JEE MAIN 2020 - 07 JAN S2 2020

(i) Electron affinity of second period p-block element is less than third period p-block element due to small size of second period p-block element.E.A. order : F < Cl(ii) Down the group electron affinity decreases due to size increases. EA. order : S > SeLi > Na

(i) $N_2+O_2 {--------▶ }^{200K}2NO$ (Redox reaction) during the reaction, oxidation of nitrogen take place from 0 to 2 and reduction of oxygen take place from 0 to -2. It means this reaction is redox reaction. (ii) $3O_2\to ^{hυ}2O_3$ (Non - redox reaction)(iii) $H_2SO_4 + 2NaOH$ $\to Na_2SO_4 + 2H_2O$ (neutralization reaction)(iv) $[Co(H_2O)6]Cl_3 + 3AgNO_3$ $\to 3A gCl ↓ + [Co(H_2O)_6](NO_3)_3$(White ppt.)

(a) $C o^{+3}$ (with strong field ligands)(b) If $\Delta_0< p;$ (c) Splitting power o f ethylenediamine (en) is greater than fluoride $(F ^{-})$ ligand therefore more energy absorbed by $[Co(en)_3]^{3+}$ as compared to $[CoF_6]^{3-}$. So wave length of light absorbed by $[Co(en)_3]^{3+}$ is lower than that of $[CoF_6]^{3-}$(d) $\Delta_t=4/9\Delta_0$So if $\Delta_0=18,000\text{cm}^{-1}$ $\Delta_t=4/9\times 18000=8000\text{cm}^{-1}$Statement (a) and (d) are incorrect.

(a) If the com plex $MA_2B_2$ is $sp^3$ hybridised then the shape of this complex is tetrahedral this structure is opticaly inactive due to the presence of plane of symmetry.Optical isomes = 0 (b) If the complex $MA_2B_2$ is $dsp^2$ hybridised then the shape of this complex is square planar. Both isomers are optically inactive due to the presence of plane of symmetry. Optical isomers = 0

{NaOH + Cl_2}_{(\text{hot} \text{and} \text{cone}.)} \to {NaClO_3}_{(A) }+{5NaCl+3H_2O} __{\text{side} \text{products}} ${Ca(OH)_2} _{\text{dry}}+ Cl_2 \to {Ca(OCl)_2}_{(B) }+{CaCl_2+2H_2O} _{\text{side} \text{products}}$

According to MOT (If z is intemuclear axis) The configuration of $CN^{-}:$ $\sigma _{1s}^{2},\sigma _{1s}^{*2},\sigma _{2s}^{2},\sigma _{2s}^{*2},\pi _{2p_x}^2$ $=\pi _{2p_x}^2,\sigma _{2p_z}^2$Bond order $=1/2(10-4)=3$ $CN ^{-}$ is diamagnetic due to absence of unpaired electron.

Option (1) is incorrect.According to Kohlrausch's law correct expression is $({\Lambda ^0}_\text{m} )_{\text{NaBr}}-({\Lambda ^0}_\text{m} )_\text{NaI}$ $=({\Lambda ^0}_\text{m} )_\text{KBr}-({\Lambda ^0}_\text{m} )_\text{KI}$The other statements are correct.

The pure solvent solution will try to maintain higher vapour pressure in the sealed container and in return the solvent vapour molecules will condense in the solution of non-volatile solute as it maintains an equilibrium with lower vapour pressure. (Lowering of vapour pressure is observed when a non volatile solute is mixed in a volatile solvent) This will eventually lead to increase in the volume of solution and decrease in the volume of solvent.

Column chromatography is a method used to separate a single compound from a mixture of compounds. $\Rightarrow$ Here we take a mixture of 3 compounds (A) benzanilide (B) aniline (C) acetophenone. $\Rightarrow$ To separate the mixture here we are using a mixture of two solvents. hexane and ethyl acetate in the ratio 20: 80 $\Rightarrow$ hexane is a non-polar solvent having $20 \%$ composition. \Rightarrow ethyl acetate \; H _{3} C - CH _{2}- O - C ^{ ‖}^{o}- CH _{3} is a polar solvent having 80 % composition. ⇒ Means nonpolar compound present in a less quality as compared to polar solvent. ⇒ Hence more polar compound present in a mixture will separate out first with the polar solvent hence. ⇒ Acetophenone is more polar among (A), (B) & (C) hence it will separate first and after that. ⇒ Benzanilide is more polar than aniline hence A will elute out after that. ⇒ Hence the answer should be (C), (A) & (B) Refer below image

$V_\mp(=\sqrt{ {2RT}/M })$ $< V_\text{av} (=\sqrt{ \frac{8RT}{\pi M} })$ $< V_\text{rms} (=\sqrt{ {3RT}/M })$

(a) $H_2O_2\to 2H_2O +O_2$ (b) $\text{KClO}_3{--------▶}^{\Delta}_{\text{MnO}_2}\text{KCl}+3/2O_2$ $\text{Pb}(\text{NO}_3)_2\to ^{\Delta}\text{PbO}+2\text{NO}_2+1/2\text{O}_2$ $\text{NaNO}_3\to ^{\Delta}\text{NaNO}_2+1/2O_2$ (c) 2-ethylanthraquinol $⇆^{\text{O}_2 (\text{air})}_{\text{H}_2/\text{Pd}}$ 2-ethylanthraquinone + $\text{H}_2\text{O}_2$ (d) $2\text{H}_3\text{B} \text{O}_3 + 2\text{NaOH} + 2\text{H}_2\text{O}_2$ $\to \text{Na}_2[\text{B}_2(\text{O}_2)(\text{OH})_4] + 4\text{H}_2\text{O}$All statements are correct

(B) with more steric crowding forms elimination product compared to substitution.$k_e(B) > k_e(A)$ $\mu _B=\frac{k_s(B)}{k_e(A)} < \mu _A= \frac{k_s(A)}{k_e(A)}$

Liquation method is used when the melting point of metal is less compare to the melting point of the associated impurity.

The stability of an anion is determined by factors that delocalize or withdraw the negative charge, such as resonance and inductive effects.

1. Structure (C) is the most stable. It is an enolate, and the negative charge on the oxygen is delocalized through resonance onto the carbon backbone ($O-C=C \leftrightarrow O=C-C^-$). This delocalization over multiple atoms is a very powerful stabilizing effect.

2. Structure (B) is more stable than (A). In (B), the oxygen is attached to a more electronegative $sp^2$ hybridized carbon. Additionally, the strongly electron-withdrawing nitro group ($-NO_2$) is attached to the double bond, stabilizing the anion through both a strong inductive effect ($-I$) and a mesomeric effect ($-M$).

3. Structure (A) is the least stable. The oxygen is attached to a less electronegative $sp^3$ carbon, and the anion is stabilized only by the inductive effect ($-I$) of the adjacent $-NO_2$ group.

Therefore, the correct order of stability is (C) > (B) > (A).

${NH_2CONH_2 }_{10 \text{mmoles} }+ 2NaOH$ $\to Na2CO_3 +{ 2NH_3}_{20\text{mmoles}}$Hence, $NH_3$ will require 20 meq.

$K_\text{eq}=k_f/k_b=\frac{[N_2][H_2O]^2}{[H_2]^2[NO]^2}$ At equilibrium $r_f = r_b$ $k_f[NO]^2[H_2]=k_b \frac{[N_2][H_2O]^2}{[H_2] }$Hence, rate expression for reverse reaction.$={k_b[N_2][H_2O]^2}/[H_2]$

$A = CrO_2Cl_2$ $B = Na_2CrO_4$ $C = C rO_5$Total number of atom in A + B + C = 18

3gm Acetic Acid + 250 ml 0.1 M HCl +Water → made to 500 ml solution.⇒ 500 ml solution has 25 meq of HCl∴ 20 ml solution has 50 meq of $CH_3COOH$, 1 meq of $HCl$, 2 meq of $CH_3COOH$We have added 2.5 meq. of NaOH $(5M,1/2ml)$Finally, NaOH & HCl are completely consumed and we are left with 0.5 meq of $CH_3COOH$ and 1.5 meq of $CH_3 COONa$ $pH=pKa+\log 1.5/0.5$= 4.75 + log 3 = 4.75 + 0.4771 = 5.2271

$2C(\text{graphite}) + 3H_2(g)\to C_2H_6(g)$ $\Delta_fH (C_2H_6) = 2\DeltaH_\text{comb} (C_\text{graphite})$ $+ 3 \DeltaH _\text{comb}(H_2) -\DeltaH_ \text{comb}(C_2H_6)$ $= -(286\times 2)-(393.5\times 3)-(-1560)$ $= -572 -1180.5 +1560$ = -192.5 kJ/mole

1 L solution requires 30 m.mol HCl 250 ml sol. will require 7.5 m.mol HCl or 3.75 m.mol $H_2SO_4$ $\Rightarrow {3.75\times 98}/100\text{gm} H_2SO_4$= 0.3675 gm $H_2SO_4$

no. of $sp^2$-carbon → 9

Put x = sin θ, y = sin α $y\sqrt{1-x^2}=k-x\sqrt{1-y^2}$ $\sin \alpha . \cos\theta + \cos\alpha . \sin\theta = k$ $\sin(\alpha + \theta ) = k$ $\alpha + \theta = \sin ^{-1} k$ $\sin^{-1}x + \sin^{-1} y = \sin^{-1}k$ $1/\sqrt{1-x^2}+1/\sqrt{1-y^2}\frac{dy}{dx}=0$ at $x=1/2,y=-1/2$ $\frac{dy}{dx}=-\sqrt5/2$

$4x^2 - y \le 0$ and $8x - y + 12 \ge 0$On solving $y=4x^2$ and $y = 8x + 12$We get A (-1, 4) & B(3, 36)Required area = area of the shaded region $=∫^{3}_{-1}(8x+12-4x^2)dx=128/3$

$\vec{a} + \vec{b} + \vec{c} =0$ $\Rightarrow {|\vec{a}|}^{2+}{|\vec{b}|}^{2+}{|\vec{c}|}^{2+}2(\vec{a}.\vec{b})$ $+2(\vec{b}.\vec{c})+2(\vec{c}.\vec{a})=0$ $\lambda =\vec{a} . \vec{b} +\vec{b} .\vec{c} + \vec{c} .\vec{a}$ $=-3/2$ $\vec{d} =\vec{a} \times \vec{b} +\vec{b} \times \vec{c} + \vec{c}\times \vec{a}$ $\vec{a} + \vec{b} + \vec{c} =0$ $\Rightarrow \vec{a} \times \vec{b} =\vec{b} \times \vec{c} = \vec{c}\times \vec{a}$ $\Rightarrow \vec{d}=3(\vec{a}\times \vec{b})$

Sum o f the 40 terms of 3 + 4 + 8 + 9+13 + 14+18+ 19...= (3 + 8 + 13 +...upto 20 term) + [4 + 9 + 15 +... upto 20 terms]= 10 [{6 + 19 x 5} + {8 + 19 x 5}]= 10 x 204 = 20 x 102

$f(0) = 11$ $f(1) = 16$ $\frac{f(1)-f(0)}{1-0}=3c^{2-}8c+8$ $\Rightarrow 3c^{2-}8c+8=5$ $\Rightarrow 3c^{2-}8c+3=0$ $c \in [0, 1] \Rightarrow c={4-\sqrt7}/3$

$2 \cot ^{2} \theta -\;\frac{5}{\sin \theta }+4=0$ $\;\frac{2 \cos ^{2} \theta }{\sin ^{2} \theta }-\;\frac{5}{\sin \theta }+4=0$ $2 \cos ^{2} \theta -5 \sin \theta +4 \sin ^{2} \theta =0$ $2 \sin ^{2} \theta -5 \sin \theta +2=0$ $(2 \sin \theta -1)(\sin \theta -2)=0$ $\Rightarrow 2 \sin \theta -1=0$ or $\sin \theta \ne 2$ $\Rightarrow \sin \theta =\;\frac{1}{2}$ $\Rightarrow \theta =\;\frac{\pi }{6}, \;\frac{5 \pi }{6}$ as $\theta$ lies in $(0,2 \pi )-\{\pi \}$ $\therefore \theta _{1}=\;\frac{\pi }{6}, \theta _{2}=\;\frac{5 \pi }{6}$ $\therefore ∫_{\theta _{1}}^{\theta _{2}} \cos ^{2} 3 \theta =∫_{\pi / 6}^{5 \pi / 6} \;\frac{1+\cos 6 \theta }{2} d \theta$ $=\;\frac{1}{2}[\theta +\;\frac{\sin 6 \theta }{6}]_{\pi / 6}^{5 \pi / 6}$ $=\;\frac{1}{2}[\;\frac{5 \pi }{6}+\;\frac{\sin 5 \pi }{6}-\;\frac{\pi }{6}-\;\frac{\sin \pi }{6}]=\;\frac{1}{2}[\;\frac{2 \pi }{3}]$ $=\;\frac{\pi }{3}$

$6\times ^35C_r = (k^2 - 3)\times ^36C_{r +1}$ $k^{2-}3 > 0 \Rightarrow k^2 > 3$ $k^{2-}3=\frac{6\times ^35C_r}{^36C_{r+1}}={r+1}/6$ Possible values of r for integral values of k, are r = 5, 35number of ordered pairs are 4 : (5, 2), (5, -2), (35, 3), (35, -3)

$b _{ij}= (3) ^(i+j-2)a_{ij}$ $B=\begin{bmatrix}a_11 & 3a_12 & 3^2a_13 \\ 3a_21 & 3a_22 & 3a_23 \\ 3^2a_31 & 3^2a_32 & 3^2a_33\end{bmatrix}$ $\Rightarrow |B|=3\times 3^2\begin{vmatrix}a_11 & a_12 & a_13 \\ 3a_21 & 3a_22 & 3a_23 \\ 3^2a_31 & 3^2a_32 & 3^2a_33\end{vmatrix}=3^6{|A|}$ $\Rightarrow |A|=81/{27\times 27}=1/9$

$a_1+a_2=4$ $r^2a_1+r^2a_2=16$ $\Rightarrow r^2=4\Rightarrow r=-2\text{as} a_1 < 0$ $a_1+a_1(-2)=4$ $\Rightarrow a_1=-4$ $4\lambda =(-4)(\frac{ (-2)^{9-}1}{-2-1})$ $\lambda =-171$

Contrapositive of p → q is ~q → ~p( A ⊆ B ) \wedge ( B ⊆ D ) → (A ⊆ C)Contrapositive is ~(A ⊆ C ) → ~(A ⊆ B) \vee ~(B ⊆ D)A ⊈ C → ( A ⊈ B ) \vee ( B ⊈ D )

$3x + 4y = 12V\sqrt12$ is tangent to $x^2/a^{2+}y^2/9=1$ $c^2=m^2a^{2+}b^2$ $\Rightarrow a^2=16$ $e=\sqrt{1-9/16}=\sqrt7/4$Distance between focii $= 2ae = 2\sqrt7$

4\alpha \[∫^{0}{-1}e^{αx}dx+∫^{2}{0}e^{-αx}dx]$=5$$\Rightarrow 4\alpha ([ e^{\alpha x}/\alpha ]^0_{-1}+$[e^\frac{-αx}{-α}]$)=5$$\Rightarrow 4e^{-2\alpha } + 4e ^{-\alpha } - 3 = 0$Let$e^{-\alpha }=t,4t^{2+}4t-3=0,$$t=1/2,-3/2$(Rejected)$e^{-\alpha }=1/2;\alpha =\ln2$

Coefficient of $x^7$ is $^10C_7+^9C_6+^8C_5+$ $...+^4C_1+^3C_0$ ${^4C_0+^4C_1}_{^5C_1}+^5C_2+$ $...+^10C_7=^11C_7=330$

$\alpha +\beta =1,\alpha \beta =-1$ $p_k=\alpha ^k+\beta ^k$ $\alpha ^{2-}\alpha -1=0$ $\Rightarrow a^k-\alpha ^{k-1}-\alpha ^{k-2}=0$ & $\beta ^k-\beta ^{k-1}-\beta ^{k-2}=0$ $\Rightarrow p_k=p_{k-1}+p_{k-2}$ $p_1=\alpha +\beta =1$ $p_2=(\alpha +\beta )^{2-}2\alpha \beta =1+2=3$ $p_3=4$ $p_4=7$ $p_5=11$

$\frac{\alpha -\beta }{2\alpha -\beta }=-1$ $3\alpha =2\beta$ $h={2\alpha +\beta }/2$ $2h={7\alpha }/2$ $k={\alpha +\beta }/2$ $2k={5\alpha }/2$ $h/k=7/5$ $5x=7y$

$\frac{3+\text{isin} \theta }{4-\text{icos}\theta }$ is a real number$\Rightarrow 3\cos\theta +4\sin\theta =0$ $\Rightarrow \tan\theta =-3/4$argument of sinθ + icosθ $=\pi -\tan^{-1}4/3$

$(y^{2-} x ) \frac{dy}{dx}= 1$ $\Rightarrow \frac{dx}{dy}+x=y^2$ $\text{I}.\text{F}.=e^{∫dy}=e^y$Solution is given by$xe^y=∫y^2e^ydy+C$ $\Rightarrow xe^y=(y^{2-}2y+2)e^y+C$x = 0, y = 1, gives C = -eIf y = 0, then x = 2 - e