -\frac{\lambda^2}{2} + \lambda + 3 = 0$.
$\lambda^2 - 2\lambda - 6 = 0$, $\lambda = 1 \pm \sqrt{7}$, i.e. about $3.65$ and $-1.65$. Neither lies in $[-1, 3]$.
Answer: (A) $0$. | ||||
| 14 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctEasy | |
A easy single correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry The line passes through $A(1, -2, -3)$ with direction $\vec{d} = (2, -1, 2)$, $|\vec{d}| = 3$.
$\vec{AP} = (1, -8, 4)$, $|\vec{AP}|^2 = 81$. Projection on the line: $\frac{\vec{AP}\cdot\vec{d}}{|\vec{d}|} = \frac{2 + 8 + 8}{3} = 6$.
Distance$^2 = 81 - 36 = 45$, distance $= 3\sqrt{5}$.
Answer: (C) $3\sqrt{5}$. | ||||
| 15 | Three Dimensional GeometryEquation of a line | Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry | Single correctMedium | |
A medium single correct question on equation of a line, from Three Dimensional Geometry. NCERT: Class 11: Introduction to Three Dimensional Geometry; Class 12: Three Dimensional Geometry $Q = (-2 + 3t, -1 + 2t, 3 + 2t)$, $t \neq 0$.
$QR^2 = (3 - 3t)^2 + (4 - 2t)^2 + (2t)^2 = 17t^2 - 34t + 25 = 25$, so $t = 2$ and $Q = (4, 3, 7)$.
$\vec{PQ} = (6, 4, 4)$, $\vec{PR} = (3, 4, 0)$, $\vec{PQ}\times\vec{PR} = (-16, 12, 12)$, $|\vec{PQ}\times\vec{PR}|^2 = 544$.
(Area)$^2 = \frac{544}{4} = 136$.
Answer: (A) $136$. | ||||
| 17 | Applications of DerivativesMaxima and minima of one variable | Class 12: Application of Derivatives | Single correctMedium | |
A medium single correct question on maxima and minima of one variable, from Applications of Derivatives. NCERT: Class 12: Application of Derivatives $f'(x) = 2x \cdot \frac{x^4 - 8x^2 + 15}{e^{x^2}} = 2x(x^2 - 3)(x^2 - 5)e^{-x^2}$.
Critical points: $0, \pm\sqrt{3}, \pm\sqrt{5}$. Sign of $f'$ on the intervals from left to right:
$(-\infty, -\sqrt{5})$: $-$; $(-\sqrt{5}, -\sqrt{3})$: $+$; $(-\sqrt{3}, 0)$: $-$; $(0, \sqrt{3})$: $+$; $(\sqrt{3}, \sqrt{5})$: $-$; $(\sqrt{5}, \infty)$: $+$.
Minima at $-\sqrt{5}, 0, \sqrt{5}$ (sign $-$ to $+$); maxima at $-\sqrt{3}, \sqrt{3}$.
2 local maxima and 3 local minima.
Answer: (B) 2 and 3. | ||||
| 18 | Integral CalculusArea under simple curves | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on area under simple curves, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Put $u = x - 2$: the curves are $y = u^2$ and $y^2 = -8u$.
Intersections: $u^4 = -8u$, so $u = 0$ ($y = 0$) or $u = -2$ ($y = 4$).
For $0 \le y \le 4$ the region lies between $u = -\sqrt{y}$ (left) and $u = -\frac{y^2}{8}$ (right).
Area $= \int_0^4\left(\sqrt{y} - \frac{y^2}{8}\right)dy = \frac{2}{3}\cdot 8 - \frac{64}{24} = \frac{16}{3} - \frac{8}{3} = \frac{8}{3}$.
Answer: (D) $\frac{8}{3}$. | ||||
| 19 | Differential EquationsLinear differential equations | Class 12: Differential Equations | Single correctMedium | |
A medium single correct question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations Rewrite with $x$ as a function of $y$: $\frac{dx}{dy} + \frac{x}{1 + y^2} = \frac{2e^{\tan^{-1}y}}{1 + y^2}$.
Integrating factor $e^{\tan^{-1}y}$: $\frac{d}{dy}\left(x e^{\tan^{-1}y}\right) = \frac{2e^{2\tan^{-1}y}}{1 + y^2}$, so $x e^{\tan^{-1}y} = e^{2\tan^{-1}y} + C$.
$y = 0$, $x = 1$: $C = 0$. Hence $x = e^{\tan^{-1}y}$ and $f\left(\frac{1}{\sqrt{3}}\right) = e^{\pi/6}$.
Answer: (A) $e^{\pi/6}$. | ||||
| 20 | Integral CalculusIntegral as an antiderivative | Class 12: Integrals; Class 12: Application of Integrals | Single correctMedium | |
A medium single correct question on integral as an antiderivative, from Integral Calculus. NCERT: Class 12: Integrals; Class 12: Application of Integrals Let $h(x) = \frac{x\sin^{-1}x}{\sqrt{1 - x^2}}$. Then
$h'(x) = \frac{\sin^{-1}x}{\sqrt{1 - x^2}} + \frac{x}{1 - x^2} + \frac{x^2\sin^{-1}x}{(1 - x^2)^{3/2}} = \frac{\sin^{-1}x}{(1 - x^2)^{3/2}} + \frac{x}{1 - x^2}$.
So the integrand is $e^x\left(h + h'\right)$, whose integral is $e^x h(x) + C$: $g(x) = \frac{x e^x\sin^{-1}x}{\sqrt{1 - x^2}}$.
$g\left(\frac{1}{2}\right) = \frac{\frac{1}{2}\sqrt{e}\cdot\frac{\pi}{6}}{\frac{\sqrt{3}}{2}} = \frac{\pi}{6}\cdot\frac{\sqrt{e}}{\sqrt{3}} = \frac{\pi}{6}\sqrt{\frac{e}{3}}$.
Answer: (C) $\frac{\pi}{6}\sqrt{\frac{e}{3}}$. | ||||
| 21 | Binomial TheoremSimple applications | Class 11: Binomial Theorem | NumericalMedium | |
A medium numerical question on simple applications, from Binomial Theorem. NCERT: Class 11: Binomial Theorem $\frac{{}^{30}C_r}{{}^{30}C_{r-1}} = \frac{31 - r}{r}$, so each term is $r^2\cdot{}^{30}C_r\cdot\frac{31 - r}{r} = r(31 - r)\,{}^{30}C_r$.
$\sum r\,{}^{30}C_r = 30\cdot2^{29}$ and $\sum r^2\,{}^{30}C_r = 30\cdot31\cdot2^{28}$.
Sum $= 31\cdot30\cdot2^{29} - 30\cdot31\cdot2^{28} = 30\cdot31\cdot2^{28} = 465\cdot2^{29}$.
Answer: 465. | ||||
| 22 | Straight LinesCentroid, orthocentre and circumcentre | Class 11: Straight Lines | NumericalMedium | |
A medium numerical question on centroid, orthocentre and circumcentre, from Straight Lines. NCERT: Class 11: Straight Lines All three vertices are at distance 10 from the origin, so the circumcentre is $O(0, 0)$. Then the orthocentre is $A + B + C$ (as vectors) and the centroid is $\frac{A + B + C}{3}$.
$A + B + C = \left(6 + 10(\cos\alpha - \sin\alpha),\ 8 + 10(\cos\alpha - \sin\alpha)\right)$.
$8 + 10(\cos\alpha - \sin\alpha) = 9$ gives $\cos\alpha - \sin\alpha = 0.1$, so $a = 7$.
$G = \left(\frac{7}{3}, 3\right)$: $h = \frac{7}{3}$, $k = 3$.
$(\cos\alpha - \sin\alpha)^2 = 1 - \sin2\alpha = 0.01$, so $100\sin2\alpha = 99$.
$5a - 3h + 6k + 100\sin2\alpha = 35 - 7 + 18 + 99 = 145$.
Answer: 145. | ||||
| 23 | Differential EquationsLinear differential equations | Class 12: Differential Equations | NumericalMedium | |
A medium numerical question on linear differential equations, from Differential Equations. NCERT: Class 12: Differential Equations Integrating factor: $e^{\int\frac{x}{x^2 - 1}dx} = \sqrt{1 - x^2}$ on $(-1, 1)$.
$\frac{d}{dx}\left(y\sqrt{1 - x^2}\right) = x^6 + 4x$, so $y\sqrt{1 - x^2} = \frac{x^7}{7} + 2x^2 + C$; $f(0) = 0$ gives $C = 0$.
$f(x) = \frac{x^7/7}{\sqrt{1 - x^2}} + \frac{2x^2}{\sqrt{1 - x^2}}$. The first part is odd and integrates to 0 over $[-\frac{1}{2}, \frac{1}{2}]$.
$\int_{-1/2}^{1/2}\frac{2x^2}{\sqrt{1 - x^2}}dx = 4\int_0^{\pi/6}\sin^2\theta\,d\theta = 4\left(\frac{\pi}{12} - \frac{\sqrt{3}}{8}\right) = \frac{\pi}{3} - \frac{\sqrt{3}}{2}$ (with $x = \sin\theta$).
$6\int f = 2\pi - 3\sqrt{3}$, so $\alpha = 3\sqrt{3}$ and $\alpha^2 = 27$.
Answer: 27. | ||||
| 24 | Straight LinesCartesian coordinates and distance formula | Class 11: Straight Lines | NumericalMedium | |
A medium numerical question on cartesian coordinates and distance formula, from Straight Lines. NCERT: Class 11: Straight Lines Take the lines $y = 0$ and $y = 5$ and $P = (0, 1)$. Let the side be $s$, with $\vec{PQ} = s(\cos\theta, \sin\theta)$ and $\vec{PR} = s(\cos(\theta + 60^\circ), \sin(\theta + 60^\circ))$.
Q on $y = 0$: $s\sin\theta = -1$. R on $y = 5$: $s\sin(\theta + 60^\circ) = 4$, i.e. $\frac{1}{2}s\sin\theta + \frac{\sqrt{3}}{2}s\cos\theta = 4$, so $s\cos\theta = \frac{9}{\sqrt{3}} = 3\sqrt{3}$.
$s^2 = (s\cos\theta)^2 + (s\sin\theta)^2 = 27 + 1 = 28$. (The other orientations, or Q on $y = 5$, give the same value.)
$QR^2 = s^2 = 28$.
Answer: 28. | ||||
| 25 | Sets, Relations and FunctionsTypes of relations | Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions | NumericalMedium | |
A medium numerical question on types of relations, from Sets, Relations and Functions. NCERT: Class 11: Sets; Class 11: Relations and Functions; Class 12: Relations and Functions Reflexivity and transitivity force $R_0 = \{(1,1), (2,2), (3,3), (1,2), (2,3), (1,3)\}$. The remaining pairs that may be added are $(2,1)$, $(3,2)$, $(3,1)$.
Check each choice for transitivity:
nothing added: transitive;
$(2,1)$ alone: transitive;
$(3,2)$ alone: transitive;
$(3,1)$ alone: $(3,1),(1,2)$ needs $(3,2)$ — fails;
$(2,1),(3,2)$: $(3,2),(2,1)$ needs $(3,1)$ — fails;
$(2,1),(3,1)$: $(3,1),(1,2)$ needs $(3,2)$ — fails;
$(3,2),(3,1)$: $(2,3),(3,1)$ needs $(2,1)$ — fails;
all three: the full relation, which is symmetric — excluded.
The three transitive ones are not symmetric (each lacks some reverse pair).
Answer: 3. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 26 | Physical World and MeasurementDimensional analysis | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on dimensional analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement Capacitance $= \frac{Q}{V}$ and potential $V = \frac{W}{Q}$, so capacitance $= \frac{Q^2}{W}$.
$[W] = ML^2T^{-2}$, so $[F] = \frac{C^2}{ML^2T^{-2}} = C^2M^{-1}L^{-2}T^2$.
Answer: (D) $[F] = [C^2M^{-1}L^{-2}T^2]$. | ||||
| 27 | Physical World and MeasurementError analysis | Class 11: Units and Measurement | Single correctEasy | |
A easy single correct question on error analysis, from Physical World and Measurement. NCERT: Class 11: Units and Measurement $\rho = \frac{m}{\pi r^2 L}$, so $\frac{\Delta\rho}{\rho} = \frac{\Delta m}{m} + 2\frac{\Delta r}{r} + \frac{\Delta L}{L}$.
$= \frac{0.003}{0.60} + 2\times\frac{0.01}{0.50} + \frac{0.05}{10.00} = 0.005 + 0.04 + 0.005 = 0.05$, i.e. $5\%$.
Answer: (C) $5$. | ||||
| 28 | Motion of System of Particles and Rigid BodyTorque and rotational dynamics | Class 11: Systems of Particles and Rotational Motion | Single correctEasy | |
A easy single correct question on torque and rotational dynamics, from Motion of System of Particles and Rigid Body. NCERT: Class 11: Systems of Particles and Rotational Motion $\vec{\tau} = \vec{r}\times\vec{F} = (\hat{i} + \hat{j} + \hat{k})\times(2\hat{i} + \hat{j} + 2\hat{k})$
$= \hat{i}(1\cdot2 - 1\cdot1) - \hat{j}(1\cdot2 - 1\cdot2) + \hat{k}(1\cdot1 - 1\cdot2) = \hat{i} - \hat{k}$.
Answer: (A) $\hat{i} - \hat{k}$. | ||||
| 29 | KinematicsProjectile motion | Class 11: Motion in a Straight Line; Class 11: Motion in a Plane | Single correctEasy | |
A easy single correct question on projectile motion, from Kinematics. NCERT: Class 11: Motion in a Straight Line; Class 11: Motion in a Plane Initial kinetic energy $= \frac{1}{2}(0.1)(20)^2 = 20$ J.
At the highest point only the horizontal component remains: $20\cos60^\circ = 10$ m/s, kinetic energy $= \frac{1}{2}(0.1)(10)^2 = 5$ J.
Decrease $= 20 - 5 = 15$ J.
Answer: (B) 15 J. | ||||
| 31 | Work, Energy and PowerWork-energy theorem | Class 11: Work, Energy and Power | Single correctEasy | |
A easy single correct question on work-energy theorem, from Work, Energy and Power. NCERT: Class 11: Work, Energy and Power $W = \vec{F}\cdot\vec{s} = 2(1) + b(-2) + 1(-1) = 1 - 2b = 0$, so $b = \frac{1}{2}$.
Answer: (A) $\frac{1}{2}$. | ||||
| 32 | Properties of Bulk MatterFluid mechanics | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctEasy | |
A easy single correct question on fluid mechanics, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter Equation of continuity: $A_1v_1 = A_2v_2$, i.e. $\pi r_1^2v_1 = \pi r_2^2v_2$.
$v_2 = v_1\left(\frac{r_1}{r_2}\right)^2 = 2\times\left(\frac{2}{1}\right)^2 = 8$ m/s.
Answer: (B) 8 m/s. | ||||
| 33 | Properties of Bulk MatterViscosity | Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter | Single correctEasy | |
A easy single correct question on viscosity, from Properties of Bulk Matter. NCERT: Class 11: Mechanical Properties of Solids; Class 11: Mechanical Properties of Fluids; Class 11: Thermal Properties of Matter At terminal velocity the net force is zero: weight $=$ buoyant force $+$ viscous force.
Buoyant force $= V\rho_{gly}g = V\frac{\rho_{ball}}{2}g = \frac{Mg}{2}$.
Viscous force $= Mg - \frac{Mg}{2} = \frac{Mg}{2}$.
Answer: (D) $\frac{Mg}{2}$. | ||||
| 34 | Behaviour of Perfect Gas and Kinetic TheoryDegrees of freedom | Class 11: Kinetic Theory | Single correctEasy | |
A easy single correct question on degrees of freedom, from Behaviour of Perfect Gas and Kinetic Theory. NCERT: Class 11: Kinetic Theory Rigid diatomic: 5 degrees of freedom, $C_v = \frac{5}{2}R$, $\gamma_1 = \frac{7}{5} = 1.4$.
With the vibrational mode active (2 more quadratic terms): $C_v = \frac{7}{2}R$, $\gamma_2 = \frac{9}{7} \approx 1.29$.
More degrees of freedom lower $\gamma$: $\gamma_2 < \gamma_1$.
Answer: (B) $\gamma_2 < \gamma_1$. | ||||
| 35 | ThermodynamicsThermodynamic processes | Class 11: Thermodynamics | StatementsEasy | |
A easy statements question on thermodynamic processes, from Thermodynamics. NCERT: Class 11: Thermodynamics Extensive variables depend on the amount of the system (internal energy, volume, mass); intensive ones do not (pressure, temperature, density).
(A) true. (B) true. (C) false: volume is extensive. (D) false: temperature is intensive.
Answer: (A) (A) and (B) Only. | ||||
| 36 | GravitationAcceleration due to gravity | Class 11: Gravitation | Assertion and reasonEasy | |
A easy assertion and reason question on acceleration due to gravity, from Gravitation. NCERT: Class 11: Gravitation On the planet $g' = \frac{G(4M)}{(2R)^2} = \frac{GM}{R^2} = g$, so $T = 2\pi\sqrt{\frac{l}{g}}$ is unchanged: (A) is true.
Mass is a property of the bob and does not change from place to place: (R) is true.
But the period of a simple pendulum does not depend on the bob's mass at all; it is the same because $g$ is the same. (R) does not explain (A).
Answer: (B). | ||||
| 38 | Electromagnetic Induction and Alternating CurrentsFaraday's law | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on faraday's law, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current While the loop is partly inside, the flux is $\Phi = B\,l\,x$, where $l$ is the side perpendicular to the motion and $x$ the length still inside. $x$ decreases at the constant rate $v$.
$\varepsilon = \left|\frac{d\Phi}{dt}\right| = Blv$, which is constant in time.
The graph is a horizontal line.
Answer: (A). | ||||
| 39 | Electromagnetic Induction and Alternating CurrentsLCR circuits and resonance | Class 12: Electromagnetic Induction; Class 12: Alternating Current | Single correctEasy | |
A easy single correct question on lcr circuits and resonance, from Electromagnetic Induction and Alternating Currents. NCERT: Class 12: Electromagnetic Induction; Class 12: Alternating Current At resonance $X_L = X_C$ and the impedance is $Z = R$, so $I_0 = \frac{E_0}{R}$.
Doubling $R$ (the resonant frequency $\frac{1}{2\pi\sqrt{LC}}$ is unchanged) halves the amplitude: $\frac{I_0}{2}$.
Answer: (B) $\frac{I_0}{2}$. | ||||
| 49 | Magnetic Effects of Current and MagnetismForce on moving charges and currents | Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter | NumericalEasy | |
A easy numerical question on force on moving charges and currents, from Magnetic Effects of Current and Magnetism. NCERT: Class 12: Moving Charges and Magnetism; Class 12: Magnetism and Matter Circular motion in B alone: $r = \frac{mv}{qB}$, so $B = \frac{mv}{qr} = \frac{1.6\times10^{-27}\times2\times10^5}{1.6\times10^{-19}\times0.02} = 0.1$ T.
Undeflected motion: $qE = qvB$, so $E = vB = 2\times10^5\times0.1 = 2\times10^4$ N/C.
Answer: 2. | ||||
| Q | Chapter and topic | NCERT | Kind and level | Explanation |
|---|---|---|---|---|
| 55 | SolutionsConcentration terms | Class 12: Solutions | Single correctEasy | |
A easy single correct question on concentration terms, from Solutions. NCERT: Class 12: Solutions Take 1 L of solution: mass $= 1000\times1.25 = 1250$ g.
NaCl: 3 mol $= 3\times58.5 = 175.5$ g. Solvent $= 1250 - 175.5 = 1074.5$ g.
Molality $= \frac{3}{1.0745} = 2.79$ m.
Answer: (B) 2.79 m. | ||||
| 56 | Redox ReactionsOxidation number | Class 11: Redox Reactions | Single correctEasy | |
A easy single correct question on oxidation number, from Redox Reactions. NCERT: Class 11: Redox Reactions Disproportionation needs an element in an intermediate oxidation state, so that it can be both oxidised and reduced.
Chlorine is $+1$ in $\mathrm{ClO^-}$, $+3$ in $\mathrm{ClO_2^-}$, $+5$ in $\mathrm{ClO_3^-}$ — all can disproportionate (e.g. $3\mathrm{ClO^-} \rightarrow 2\mathrm{Cl^-} + \mathrm{ClO_3^-}$).
In $\mathrm{ClO_4^-}$ chlorine is $+7$, its highest state; it cannot be oxidised further, so it cannot disproportionate.
Answer: (D) $\mathrm{ClO_4^-}$. | ||||
| 57 | Chemical KineticsArrhenius equation | Class 12: Chemical Kinetics | Single correctEasy | |
A easy single correct question on arrhenius equation, from Chemical Kinetics. NCERT: Class 12: Chemical Kinetics The product lies $E_2$ above the reactant (endothermic), and the activated complex lies $E_1$ above the product.
Forward activation energy (reactant to peak) $= E_1 + E_2$; backward activation energy (product to peak) $= E_1$.
The product has the higher energy, so it is less stable than the reactant.
(B) and (D) say the product is more stable — wrong; (C) gives $E_1 + E_2$ for the backward reaction — wrong.
Answer: (A). | ||||
| 58 | Chemical Bonding and Molecular StructureDipole moment and Fajans' rule | Class 11: Chemical Bonding and Molecular Structure | Single correctMedium | |
A medium single correct question on dipole moment and fajans' rule, from Chemical Bonding and Molecular Structure. NCERT: Class 11: Chemical Bonding and Molecular Structure In $\mathrm{NF_3}$ the lone pair's moment opposes the N–F bond moments, leaving a very small dipole (about 0.24 D).
HBr: about 0.8 D. $\mathrm{H_2S}$ (bent): about 0.95 D. $\mathrm{CHCl_3}$: about 1.0 D.
Increasing order: $\mathrm{NF_3 < HBr < H_2S < CHCl_3}$.
Answer: (A). | ||||
| 59 | Classification of Elements and PeriodicityPeriodic trends | Class 11: Classification of Elements and Periodicity in Properties | StatementsEasy | |
A easy statements question on periodic trends, from Classification of Elements and Periodicity. NCERT: Class 11: Classification of Elements and Periodicity in Properties Elements at the extreme left (alkali metals) form basic oxides, e.g. $\mathrm{Na_2O + H_2O \rightarrow 2NaOH}$. Statement I is false.
The reactive elements at the extreme right (halogens; the noble gases are unreactive) form acidic oxides, which give acids with water, e.g. $\mathrm{Cl_2O_7 + H_2O \rightarrow 2HClO_4}$. Statement II is true.
Answer: (D). | ||||
| 60 | p-Block ElementsGroup 15 elements | No matching NCERT chapter | Single correctMedium | |
A medium single correct question on group 15 elements, from p-Block Elements. NCERT: No matching NCERT chapter The non-metals of group 15 are nitrogen and phosphorus. The N–N single bond is weaker than the P–P single bond, because the lone pairs on the small nitrogen atoms repel each other strongly at the short bond length.
So E is nitrogen. Having no d orbitals, nitrogen can use only its four valence orbitals (one s, three p): its maximum covalency is 4 (as in $\mathrm{NH_4^+}$).
Answer: (B) $4$. | ||||
| 61 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | Single correctMedium | |
A medium single correct question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Homoleptic means only one kind of ligand; (A) has CN$^-$ and NO, so it is heteroleptic and excluded.
(B) $\mathrm{Co^{3+}}$, $d^6$ with weak-field F$^-$: high spin.
(C) $\mathrm{Fe^{2+}}$, $d^6$ with strong-field CN$^-$: $t_{2g}^6$, low spin.
(D) $\mathrm{Co^{3+}}$, $d^6$ with NH$_3$: $t_{2g}^6$, low spin (diamagnetic).
(E) $\mathrm{Cr^{2+}}$, $d^4$ with weak-field H$_2$O: high spin.
Low-spin homoleptic: (C) and (D).
Answer: (A) (C) and (D) only. | ||||
| 62 | Coordination CompoundsCrystal field theory | Class 12: Coordination Compounds | Single correctMedium | |
A medium single correct question on crystal field theory, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Compare the magnitudes of CFSE.
$\mathrm{[Co(NH_3)_4]^{2+}}$: tetrahedral $d^7$ ($e^4t_2^3$), CFSE $= 1.2\Delta_t \approx 0.53\Delta_o$ — the smallest (and $\Delta_t \approx \frac{4}{9}\Delta_o$).
$\mathrm{[Co(NH_3)_6]^{2+}}$: octahedral $d^7$ high spin ($t_{2g}^5e_g^2$), CFSE $= 0.8\Delta_o$, with the small $\Delta_o$ of a $+2$ ion.
$\mathrm{[Co(NH_3)_6]^{3+}}$: $d^6$ low spin ($t_{2g}^6$), CFSE $= 2.4\Delta_o$, and $\Delta_o$ is larger for $\mathrm{Co^{3+}}$.
$\mathrm{[Co(en)_3]^{3+}}$: also $t_{2g}^6$, but en is a stronger-field ligand than NH$_3$ (chelate), so $\Delta_o$ and the CFSE are the largest.
Order: $\mathrm{[Co(NH_3)_4]^{2+} < [Co(NH_3)_6]^{2+} < [Co(NH_3)_6]^{3+} < [Co(en)_3]^{3+}}$.
Answer: (C). | ||||
| 63 | Organic Chemistry: Some Basic Principles and TechniquesQualitative and quantitative analysis | Class 11: Organic Chemistry – Some Basic Principles and Techniques | StatementsEasy | |
A easy statements question on qualitative and quantitative analysis, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Statement I is the standard description of Lassaigne's test: nitrogen, sulphur, halogens and phosphorus in an organic compound are detected through the sodium fusion extract. True.
In Lassaigne's test the compound is fused with sodium metal, not magnesium; sodium converts the covalently bound elements into ionic NaCN, Na$_2$S, NaX (and phosphate after oxidation). Statement II is false.
Answer: (C). | ||||
| 64 | HydrocarbonsAlkanes | Class 11: Hydrocarbons | Single correctHard | |
A hard single correct question on alkanes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons Bromination in sunlight is a free-radical substitution, and bromine is highly selective: the major product comes from the most stable radical, i.e. abstraction of a tertiary hydrogen.
sec-Butylcyclohexane, $\mathrm{C_6H_{11}{-}CH(CH_3){-}CH_2CH_3}$, has two tertiary hydrogens: on the side-chain CH and on the ring carbon that carries the side chain. Options (C) and (D) come from secondary hydrogens and are minor.
The side-chain tertiary radical has six $\alpha$-hydrogens for hyperconjugation (CH$_3$: 3, CH$_2$: 2, ring CH: 1), the ring one only five (two ring CH$_2$: 4, side-chain CH: 1), so the side-chain radical is the more stable and gives the major product: Br on the side-chain carbon.
Answer: (A). | ||||
| 65 | Organic Chemistry: Some Basic Principles and TechniquesReaction intermediates | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctEasy | |
A easy single correct question on reaction intermediates, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques All four are benzyl cations, stabilised by resonance with the ring; substituents change this.
para-OMe donates electrons by resonance (+R) straight into the carbon bearing the positive charge — the strongest stabilisation.
para-CH$_3$ stabilises by hyperconjugation and +I, less than OMe.
meta-OMe cannot donate by resonance to the benzylic position and withdraws by $-$I, so it is less stable than the unsubstituted benzyl cation.
Most stable: p-methoxybenzyl cation.
Answer: (B). | ||||
| 66 | Organic Chemistry: Some Basic Principles and TechniquesIUPAC nomenclature | Class 11: Organic Chemistry – Some Basic Principles and Techniques | Single correctMedium | |
A medium single correct question on iupac nomenclature, from Organic Chemistry: Some Basic Principles and Techniques. NCERT: Class 11: Organic Chemistry – Some Basic Principles and Techniques Secondary hydrogens sit on CH$_2$ groups bonded to two carbons. Count the CH$_2$ groups.
(A) $\mathrm{CH_3CH_2CH(CH_3)C(CH_3)(C_2H_5)CH_2CH_2CH_2CH_3}$: CH$_2$ at C-2, C-5, C-6, C-7 and in the ethyl group — 10 secondary H.
(B) $\mathrm{(CH_3)_3C{-}CH_2{-}C(CH_3)_2{-}CH_2CH_3}$: two CH$_2$ — 4 secondary H.
(C) $\mathrm{(CH_3)_3C{-}C(CH_3)_2{-}CH_2CH_3}$: one CH$_2$ — 2 secondary H.
(D) $\mathrm{(CH_3)_3C{-}CH_2{-}CH(CH_3){-}CH(CH_3){-}CH_2CH_3}$: two CH$_2$ — 4 secondary H.
Answer: (C) 2,2,3,3-Tetramethylpentane. | ||||
| 67 | HydrocarbonsAlkanes | Class 11: Hydrocarbons | Single correctEasy | |
A easy single correct question on alkanes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons RMgBr + H$_2$O gives RH, so R–Br must be a monobromo derivative of 2-methylbutane, $\mathrm{(CH_3)_2CH{-}CH_2{-}CH_3}$, with Br on any carbon.
The distinct positions give: 1-bromo-2-methylbutane, 2-bromo-2-methylbutane, 2-bromo-3-methylbutane and 1-bromo-3-methylbutane (the two methyls on C-2 are equivalent).
Four structural isomers.
Answer: (B) $4$. | ||||
| 68 | Aldehydes, Ketones and Carboxylic AcidsNucleophilic addition | Class 12: Aldehydes, Ketones and Carboxylic Acids | Single correctMedium | |
A medium single correct question on nucleophilic addition, from Aldehydes, Ketones and Carboxylic Acids. NCERT: Class 12: Aldehydes, Ketones and Carboxylic Acids (i) and (ii) are the Etard reaction: chromyl chloride forms the chromium complex $\mathrm{C_6H_5CH(OCrOHCl_2)_2}$, which acid hydrolysis turns into benzaldehyde (excess toluene stays unchanged).
(iii) Benzaldehyde adds sodium bisulphite: $\mathrm{C_6H_5CHO + NaHSO_3 \rightarrow C_6H_5CH(OH)SO_3Na}$, a crystalline solid that is filtered off as the residue (A), leaving toluene behind.
Dilute HCl decomposes the bisulphite adduct back into the aldehyde: (B) is benzaldehyde.
Answer: (D). | ||||
| 69 | AminesClassification and identification of amines | Class 12: Amines | MatchingMedium | |
A medium matching question on classification and identification of amines, from Amines. NCERT: Class 12: Amines (A) An amide is reduced to an amine by $\mathrm{LiAlH_4}$ followed by water: $\mathrm{RCONH_2 \rightarrow RCH_2NH_2}$ — (III).
(B) Nitrobenzene is reduced to aniline by Sn and HCl — (IV).
(C) A nitrile is reduced to a primary amine by catalytic hydrogenation, $\mathrm{H_2/Ni}$ — (II).
(D) The N-alkylphthalimide gives the primary amine $\mathrm{RNH_2}$ on hydrolysis with aqueous NaOH (Gabriel synthesis) — (I).
A-III, B-IV, C-II, D-I.
Answer: (B). | ||||
| 70 | BiomoleculesCarbohydrates | Class 12: Biomolecules | Single correctMedium | |
A medium single correct question on carbohydrates, from Biomolecules. NCERT: Class 12: Biomolecules A sugar is correlated to D-glyceraldehyde (D-series) when, with CHO at the top of the Fischer projection, the OH on the chiral carbon farthest from CHO (next to the terminal CH$_2$OH) is on the right.
(A) CHO, then OH right, OH right, CH$_2$OH: the last chiral OH is on the right — D.
(B) CHO, OH right, OH left, OH right, CH$_2$OH: last chiral OH right — D.
(C) CHO, OH right, OH left, OH left, CH$_2$OH: last chiral OH left — L.
(D) CHO is at the bottom. Turning the projection through $180^\circ$ in the plane (allowed) puts CHO on top and moves every group to the other side. The chiral carbon next to CH$_2$OH (second from the top as drawn) has OH on the left as drawn, so on the right after turning — D.
Three structures: (A), (B), (D).
Answer: (C) three. | ||||
| 71 | Some Basic Concepts of ChemistryMole concept | Class 11: Some Basic Concepts of Chemistry | NumericalEasy | |
A easy numerical question on mole concept, from Some Basic Concepts of Chemistry. NCERT: Class 11: Some Basic Concepts of Chemistry Moles of NaOH: $0.020\times2 + 0.400\times0.5 = 0.04 + 0.20 = 0.24$ mol in $0.420$ L (volumes taken as additive).
$M = \frac{0.24}{0.42} = 0.5714$ M $= 57.14\times10^{-2}$ M, nearest integer 57.
Answer: 57. | ||||
| 72 | ThermodynamicsHess law | Class 11: Thermodynamics | NumericalEasy | |
A easy numerical question on hess law, from Thermodynamics. NCERT: Class 11: Thermodynamics Formation: $\mathrm{2C(graphite) + 3H_2(g) \rightarrow C_2H_6(g)}$ $= 2\times(2) + 3\times(3) - (1)$.
$\Delta H_f^\circ = 2(-393.5) + 3(-286) - (-1550) = -787 - 858 + 1550 = -95$ kJ mol$^{-1}$.
Magnitude: 95.
Answer: 95. | ||||
| 73 | Structure of AtomElectronic configuration | Class 11: Structure of Atom | NumericalMedium | |
A medium numerical question on electronic configuration, from Structure of Atom. NCERT: Class 11: Structure of Atom Both have anomalous ground-state configurations with a single 5s electron.
Nb ($Z = 41$): $\mathrm{[Kr]\,4d^4\,5s^1}$, so $x = 4$.
Ru ($Z = 44$): $\mathrm{[Kr]\,4d^7\,5s^1}$, so $y = 7$.
$x + y = 11$.
Answer: 11. | ||||
| 74 | HydrocarbonsAlkenes and alkynes | Class 11: Hydrocarbons | NumericalMedium | |
A medium numerical question on alkenes and alkynes, from Hydrocarbons. NCERT: Class 11: Hydrocarbons $\mathrm{C_6H_6}$ has 4 degrees of unsaturation. Taking up 4 mol $\mathrm{H_2}$ means all four are $\pi$ bonds (no ring), so it is not benzene (which takes 3 mol).
Only one monobromo derivative means all six H are equivalent: hexa-2,4-diyne, $\mathrm{CH_3{-}C{\equiv}C{-}C{\equiv}C{-}CH_3}$.
Two triple bonds hold 4 $\pi$ bonds, i.e. 8 $\pi$ electrons.
Answer: 8. | ||||
| 75 | Coordination CompoundsImportance of coordination compounds | Class 12: Coordination Compounds | NumericalMedium | |
A medium numerical question on importance of coordination compounds, from Coordination Compounds. NCERT: Class 12: Coordination Compounds Dimethylglyoxime is $\mathrm{CH_3C(=NOH)C(=NOH)CH_3}$, $\mathrm{C_4H_8N_2O_2}$. Each ligand loses one oxime H to bind $\mathrm{Ni^{2+}}$, giving the red complex $\mathrm{[Ni(dmgH)_2]}$, $\mathrm{Ni(C_4H_7N_2O_2)_2}$ (the remaining two oxime H form the intramolecular O–H···O bonds).
H atoms: $2\times7 = 14$.
Answer: 14. | ||||
Sit this paper free.
Timed like the exam or chapter by chapter, with a solution for each question. Run by a non-profit trust.