JEE MAIN 2020 - 09 JAN S2 2020

Complex (I) has the central metal ion as $Fe ^{2+}$ with strong field ligands. Configuration of $Fe ^{2+}=[ Ar ] 3 d ^{6}$ Strong field ligands will pair up all the electrons and hence the magnetic moment will be zero. Complex(II) has the central metal ion as $Cr ^{2+}$ with weak field ligands. Configuration of $Cr ^{2+}=[ Ar ] 3 d ^{4}$ As weak field ligands are present, pairing does not take place. There will be 4 unpaired electrons and hence the magnetic moment $=\sqrt24$ B.M. Complex (III) has the central metal ion as $Co ^{2+}$ with weak field ligands. Configuration of $Co ^{2+}=[ Ar ] 3 d ^{7}$ As weak field ligands are present no pairing can occur. There will be 3 unpaired electrons and hence the magnetic moment $=\sqrt{15}$ B.M. Complex (IV) has the central metal ion as $Fe ^{3+}$ with strong field ligands. Configuration of $Fe ^{3+}=[ Ar ] 3 d ^{5}$ Strong field ligands will pair up the electrons but as we have a [Ar] $3 d ^{5}$ configuration, one electron will remain unpaired and hence the magnetic moment will be $\sqrt{3}$ B.M.

IE values indicate, that the metal belongs to Ist group since second IE is very high (∵ only one valence electron)Metal hydroxide will be of type, MOH.${MOH }_{1\text{mol}}+ {HCl}_{1\text{mol}} \to MCI + H_2O$ ${MOH}_{1\text{mol}}+{1/2H_2SO_4}_{1/2\text{mol}}\to 1/2M_2SO_4+H_2O$So one mole of HCl required to react with one mole MOH. So $1/2$ mole of $H_2SO_4$ required to react with one mole MOH.

(No chiral centre, so no racemisation possible)

Adsorption increase → decrease pressure → With time pores of charcoal decrease. → So initial it will decrease fast but after some time its decrease rate will be slow. Refer the image. $p = k(x/m)^{1/n}$ As 'x' increases , 'p' decreases

A chiral monomer is a molecule that contains at least one chiral center, which is a carbon atom bonded to four different groups. Let's examine the monomers of each polymer. The monomers of Buna-N (1,3-butadiene, acrylonitrile), Nylon 6,6 (adipic acid, hexamethylenediamine), and Neoprene (chloroprene) are all achiral as they lack a chiral center. PHBV (Poly(3-hydroxybutyrate-co-3-hydroxyvalerate)) is a copolymer made from 3-hydroxybutanoic acid and 3-hydroxypentanoic acid. The third carbon ($C_3$) in 3-hydroxybutanoic acid, $CH_3-\stackrel{*}{C}H(OH)-CH_2-COOH$, is a chiral center. Similarly, the $C_3$ in 3-hydroxypentanoic acid, $CH_3CH_2-\stackrel{*}{C}H(OH)-CH_2-COOH$, is also chiral. Thus, PHBV is formed from chiral monomers.

Biochemical oxygen demand (BOD) is amount of oxygen required by bacteria to break down organic waste in a certain volume of water sample.

Lithium has highest hydration enthalpy among alkali metals due to its small size.LiCl is soluble in pyridine because LiCl have more covalent character.Li does not form ethynide with ethyne. Both Li and Mg reacts slowly with $H_2O$

Distilled water have lowest ionic conductance.

Due to -M effect of -$NO_2$ and + M effect of Cl more D.B. character between C - Cl bond. So shortest bond length.

(1 or Bonus)(no reaction is given)$A ⇌ B$ (Assume reaction)$K=[B]/[A]=11/6 \approx 2$

Basicity $\propto$ electron pair intensity on nitrogen. (III) Refer image 1 No resonance and + I effect from both sides $\therefore$ most basic. (II) Refer image. 2 Nitrogen do not involve in resonance $\therefore$ more basic. (I) Refer image.3 Nitrogen lone pair is involved in resonance. Therefore less available for donation. $\therefore$ less basic. (IV) Refer image.4 $6 \pi {e}^{-},$ Aromatic nitrogen lone pair participating in aromaticity $\therefore$ less available for donation. : least basic. ∴ III > II > I > IV (Basicity order)

$Cr(OH)_3(s) ⇌ {Cr^{3+}(aq.) }_{s}+ {3OH^{-}(aq.)}_{3s}$ $k _{sp} = 27 (s)^4= 6\times 10^{-31}$ $\Rightarrow [3(s)]^4 = 18 \times 10^{-31}$ [OH^{-}] = 3(s) = \[18× 10^{-31}]^{1/4}$

$Zn + 2HCl\to ZnCl_2 + H_2$ $Zn + 2NaOH\to Na_2ZnO_2 + H_2$The ratio of the volume of $H_2$ is 1 : 1

Alanine does not show Biuret test because Biuret test is used for deduction of peptide linkage & alanine is amino acid. Albumine is protein so have paptide linkage so it gives positive Biuret test.Positive B arfoed test is shown by monosaccharide but not disaccharide. Positive M olisch's test is shown by glucose.

The starting compound (A), $H_3N_3B_3Cl_3$, is trichloroborazine. The first reaction involves the reduction of (A) with lithium borohydride ($LiBH_4$), a source of hydride ions ($H^-$). The hydride ions replace the chlorine atoms on the boron ring to form $B_3N_3H_6$, which is borazine, commonly known as inorganic benzene. Therefore, compound (B) is borazine.

The second reaction involves the conversion of (A) to $H_3N_3B_3(Me)_3$. This is a substitution reaction where the chlorine atoms are replaced by methyl ($Me$) groups. This requires a strong nucleophilic source of methyl groups to attack the electrophilic boron atoms. A Grignard reagent, such as methylmagnesium bromide ($MeMgBr$), is an excellent nucleophile for this purpose. Thus, compound (C) is $MeMgBr$.

$[Co(NH_3)_4Cl_2]$ has 2 geometrical isomerscis isomer has Cl-Co-Cl angle of 90°

Each carbon atom is $sp^2$ hybridizedTherefore each carbon has 3 $sp^2$ hybrid orbitals.Hence total $sp^2$ hybrid orbitals are 18.

$ds=∫q_\text{rev}/T$

for gas P $={0.1\times 0.08\times 272}/1$ P = 2.176 atm$P_1V_1 = P_2V_2$ $2.176\times 1 = 1 \times V_2$ $V_2 =$ 2.176 litre

ppm $={10.3\times 10^{-3}}/1030\times 10^6=10$

$\ln(t_1/t_2)={-Ea}/R[1/T_2-1/T_1]$ $\ln(60/40)={-Ea}/8.3[1/400-1/300]$E = 0.4 × 1200 × 8.3 = 3.984 kJ/mole

Total number of bonds between chromium and oxygen in both structures are 18.Note But answer of NTA is 12. They consider only linkages between Chromium and Oxygen but in question total no. of bonds are asked so σ and π bonds must be considered separately.

C ⇒ 12 × 4 = 48H ⇒ 8 × 1 = 8O ⇒ 16 × 1 = 16Total = 72% of C $=48/72\times 100=66.66$

$A=\lim_{x\to 0}x[4/x]$ $=\lim_{x\to 0}x(4/x)-x\{4/x\}=4$ f(x) = \[x^2]$\sin(\pi x)$will be discontinuous at non integers$\therefore x=\sqrt{A+1}$i.e$\sqrt5$

7x + 6y - 2z = 0 .... (1)3x + 4y + 2z = 0 .... (2)x - 2y - 6z = 0 .... (3)$\Delta=\begin{vmatrix}7 & 6 & -2 \\ 3 & 4 & 2 \\ 1 & -2 & -6\end{vmatrix}=0$ ⇒ infinite solutionsNow (1) + (2) ⇒ y = -x put in (1), (2) & (3) all will lead to x = 2z

(Bonus) $x = 2\sin\theta - \sin2\theta$ $\Rightarrow \frac{dx}{d\theta }=2\cos\theta -2\cos2\theta$ $=4\sin(\theta /2)\sin({3\theta }/2)$ $y = 2\cos\theta - cos2\theta$ $\Rightarrow \frac{dy}{d\theta }=-2\sin\theta +2\sin2\theta =4\sin\theta /2\cos{3\theta }/2$ $\Rightarrow \frac{dy}{dx}=\cot({3\theta }/2)$ $\Rightarrow \frac{d^2y}{dx^2}= \frac{-3/2\text{cosec}^2({3\theta }/2) }{4\sin(\theta /2)\sin{3\theta }/2}$ $\Rightarrow (\frac{d^2y}{dx^2})_{\theta =\pi }=3/8$

Let $x^2/a^{2+}y^2/b^2=1;a > b;$ $2b =4/\sqrt3$ $\Rightarrow b=2/\sqrt3\Rightarrow b^2=4/3$tangent $y=-x/6+4/3$ compare with $y=mx±\sqrt{a^2m^{2+}b^2}$ $\Rightarrow m=-1/6\Rightarrow \sqrt{a^2/36+4/3}=4/3$ $\Rightarrow a=4$ $e=\sqrt{1-b^2/a^2}=1/2\sqrt{11/3}$

$ax^2 - 2bx + 5 = 0$ $\Rightarrow \alpha =b/a;\alpha ^2=5/a\Rightarrow b^2=5a$ $x^2 - 2bx - 10 = 0$ $\Rightarrow \alpha ^{2-}2b\alpha -10=0$ $\Rightarrow a=1/4\Rightarrow \alpha ^2=20$; $\alpha \beta =-10\Rightarrow \beta ^2=5$ $\Rightarrow \alpha ^{2+}\beta ^2=25$

Required area = Area of trepezium ABCD - Area of parabola between $x =1/2$ & $x=\sqrt3/2$ $A=1/2(\sqrt3/2-1/2)(1/2+1-\sqrt3/2)$ $-∫^{\sqrt3/2}_{1/2}(x-1/2)^2dx$ $=\sqrt3/4-1/3$

∑ P(X) = 1 $\Rightarrow K^2 + 2K + K + 2K + 5K^2 = 1$ $\Rightarrow 6K^2 + 5K - 1 = 0$ $\Rightarrow (6K - 1) (K + 1) = 0$ $\Rightarrow K = -1$ (rejected) $\Rightarrow K =1/6$ $P(X > 2) = K + 2K + 5K^2=23/36$

$x= ∑^{\infty }_{n=0}(-1)^n\tan^{2n}\theta$ $=1-\tan^2\theta +\tan^4\theta +..$ $\Rightarrow x=\cos^2\theta$ $y= ∑^{\infty }_{n=0}\cos^{2n}\theta$ $=1+\cos^2\theta +\cos^4\theta +...$ $\Rightarrow y=1/{\sin^2\theta }\Rightarrow y=1/{1-x}$ $\Rightarrow y(1- x) = 1$

$F'(x )=x^2g(x)$ $=x^2 ∫^{x}_{1}f(u)du$ $\Rightarrow F'(1)=0$ $F"(x)=x^2f(x)-2x∫^{x}_{1}f(u)du$ $F"(1) = 1.f(1) - 2 \times 0$ $F"(1) = 3$F'(1) = 0 and F"(1) = 3 > 0 So, Minima

$y^2 = 8x$ $4t_1=-2\Rightarrow t_1=-1/2$ $t_1.t_2=-1$ $t_2=-1/t_1$ $\Rightarrow t_2=2$So coordinate of B is (8, 8)∴ Equation of tangent at B is8y = 4(x + 8) ⇒ 2y = x + 8

(BONUS)10 different balls in 4 different boxes. $1/4^10(4!\times 10!/{2!\times 3!\times 0\times 5!}+4!\times 10!/{2!\times 3!\times 1!\times 4!}+4!\times 10!/{(2!)^2\times 2!\times (3!)^2\times 2!})$ $={17\times 945}/2^15$

A : x ∈ (-2, 2); B : x ∈ (-∞, -1] ∪ [5, ∞)⇒ B - A = R - (-2, 5)

$\frac{dy}{dx}=\frac{xy}{x^{2+}y^2}$Let $y = vx$ $\frac{dy}{dx}=v+x\frac{dv}{dx}$ $v+x\frac{dv}{dx}=\frac{xvx}{x^{2+}v^2x^2}$ $=v/{1+v^2}$ $x\frac{dv}{dx}=-v^3/{1+v^2}$ $∫{1+v^2}/v^3dv=∫-{dx}/x$ $\Rightarrow ∫v^{-3}dv+∫1/vdv=-∫{dx}/x$ $\Rightarrow v^\frac{-2}{-2}+\ln v=-\ln x+\lambda$ $\Rightarrow -1/2x^2/y^{2+}\ln y-\ln x=-\ln x+\lambda$ $\Rightarrow -1/2+0=\lambda \Rightarrow \lambda =-1/2$ $\Rightarrow -1/2x^2/y^{2+}\ln y +1/2=0$ at $y=e$ $\Rightarrow -1/2x^2/e^{2+}1+1/2=0$ $\Rightarrow x^2/{2e^2}=3/2\Rightarrow x^2=3e^2$ $\therefore x=\sqrt3e$

$I=∫\frac{d\theta }{\cos^2\theta (\tan2\theta +\sec2\theta )}$ $=∫\frac{\sec^2\theta }{\frac{2\tan\theta }{1-\tan^2\theta }+\frac{1+\tan^2\theta }{1-\tan^2\theta }}$ $=∫{(1-\tan^2\theta )\sec^2\theta d\theta }/(1+\tan\theta )^2$ $\tan\theta =t\Rightarrow \sec^\theta d\theta =dt$ $I=∫{1-t^2}/(1+t)^2dt$ $=∫{({1-t})(1+t)}/(1+t)^2dt$ $=∫1/{1+t}-t/{1+t}dt$ $=\ln |1+t|-∫(\frac{1+t}{1+t}-1/{1+t})dt$ $=\ln |1+t|-t+\ln |1+t|$ $=2\ln|1+t|-t+C$ $\lambda =-1,f(\theta )=1+\tan\theta$

$z=x+iy$ $|x|+|y|=4$ $|z|=\sqrt{x^{2+}y^2}$ $\Rightarrow |z|_{\min}=\sqrt8$ & $|z|_{\max}=4=\sqrt16$So |z| cannot be $\sqrt7$

p → (p \wedge ~q) is F⇒ p is T & p \wedge ~ q is F ⇒ q is T∴ p is T, q is T

$R_1 \to R_1 + R_3 - 2R_2$ $f(x)=\begin{vmatrix}a+c-2b & 0 & 0 \\ x+b & x+3 & x+2 \\ x+c & x+4 & x+3\end{vmatrix}$ $= (a + c - 2b) ((x + 3)^2 - (x + 2)(x + 4))$ $= x^2 + 6x + 9 - x^2 - 6x - 8 = 1$ $\Rightarrow f(x) = 1 \Rightarrow f(50) = 1$

$T_{r+1}=^16C_r(x/{\cos\theta })^{16-r} (1/{x\sin\theta })^r$ $=^16C_r(x)^{16-2r}\times 1/{ (\cos\theta )^{16-r}(\sin\theta )^r}$For independent of x ; 1$6-2r = 0\Rightarrow r = 8$ $\Rightarrow T_9=^16C_8 1/{\cos^8\theta \sin^8\theta }$ $=^16C_8 2^8/{(\sin2\theta )^8}$for $\theta \in [\pi /8,\pi /4]$ $l_1$ is least for $\theta _1=\pi /4$for $\theta \in [\pi /16,\pi /8]$ $l_2$ is least for $\theta _2=\pi /8$ $l_2/l_1=(\sin2\theta _1)^8/(\sin2\theta _2)^8$ $=(\sqrt2)^8=16/1$

$∑^{100}_{n=1}a_{2n+1}=200$ $\Rightarrow a_3+a_5+a_7+....+a_201=200$ $\Rightarrow ar^2(r^{200-}1)/(r^{2-}1)=200$ $∑^{100}_{n=1}a_{2n}=100$ $\Rightarrow a_2+a_4+a_6+...+a_200=100$ $\Rightarrow {ar(r^{200-}1)}/(r^{2-}1)=100$On dividing r = 2on adding $a_2 + a_3 + a_4 + a_5$ $+ ... +a_200 + a_201 = 300$ $\Rightarrow r(a_1+a_2+a_3+...+a_200)=300$ $\Rightarrow ∑^{200}_{n=1}a_n=150$

f(g(x)) = xf'(g(x)) g'(x) = 1put x = a⇒ f'(b) g'(a) = 1$f'(b)=1/5$

Common term are : 23, 51, 79, ... $T_n$ $T_n \le 407$ $\Rightarrow 23 + (n - 1)28 \le 407$ $\Rightarrow n \le 14.71$ $n = 14$

$\vec{b} . \vec{c} = 10$ $\Rightarrow 5|\vec{c}|\cos\pi /3=10$ $\Rightarrow |\vec{c}|=4$ $|\vec{a}\times (\vec{b}\times \vec{c})|=||\vec{a}||\vec{b}\times \vec{c}|$ $=\sqrt3 . 5 . 4 . \sin\pi /4=30$

If λ = -7, then planes will be parallel & distance between them will be $3/\sqrt633$ $\Rightarrow k=3$But if $\lambda \ne -7$, then planes will be intersecting & distance between them will be 0