NEET 2021 Chemistry Questions with Answers Key Solutions

Solution:

A hexagonal primitive unit cell contains 6 atoms in the crystal structure i.e. n = 6
We know that, for hcp,

The number of octahedral voids will be equal to the number of atoms in the crystal structure.
No. of octahedral voids = 6.

The number of tetrahedral voids will be equal to two times the number of atoms in the crystal structure .
No. of tetrahedral voids = 2 x 6 = 12.

Solution:

$Zr$ and $Hf$ have same size due to Lanthanoid contraction

Solution:

It is an exothermic reaction
For exothermic reaction, energy of reactants is greater than energy of products. By inspection option $3$ is correct.

Solution:

$_1H^3 \to \,_2He^3 + \, _{-1}e^0 (\beta-$ ray)

Solution:

Deficiency of vitamin $B_{12}$ causes Megaloblastic Anaemia/Pernicious anaemia

Solution:

$\lambda_{ M }^{ CH _{3} COOH }=\lambda_{ CH _{3} COO^- }^{ M }+\lambda_{ Na ^{+}}^{ M }+\lambda_{ H ^{+}}^{ M }+\lambda_{ Cl^- }^{ M }-\lambda_{ Na ^{+}}^{ M }-\lambda_{Cl^-}^{ M }$
$=\lambda_{ CH _{3} COONa }^{ M }+\lambda_{ HCl }^{ M }-\lambda_{ NaCl }^{ M }$
$=91+426.16-126.45$
$=390.71 \,S\,cm ^{2} mole ^{-1}$

Solution:

Solution:

Temperature about $2200 \,K$. This temperature is attained at the bottom near tuyers

Solution:

$\underset{1^{\circ}\text{amine}}{R - NH - H } + \underset{\text{Hinsberg reagent}}{SO_2C_6H_5} \to \underset{\text{N - alkyl benzene Sulphonamide alkali soluble}}{R - NHSO_2C_6H_5} + HCl$
$H$ - atom attached to nitrogen is highly acidic because it is attached to strong electron withdrawing $- SO _{2} C _{6} H _{5}$ group

Solution:

- Osmotic pressure $(\pi)=i C R T$
where $C$ is molar concentration of the solution
- With increase in molar concentration of solution osmotic pressure increases.
- Since, weight of all solutes and its solution volume are equal, so higher will be the molar mass of solute, smaller will be molar concentration and smaller will be the osmotic pressure.
- Order of molar mass of solute decreases as
Sucrose > Glucose > Urea
- So, correct order of osmotic pressure of solution is $P _{3}< P _{1}< P _{2}$

Solution:

Solution:

Aspirin and Paracetamol are non - narcotic where as Morphine and Heroin are narcotic analgesics

Solution:

Correct sequence of bond enthalpy of $C-x$ bond is
$ CH _{3}- F > CH _{3}- Cl > CH _{3}- Br > CH _{3}-1 $

Solution:

Solution:

For an ideal gas, $C_P - C_V = R$

Solution:

Due to high polarising power of $Be ^{+2}$ ion, all beryllium halides are predominantly covalent. They are soluble in organic solvents

Solution:

$ \% C =78$
$ \% H =22 $
$ C : H =\frac{78}{12}: \frac{22}{1}$
$=6.5: 22 $
$ \frac{6.5}{6.5} : \frac{22}{6.5}=1: 3.3 $
$ C : H = CH _{3} $

Solution:

$CH_3 - CH(Br) - CH_2 - CH_2 - CH_3 \xrightarrow{alc. KOH}$
$\underset{\text{1 - pentene (minor)}}{CH_2 = CH - CH_2 - CH_2 - CH_3} + \underset{\text{2 - pentene(major)}}{CH_3 - CH = CH - CH_2 - CH_3}$
During dehydrohalogenation most stable alkene is formed by Saytzeff rule. According to Saytzeff’s rule an alkene with more number of hyperconjugated
hydrogens is the major product

Solution:

Solution:

Noble gases have low melting point and boiling point due to weak London dispersion forces

Solution:

For a salt of weak acid and weak base $P^{H}$ does not depend on concentration of salt.
$ P ^{ H }=7+\frac{1}{2}\left[ P ^{ k _{ a }}- P ^{ K _{ b }}\right] $
$=7+\frac{1}{2}[4.77-3.27]=7.75$

Solution:

Colloidal sol shows Tyndall effect. Starch is a colloid

Solution:

Down the group acidic strength increases as bond length increases and bond strength decreases due to which ease of release of $ H ^+$ increases. Acidic strength increases

Solution:

Solution:

$P \propto \frac{1}{ V }$
at constant temperature $PV = K$
Greater the temperature greater the magnitude of $PV$

Solution:

Beryllium chloride has chain structure in solid state. Vapour phase forms chlorobridged dimer

Solution:

Only metal stable in liquid state at room temperature is $Hg$. It has non volatile impurity. Therefore, it is purified by distillation

Solution:

With the formula $C _{4} H _{10} O$ the possible ethers are
(1) diethyl ether
(2) methyl $n$ - propyl ether
(3) iso propyl methyl ether
(1) and (2), (1) and (3) are metamers

Solution:

Cubic, Tetragonal, Orthorhombic

Solution:

Teflon is formed by addition polymerisation of Tetrafluoro ethylene
$nCF_2 = CF_2 \xrightarrow[\text{Under pres}]{\text{Peroxide}}-(-CF_2 - CF_2 -)_n-$

Solution:

Wavelength $(\lambda)=\frac{c}{v}$
$c$ is speed of light
$v$ is frequency
$ \lambda=\frac{3 \times 10^{8}}{1368 \times 10^{3}}=219.298 m =219.3 m $

Solution:

More electropositive element displaces less electropositive metal

Solution:

Most of Lanthanoids in trivalent state are coloured due to unpaired electrons in $f$ - subshell. $La^{+3}, Lu^{+3}$ are colourless

Solution:

Eclipsed conformer is highly unstable Dihedral angle is zero

Solution:

$PCl _{5}$ - Trigonal bipyramidal ( $5$ bond pairs)
$SF _{6}$ - Octahedral ( $6$ bond pairs)
$BrF _{5}$ - Square pyramidal ( $5$ bond pairs $+ 1$ lone pair)
$BF _{3}$ - Trigonal planar ( $3$ bond pairs)

Solution:

$CH_3CH_2COONa + NaOH \xrightarrow[\Delta]{CaO}CH_3 - CH_3 + Na_2CO_3$
This is a decarboxylation reaction

Solution:

Solution:

(a) $2 SO _{2}( g )+ O _{2}( g ) \rightarrow 2 SO _{3}( g ) \ldots$ Tropospheric pollution
(b) $HOCl ( g ) \xrightarrow{ hv } H \dot{ O }+\dot{ C } l \ldots$ Ozone depletion
(c) $CaCO _{3}+ H _{2} SO _{4} \longrightarrow CaSO _{4}+ H _{2} O + CO _{2} \ldots$ Acid rain
(d) $NO _{2}( g ) \xrightarrow{ hv } NO ( g )+ O ( g ) \ldots$ Photo chemical smog

Solution:

Solution:

Solution:

Solution:

$NaBH_4/ C_2 H_5 OH$ is a weak reducing agent. It cannot reduce $- COOR$ group
It reduces $- \overset{\underset{||}{O}}{C}-$ to $- COOH-$ group

Solution:

Isoelectronic refers to two atoms, ions, or molecules that have the same number of total electrons and the same number of valence electrons. (a) $O ^{2-}$ and $F ^{-}$both have 10 electrons (b) $Na ^{+}$and $Mg ^{2+}$ both have 10 electrons (c) $Mn ^{2+}$ and $Fe ^{3+}$ both have 23 electrons (d) $Fe ^{2+}$ has 24 electrons. $Mn ^{2+}$ has 23 electrons.

Solution:

Degree of dissociation $(\alpha)$ of $CH _{3} COOH$
$=\frac{\Lambda_{ C }}{\Lambda_{0}}=\frac{20}{350+50}=\frac{1}{20}$
$ K _{ a }=\frac{ C \alpha^{2}}{1-\alpha} $
$1-\alpha \approx 1 $
$K _{ a }= C \alpha^{2} $
$ K _{ a }=0.007 \times \frac{1}{20} \times \frac{1}{20} $
$ K _{ a }=1.75 \times 10^{-5}$

Solution:

$n = n _{ O _{2}}+ n _{ H _{2}}$
$n =\frac{4}{32}+\frac{2}{2}$
$n =\frac{9}{8}$
$P =\frac{ nRT }{ V }$
$P =\frac{9}{8} \times \frac{0.0821 \times 273}{1}$
$=25.18\, atm$

Solution:

Applying Raoult's law,
$P_{\text {total }}=P_{A}^{0} . X_{A}+P_{B}^{0} . X_{B}$
$X_{A}=\frac{3}{5}, X_{B}=\frac{2}{5}$
$P_{\text {total }}=280 \times\left(\frac{3}{5}\right)+420 \times\left(\frac{2}{5}\right)$
$=336 \,mm$ of $ Hg$

Solution:

For an isothermal process, $\Delta T =0$
$ \therefore \Delta U =0$
For an irreversible expansion, $\Delta S _{\text {total }} \neq 0$

Solution:

dipole moment $(\mu) = 0$

Solution:

From $H _{2} O$ to $H _{2}$ Te as $X - H$ bond enthalpy decreases, acidic strength increases ( $K _{ a }$ increases)
$\therefore H _{2} O < H _{2} S < H _{2} Se < H _{2} \text { Te....K } K _{ a } $
$ H _{2} O > H _{2} S > H _{2} Se > H _{2} \text { Te....P }^{ K _{ a }}$

Solution:

$K = Ae ^{-\frac{E_{a}}{R T}}$
$\underset{y}{\ln K} = \underset{C}{\ln A} -\underset{m}{\frac{ E _{ a }}{ R }} \cdot \underset{x}{\frac{1}{ T }}$
Slope $=-\frac{ E _{ a }}{ R }$
$-\frac{ E _{ a }}{ R }=-5 \times 10^{3} K$
$E _{ a }=5 \times 8.314 \,J K ^{-1} mole ^{-1} \times 10^{3} K$