NEET 2020 Chemistry Questions with Answers Key Solutions

Solution:

Species in which bond order is zero do not exist.
(1) $O_2 = (KK)(\sigma 2s^2)(\sigma^* 2s^2)(\sigma 2p_z^2)$
$(\pi 2p_x^2 = \pi 2p_y^2)( \pi^* 2p_x^1 = \pi^* 2p_y^1)$
Bond Order $ = \frac{10 - 6}{2} = 2$
(2) $He_2 = (\sigma 1s^2)(\sigma^* 1s^2)$
Bond Order $= 0$
(3) $Li_2 = (\sigma1s^2)(\sigma^* 1s^2)(\sigma^* 2s^2)$
Bond Order $= 1$
(4) $C_2 = (\sigma 1s^2)(\sigma^* 1s^2)(\sigma 2s^2)(\sigma^* 2s^2)$
$(\pi 2p_x^2 = \pi 2p_y^2)$
Bond Order $= 2$
As, bond order of $He_2$ is zero it does not exist.

Solution:

$Ni(OH)_2 \rightleftharpoons \underset{(S)}{Ni^{2+}} + \underset{(2S)}{2OH^-} S =$ molar conc. of $Ni(OH)_2$
$NaOH \rightarrow Na^+ +\underset{(0.1\,M}{OH^-}$
$K_{SP} = [Ni^{2+}][OH^-]^2$
$ = (S)(2S + 0.1)^2$
$=(S)(4S^2 + 0.01 + 0.4S)$
$= 4S^3 + 0.01S + 0.4 S^2$
Neglecting higher power of $S$,
$2 \times 10^{-15} = 0.01S$
$S = 2 \times 10^{-13}M$

Solution:

(1) $CO_2(s)$ (dry ice) [not $CO_2(g)$] is used as refrigerant for ice cream and frozen food.
(2) The structure of $C_{60}$ contains $20$ six carbon rings and $12$ five carbon rings.
(3) $ZSM-5$ (a type of zeolite) is used to convert alcohols directly into gasoline.
(4) $CO(g)$ is colourless and adourless gas

Solution:

$\Delta G^{\circ} = -RT\,ln\,K_C$
$\Delta G^{\circ} = -8.314 \times 300 \,ln ( 2 \times 10^{13})$

Solution:

Solution:

Atomic number	Name	IUPAC offical name
101	Unnilunium	Mendelevium
103	Unniltrium	Lawrencium
106	Unnilhexium	Seaborgium
111	Unununnium	Roentgenium

Element $(Z = 110)$ is ununnilium and its IUPAC offical name is Darmstadtium.

Solution:

In bcc unit cell, atoms lying along a body diagonal touch each other
Length of body diagonal $=\sqrt{3}a$
$\sqrt{3}a=4r$

$r=\frac{\sqrt{3}}{4}a$
As $a=288\,pm$
$\therefore r=\frac{\sqrt{3}}{4}\times288Pm$

Solution:

The species with polar bonds and shall be non polar i.e., have zero dipole moment if resultant of all dipoles is cancelled due to symmetrical structure.

Solution:

Dissociation of sulfuric acid,
$H _{2} SO _{4} \rightarrow 2 H ^{+}( aq )+ SO _{4}^{-2}( aq )$
Disssociation of water,
$H _{2} O \rightarrow H ^{+}( aq )+ OH ^{-}( aq )$
At Anode : $4 OH ^{-} \rightarrow 2 H _{2} O ^{+}( l )+ O _{2}( g )+4 e ^{-}$
At Cathode: $2 H ^{+}( aq )+2 e ^{-} \rightarrow H _{2}( g )$

Solution:

Solution:

	Oxyacid	Structure
(1)	$H_{2}S_{2}O_{7}$
(2)	$H_{2}SO_{3}$ (Sulphurous acid)
(3)	$H_{2}SO_{4}$ (Sulphuric acid)
(4)	$H_{2}S_{2}O_{8}$

Solution:

The Carbylamine test is a test for the detection of primary amines. (both aliphatic and aromatic amines)

In this test, the analyte is heated with alcoholic potassium hydroxide and chloroform.

If a primary amine is present, the foul smelling compound isocyanide is formed.
$
R - NH _{2}+ CHCl _{3}+3 KOH \rightarrow RN ^{+} \equiv C ^{-}+3 KCl +3 H _{2} O
$

Solution:

Electronic configuration of $Cr=[Ar]3d^{5}\,4s^{1}$
Electronic configuration of $Cr^{+2}=[Ar]3d^{4}$
Number of unpaired electrons $= 4$
Spin only magnetic moment $=\sqrt{n\left(n+2\right)}BM $ as $n=4$
$\therefore \mu=\sqrt{4\left(4+2\right)}$
$=\sqrt{24}$
$=4.90\,BM$

Solution:

For free expansion of ideal gas $P _{ e x t }= 0$
$W _{ p v }= 0$
$q = 0$ (adiabatic process)
$\Delta E = q + w$
$\Delta E = 0$
$\Delta E = n C _{ v m } \Delta T = 0$
$q = 0 , \Delta \mathbf { T } = 0 , w = 0$

Solution:

$\Delta\,T_{f}=iK_{f}m$
As solute is non-electrolyte, $i = 1$
$\Delta\,T_{f}=1\times 5.12\times 0.078$
$=0.40\,K$

Solution:

$ \begin{array}{l} w = ZQ \\ Q =\frac{ w }{ Z }=\frac{20}{40 / 2} F =1 F \end{array} $
where $Z=\frac{E}{F}$

Solution:

This reactions carried out between two different aldehydes and/or ketones is called cross aldol condensation

Solution:

Paper Chromatography is a type of partition chormatography. Chormatography paper contains water trapped in it, which acts as the stationary phase. The paper selectively retains different components according to their differing partition in two phases.

Solution:

Collision frequency $\propto$ number of reactant molecules per unit volume
As concentraction o f reactants o f a reaction increase the number of reactant molecules per unit volume increase which increases the collision frequency

Solution:

Partial presure of $N_2 =$ mole fraction $\times P_{Total}$ of $N_2$
$X_{N_2} = \frac{\text{moles of }N_2}{\text{ total moles}}$
moles of $N_2 = \frac{7}{28} = \frac{1}{4};$
moles of $Ar = \frac{8}{40} = \frac{1}{5}$
$X_{N_2} = \frac{(\frac{1}{4})}{\frac{1}{4} + \frac{1}{5}} = \frac{5}{9}$
$P_{N_2} = \frac{5}{9} \times 27 = 15 $ bar

Solution:

(1) Pig iron contains about 4% carbon and many impurities in smaller amount (e.g., S, P, Si, Mn). It can be moulded into variety o f shapes
(2) Wrought iron is the purest form o f carbon
(3) Blister copper has blistered appearance due to evolution of $SO_{2}$
(4) Vapour phase refining is carried out for nickel by Mond’s process

Solution:

Tertiary carbocations are stable due to inductive effect (+I effect of alkyl groups) and hyperconjugation.
Tert. carbocation has greater number of alpha H in comparison to sec carbocation. This increases hyperconjugation which results to its greater stability.

Solution:

Sodium dodecyl benzene sulphonate (anionic detergent)
(2) $CH_{3}(CH_{2})_{10}CH_{2}OSO^{-}_{3}Na^{+}$
Sodium lauryl sulphate (anionic detergent)
(3) $C_{17}H_{35}COONa$
Sodium stearate (Soap)

Cetyltrimethylammonium bromide (Cationic detergent)

Solution:

This reaction is an example of $\beta$-elimination.Hydrogen is removed from $\beta-$ carbon and halogen from $\alpha$-carbon, hence dehydrohalogenation reaction. Pent-2-ene is major product known as Saytzeff's product and it is more stable alkene than Pent-1-ene.
Generally, in $E _{2}$ reaction Zaitsev alkene is formed as a major product (more stable alkene).

Solution:

	Solutions	Type
(1)	Chloroethane + Bromoethane	Ideal Solutions
(2)	Ethanol + Acetone	Nonideal (+ve deviation)
(3)	Benzene + Toluene	Ideal solutions
(4)	Acetone+$CHCl_{3}$	Nonideal (-ve deviation)

Solution:

Arrangement o f different ligands in increasing order of field strength called spectrochemical series is as follows:
$I^{-}<\,SCN^{-}<\,Cl^{-}<\,S^{2-}<\,F^{-}<\,OH^{-}<\,C_{2}O^{2-}_{4}<\,H_{2}O<\,NH_{3}<\,CN^{-}<\,CO$

Solution:

Amine	Structure	Nature
Lysine		Basic (As it contains more number of - $NH_{2}$ groups as composed to $-COOH$ group)
Serine		Neutral
Alanine		Neutral
Tyrosine		Neutral

Solution:

On passing $HCl$ through aqueous solution of $CaCl_2, MgCl_2$ and $NaCl$, crystals of pure $NaCl$ seperate out. Calcium and magnesium chloride being more soluble than $NaCl$, remain in solution.

Solution:

Solution:

$CO$ is formed by incomplete combustion. It forms stable complex with $O_2$ called carboxyhaemoglobin which is more stable than oxyhaemoglobin and it reduces oxygen carrying ability of blood.

Solution:

$C_{12}H_{22}O_{11}+H_{2}O \to\underset{\text{D(+)Glucose}}{{C_{6}H_{12}O_{6}}}+\underset{\text{D(-)fructose}}{{C_{6}H_{12}O_{6}}}$
Sucrose contains two monosaccharides i.e., $\alpha-D$ glucose and $\beta-D-$ fructose which are obtained on its hydrolysis

Solution:

Potassium ions are the most abundant cations within cell fluids where they activate many enymes, participate in the oxidation of glucose to produce ATP and with sodium are responsible for transmission of nerve signals.

Solution:

Number of atoms in $1\,g$ of $Li = \frac{1}{7} \times N_A$
$ = 0.86 \times 10^{23}$
Number of atoms in $1\, g$ of $Ag = \frac{1}{108} \times N_A$
$ = 0.056 \times 10^{23}$
Number of atoms in $1\,g$ of $Mg = \frac{1}{24} \times N_A$
$= 0.25 \times 10^{23}$
Number of atoms in $1\,g$ of $O_2 = \frac{1}{32} \times N_A \times 2 $
$= \frac{N_A}{16} = 0.37 \times 10^{23}$
Maximum number of atoms are present in$1\,g$ of $Li$.

Solution:

$^{175} Lu_{71} $ Number of electrons
= Number of protons $= 71$
Number of neutrons $= 175 - 71 = 104$

Solution:

Oxidation number of $C$ in $CH_4$ is $- 4$
as $x + 4 = 0$$\,\, x = - 4 (x$ is oxidation state of carbon)
oxidation of $C$ in $CCl_4$ is $+ 4$
as $x - 4 = 0 \,\,\,x = + 4$
$\therefore $ Change in oxidation state of carbon is from $-4$ to $+4$

Solution:

Chromate $\left( CrO _{4}^{-2}\right) \Rightarrow$ oxidation state $=+6$
Dichromate $\left( Cr _{2} O _{7}^{-2}\right) \Rightarrow$ oxidation state $=+6$
Oxidation states are same.
All other statements are correct except on oxidation states of chromium.

Solution:

$2 C l(g) \longrightarrow C l_2(g)$
As the process involves bond formation $\Delta H=-v e$ and as the number of gaseous particles are decreasing in the reaction $\Delta S$ is also $-v e$
i.e., $\Delta_r H< 0$ and $\Delta_r S< 0$

Solution:

Greater the Zeta potential more will be the stability of colloidal particles as there will be a balance between the repulsive interactions between the charges of electrical double layer and attractive van der Waal interactions between molecules

Solution:

$NH_{2}CONH_{2}+H_{2}O \to$
$\underset{(A)}{(NH_{4})_{2}CO_{3}} \rightleftharpoons \underset{(B)}{2NH_{3}}+H_{2}O+CO_{2}$
$\underset{(B)}{4NH_{3}}+\overset{2+}{Cu(aq)} \to \underset{\text{(deep blue (C)}}{\left[Cu\left(NH_{3}\right)_{4}\right]_{\left(aq\right)}^{2+}}$

Solution:

a	$CO + H_2$	iii	Synthesis gas
b	Temporary hardness of water	i	$Mg(HCO_3)_2 + Ca(HCO_3)_2$
c	$B_2H_6$	ii	An electron deficient hydride
d	$H_2O_2$	iv	Non planar structure

Solution:

Oxide		Nature
(a)	CO	(ii)	Neutral
(b)	BaO	(i)	Basic
(c)	$Al_{2}O_{3}$	(iv)	Amphoteric
(d)	$Cl_{2}O_{7}$	(iii)	Acidic

Solution:

$K=\frac{2.303}{t}log\, \frac{a}{a-x}$
$t=\frac{2.303}{4.606\times 10^{-3}}$
$log\,\frac{2}{0.2}$
$=\frac{1000}{2} \,\log\,10$
$=500\,\sec$
(as log $10=1)$

Solution:

Solution:

$n$-butane, $n$-hexane, $2,3$-dimethylbutane are symmetrical alkanes. Such alkanes are prepared in good yield by Wurtz reaction as these require single alkyl halide for their preparation.
$CH_3CH_2Cl \xrightarrow[\text{dry ether}]{+2\,Na} \underset{\text{(n-butane)}}{CH_3CH_2CH_2CH_3}$
$CH_3CH_2CH_2Cl \xrightarrow[\text{dry ether}]{+2\,Na}$
$ \underset{\text{(n-hexane)}}{CH_3CH_2CH_2CH_2CH_2CH_3}$
$CH_3- \underset{\overset{|}{CH_3}}{CH}-Cl \xrightarrow[\text{dry ether}]{+2\,Na}\underset{\text{(2,3 -Dimethyl butane)}}{CH_3-\underset{\overset{|}{CH_3}}{CH}-\underset{\overset{|}{CH_3}}{CH} - CH_3}$
$n$-heptane is an unsymmetrical alkane that requires two different alkyl halides for its preparation. This will form three different alkanes, thus, lowering the yield of required alkane.
$CH_3CH_2CH_2Cl + CH_3CH_2CH_2CH_2Cl \xrightarrow[\text{dry ether}]{+2\,Na}$
$CH_3CH_2CH_2CH_2CH_2CH_3 $
$+ CH_3CH_2CH_2CH_2CH_2CH_2CH_3$
$+CH_3CH_2CH_2CH_2CH_2CH_2CH_2CH_2CH_3$

Solution:

Reaction is $S_{N}2$ reaction in which $I^{-}$ ttacks at alkyl group with lesser steric hindrance and weak $C-O$ bond