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NEET SECTION - A [CHEMISTY - MCQ]

Model Paper 8 · Gujarat Board (GSEB) STD 11 Science (GSEB - English Medium). 45 questions.

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SECTION - A [CHEMISTY - MCQ]

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45 questions · 3 auto-graded MCQ + 42 self-marked written.

MCQ 24 Marks
During the fusion of an organic compound with sodium metal, nitrogen of the organic compound is converted into
  • A
    NaCN
  • B
    $NaNH _2$
  • C
    $NaNO _2$
  • D
    NaNC
Answer
A. NaCN
Explanation:
During the fusion of an organic compound with sodium metal, nitrogen of the organic compound is converted into NaCN.
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MCQ 34 Marks
Consider $Cu ^{2+} \mid Cu$ electrode. What is the electrode potential if $0.01 M Cu ^{2+}$ solution is used?
$
\left(T=298 K, E_{\frac{C u^{2+}}{C u}}^o=0.34 V\right)
$
  • A
    0.399 V
  • B
    0.281 V
  • C
    0.591 V
  • D
    0.222 V
Answer
B. 0.281 V
Explanation:
$\begin{array}{l}E_{\left(\frac{C_u{ }^{2+}}{C_u}\right)}=E_{\left(\frac{C u{ }^{2+}}{C_u}\right)}^o-\frac{0.059}{n} \log \left[\frac{1}{\left[C u^{2+}\right]}\right. \\=0.34-\frac{0.059}{2} \log \frac{1}{0.01} \\=0.3400-0.059=0.281 V\end{array}$
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MCQ 54 Marks
Which of the following statement is not true about sucrose?
  • A
    On hydrolysis, it produces glucose and fructose
  • B
    It is also named as invert sugar
  • C
    The glycosidic linkage is present between $C _1$ of $\alpha$-glucose and $C _1$ of $\beta$-fructose
  • D
    It is a non-reducing sugar
Answer
C. The glycosidic linkage is present between $C _1$ of $\alpha$-glucose and $C _1$ of $\beta$-fructose
Explanation:
Statement-(b) is not true for sucrose. It is linked through a glycosidic linkage between C-1 of $\alpha$-glucose and C-2 of $\beta$-fructose. Since, the reducing groups of glucose and fructose are involved in glycosidic bond formation, sucrose is a non-reducing sugar.
Image
On hydrolysis with acids or enzyme, sucrose gives equimolar mixture of D-(+)-glucose and D-(-)-fructose.
Image
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MCQ 64 Marks
Cobalt as a rare element is essential in the synthesis of this:
  • A
    Vitamin D
  • B
    Vitamin $B_1$
  • C
    Vitamin C
  • D
    Vitamin $B _{12}$
Answer
D. Vitamin $B _{12}$
Explanation:
Because cobalt is also an essential trace element for humans and found at the center of $B _{12}$ and produces red blood cells.
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MCQ 94 Marks
$
R-O^{-}+H-O-H \longrightarrow R-O-H^{}+ OH^{-}
$
With respect to above reaction, choose an incorrect statement from the following:
  • A
    The alkoxide ions act as a Bronsted base.
  • B
    Alcohol is a better proton donor than the conjugate acid of hydroxide ions.
  • C
    Water is a stronger acid as compared to alcohol.
  • D
    The conjugate base of water is weaker than alkoxide ions.
Answer
B.Alcohol is a better proton donor than the conjugate acid of hydroxide ions.
Explanation:
The conjugate acid of hydroxide ions (i.e., water) is a better proton donor than alcohol.
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MCQ 114 Marks
Which one is most reactive towards $S _{ N } 1$ reaction?
  • A
    $C _6 H _5 CH _2 Br$
  • B
    $C _6 H _5 CH \left( C _6 H _5\right) Br$
  • C
    $C _6 H _5 C \left( CH _3\right)\left( C _6 H _5\right) Br$
  • D
    $C _6 H _5 CH \left( CH _3\right) Br$
Answer
C. $C _6 H _5 C \left( CH _3\right)\left( C _6 H _5\right) Br$
Image
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MCQ 124 Marks
Which of the following is correct for the complex $\left[ NiBr _2\left( PPh _3\right)_2\right]$ ?
  • A
    It is tetrahedral with two unpaired electrons.
  • B
    It is tetrahedral with one unpaired electron.
  • C
    It is square planar and diamagnetic
  • D
    It is square planar with one unpaired electron.
Answer
B. It is tetrahedral with one unpaired electron.
Explanation:
Image
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MCQ 134 Marks
Which of the following is a didentate ligand?
  • A
    en
  • B
    EDTA $^{4-}$
  • C
    dien
  • D
    trien
Answer
A. en
Explanation:
en (ethylenediamine) is a bidenate ligand.
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MCQ 144 Marks
Gadolinium belongs to 4 f series, Its atomic number is 64 . Which of the following is the correct electronic configuration of gadolinium?
  • A
    $[ Xe ] 4 f ^8 6 d^2$
  • B
    $[ Xe ] 4 f ^7 5 d^1 6 s^2$
  • C
    $[ Xe ] 4 f ^6 5 d^2 6 s^2$
  • D
    $[X e] 4 f^9 5 s^1$
Answer
B. $[ Xe ] 4 f ^7 5 d^1 6 s^2$
Explanation:
$
[Xe] 4 f^7 5 d^1 6 s^2
$
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MCQ 154 Marks
The correct statement among the following is
  • A
    $\left( SiH _3\right)_3 N$ is planar and less basic than $\left( CH _3\right)_3 N$
  • B
    $\left( SiH _3\right)_3 N$ is pyramidal and less basic than $\left( CH _3\right)_3 N$
  • C
    $\left( SiH _3\right)_3 N$ is pyramidal and more basic than $\left( CH _3\right)_3 N$
  • D
    $\left( SiH _3\right)_3 N$ is planar and more basic than $\left( CH _3\right)_3 N$
Answer
A. $\left( SiH _3\right)_3 N$ is planar and less basic than $\left( CH _3\right)_3 N$
Explanation:
The correct statement is that $\left( SiH _3\right)_3 N$ is planar and less basic than $\left( CH _3\right)_3 N$. The compounds trimethylamine $\left( CH _3\right)_3 N$ and triethylamine $\left( SiH _3\right)_3 N$ have similar formulae but have totally different structures. In trimethylamine the arrangement of electrons is as follows:
Image
In trisilylamine, three $sp ^2$ orbitals are used for a-bonding, giving a plane triangular structure.
Image
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MCQ 164 Marks
Name the gas that can readily decolourise acidified $KMnO _4$ solution:
  • A
    $P _2 O _5$
  • B
    $SO _2$
  • C
    $NO _2$
  • D
    $CO _2$
Answer
B. $SO _2$
Explanation:
When $SO _2$ gas is passed through acidified $KMnO _4$ solution, the solution turns colourless.
$
2 KMnO_4+5 SO_2+2 H_2 O \rightarrow K_2 SO_4+2 MnSO_4+2 H_2 SO_4
$
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MCQ 174 Marks
For the reaction, $H _2(g)+ Br _2(g)=2 HBr ( g )$, the reaction rate $= K \left[ H _2\right]\left[ Br _2\right]^{1 / 2}$. Which statement is true about this reaction?
  • A
    The unit of K is $sec ^{-1}$
  • ✓
    Molecularity of the reaction is 2
  • C
    The reaction is of second order
  • D
    Molecularity of the reaction is $\frac{3}{2}$
Answer
Correct option: B.
Molecularity of the reaction is 2
B
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MCQ 184 Marks
If for a certain chemical reaction, $A + B \longrightarrow C + D$, the unit of its rate constant is $M ^{-2}$ $s ^{-1}$, which of the following could be a possible rate law for the reaction?
  • A
    Rate $= k [ A ][ B ]^2$
  • B
    Rate $=k[A][B]$
  • C
    Rate $= k [ A ][ B ]^3$
  • D
    Rate $=k[A]^2[B]^2$
Answer
A. Rate $= k [ A ][ B ]^2$
Explanation:
In general, for nth order reaction, the unit of rate constant
$=(\text { Concentration })^{1-n}(\text { time })^{-1}$
So, $1- n =-2 \Rightarrow n =3$
Thus, the unit of rate constant $M ^{-2} s^{-1}$ indicates that the given reaction must be third order.
Rate $= k [ A ][ B ]^2$ is third order overall.
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MCQ 194 Marks
In an electrolytic cell of $Ag \left| AgNO _3\right| Ag$, when current is passed the concentration of $AgNO _3$ :
  • A
    increases
  • B
    Rises
  • C
    decrease
  • D
    remains same
Answer
D. remains same
Explanation:
In the case of attacked electrodes, the metal dissolves at anode and deposites at the cathode.
$
\begin{array}{l}
{\left[Ag \longrightarrow Ag^{+}+e\right. \text { (anode); }} \\
\left.Ag^{+}+e \longrightarrow Ag \text { (cathode) }\right]
\end{array}
$
Thus the concentration of salt does not change.
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MCQ 204 Marks
If $E_{ Fe ^{2+} / Fe }^{\circ}=-0.441 V$ and $E_{ Fe ^{3+} / Fe ^{2+}}^{\circ}=0.771 V$, the standard e.m.f. of the reaction, Fe $+2 Fe ^{3+} \rightarrow 3 Fe ^{2+}$ will be:
  • A
    0.111 V
  • B
    1.212 V
  • C
    0.330 V
  • D
    1.653 V
Answer
B.1.212 V
Explanation:
$E_{Fe^{2+} / Fe}^{\circ}=-0.441 V $
$E_{Fe^{3+} / Fe}^{\circ}=-0.771 V $
$E_{\text {cell }}^{\circ}=E_{OP_{Fe / Fe^{2+}}}^{\circ}+E_{R P_{Fe^{3+} / Fe^{2+}}}^{\circ} \text { (see redox change) }$
$=+0.441+0.771=1.212 V$
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MCQ 214 Marks
Kohlrausch's law states that at:
  • A
    infinite dilution, each ion makes definite contribution to conductance of an electrolyte whatever be the nature of the other ion of the electroly.
  • B
    infinite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte.
  • C
    infinite dilution each ion makes definite contribution to equivalent conductance of an electrolyte depending on the nature of the other ion of the electrolyte.
  • D
    finite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte.
Answer
B. infinite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte.
Explanation:
Kohlrausch's law states that at infinite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte.
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MCQ 224 Marks
Isotonic solutions are those which have the:
  • A
    same density
  • B
    same normality
  • C
    same molarity
  • D
    same osmotic pressure
Answer
D. same osmotic pressure
Explanation:
A solution of $\frac{M}{2} NaCl$ is isotonic with M glucose. The required condition is $\pi_1=\pi_2$.
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MCQ 234 Marks
Which of the following colligative property can provide molar mass of proteins (or polymers or colloids) with greatest precision?
  • A
    Osmotic pressure
  • B
    Depression in freezing point
  • C
    Relative lowering of vapour pressure
  • D
    Elevation in boiling point
Answer
A.Osmotic pressure
Explanation:
Osmotic pressure is a colligative property which is used for finding the molecular weight of polymer because other colligative properties give very low measurement values which is difficult to measure accurately.
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MCQ 244 Marks
The observed depression in freezing point of a solution will be __________ times if the weight of the solute dissolved is four times higher, and the weight of solvent taken is halved.
  • A
    eight
  • B
    four
  • C
    two
  • D
    one-half
Answer
A. eight
Explanation:
Depression in freezing point will be eight times by the formula,
$
\begin{array}{l}
\Delta T_{f}=\frac{1000 K_f \times W_B}{M_B W_A} \\
\Delta T_{f} \propto \frac{W_B}{W_A}
\end{array}
$
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MCQ 254 Marks
A solution at $20^{\circ} C$ is composed of 1.5 mol of benzene and 3.5 mol of toluene. If the vapour pressure of pure benzene and pure toluene at this temperature are 74.7 torr and 22.3 torr, respectively, then the total vapour pressure of the solution and the benzene mole fraction in equilibrium with it will be, respectively:
  • A
    35.8 torr and 0.280
  • B
    38.0 torr and 0.589
  • C
    30.5 torr and 0.389
  • D
    30.5 torr and 0.480
Answer
B. 38.0 torr and 0.589
Explanation:
Total V.P. of solution $=P_A^{\circ} X_A+P_B^{\circ} X_B$
Given, $P_A^{\circ}=74.7$ torr, $P_B^{\circ}=22.3$ torr
$n _{\text {benzene }}=1.5 mol, n _{\text {toluene }}=3.5 mol$
$n _{\text {solution }}=1.5+3.5=5 mol$
Image
Total V.P. of solution $=(0.3 \times 74.7+0.5 \times 22.3)$ torr
$=(22.4+15.6)$ torr $=38$ torr
Mole fraction of benzene in vapour form $=\frac{22.4}{38}=0.589$
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MCQ 264 Marks
$
Ph-CH=CH-Ph \xrightarrow[CC l_4]{Cl_2} X \xrightarrow{2 NaNH_2} Y \xrightarrow{H_2, Pd-CaCO_3} Z
$
Identify product $(Z)$ of the reaction.
  • A

    Image
  • B
    $Ph - C \equiv C - Ph$
  • C

    Image
  • D

    Image
Answer
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MCQ 274 Marks
The following compounds differ in:
Image
  • A
    configuration
  • B
    chirality
  • C
    conformation
  • D
    structure
Answer
D. structure
Explanation
structure
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MCQ 284 Marks
Pick the CORRECT match for the molecule in column A with the number of $C - C \sigma$, $C - H \sigma$ and $C - C \pi$ bonds in column B.
Column AColumn B
i. $HC \equiv C - CH = CH - CH _3$a. 2,3 and 0
ii. $CH _2= C = CH - CH _3$b. 4,6 and 2
iii. $CH _2- CH - C \equiv N$c. 4,6 and 3
iv. Imaged. 3, 6 and 2
e. 3, 6 and 2
  • A
    i - b, ii - d , iii - c, iv - e
  • B
    i - c, ii - e , iii - a , iv - d
  • C
    i - b , ii - e , iii - d , iv - c
  • D
    i - c, ii - b , iii - e , iv - b
Answer
B. i - c, ii - e , iii - a, iv - d
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MCQ 294 Marks
Which of the following is incorrect statement?
  • A
    $PbF _4$ is covalent in nature
  • B
    $GeX _4( X = F , Cl , Br , I )$ is more stable than $GeX _2$
  • C
    $SnF _4$ is ionic nature
  • D
    $SiCl _4$ is easily hydrolysed
Answer
A. $PbF _4$ is covalent in nature
Explanation:
$PbF _4$ and $SnF _4$ are ionic in nature.
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MCQ 304 Marks
A compound $X$ upon reaction with $H _2 O$ produces a colourless gas $Y$ with rotten fish smell. Gas $Y$ is absorbed in a solution of $CuSO _4$ to give $Cu _3 P _2$ as one of the products. Predict the compound $X$.
  • A
    $Ca \left( PO _4\right)_2$
  • B
    $NH _4 Cl$
  • C
    $As _2 O _3$
  • D
    $Ca _3 P _2$
Answer
D. $Ca _3 P _2$
Explanation:
Image
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MCQ 314 Marks
The reaction between cadmium oxide and carbon monoxide produces cadmium and carbon dioxide. Which element displays the highest oxidation number in this reaction?
  • A
    C
  • B
    O
  • C
    Both Cd and C
  • D
    Cd
Answer
A. C
Explanation:
Image
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MCQ 324 Marks
Identify disproportionation reaction.
i. $CH _4+2 O _2 \longrightarrow CO _2+2 H _2 O$
ii. $CH _4+4 Cl _2 \longrightarrow CCl _4+4 HCl$
iii. $2 F_2+2 OH ^{-} \longrightarrow 2 F+ OF _2+ H _2 O$
iv. $2 NO _2+2 OH ^{-} \longrightarrow NO _2^{-}+ NO _3^{-}+ H _2 O$
  • A
    Only (ii)
  • B
    Only (i)
  • C
    Only (iii)
  • D
    Only (iv)
Answer
D. Only (iv)
Explanation:
$
2 NO_2+2 OH^{-} \longrightarrow NO_2^{-}+NO_3^{-}+H_2 O
$
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MCQ 334 Marks
The equivalent conductance of NaCl at concentration C and at infinite dilution are $\lambda_{ c }$ and $\lambda_{\infty}$ relationship. The correct relationship between $\lambda_{ c }$ and $\lambda_{\infty}$ is given as (where, the constant B is positive)
  • A
    $\lambda_{ c }=\lambda_{\infty}+( B ) C$
  • B
    $\lambda_{ c }=\lambda_{\infty}+( B ) \sqrt{C}$
  • C
    $\lambda_{ c }=\lambda_{\infty}-( B ) \sqrt{C}$
  • D
    $\lambda_{ c }=\lambda_{\infty}-( B ) C$
Answer
C. $\lambda_{ c }=\lambda_{\infty}-( B ) \sqrt{C}$
Explanation:
According to Debye Huckel Onsager equation,
$
\lambda_{c}=\lambda_{\infty}-(B) \sqrt{C}
$
where, $\lambda_{ c }=$ limiting equivalent conductivity at concentration C
$\lambda_{\infty}=$ limiting equivalent conductivity at infinite dilution
$C =$ concentration
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MCQ 344 Marks
The conjugate base of $H _2 PO _4^{-}$is:
  • A
    $P _2 O _5$
  • B
    $HPO _4^{2-}$
  • C
    $PO _4^{3-}$
  • D
    $H _3 PO _4$
Answer
B. $HPO _4^{2-}$
Explanation:
$
H P O_4^{2-}
$
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MCQ 354 Marks
The covalent bond energy of A - Z, B - Z, C - Z and D - Z and 240, 382, 276, 486 respectively. The shortest bond length exist for:
  • A
    $A-Z$
  • B
    $B - Z$
  • C
    $D-Z$
  • D
    $C - Z$
Answer
C. $D - Z$
Explanation:
Shorter the bond, more is bond energy.
The bond $D - Z$ has maximum covalent bond energy which is equal to 486 .
Hence, $D - Z$ bond has the shortest bond length.
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MCQ 364 Marks
For the chemical reaction, $X \rightleftharpoons Y$, the standard reaction Gibbs energy depends on temperature T (in K ) as
$
\Delta_r G^{o}\left(\text { in } kJ mol^{-1}\right)=120-\frac{3}{8} T
$
The major component of the reaction mixture at T is
  • A
    Y if $T =280 K$
  • B
    X if $T =315 K$
  • C
    Y if $T =300 K$
  • D
    X if $T =350 K$
Answer
B. X if $T =315 K$
Explanation:
For a given value of T,
i. If $\triangle_r G ^{\circ}$ becomes $<0$, the forward direction will be spontaneous and then the major and minor components will be Y and X respectively.
ii. If $\triangle_r G^{\circ}$ becomes $>0$, the forward direction will be non-spontaneous and then the major and minor components will be X and Y respectively.
$\triangle_{ r } G ^{\circ}=120-\frac{3}{8} \times 315=1.875$
i.e $\triangle_r G ^{\circ}>0$, major component $= X$
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MCQ 374 Marks
In which of the following reactions the hybridisation state of underlined atom is same in reactants and products?
  • A
    $\underline{ N } H _3+ H ^{+} \rightarrow NH _4^{+}$
  • B
    $\underline{ Al}Cl _3+ Cl ^{-} \rightarrow AlCl _4^{-}$
  • C
    $\underline{N}O _3^{-}+ Zn + OH ^{-} \rightarrow$ $NH_{3} + ZnO^{2-}_{2}$
  • D
    $\underline{ B } F_3+ F ^{-} \rightarrow BF _4^{-}$
Answer
A. $\underline{ N } H _3+ H ^{+} \rightarrow NH _4^{+}$
Explanation:
$
NH_3\left(s p^3\right) \longrightarrow NH_4^{+}\left(s p^3\right)
$
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MCQ 384 Marks
Find the ions which do not have X-O-X type of linkage:
  • A
    $H _2 P _2 O _5^{2-}$
  • B
    $Si _2 O _7^{6-}$
  • C
    $\left( P _3 O _9\right)^{3-}$
  • D
    $S _2 O _3^{2-}$
Answer
D. $S _2 O _3^{2-}$
Explanation:
Image
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MCQ 394 Marks
The geometry of $XeOF _4$ by VSEPR theory is:
  • A
    Square pyramidal
  • B
    Pentagonal planar
  • C
    Trigonal bipyramidal
  • D
    Octahedral
Answer
A. Square pyramidal
Explanation:
Image
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MCQ 404 Marks
Lanthanide contraction means:
  • A
    Contraction of atoms of elements after lanthanum due to poor shielding of the $f$-subshell electron.
  • B
    Contraction of atoms of elements before lanthanum due to poor shielding of the f subshell electron.
  • C
    Contraction of an atom of lanthanum element due to poor shielding d-subshell electron.
  • D
    Contraction of an atom of lanthanum element due to high shielding of the d-subshell electron.
Answer
A. Contraction of atoms of elements after lanthanum due to poor shielding of the $f$-subshell electron.
Explanation:
Contraction of atoms of elements after lanthanum due to poor shielding of the f-subshell electron.
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MCQ 414 Marks
The wave number of the first emission line in the Balmer series of H-Spectrum is: ( R $=$ Rydberg constant)
  • A
    $\frac{9}{400} R$
  • B
    $\frac{7}{6} R$
  • C
    $\frac{3}{4} R$
  • D
    $\frac{5}{36} R$
Answer
D.$\frac{5}{36} R$
Explanation:
$
\begin{array}{l}
\bar{v}=R Z^2\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \\
=R\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{5 R}{36}
\end{array}
$
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MCQ 434 Marks
An ion which has 18 electrons in the outermost shell is:
  • A
    $K ^{+}$
  • B
    $Cs ^{+}$
  • C
    $Th ^{4+}$
  • D
    $Cu ^{+}$
Answer
D. $Cu ^{+}$
Explanation:
$Cu ^{+}$has $3 d^{10}$ configuration.
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MCQ 444 Marks
The normality of $10 \%$ (mass/volume) acetic acid is:
  • A
    0.83 N
  • B
    10 N
  • ✓
    1.7 N
  • D
    1 N
Answer
Correct option: C.
1.7 N
C
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MCQ 454 Marks
How many moles of electrons weigh one kilogram?
  • A
    $\frac{1}{2.108} \times 10^{31}$
  • B
    $6.023 \times 10^{23}$
  • C
    $\frac{6.023}{9.108} \times 10^{54}$
  • D
    $\frac{1}{9.108 \times 6.023} \times 10^8$
Answer
D.$\frac{1}{9.108 \times 6.023} \times 10^8$
Explanation:
$9.108 \times 10^{-31} kg=1$ electron
$\therefore 1 kg=\frac{1}{9.108 \times 10^{-31}}$ electron
$=\frac{1}{9.108 \times 10^{-31}} \times \frac{1}{6.023 \times 10^{23}}$ mole electron
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