MCQ 11 Mark
The number of alkynes possible with molecular formula $C _5 H _8$ is:
- A5
- B4
- C2
- ✓3
Answer
View full question & answer→Correct option: D.
3
(d) 3
Questions
M.C.Q (1 Marks)
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12 questions · 10 auto-graded MCQ + 2 self-marked written.
MCQ 21 Mark
The equilibrium constant for a reaction is 10 . What will be the value of $\Delta G^0$ ? $R =8.314 JK ^{-1}, T=300 K, T =$ 300 K ?
- ✓$-5.744 kJ mol ^{-1}$
- B$-5.456 kJ mol ^{-1}$
- C$-6.132 kJ mol ^{-1}$
- D$-5.978 kJ mol ^{-1}$
Answer
View full question & answer→Correct option: A.
$-5.744 kJ mol ^{-1}$
(a) $-5.744 kJ mol ^{-1}$
Explanation: $\Delta_{ r } G ^0=- RT \ln K _{ C }$.
$
\Delta G^0=-2.303 \times 8.314 \times 300 \times \log 10=-5744.14 Jmol^{-1}=-5.74414 kJmol^{-1}
$
Explanation: $\Delta_{ r } G ^0=- RT \ln K _{ C }$.
$
\Delta G^0=-2.303 \times 8.314 \times 300 \times \log 10=-5744.14 Jmol^{-1}=-5.74414 kJmol^{-1}
$
MCQ 31 Mark
Choose one of the following in the order of increasing radii:
- A$I ^{-}< I ^{+}< I$
- B$I < I ^{+}< I ^{-}$
- C$I ^{+}< I ^{-}< I$
- ✓$I ^{-}> I > I ^{+}$
Answer
View full question & answer→Correct option: D.
$I ^{-}> I > I ^{+}$
(d) $I ^{-}>$I $>$I $^{+}$
Explanation: The size of the anion is greater than the size of the parent atom and the size of the cation. As the -ve charge of anion increases ionic radii increases due to a decrease in the effective nuclear charge and vice versa.
Explanation: The size of the anion is greater than the size of the parent atom and the size of the cation. As the -ve charge of anion increases ionic radii increases due to a decrease in the effective nuclear charge and vice versa.
MCQ 41 Mark
The synthesis of $3-$octyne is achieved by subsequent stepwise reactions of sodium amide with an alkyne, and a bromoalkane. The bromoalkane and the other alkyne respectively are:
- A$BrCH_2CH_2CH_3$ and $CH_3 CH_2 CH_2C ≡ CH$
- B$BrCH_2CH_2CH_2CH_2CH_3$ and $CH_3C ≡ CH$
- C$BrCH_2CH_2CH_2CH_3$ and $CH_3 CH_2C ≡ CH$
- D$BrCH_2CH_2CH_2CH_2CH_3$ and $CH_3 CH_2C ≡ CH$
Answer
View full question & answer→$ 3 -$ Octyne can be synthesized as per the following stepwise conversion reactions:
Step $1.$ Formation of a Sodium alkynide, by the reaction of $1 -$butyne with sodium amide
$CH_3 CH_2 C \equiv CH+NaNH_2 \rightarrow CH_3 CH_2 C \equiv CNa^{+}+NH_3$
$1-$butyne sodium amide sodium butynide
Step $2.$ Reaction of the alkynide $($ sodium butynide$)$ with $1 -$ bromobutane
$CH_3 CH_2 C \equiv CNa^{+}+CH_3 CH_2 CH_2 CH_2 Br \rightarrow CH_3 CH_2 C \equiv CCH_2 CH_2 CH_2 CH_3+NaBr$
Sodium Butynide $1-$bromobutane $3 -$ Octyne
Step $1.$ Formation of a Sodium alkynide, by the reaction of $1 -$butyne with sodium amide
$CH_3 CH_2 C \equiv CH+NaNH_2 \rightarrow CH_3 CH_2 C \equiv CNa^{+}+NH_3$
$1-$butyne sodium amide sodium butynide
Step $2.$ Reaction of the alkynide $($ sodium butynide$)$ with $1 -$ bromobutane
$CH_3 CH_2 C \equiv CNa^{+}+CH_3 CH_2 CH_2 CH_2 Br \rightarrow CH_3 CH_2 C \equiv CCH_2 CH_2 CH_2 CH_3+NaBr$
Sodium Butynide $1-$bromobutane $3 -$ Octyne
MCQ 51 Mark
The order of decreasing priority for some functional groups in the naming of an organic compound is:
- A$- COOR (R = alkyl \ group \ right ), - COCl, CONH_2 , - CN, - HC = O, > C = O, - OH, H_2$
- B$- CN, - HC = O, > C = O, - OH, - NH_2, COOR (R = alkyl \ group ), - COCl, - CONH_2$
- C$- CONH_2, - CN, - HC = O, > C = O, - OH, NH_2, - COOR (R = alkyl \ group), - COCl$
- D$- COCl, - CONH_2, - CN, - HC = O, > C = O, - OH, - NH_2, - COOR (R = alkyl \ group)$
Answer
View full question & answer→$COOR ( R =$ alkyl group $) > - COCl ,> - CONH _2 > - CN > - HC = O >> C = O > - OH ,- NH _2$ is the decreasing order of priority
ester $>$ acyl halide $>$ acyl amide $>$ nitrile $>$ aldehyde $>$ ketones $>$ alcohol $>$ amine
ester $>$ acyl halide $>$ acyl amide $>$ nitrile $>$ aldehyde $>$ ketones $>$ alcohol $>$ amine
MCQ 61 Mark
Choose the correct explanation regarding half-reaction such as $Cr _2 O _7^{2-} \rightarrow Cr ^{3+}$ from the following.
- A$Cr _2 O _7^{2-}$ is a good reducing agent
- BIt is oxidation half-reaction
- ✓Chromium being reduced
- DChromium being oxidised
Answer
View full question & answer→Correct option: C.
Chromium being reduced
(c) Chromium being reduced
MCQ 71 Mark
Orbital angular momentum depends on ____________ .
- Am and s
- Bn and m
- ✓l
- Dn and l
Answer
View full question & answer→Correct option: C.
l
(c) l
Explanation: Azimuthal quantum number. 'I' also known as orbital angular momentum or subsidiary quantum number. It defines the three-dimensional shape of the orbital. For a given value of $n, 1$ can have $n$ values ranging from 0 to $n-1$, that is, for a given value of $n$, the possible values of 1 are: $1=0,1,2, \ldots \ldots . .(n-1)$.
Explanation: Azimuthal quantum number. 'I' also known as orbital angular momentum or subsidiary quantum number. It defines the three-dimensional shape of the orbital. For a given value of $n, 1$ can have $n$ values ranging from 0 to $n-1$, that is, for a given value of $n$, the possible values of 1 are: $1=0,1,2, \ldots \ldots . .(n-1)$.
MCQ 81 Mark
For a cyclic process, the change in internal energy of the system is _______________ .
- Aalways + ve
- Balways - ve
- Cnone of these
- ✓zero
Answer
View full question & answer→Correct option: D.
zero
(d) zero
Explanation: For a cyclic process, the initial state is the same as that of the final state. As internal energy is a state function, its value at the initial point is the same as that at the final point and enthalpy change is zero.
Explanation: For a cyclic process, the initial state is the same as that of the final state. As internal energy is a state function, its value at the initial point is the same as that at the final point and enthalpy change is zero.
MCQ 91 Mark
The product of the uncertainties of the exact position and exact momentum (or velocity) of an electron. The product of their uncertainties is always
- Aequal to or greater than $h / 3 \pi$
- ✓equal to or greater than $h / 4 \pi$
- Cequal to or greater than $h / 2 \pi$
- Dequal to or greater than $h / \pi$
Answer
View full question & answer→Correct option: B.
equal to or greater than $h / 4 \pi$
(b) equal to or greater than $h / 4 \pi$.
Explanation: The uncertainty principle says that we cannot measure the position ( x ) and the momentum ( p ) of a particle with absolute precision. The more accurately we know one of these values, the less accurately we know the other. Multiplying together the errors in the measurements of these values has to give a number greater than or equal to half of a constant called "h-bar". This is equal to Planck's constant (usually written as h) divided by $2 \pi$.
$
\Delta x \cdot \Delta p \geq \frac{h}{4 \pi}
$
Explanation: The uncertainty principle says that we cannot measure the position ( x ) and the momentum ( p ) of a particle with absolute precision. The more accurately we know one of these values, the less accurately we know the other. Multiplying together the errors in the measurements of these values has to give a number greater than or equal to half of a constant called "h-bar". This is equal to Planck's constant (usually written as h) divided by $2 \pi$.
$
\Delta x \cdot \Delta p \geq \frac{h}{4 \pi}
$
MCQ 101 Mark
No work is done on the system, but q amount of heat is taken out from the system and given to the surrounding. The change in internal energy of a system is
- ✓$\Delta U=- q$
- B$\Delta U= W _{ ad }$
- C$\Delta U=+ q$
- D$\Delta U= q - W$
Answer
View full question & answer→Correct option: A.
$\Delta U=- q$
(a) $\Delta U=- q$
Explanation: $\Delta U=- q$
Explanation: $\Delta U=- q$
MCQ 111 Mark
Cathode rays are deflected by:
- ✓electric and magnetic fields
- Bmagnetic field only
- Cthin metal foils only
- Delectric field only
Answer
View full question & answer→Correct option: A.
electric and magnetic fields
(a) electric and magnetic fields
Explanation: When an electric field is applied to a stream of cathode rays, they get deflected towards the positive plate. On the application of a magnetic field perpendicular to the path of the cathode rays, they get deflected in the direction expected of negative particles.
Explanation: When an electric field is applied to a stream of cathode rays, they get deflected towards the positive plate. On the application of a magnetic field perpendicular to the path of the cathode rays, they get deflected in the direction expected of negative particles.
MCQ 121 Mark
30 mg is the same mass as:
- A0.0003 kg
- ✓0.03 g
- C300 decigrams
- D0.3 grams
Answer
View full question & answer→Correct option: B.
0.03 g
(b) 0.03 g
Explanation: since $1000 mg =1 g$
$
\begin{aligned}
& \therefore 30 mg \\
& =[(1 / 1000) \times 30] g \\
& =0.03 g
\end{aligned}
$
Explanation: since $1000 mg =1 g$
$
\begin{aligned}
& \therefore 30 mg \\
& =[(1 / 1000) \times 30] g \\
& =0.03 g
\end{aligned}
$