JEE Main 2017 Chemistry Question Paper with Answer and Solution

(B) To determine the shape of a molecule,we calculate the hybridization of the central atom using the formula: $H = \frac{1}{2}(V + M - C + A)$,where $V$ is the number of valence electrons,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.
$1$. For $CO_3^{2-}$: $H = \frac{1}{2}(4 + 0 - 0 + 2) = 3$ ($sp^2$ hybridization,triangular planar).
$2$. For $NO_3^-$: $H = \frac{1}{2}(5 + 0 - 0 + 1) = 3$ ($sp^2$ hybridization,triangular planar).
$3$. For $SO_3$: $H = \frac{1}{2}(6 + 0 - 0 + 0) = 3$ ($sp^2$ hybridization,triangular planar).
All species in option $B$ have $sp^2$ hybridization with no lone pairs on the central atom,resulting in a triangular planar geometry.

(D) The reactivity of a compound toward an aqueous solution of sodium carbonate $(Na_2CO_3)$ depends on its acidity. Compounds that are sufficiently acidic to donate a proton to the carbonate ion $(CO_3^{2-})$ or bicarbonate ion $(HCO_3^-)$ will react.
In this case,we evaluate the acidity of the hydrocarbons by looking at the stability of their conjugate bases (the corresponding carbanions formed after deprotonation).
$(a)$ Cyclopropene: Forms an antiaromatic $(4\pi e^-)$ carbanion,which is highly unstable.
$(b)$ Cyclobutene: Forms an antiaromatic $(4\pi e^-)$ carbanion,which is highly unstable.
$(c)$ $1,3-$Cyclohexadiene: Forms a homoaromatic carbanion,which is unstable due to the lack of continuous delocalization.
$(d)$ $1,3-$Cyclopentadiene: Upon losing a proton,it forms the cyclopentadienyl anion,which is aromatic $(6\pi e^-)$ and highly stable.
Because the conjugate base of $1,3-$cyclopentadiene is aromatic and exceptionally stable,$1$,$3$-cyclopentadiene is the most acidic among the given options and thus the most reactive toward an aqueous solution of sodium carbonate.

(B) For geometrical isomerism to occur,each carbon atom of the double bond must be attached to two different groups.
At site $I$,the carbon atom is attached to two identical methyl groups,so it cannot show geometrical isomerism.
At site $II$,the carbon is attached to a hydrogen and a carbon chain,and the other carbon is attached to a methyl and a carbon chain; thus,it shows geometrical isomerism.
At site $III$,the carbon is attached to a methyl and a carbon chain,and the other carbon is attached to a methyl and a carbon chain; thus,it shows geometrical isomerism.
At site $IV$,the carbon is attached to a methyl and an ethyl group,and the other carbon is attached to a methyl and a hydrogen; thus,it shows geometrical isomerism.
Therefore,geometrical isomerism is not possible only at site $I$.

(B) The reaction of $1$-methylcyclohex-$1$-ene with $Br_2$ in the presence of light $(h\nu)$ is a free radical substitution reaction (allylic bromination).
In this reaction,the bromine radical $(Br^\bullet)$ abstracts a hydrogen atom from the allylic position.
The most stable free radical formed is the tertiary $(3^\circ)$ allylic radical at the $C-1$ position.
This radical then reacts with $Br_2$ to form $1$-bromo-$1$-methylcyclohex-$2$-ene as the major product.

(A) The dipole moment of a molecule is directly related to the polarity of its bonds and the contribution of its resonance structures.
For compound $I$ (cyclopropenone),the resonance structure involves the transfer of electrons from the $C=O$ bond to the oxygen atom,resulting in a positive charge on the three-membered ring.
This resonance structure is aromatic because the ring contains $2 \pi$ electrons ($H$ückel's rule,$4n+2$ where $n=0$),which provides significant stability.
Due to this aromatic character,the dipolar resonance structure makes a major contribution to the overall electronic structure,leading to a very high dipole moment compared to the other compounds where such aromatic stabilization of the dipolar form is not present.

(C) Let $p$ be the statement "Two numbers are not equal" and $q$ be the statement "Their squares are not equal".
The given statement is in the form $p \to q$.
The contrapositive of $p \to q$ is defined as $\sim q \to \sim p$.
Here,$\sim q$ is "The squares of two numbers are equal" and $\sim p$ is "The numbers are equal".
Therefore,the contrapositive is "If the squares of two numbers are equal,then the numbers are equal".

(C) For the diverging lens,the incident light is parallel,so the image is formed at its focus. Since it is a diverging lens,$f_1 = -25 \, cm$. The image is virtual and formed at $v_1 = -25 \, cm$ from the diverging lens.
This image acts as an object for the converging lens. The distance between the lenses is $d = 15 \, cm$. The object distance for the converging lens is $u_2 = -(25 + 15) \, cm = -40 \, cm$.
For the converging lens,$f_2 = +20 \, cm$. Using the lens formula $\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2}$:
$\frac{1}{v_2} - \frac{1}{-40} = \frac{1}{20}$
$\frac{1}{v_2} = \frac{1}{20} - \frac{1}{40} = \frac{2-1}{40} = \frac{1}{40}$
$v_2 = +40 \, cm$.
Since $v_2$ is positive,the final image is real and formed at a distance of $40 \, cm$ from the converging lens.

(C) Let the velocity of particle $A$ after the collision be $v_{1}$ and the velocity of particle $B$ be $v_{2}$.
Using the law of conservation of linear momentum:
$m v = m v_{1} + (m/2) v_{2}$
$2v = 2v_{1} + v_{2} \quad \dots (i)$
For an elastic head-on collision,the coefficient of restitution $e = 1$:
$e = \frac{v_{2} - v_{1}}{v - 0} = 1$
$v = v_{2} - v_{1} \implies v_{2} = v + v_{1} \quad \dots (ii)$
Substituting $(ii)$ into $(i)$:
$2v = 2v_{1} + (v + v_{1})$
$v = 3v_{1} \implies v_{1} = \frac{v}{3}$
$v_{2} = v + \frac{v}{3} = \frac{4v}{3}$
The de-Broglie wavelength is $\lambda = \frac{h}{p} = \frac{h}{mv}$.
Ratio of wavelengths: $\frac{\lambda_{A}}{\lambda_{B}} = \frac{p_{B}}{p_{A}} = \frac{(m/2) v_{2}}{m v_{1}}$
$\frac{\lambda_{A}}{\lambda_{B}} = \frac{(m/2) \times (4v/3)}{m \times (v/3)} = \frac{2mv/3}{mv/3} = 2$.

(C) For geometrical isomerism to occur at a double bond,each carbon atom of the double bond must be attached to two different groups.
$I$: The carbon atom is attached to two identical methyl groups $(a, a)$,so geometrical isomerism is not possible.
$II$: Both carbons are attached to different groups,so it shows geometrical isomerism.
$III$: The carbon atom is attached to two identical methyl groups,so geometrical isomerism is not possible.
$IV$: Both carbons are attached to different groups,so it shows geometrical isomerism.
Therefore,geometrical isomerism is not possible at sites $I$ and $III$.

(A) According to Le Chatelier's principle,the equilibrium position is affected by changes in the concentration of gaseous or aqueous species.
In the given reaction,$Fe_2O_{3(s)}$ is a solid.
The active mass of a pure solid or pure liquid is taken as unity $(1)$ and remains constant regardless of the amount present.
Therefore,the addition or removal of $Fe_2O_{3(s)}$ does not change the concentration of the reactants or products involved in the equilibrium expression,and thus it will not disturb the equilibrium.