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^{th}**Ionic Equilibrium**Theory

**Fundamentals of Acids, Bases & Ionic Equilibrium**

**Acids & Bases**When dissolved in water, acids release $\mathrm{H}^{+}$ ions, base release $\mathrm{OH}^{-}$ ions.

**Arrhenius Theory**When dissolved in water, the substances which release (i) $\mathrm{H}^{+}$ ions are called acids (ii) $\mathrm{OH}^{-}$ ions are called bases

**Bronsted & Lowry Concept**Acids are proton donors, bases are proton acceptors Note that as per this definition, water is not necessarily the solvent.

*When a substance is dissolved in water, it is said to react with water e.g.*$\mathrm{HCl}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{Cl}^{-}$

*; HCl donates H+ to water, hence acid.*$\mathrm{NH}_{3}+\mathrm{H}_{2} \mathrm{O} \quad \rightarrow \mathrm{NH}_{4}^{+}+\mathrm{OH}^{-}$

*; NH3 takes H+ from water, hence base.*

*For the backward reaction, $N H_{4}^{+}$ donate $H^{+}$ , hence it is an acid; $O H^{-}$ accepts H+, hence it is base. $N H_{3}$ (base) & $N H_{4}^{+}$ (acid) from conjugate acid base pair.*

**To get conjugate acid of a given species add $\mathrm{H}^{+}$ to it. e.g. conjugate acid of N2H4 is N2H5+. To get conjugate base of any species subtract H+ from it. e.g. Conjugate base of $\mathrm{NH}_{3}$ is $\mathrm{NH}_{2}^{-}$.**

*Conjugate acid and bases***Note:**

*Although $C I^{-}$ is conjugate base of HCl, it is not a base as an independent species. In fact,anions of all strong acid like $C I, N O_{3}^{-}, C l O_{4}^{-}$ etc. are neutral anions. Same is true for cations of strong bases like $K^{+}, N a^{+}, B a^{++}$ etc. When they are dissolved in water, they do not react with water (i.e. they do not undergo hydrolysis) and*these ions

*do not cause any change in pH of water (others like $\left.C N^{-} d o\right)$.*

**Basic Anions : $\mathrm{CH}_{3} \mathrm{COO}^{-}, \mathrm{OH}^{-}, \mathrm{CN}^{-}$ (Conjugate bases of weak acids)**

*Some examples of :**Acid Anions: $H S O_{3}^{-}, H S^{-}$ etc. Note that these ions are*amphoteric

*, i.e. they can behave both as an acid and as a base. e.g. for $\mathrm{H}_{2} \mathrm{PO}_{4}^{-}:$*

*Lewis Concept :**Acids are substances which accept a pair of electrons to form a coordinate bond and bases are the substances which donate a pair of electrons to form a coordinate bond.*

**Important :**$\mathrm{Ca}+\mathrm{S} \rightarrow \mathrm{Ca}^{2+}+\mathrm{S}^{2-}$ is not a Lewis acidbase reaction since dative bond is not formed.

*Lewis Acids :**As per Lewis concept, following species can acts as Lewis Acids :*

*(i) Molecules in which central atom has*incomplete octet. \text { (e.g. }\left.\mathrm{BF}_{3}, \mathrm{AlCl}_{3} \text { etc. }\right)

*(ii) Molecules which have a central atom with empty d orbitals \text { (e.g. }\left.\operatorname{six}_{4}, \mathrm{GeX}_{4}, \mathrm{PX}_{3}, \mathrm{TiCl}_{4} \text { etc. }\right)*

*(iii)*Simple Cations:

*Though all cations can be expected to be Lewis acids,*

*\mathrm{Na}^{+}, \mathrm{Ca}^{++}, \mathrm{K}^{+} \text {etc. } show no tendency to accept electrons. However \mathrm{H}^{+}, \mathrm{Ag}^{+} etc. act as Lewis acids.*

*(iv) Molecules having multiple bond between atoms of dissimilar electronegativity.*

**Lewis bases**are typically : (i) Neutral species having at least one lone pair of electrons.

**\text { Autoprotolysis (or self-ionization) constant }\left(\mathrm{K}_{\mathrm{w}}\right)=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right] \text { Hence, } \mathrm{pH}+\mathrm{pOH}=\mathrm{pK}_{\mathrm{w}} \text { at all temperatures }**

*Autoprotolysis of water (or any solvent)*

*Condition of neutrality**\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\left[\mathrm{OH}^{-}\right] \text {(for water as solvent) }*\text { At } 25^{\circ} \mathrm{C}, \mathrm{K}_{\mathrm{W}}=10^{-14} \cdot \mathrm{K}_{\mathrm{W}} increases with increase in temperature. Accordingly, the neutral point of water \left(\mathrm{pH}=7 \text { at } 25^{\circ} \mathrm{C}\right) a

*lso shifts to a value lower than 7 with increase in temperature.*

**Important**

*: \mathrm{K}_{\mathrm{W}}=10^{-14}**s a value at \text { (i) } 25^{\circ} \mathrm{C} \text { (ii) }for water only. If the temperature changes or if some other solvent is used, autoprotolysis constant will not be same.*

*** \quad \text { For dissociation of weak acids (eg. HCN), HCN + H_{ } 2 } \mathrm{O} \text { 1 } \mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{CN}^{-} \text {the equilibrium }**

*Ionisation Constant**constant expression is written as Ka =\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{CN}^{-}\right]}{[\mathrm{HCN}]}*

** For the Polyprotic acids \left(\mathrm{e} \cdot \mathrm{g} \cdot \mathrm{H}_{3} \mathrm{PO}_{4}\right) sucessive ionisation constants are denoted \text { by } \mathrm{K}_{1}, \mathrm{K}_{2}, \mathrm{K}_{3} \text { etc. } For \mathrm{H}_{3} \mathrm{PO}_{4},*\text { Similarly, } \mathrm{K}_{\mathrm{b}} \text { denotes basic dissociation constant for a base. } \text { Also, } \mathrm{pK}_{\mathrm{a}}=-\log _{10} \mathrm{K}_{\mathrm{a}}, \mathrm{pK}_{\mathrm{b}}=-\log _{10} \mathrm{K}_{\mathrm{b}}

**\text { Some Important Results: }\**left[\mathrm{H}^{+}\right] \text {concentration of }

*Case (i) A weak acid in water**Similarly for a weak base, substitute \left[\mathrm{OH}^{-}\right] \text {and } \mathrm{K}_{\mathrm{b}} instead of \left[\mathrm{H}^{+}\right] \text {and } \mathrm{K}_{\mathrm{a}}*respectively in these expressions.

*Case (ii)**(a)*

*A weak acid and a strong acid \left[\mathrm{H}^{+}\right]**is entirely due to dissociation of strong acid*(b)

**A weak base and a strong base \left[\mathrm{H}^{+}\right]**is entirely due to dissociation of strong base

*Neglect the contribution of weak acid/base usually.*

**Condition for neglecting : \text { If } \mathrm{c}_{0}**concentration of strong acid, c_{1} = concentration of

*weak acid then neglect the contribution of weak acid if \mathrm{K}_{\mathrm{a}} \leq 0.01 \mathrm{c}_{0}^{2 /} \mathrm{c}_{1}*

**Proceed by the general method of applying two conditions (i) of electroneutrality (ii) of equilibria.**

*Case (iii) Two (or more) weak acids**The accurate treatement yields a cubic equation. Assuming that acids dissociate to*

*\text { a negligible extent }\left[\text { i.e. } c_{0}-x \approx c_{0}\right] \quad\left[\mathrm{H}^{+}\right]=\left(\mathrm{K}_{1} \mathrm{c}_{1}+\mathrm{K}_{2} \mathrm{c}_{2}+\ldots+\mathrm{K}_{\mathrm{w}}\right)^{1 / 2}*

*Case (iv) When dissociation of water becomes significant:**\text { Dissociation of water contributes significantly to }\left[\mathrm{H}^{+}\right] \text {or }\left[\mathrm{OH}^{-}\right] \text {only when for }*(i)

**strong acids (or bases) : 10^{-8} \mathrm{M}<\mathrm{c}_{0}<10^{-6} \mathrm{M}**.

*Neglecting ionisation of water at*

*10^{-6} \mathrm{M} \text { causes } 1 \% \text { error (approvable). Below } 10^{-8} \mathrm{M}, Neglecting ionisation of water at 10^{-6} \mathrm{M} \text { causes } 1 \% \text { error (approvable). } Below 10^{-8} \mathrm{M}, contribution of acid (or base) can be neglected and pH can be taken to be practically 7.*

**Weak acids (or bases) : \text { When } \mathrm{K}_{\mathrm{a}} \mathrm{c}_{0}<10^{-12}**then consider dissociation of water as well.

**HYDROLYSIS***

**Salts of strong acids and strong bases**do not undergo hydrolysis. *

**Salts of a strong acids and weak bases**give an acidic solution. e.g. \mathrm{NH}_{4} \mathrm{Cl} when

*dissolved, it dissociates to give \mathrm{NH}_{4}^{+} ions and \mathrm{NH}_{4}^{+}+\mathrm{H}_{2} \mathrm{O} \quad 1 \quad \mathrm{NH}_{3}+\mathrm{H}_{3} \mathrm{O}^{+}*

*\mathrm{K}_{\mathrm{h}}=\left[\mathrm{NH}_{3}\right]\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] /\left[\mathrm{NH}_{4}^{+}\right]=\mathrm{K}_{\mathrm{W}} / \mathrm{K}_{\mathrm{b}} of conjugate base of \mathrm{NH}_{4}^{+}*

*Important!**In general : \mathrm{K}_{\mathrm{a}}(\text { of an acid }) \mathrm{xK}_{\mathrm{b}} \text { (of its conjugate base) }=\mathrm{K}_{\mathrm{w}}*

*If the degree of hydrolysis(h) is small (<<1), \quad \mathrm{h}=\sqrt{\mathrm{K}_{\mathrm{h}} \mathrm{c}_{0}}*

*Otherwise h = \frac{-\mathrm{K}_{\mathrm{h}}+\sqrt{\mathrm{K}_{\mathrm{h}}^{2}+4 \mathrm{K}_{\mathrm{h}} \mathrm{c}_{0}}}{2 \mathrm{c}_{0}}, \quad\left[\mathrm{H}^{+}\right]=\mathrm{c}_{0} \mathrm{h}*

***

**Salts of strong base and weak acid**

*give a basic solution (\mathrm{pH}>7) when dissolved in water, e.g. NaCN,*

*\mathrm{CN}^{-}+\mathrm{H}_{2} \mathrm{O} \quad 1 \quad \mathrm{HCN}+\mathrm{OH}^{-} \quad\left[\mathrm{OH}^{-}\right]=\mathrm{c}_{0} \mathrm{h}, \mathrm{h}=\sqrt{\mathrm{K}_{\mathrm{h}} \mathrm{c}_{0}}*

***

**Salts of weak base and weak acid**

*Assuming degree of hydrolysis to be same for the both the ions,*

*\mathrm{K}_{\mathrm{h}}=\mathrm{K}_{\mathrm{w}} /\left(\mathrm{K}_{\mathrm{a}} \cdot \mathrm{K}_{\mathrm{b}}\right),\left[\mathrm{H}^{+}\right]=\left[\mathrm{K}_{\mathrm{a}} \mathrm{K}_{\mathrm{w}} / \mathrm{K}_{\mathrm{b}}\right]^{1 / 2}*

**Note:**Exact treatment of this case is difficult to solve. So use this assumption in general cases. Also, degree of anion or cation will be much higher in the case of a salt of weak acid and weak base. This is because each of them gets hydrolysed, producing \mathrm{H}^{+} \text {and } \mathrm{OH}^{-} ions. These ions combine to form water and the hydrolysis equilibrium is shifted in the forward direaction.

*Buffer Solutions**are the solutions whose pH does not change significantly on adding a small quantity of strong base or on little dilution.*

*These are typically made by mixing a weak acid (or base) with its conjugate base (or acid). e.g.*

*\mathrm{CH}_{3} \mathrm{COOH} with \mathrm{CH}_{3} \mathrm{COONa}, \mathrm{NH}_{3}(\mathrm{aq}) \text { with } \mathrm{NH}_{4} \mathrm{Cl} \text { etc. }*

*\text { If }\left.\mathrm{K}_{\mathrm{a}} \text { for acid (or } \mathrm{K}_{\mathrm{b}} \text { for base }\right) (or Kb for base) is not too high, we may write :*

*Henderson’s Equation**\mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \{[\mathrm{salt}] /[\mathrm{acid}]\} for weak acid with its conjugate base.*

*\text { or } \mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}+\log \{[\mathrm{salt}] /[\text { base }]\} for weak base with its conjugate acid.*

*Important**: For good buffer capacity, [salt] : [acid ratios should be as close to one as possible. In such a case, \mathrm{pH}=\mathrm{pK}_{\mathrm{a}} (This also is the case at midpoint of titration)*

*Buffer capacity = (no. of moles of acid (or base) added to 1L) / (change in pH)*

**Indicators.**Indicator is a substance which indicates the point of equivalence in a titration by undergoing a change in its colour. They are weak acids or weak bases.

*Theory of Indicators.**The ionized and unionized forms of indicators have different colours. If 90 % or more of a particular form (ionised or unionised) is present, then its colour can be distinclty seen.In general, for an indicator which is weak acid, HIn l H+ + In–, the ratio of ionized to unionized form can be determined from*This roughly gives the range of indicators. Ranges for some popular indicators are

*Table 1 : Indicators***Equivalence point.**The point at which exactly equivalent amounts of acid and base have been mixed.

*Acid Base Titration.**For choosing a suitable indicator titration curves are of great help. In a titration curve, change in pH is plotted against the volume of alkali to a given acid. Four cases arise.*(a)

**Strong acid vs strong base.**The curve is almost vertical over the pH range 3.5-10. This abrupt change corresponds to equivalence point. Any indicator suitable. (b)

**Weak acid vs strong base.**Final solution is basic 9 at equivalence point. Vertical region (not so sharp) lies in pH range 6.5-10. So, phenolphathlene is suitable. (c)

**Strong acid vs weak base.**Final solution acidic. Vertical point in pH range 3.8-7.2. Methyl red or methyl orange suitable.

*(d)*

*Weak acid vs weak base.**No sharp change in pH. No suitable indicator.*

**Note**

*: at midpoint of titration,*

*\mathrm{pH}=\mathrm{pK}_{\mathrm{a}}, thus by pH measurements, Ka for weak acids (or K_{b} for weak bases) can be determined.*

*Polyprotic acids and bases.*

*\text { Usually } \mathrm{K}_{2}, \mathrm{K}_{3} \text { etc. can be safely neglected and only } \mathrm{K}_{1} \text { plays a significant }*

*Solubility product**\left(\mathbf{K}_{\mathbf{s p}}\right). For sparingly soluble salts (eg.*

*\mathrm{Ag}_{2} \mathrm{C}_{2} \mathrm{O}_{4}) an equilibrium which exists is*

*Precipitation.**Whenever the product of concentrations (raised to appropriate power) exceeds the solubility product, precipitation occurs.*

*Common ion effects.**Suppression of dissociation by adding an ion common with dissociation products.*

*\text { e.g. } \mathrm{Ag}^{+} \text {or } \mathrm{C}_{2} \mathrm{O}_{4}^{2-} in the above example.*

**Simultaneous solubility.**While solving these problems, go as per general method i.e. (i) First apply condition of electroneutrality and

*(ii) Apply the equilibria conditions.*

**[JEE 2009, 4]**

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**Sol.**(A,C) $\mathrm{a}-\mathrm{a}\left(1-\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}\right)^{\frac{1}{2}}-\frac{\mathrm{x}^{2}}{4} \quad \mathrm{a}-\mathrm{a}\left(1-\frac{\mathrm{x}^{2}}{2 \mathrm{a}^{2}}-\frac{1}{8} \frac{\mathrm{x}^{4}}{\mathrm{a}^{4}}\right)-\frac{\mathrm{x}^{2}}{4}$ $\mathrm{a}=2,\left(\mathrm{coefficient} \text { of } \mathrm{x}^{2}=0\right)$ $\therefore \mathrm{L}=\frac{1}{64}$

**[JEE 2011, 3M, –1M]**

**[JEE 2012, 3M, –1M]**

**[JEE 2012, 3M, –1M]**

**[JEE(Advanced)-2014, 3]**

**[JEE(Advanced)-2016]**

**[JEE(Advanced)-2017]**

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**Sol.**(A,C)

**[JEE(Advanced)-2018]**

**[JEE(Advanced)-2018]**

**A**be the set of all 3 3 symmetric matrices all of whose entries are either 0 or 1. Five of these entries are 1 and four of them are 0. (A) The number of matrices in

**A**is – (A) 12 (B) 6 (C) 9 (D) 3 (B) The number of matrices A in

**A**for which the system of linear equations $A\left[\begin{array}{l}{x} \\ {y} \\ {z}\end{array}\right]=\left[\begin{array}{l}{1} \\ {0} \\ {0}\end{array}\right]$ has a unique solution, is – (A) less than 4 (B) at least 4 but less than 7 (C) at least 7 but less than 10 (D) at least 10 (C) The number of matrices A in

**A**for which the system of linear equations $A\left[\begin{array}{l}{x} \\ {y} \\ {z}\end{array}\right]=\left[\begin{array}{l}{1} \\ {0} \\ {0}\end{array}\right]$ is inconsistent, is – (A) 0 (B) more than 2 (C) 2 (D) 1

**[JEE 2009, 4+4+4]**

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**Sol.**( (A) A ,(b) B, (c) B )

**Note :**adj M denotes the adjoint of a square matrix M and [k] denotes the largest integer less than or equal to k].

**(C)**Let p be an odd prime number and Tp be the following set of 2 2 matrices: $\mathrm{T}_{\mathrm{p}}=\left\{\mathrm{A}=\left[\begin{array}{ll}{\mathrm{a}} & {\mathrm{b}} \\ {\mathrm{c}} & {\mathrm{a}}\end{array}\right]: \mathrm{a}, \mathrm{b}, \mathrm{c} \in\{0,1,2, \ldots \ldots, \mathrm{p}-1\}\right.$ (i) The number of $A$ in $T_{p}$ such that $A$ is either symmetric or skew symmetric or both, and det(A) divisible by p is – (A) $(\mathrm{p}-1)^{2}$ (B) 2 (p – 1) (C) $(p-1)^{2}+1$ (D) 2p –1 (ii) The number of $A$ in $T_{p}$ such that the trace of $A$ is not divisible by $p$ but det (A) is divisible by p is – [

**Note :**The trace of a matrix is the sum of its diagonal entries.] (A) $(p-1)\left(p^{2}-p+1\right)$ (B) $\mathrm{p}^{3}-(\mathrm{p}-1)^{2}$ (C) $(p-1)^{2}$ (D) $(p-1)\left(p^{2}-2\right)$ (iii) The number of $A$ in $T_{p}$ such that $\operatorname{det}(A)$ is not divisible by $p$ is – (A) $2 p^{2}$ (B) $\mathrm{p}^{3}-5 \mathrm{p}$ (C) $\mathrm{p}^{3}-3 \mathrm{p}$ (D) $\mathrm{p}^{3}-\mathrm{p}^{2}$

**[JEE 2010, 3+3+3+3+3]**

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**Sol.**($(a) A,(b) 4 ;(c)(i) D,(i i) C,(\text { iii }) D$) For the matrix $(6),(9),(10),(11)$ the system of linear equation is incosistent. (A) The given matrix system is a linear system in $\mathrm{x}, \mathrm{y}, \mathrm{z},$ hence it can have either a unique solution or no-solution or infinitely many solutions. It can never have exactly two distinct solutions.

**[JEE 2011, 4]**

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**Sol.**(Bonus)

****

*(Comment : Although 3*

*3 skew symmetric matrices can never be non-singular. Therefore the information given in question is wrong. Now if we consider only non singular skew symmetric matrices M & N, then the solution is-)***[JEE 2011, 3, (–1)]**

**[JEE 2011, 4]**

**[JEE 2011, 4]**

**[JEE 2012, 3M, –1M]**

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**Sol.**(D) $\mathrm{P}^{\mathrm{T}}=2 \mathrm{P}+\mathrm{I}$ $\Rightarrow \mathrm{P}=2 \mathrm{P}^{\mathrm{T}}+\mathrm{I}$ $\Rightarrow \mathrm{P}=2(2 \mathrm{P}+\mathrm{I})+\mathrm{I}$ $\Rightarrow P=4 P+3 I$ $\Rightarrow P=-I$ $\Rightarrow \mathrm{PX}=-\mathrm{X}$

**[JEE 2012, 4M]**

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**Sol.**(A,D)

**NOT**correct ? (A) $\mathbf{N}^{\mathrm{T}} \mathbf{M}$ N is symmetric or skew symmetric, according as M is symmetric or skew symmetric (B) MN – NM is skew symmetric for all symmetric matrices M and N (C) MN is symmetric for all symmetric matrices M and N (D) (adj M) (adj N) = adj (M N) for all invertible matrices M and N

**[JEE-Advanced 2013, 4, (–1)]**

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**Sol.**(C,D) (A) $\mathrm{B}=\mathrm{N}^{\mathrm{T}} \mathrm{MN}$ $\mathrm{B}^{\mathrm{T}}=\left(\mathrm{N}^{\mathrm{T}} \mathrm{MN}\right)^{\mathrm{T}}=\mathrm{N}^{\mathrm{T}} \mathrm{M}^{\mathrm{T}} \mathrm{N}$ Now it depands on $\mathrm{M}$ if $\mathrm{M}=\mathrm{M}^{\mathrm{T}}$ So A is true $(\mathrm{B}) \mathrm{B}=(\mathrm{MN}-\mathrm{NM})$ $\mathrm{B}=(\mathrm{MN}-\mathrm{NM})^{\mathrm{T}}$ $=\mathbf{N}^{\mathrm{T}} \mathbf{M}^{\mathrm{T}}-\mathbf{M}^{\mathrm{N}} \mathbf{N}^{\mathrm{T}}$ $=\mathrm{NM}-\mathrm{MN}=-(\mathrm{B})$ Skew symmetric B is ture $(\mathrm{C}) \mathrm{B}=\mathrm{MN}$ $\mathrm{B}^{\mathrm{T}}=(\mathrm{MN})^{\mathrm{T}}$ $\mathrm{B}^{\mathrm{T}}=\mathrm{N}^{\mathrm{T}} \mathrm{M}^{\mathrm{T}}$ $\mathrm{B}^{\mathrm{T}}=\mathrm{NM} \neq \mathrm{B}$ so wrong statement (D) Obviousely wrong because adj $(\mathrm{BA})=\operatorname{adj}(\mathrm{A}) \cdot \operatorname{adJ}(\mathrm{B})$

**[JEE(Advanced)-2014, 3]**

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**Sol.**(C,D)

**[JEE(Advanced)-2014, 3]**

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**Sol.**(A,B) $\mathrm{M}^{2}=\mathrm{N}^{4}$ $\mathrm{M}^{2}=\mathrm{N}^{4}=0(\therefore \mathrm{MN}=\mathrm{NM})$ $\left(\mathrm{M}+\mathrm{N}^{2}\right)\left(\mathrm{M}-\mathrm{N}^{2}\right)=0$ So $\left(\mathrm{M}+\mathrm{N}^{2}\right)=0$ $\mathrm{Now} \mathrm{M} \cdot\left(\mathrm{M}+\mathrm{N}^{2}\right)=0$ $\mathrm{M}^{2}+\mathrm{MN}^{2}=0$ $\left|\mathrm{M}^{2}+\mathrm{MN}^{2}\right|=0$ Option A is right So we know $A \cdot B=0$ when $\mathrm{B} \neq 0$ $\Rightarrow(\mathrm{A})=0$ So in $\mathrm{U}\left(\mathrm{M}^{2}+\mathrm{MN}^{2}\right) \mathrm{U}=0$ Because $\mathrm{U} \neq 0$ But $\left|\mathrm{M}^{2}+\mathrm{MN}^{2}\right|=0$ So option $\mathrm{B}$ is also right

**[JEE(Advanced)-2015, 4M, –2M]**

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**Sol.**(C,D) $\mathrm{x}^{\mathrm{T}}=-\mathrm{x}, \mathrm{y}^{\mathrm{T}}=-\mathrm{y}, \mathrm{z}^{\mathrm{T}}=\mathrm{z}$ (A) Let $P=y^{3} z^{4}-z^{4} y^{3}$ $\mathrm{P}^{\mathrm{T}}=\left(\mathrm{y}^{3} \mathrm{z}^{4}\right)^{\mathrm{T}}-\left(\mathrm{z}^{4} \mathrm{y}^{3}\right)^{\mathrm{T}}$ $=-z^{4} y^{3}+y^{3} z^{4}=P \Rightarrow$ symmetric (B) $\quad$ Let $\mathrm{P}=\mathrm{x}^{44}+\mathrm{y}^{44}$ $\mathrm{P}^{\mathrm{T}}=\left(\mathrm{X}^{44}\right)^{\mathrm{T}}+\left(\mathrm{y}^{44}\right)^{\mathrm{T}}=\mathrm{P} \Rightarrow$ symmetric (C) Let $P=x^{4} z^{3}-z^{3} x^{4}$ $\mathrm{P}^{\mathrm{T}}=\left(\mathrm{z}^{3}\right)^{\mathrm{T}}\left(\mathrm{x}^{4}\right)^{\mathrm{T}}-\left(\mathrm{x}^{4}\right)^{\mathrm{T}}\left(\mathrm{z}^{3}\right)^{\mathrm{T}}$ $=\mathrm{z}^{3} \mathrm{x}^{4}-\mathrm{x}^{4} \mathrm{z}^{3}=-\mathrm{P} \Rightarrow$ skew symmetric (D) Let $P=x^{23}+y^{23}$ $\mathrm{P}^{\mathrm{T}}=-\mathrm{x}^{23}-\mathrm{y}^{23}=-\mathrm{P} \Rightarrow$ skew symmetric

**[JEE(Advanced)-2016]**

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**Sol.**(B,C) $\mathrm{PQ}=\mathrm{kI}$ $|\mathrm{P}| \cdot|\mathrm{Q}|=\mathrm{k}^{3} \Rightarrow|\mathrm{P}|=2 \mathrm{k} \neq 0 \Rightarrow \mathrm{P}$ is an invertible matrix $\because \mathrm{PQ}=\mathrm{kI}$ $\therefore \mathrm{Q}=\mathrm{k} \mathrm{P}^{-1} \mathrm{I}$ $\therefore \mathrm{Q}=\frac{\mathrm{adj.P}}{2}$ $\because \mathrm{q}_{23}=-\frac{\mathrm{k}}{8}$ $\therefore \frac{-(3 \alpha+4)}{2}=-\frac{k}{8} \Rightarrow k=4$ $\therefore|\mathrm{P}|=2 \mathrm{k} \Rightarrow \mathrm{k}=10+6 \alpha \ldots(\mathrm{i})$ Put value of $k$ in (i).. we get $\alpha=-1$ $\therefore 4 \alpha-k+8=0$ $\& \operatorname{det}(\mathrm{P}(\mathrm{adj} . \mathrm{Q}))=|\mathrm{P}||\operatorname{adj} . \mathrm{Q}|=2 \mathrm{k} \cdot\left(\frac{\mathrm{k}^{2}}{2}\right)^{2}=\frac{\mathrm{k}^{5}}{2}=2^{9}$

**[JEE(Advanced)-2016]**

**[JEE(Advanced)-2017]**

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**Sol.**(A,B)

**[JEE(Advanced)-2017]**

**[JEE(Advanced)-2018, 4(–2)]**

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**Sol.**(A,C,D)

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**JEE-Mains previous year question papers from year 2009 to 2019 along with solutions.**These JEE Main previous Year Question for Chemistry plays an important role in IIT-JEE preparation. We are providing IIT-JEE Mains Previous Year Question Papers with detailed Solution. We have tried our best to provide you

**last 10 years question with solutions**. This set of question papers serves as a very important source to revise the important topics and gain an understanding into the pattern of questions asked in previous years. Practicing these papers will positively help students to gain confidence over their learning. The benefits of solving IIT-JEE previous years papers is that aspirants get to know the type of questions asked in the JEE exam. JEE aspirants can evaluate their preparation after finishing the entire syllabus, topics and chapters. They can get the experience of giving real exam while solving a past year JEE question paper to become confident for the upcoming JEE exam. While preparing for the IIT-JEE exam, aspirants should be aware about the question paper structure and the format of questions to be asked in this exam. This will help to make an effective preparation strategy for the exam. JEE Main Previous Year Question Papers are the best resources to prepare for exam. This will help an individual to understand the exam pattern of JEE. This will also enhance your level of preparation. JEE aspirants must solve multiple sample papers and analyse their performances in order to recognize their strengths and weaknesses.

**Here are the Chemistry Topic-wise Previous year question for JEE Main:**

**Chemical Kinetics – JEE Main Previous Year Questions with Solutions****Metallurgy – JEE Main Previous Year Questions with Solutions****d & f block – JEE Main Previous Year Questions with Solutions****Polymer – JEE Main Previous Year Questions with Solutions****Chemistry in Everyday Life – JEE Main Previous Year Questions with Solutions****Carboxylic Acid – JEE Main Previous Year Questions with Solutions****Carbonyl Compounds – JEE Main Previous Year Questions with Solutions****Biomolecule – JEE Main Previous Year Questions with Solutions****Alcohol & Ether – JEE Main Previous Year Questions with Solutions****Salt Analysis – JEE Main Previous Year Questions with Solutions****Surface Chemistry – JEE Main Previous Year Questions with Solutions****Solid State – JEE Main Previous Year Questions with Solutions****Nuclear Chemistry – JEE Main Previous Year Questions with Solutions****Electrochemistry – JEE Main Previous Year Questions with Solutions****Thermodynamics – JEE Main Previous Year Questions with Solutions****Thermochemistry – JEE Main Previous Year Questions with Solutions****States of Matter – JEE Main Previous Year Questions with Solutions****Redox Reaction – JEE Main Previous Year Questions with Solutions****Mole Concept – JEE Main Previous Year Questions with Solutions****Ionic Equilibrium – JEE Main Previous Year Questions with Solutions****Chemical Equilibrium – JEE Main Previous Year Questions with Solutions****Atomic Structure – JEE Main Previous Year Questions with Solutions****Structural Isomerism – JEE Main Previous Year Questions with Solutions****Oxidation and Reduction – JEE Main Previous Year Questions with Solutions****IUPAC Nomenclature – JEE Main Previous Year Questions with Solutions****Hydrocarbons – JEE Main Previous Year Questions with Solutions****Halogen Derivatives – JEE Main Previous Year Questions with Solutions****General Organic Chemistry – JEE Main Previous Year Questions with Solutions****Environmental Chemistry – JEE Main Previous Year Questions with Solutions****Aromatic Compound – JEE Main Previous Year Questions with Solutions****P-bock – JEE Main Previous Year Questions with Solutions****s-block JEE Main Previous Year Questions with Solutions****Hydrogen – JEE Main Previous Year Questions with Solutions ‘****Periodic Table – JEE Main Previous Year Questions with Solutions****Chemical Bonding – JEE Main Previous Year Questions with Solutions**

### Click Here to Download Math Topic-wise JEE Main Previous Year Question with Solutions

### Click Here to Download Physics Topic-wise JEE Main Previous Year Question with Solutions

**Click Here to Download JEE Advanced Previous Year Questions with solutions. **

We believe that your preparation for the JEE Mains Exam will be simple as long as an organized and consistent plan is followed. Solving the JEE Main Previous Year Papers will absolutely help you in your JEE Mains Exam preparation. As a result, no stone has been left unturned by eSaral to ensure that these IIT-JEE mock tests are similar to the actual exam.
**For free study material and video tutorials, Download eSaral app.**

**JEE Main Previous Year Papers Questions of Chemistry with Solutions are available at eSaral. Practicing JEE Main chapter wise questions of Chemistry will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas.**

**Simulator**

**Previous Years AIEEE/JEE Main Question**

**[AIEEE-2009]**

**incorrect ? (1) Ferrous compounds are more easily hydrolysed than the corresponding ferric compounds. (2) Ferrous oxide is more basic in nature than the ferric oxide. (3) Ferrous compounds are relatively more ionic than the corresponding ferric compounds. (4) Ferrous compounds are less volatile than the corresponding ferric compounds.**

**[AIEEE-2012]**

**[JEE MAIN-2013]**

**[JEE MAIN-2013]**

**[JEE MAIN-2013, Online]**

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**Sol.**(4) Explanation. $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+6 \mathrm{H}_{2} \mathrm{SO}_{4}+4 \mathrm{NaCl} \longrightarrow 2 \mathrm{KHSO}_{4}+4 \mathrm{NaHSO}_{4}+2 \mathrm{CrO}_{2} \mathrm{Cl}_{2}+3 \mathrm{H}_{2} \mathrm{O}$

**[JEE MAIN-2013, Online]**

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**Sol.**(3) Explanation $\mathrm{Mn} \longrightarrow \mathrm{Mn}^{+7}$

**[JEE MAIN-2013, Online]**

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**Sol.**(3) Explanation

**[JEE MAIN-2014]**

**[JEE MAIN-2014]**

**not**formed when $\mathrm{H}_{2} \mathrm{S}$ reacts with acidic $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$solution ? ( 1) $\mathrm{K}_{2} \mathrm{SO}_{4}$ (2) $\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}$ (3) $\mathrm{S}$ (4) $\mathrm{CrSO}_{4}$

**[JEE MAIN-2014, Online]**

**[JEE MAIN-2014, Online]**

**[JEE MAIN-2014, Online]**

**[JEE MAINS-2014,Online]**

**[JEE MAIN-2014, Online]**

**[JEE MAIN-2015]**

**[JEE MAIN-2015, Online]**

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*Simulator***Previous Years AIEEE/JEE Mains Questions**

**[AIEEE 2012]**

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**Sol.**(4) Very pure hydrogen (99.9%) can be made by electrolysis of water.

**[JEE(Main) 2014]**

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**Sol.**(2) When $\mathrm{H}_{2} \mathrm{O}_{2}$ act as reducing agent then it evolve.

**[JEE(Main) 2014]**

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**Sol.**(3) $\mathrm{Na}_{2} \mathrm{O}_{2}$ is peroxide of sodium

**[JEE(Main)Online-2015]**

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**Sol.**(3) Permanent hardness in water cannot cured by boiling of water

**[JEE(Main)Online-2015]**

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**Sol.**(3) $\mathrm{H}_{2} \mathrm{O}_{2}$ can act as oxidizing as well as reducing agent depend on condition.

**[JEE(Main)Online-2018]**(1) $\left(\mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}\right)$ and $\left(\mathrm{H}_{2} \mathrm{O}+\mathrm{OH}^{-}\right)$ (2) $\mathrm{H}_{2} \mathrm{O}$ and $\left(\mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}\right)$ (3) $\mathrm{H}_{2} \mathrm{O}$ and $\left(\mathrm{H}_{2} \mathrm{O}+\mathrm{OH}^{-}\right)$ (4) $\left(\mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}\right)$ and $\mathrm{H}_{2} \mathrm{O}$

*Simulator***Previous Years AIEEE/JEE Mains Questions**

**[AIEEE-2009]**

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**Sol.**(1) Due to back bonding in BF3 molecule all B–F bond having partial double bond character.

**[AIEEE-2009]**

**[AIEEE-2011]**

**[AIEEE-2011]**

**[AIEEE-2011]**

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**Sol.**(1) $\mathrm{Al}^{+3}$ having highest polarizing power than other : having greater covalent character

**[AIEEE-2011]**

**[AIEEE-2011]**

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**Sol.**(1) $\mathrm{Ca}^{+2}(\mathrm{C} \equiv \mathrm{C})^{2-}$

**[AIEEE-2012]**

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**Sol.**(4) Bond angle order $\mathrm{NCl}_{3}>\mathrm{PCl}_{3}>\mathrm{AsCl}_{3}>\mathrm{SbCl}_{3}$

**[AIEEE-2012]**

**[JEE-MAINS(Online) 2012]**

**[JEE-MAINS(Online) 2012]**

**[JEE-MAINS(Online) 2012]**

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**Sol.**(Bonus)

**[JEE-MAINS(Online) 2012]**

**[JEE-MAINS(Online) 2012]**

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**Sol.**(4) Highest melting is of NaCl

**[JEE-MAINS(Online) 2012]**

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**Sol.**(3) Hybridisation $\mathrm{Sp}^{3} \mathrm{d}^{2}$ Shape – square planar

**[JEE-MAINS(Online) 2012]**

**[JEE-MAINS(Online) 2012]**

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**Sol.**(3) $\mathrm{CCl}_{4}<\mathrm{CHCl}_{3}<\mathrm{CH}_{2} \mathrm{Cl}_{2}<\mathrm{CH}_{3} \mathrm{Cl}(\text { Dipolar moment })$

**[JEE-MAINS(Online) 2012]**

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**Sol.**(1) $\mathrm{CH}_{2} \mathrm{Cl}_{2}$ has lowest dipole moment

**[AIEEE-2013]**

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**Sol.**(2) $\mathrm{N}_{2}$ is diamagnetic

**[JEE-maIN 2013]**

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**Sol.**(1) Number of electron in ONCl an $\mathrm{ONO}^{-}$ is 32 & 24 respectively.

**[JEE-main 2013]**

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**Sol.**(3) Bond order of $\mathrm{H}_{2}^{2+} \& \mathrm{He}_{2}$ is zero i.e. these molecule do not exist.

**[JEE-main 2013]**

**[JEE-MAINS(Online) 2013]**

**[JEE-MAINS(Online) 2013]**

**[JEE-MAINS(Online) 2013]**

**[JEE-MAINS(Online) 2013]**

**[JEE-MAINS(Online) 2013]**

**[JEE-MAINS(Online) 2013]**

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**Sol.**(1) Solubility order LiF < NaF < KF < RbF

**[JEE-MAINS(Online) 2013]**

**[JEE-MAINS(Online) 2013]**

**[JEE-MAINS(Online) 2013]**

**[JEE-MAINS(Online) 2013]**

**[JEE-MAINS(Online) 2013]**

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**Sol.**(2) Catenation/Bond energy order C – C > Si – Si > Ge – Ge

**[JEE-MAINS(Online) 2013]**

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**Sol.**(3) $\mathrm{BF}_{4}^{-}, \mathrm{CC}_{4}, \mathrm{NH}_{4}^{+}, \mathrm{PCl}_{4}^{+}$ are terahedral

**[JEE-MAINS(Online) 2013]**

**[JEE-main 2014]**

**[JEE-main 2014]**

**[JEE-MAINS(Online) 2014]**

**[JEE-MAINS(Online) 2014]**

**[JEE-MAINS(Online) 2014]**

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**Sol.**(1) $\mathrm{O}_{2}^{-}$ ha one unpaired electron

**[JEE-MAINS(Online) 2014]**

**[JEE-MAINS(Online) 2014]**

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**Sol.**(2) $\mathrm{BrF}_{5}$ have square pyramidal shape.

**[JEE-MAINS(Online) 2014]**

**[JEE-MAINS(Online) 2014]**

**[JEE-MAINS(Online) 2014]**

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**Sol.**(1) NO has unpaired e $^{-} \therefore$ paramagnetic is nature

**[JEE-MAINS(Online) 2014]**

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**Sol.**(4) RbCl has highest ionic & and $\mathrm{BeCl}_{2}$ is most covalent

**[JEE-MAINS 2016]**

**[JEE-MAINS 2017]**

**[JEE-MAINS 2018]**

**[JEE-MAINS 2018]**

**[JEE-MAINS 2018]**

**Click Here for Complete Physics Revision Series by Saransh Gupta Sir (AIR-41)**

**About Saransh Gupta Sir***Computer Science Graduate from IIT Bombay. Cracked IIT-JEE with AIR-41 and AIEEE (JM) with AIR-71 in 2006. He is an author of JEE Mentorship Book “StrateJEE”.*

*He taught Physics for 5 years at Allen and was loved immensely by all the students. Many of his students bagged success with flying colours in JEE & NEET Exams.*eSaral brings you detailed Class 11th Physics study material. eSaral provides a series of detailed chapter wise notes for all the Subjects of class 11th and 12th. These notes will also help you in your IIT JEE & NEET preparations. We hope these Physics Notes for Class 11 will help you understand the important topics and remember the key points for the exam point of view. Get

**Complete Physics Notes for Physics Class 11**for easy learning and understanding. For free video lectures and complete study material, Download

**eSaral APP.**

*About eSaral**At eSaral we are offering a complete platform for IIT-JEE & NEET preparation. The main mission behind eSaral is to provide education to each and every student in India by eliminating the Geographic and Economic factors, as a nation’s progress and development depends on the availability of quality education to each and every one. With the blend of education & technology, eSaral team made the learning personalized & adaptive for everyone.*

This article contains Work Energy and Power Class 11 Physics Notes.

Work Energy and Power can never be ignored as it covers basic concepts which provides a smooth access to the next some topics in Mechanics. This chapter is a key to prepare for entrance examinations like IIT-JEE & NEET. This is one of the most important topic in mechanics for IIT-JEE, NEET and CBSE perspectives. The terms work, energy and power have specific meanings in Physics. Work is done when a force produces motion. To do work, energy is required. This energy we acquire from the food which we eat and if work is done by machine, then energy is supplied by fuels or by current.

There are various forms of energy – kinetic energy, potential energy, heat energy, chemical energy, electrical energy, nuclear energy etc. Work Energy and Power Class 11 Physics notes are given here:

Click Here to Study**Newton Laws of Motion Class 11th**To watch free video tutorials on Physics topics by Saransh Sir, AIR-41 Install eSaral App eSaral have already come up with an amazing revision series of Physics where you can easily revise your chapter within minutes with all the important formulae and key points. This revision series is free for all.

**Click Here for Complete Physics Revision Series by Saransh Gupta Sir (AIR-41)**

**About Saransh Gupta Sir***Computer Science Graduate from IIT Bombay. Cracked IIT-JEE with AIR-41 and AIEEE (JM) with AIR-71 in 2006. He is an author of JEE Mentorship Book “StrateJEE”.*

*He taught Physics for 5 years at Allen and was loved immensely by all the students. Many of his students bagged success with flying colours in JEE & NEET Exams.*eSaral brings you detailed Class 11th Physics study material. eSaral provides a series of detailed chapter wise notes for all the Subjects of class 11th and 12th. These notes will also help you in your IIT JEE & NEET preparations. We hope these Physics Notes for Class 11 will help you understand the important topics and remember the key points for the exam point of view. Get

**Complete Physics Notes for Physics Class 11**for easy learning and understanding. For free video lectures and complete study material, Download

**eSaral APP.**

*About eSaral**At eSaral we are offering a complete platform for IIT-JEE & NEET preparation. The main mission behind eSaral is to provide education to each and every student in India by eliminating the Geographic and Economic factors, as a nation’s progress and development depends on the availability of quality education to each and every one. With the blend of education & technology, eSaral team made the learning personalized & adaptive for everyone.*

To get into IIT is the dream of every **JEE** Aspirant. Most of them started preparing for JEE from foundation classes while some realises their interest and inclination towards engineering after class 10^{th}. Now for the aspirants who are preparing for 2020, few months left for JEE-Mains 2019. Many students started worrying on **how to score good marks in JEE mains exam.** During this time of preparation, aspirants have many more questions in their mind, Such as:

*How to improve marks during mock test**How to score above 300 marks in JEE mains**How to attempt questions to score well in JEE mains**Where to practice JEE mains mock Test.**Where to get online test series for JEE Mains.*

If you are seeking answer for any of the question above, then this article is for you!!

From last year, NTA (National Testing Agency) is conducting JEE Exam. From 2019 JEE mains exam held twice in year. In January and April.

Lot of hard work and practice is required to get a good score in JEE.

To score well in JEE mains, focus more on your learning & basic understanding of the concepts of Physics, Chemistry, and Mathematics. Click to watch Exam Pattern of JEE mains.

**How to Study Physics for JEE Mains**

Start with the basic concepts of Physics. Then memorize all the important Physics formulas for JEE Mains.

Practice makes everyone perfect so, practice questions more and more. Also practice previous year’s questions to make an idea on the question patterns that will appear in JEE mains. By practicing previous year questions, you will also get to know the most important chapters in JEE mains Physics. This will also help in strengthening your concepts and will make you familiar with a variety of questions asked in the exam.

Practice Mock Test on regular basis to improve score.

**Don’t be overconfident. **Sometimes it make mistakes if the question seems easy. Mostly mistakes are made in unit system where student ignore the difference between CGS and MKS systems. So, be easy-going and have the presence of mind while solving the problems.

**How to Study Chemistry for JEE Mains**

Chemistry comprises of three sections i.e., Physical Chemistry, Inorganic Chemistry and Organic Chemistry. The numerical questions are asked only from Physical Chemistry. Chemistry is a scoring subject in JEE mains Exam as the difficulty level of the questions in Chemistry is easier than the other two subjects i.e., Physics and Mathematics. That’s why, Chemistry is the easiest part of the JEE Main Examination for many students.

Learn all important formulas in Physical Chemistry which will help you to solve numerical problems in less time in JEE Main Examination 2020.

Don’t be overconfident. Chemistry is the easiest part of the JEE Examination. Therefore, some students did not give sufficient time to Chemistry subject. As a result, they score less marks in JEE Main Examination. They think that they can learn the complete subject within a month which is not possible.

**How to study Mathematics for JEE Mains**

For most of the aspirants, Math is one of the toughest subject among all three. Mostly, performance of JEE mains aspirants are affected by fear of mathematics.

To cope up with the fear of math, there is only one solution. Practice! Practice! And Practice! You can also make an actionable study plan to conquer the** JEE Mains Mathematics Preparation.**

An Ideal rule of solving mathematics problem is that never start with a challenging problem. If you can’t solve few basic problems, it will make you nervous & hence it will distress your complete performance in JEE Mains paper.

Solve the easier mathematical problems first & then progress towards the challenging ones.

The more you practice Mathematics, the less you make mistakes while solving problems and hence you will get good score in the JEE Mains.

Take mock tests and practice previous year questions to get first real look and feel of the examination. JEE Mains mock tests provide good practice and are a guide to judge your abilities. You can get mock test at eSaral app. eSaral also provides Topic-wise tests and offers** more than 5 lakh questions to practice**. For more details Click here.

It has been observed from the toppers talk that **mock tests during exam preparation helped them a lot in managing time as well as scoring well in JEE mains exam paper. **Hence Mock tests plays an important role in scoring good marks.

Finally, I would like to say have faith in yourself and be confident on your learning and preparation for JEE exam. Focus on every subject individually to score well.

Hope you liked reading the above post. Please share it with your friends to help them in getting the relevant information. **Stay motivated and keep learning from eSaral!!!**

**Continuing sleep with dreams.**OR

**Wake up early to chase your dreams.**The choice is yours only. Waking up early in the morning will help you in maintaining the sleep cycle. Thus, improves health. But also a good night’s sleep is extremely important for students to produce improved study outputs. Some other

**benefits to wake up early in the morning for students are:**

- Better concentration power.
- The stress level is minimum.
- Better Study Output.
- Improves mental health.
- Improves brain efficiency.
- Less traffic, more peace.

**some mistakes that are making difficult to wake up early**for them. Such as:

- Not having a regular sleep schedule.
- Sleeping late in the night.
- Drinking Caffeine during study hours.
- Taking long naps instead of short naps.
- Eating late at night.
- Hitting snooze.

**tips to wake up early in the morning**. He mentored many students and the tips are proved to be useful in getting up early. So Guys, watch out the Video on How to wake up early in the morning by Saransh Sir.

**some tips to help you in the fight & wake up early in the morning:**

**Get excited to wake up early:**find a strong purpose to get up early morning.**Don’t put the alarm on the bed**or near to the bed. Keep it at distance.- You can also
**find a wake-up partner**to help you to wake up early in the morning. - Sleep at the right time to
**get proper sleep of 6-7 hours.** **Do stretch your body after waking up**. This will help you in activating blood circulation and also helps in reducing laziness.**Give 10-15 minutes to meditation or exercise**to maintain the oxygen level.**Hydrate your body by drinking 2-3 glasses of water**just after waking up.