ChemistryQ1–29 of 29 questions
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1
ChemistryMediumMCQAIIMS · 1983
Which one of the following explains light both as a stream of particles and as wave motion?
A
Diffraction
B
$\lambda = h/p$
C
Interference
D
Photoelectric effect
Solution
(B) The de-Broglie equation,$\lambda = \frac{h}{p} = \frac{h}{mv}$,relates the wavelength (wave property) to the momentum (particle property) of a moving object,including photons. This equation provides the mathematical basis for the wave-particle duality of light and matter.
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2
ChemistryMediumMCQAIIMS · 1983
In which one of the following pairs of experimental observations and phenomena does the experimental observation correctly account for the phenomenon?
A
$X$-ray spectra: Charge on the nucleus
B
$\alpha$-particle scattering: Quantized electron orbit
C
Emission spectra: The quantization of energy
D
The photoelectric effect: The nuclear atom
Solution
(C) The correct pair is $(C)$.
Emission spectra consist of discrete lines at specific wavelengths,which directly provide evidence for the quantization of energy levels within an atom.
$X$-ray spectra relate to the atomic number $(Z)$,$\alpha$-particle scattering demonstrates the existence of a small,dense,positively charged nucleus,and the photoelectric effect demonstrates the particle nature of light (photons).
Emission spectra consist of discrete lines at specific wavelengths,which directly provide evidence for the quantization of energy levels within an atom.
$X$-ray spectra relate to the atomic number $(Z)$,$\alpha$-particle scattering demonstrates the existence of a small,dense,positively charged nucleus,and the photoelectric effect demonstrates the particle nature of light (photons).
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3
ChemistryEasyMCQAIIMS · 1983
For the energy levels in an atom,which one of the following statements is correct?
A
There are seven principal electron energy levels
B
The second principal energy level can have four sub-energy levels and contains a maximum of eight electrons
C
The $M$ energy level can have a maximum of $32$ electrons
D
The $4s$ sub-energy level is at a higher energy than the $3d$ sub-energy level
Solution
(B) The correct statement is $(b)$.
The second principal energy level $(n=2)$ consists of $2s$ and $2p$ subshells.
The $2s$ subshell has $1$ orbital and the $2p$ subshell has $3$ orbitals,totaling $4$ orbitals.
Each orbital can hold $2$ electrons,so the maximum capacity is $4 \times 2 = 8$ electrons.
Option $(a)$ is incorrect as there are theoretically infinite energy levels.
Option $(c)$ is incorrect because the $M$ shell $(n=3)$ can hold $2n^2 = 2(3)^2 = 18$ electrons.
Option $(d)$ is incorrect because according to the $(n+l)$ rule,$4s$ $(4+0=4)$ has lower energy than $3d$ $(3+2=5)$.
The second principal energy level $(n=2)$ consists of $2s$ and $2p$ subshells.
The $2s$ subshell has $1$ orbital and the $2p$ subshell has $3$ orbitals,totaling $4$ orbitals.
Each orbital can hold $2$ electrons,so the maximum capacity is $4 \times 2 = 8$ electrons.
Option $(a)$ is incorrect as there are theoretically infinite energy levels.
Option $(c)$ is incorrect because the $M$ shell $(n=3)$ can hold $2n^2 = 2(3)^2 = 18$ electrons.
Option $(d)$ is incorrect because according to the $(n+l)$ rule,$4s$ $(4+0=4)$ has lower energy than $3d$ $(3+2=5)$.
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4
ChemistryMediumMCQAIIMS · 1983
Which of the following statements is not correct?
A
Hybridization is the mixing of atomic orbitals prior to their combining into molecular orbitals.
B
$sp^2$ hybrid orbitals are formed from two $p$ atomic orbitals and one $s$ atomic orbital.
C
$d^2sp^3$ hybrid orbitals are directed towards the corners of a regular octahedron.
D
$dsp^3$ hybrid orbitals are all at $90^o$ to one another.
Solution
(D) is the correct answer because $dsp^3$ hybridization results in a trigonal bipyramidal geometry.
In this geometry,the bond angles are $120^o$ (equatorial) and $90^o$ (axial),not all at $90^o$.
In this geometry,the bond angles are $120^o$ (equatorial) and $90^o$ (axial),not all at $90^o$.
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5
ChemistryMediumMCQAIIMS · 1983
The bond order is maximum in
A
$O_2$
B
$O_2^{-}$
C
$O_2^{+}$
D
$O_2^{2-}$
Solution
(C) The bond order $(B.O.)$ is calculated using the formula: $B.O. = \frac{1}{2} (N_b - N_a)$.
For $O_2$ ($16$ electrons): $B.O. = \frac{1}{2} (10 - 6) = 2.0$.
For $O_2^{-}$ ($17$ electrons): $B.O. = \frac{1}{2} (10 - 7) = 1.5$.
For $O_2^{+}$ ($15$ electrons): $B.O. = \frac{1}{2} (10 - 5) = 2.5$.
For $O_2^{2-}$ ($18$ electrons): $B.O. = \frac{1}{2} (10 - 8) = 1.0$.
Comparing these values,the bond order is maximum in $O_2^{+}$.
For $O_2$ ($16$ electrons): $B.O. = \frac{1}{2} (10 - 6) = 2.0$.
For $O_2^{-}$ ($17$ electrons): $B.O. = \frac{1}{2} (10 - 7) = 1.5$.
For $O_2^{+}$ ($15$ electrons): $B.O. = \frac{1}{2} (10 - 5) = 2.5$.
For $O_2^{2-}$ ($18$ electrons): $B.O. = \frac{1}{2} (10 - 8) = 1.0$.
Comparing these values,the bond order is maximum in $O_2^{+}$.
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6
ChemistryMediumMCQAIIMS · 1983
Which of the following species has the maximum number of unpaired electrons?
A
$O_2$
B
$O_2^+$
C
$O_2^-$
D
$O_2^{2-}$
Solution
(A) According to Molecular Orbital Theory $(MOT)$,the electronic configuration of the species are as follows:
$O_2$ ($16$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$. It has $2$ unpaired electrons.
$O_2^+$ ($15$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$. It has $1$ unpaired electron.
$O_2^-$ ($17$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^1$. It has $1$ unpaired electron.
$O_2^{2-}$ ($18$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$. It has $0$ unpaired electrons.
Therefore,$O_2$ has the maximum number of unpaired electrons.
$O_2$ ($16$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$. It has $2$ unpaired electrons.
$O_2^+$ ($15$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1$. It has $1$ unpaired electron.
$O_2^-$ ($17$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^1$. It has $1$ unpaired electron.
$O_2^{2-}$ ($18$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$. It has $0$ unpaired electrons.
Therefore,$O_2$ has the maximum number of unpaired electrons.
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7
ChemistryMediumMCQAIIMS · 1983
For which of the following reactions does the equilibrium constant depend on the units of concentration?
A
$NO_{(g)} \rightleftharpoons \frac{1}{2}N_{2(g)} + \frac{1}{2}O_{2(g)}$
B
$Zn_{(s)} + Cu^{2+}_{(aq)} \rightleftharpoons Cu_{(s)} + Zn^{2+}_{(aq)}$
C
$C_2H_5OH_{(l)} + CH_3COOH_{(l)} \rightleftharpoons CH_3COOC_2H_{5(l)} + H_2O_{(l)}$ (Reaction carried in an inert solvent)
D
$COCl_{2(g)} \rightleftharpoons CO_{(g)} + Cl_{2(g)}$
Solution
(D) The equilibrium constant $K_c$ depends on the units of concentration if the change in the number of moles of gaseous species,$\Delta n_g$,is non-zero.
For option $A$: $\Delta n_g = (0.5 + 0.5) - 1 = 0$.
For option $B$: $\Delta n_g = 0$ (as only aqueous species are involved in the expression).
For option $C$: $\Delta n_g = 0$ (as liquid species are not included in the $K_c$ expression).
For option $D$: $\Delta n_g = (1 + 1) - 1 = 1$.
Since $\Delta n_g \neq 0$ for reaction $D$,the equilibrium constant $K_c$ depends on the units of concentration.
For option $A$: $\Delta n_g = (0.5 + 0.5) - 1 = 0$.
For option $B$: $\Delta n_g = 0$ (as only aqueous species are involved in the expression).
For option $C$: $\Delta n_g = 0$ (as liquid species are not included in the $K_c$ expression).
For option $D$: $\Delta n_g = (1 + 1) - 1 = 1$.
Since $\Delta n_g \neq 0$ for reaction $D$,the equilibrium constant $K_c$ depends on the units of concentration.
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8
ChemistryMediumMCQAIIMS · 1983
At constant $T$ and $P$,which one of the following statements is correct for the reaction,$CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)}$?
A
$\Delta H$ is independent of the physical state of the reactants of that compound
B
$\Delta H > \Delta E$
C
$\Delta H < \Delta E$
D
$\Delta H = \Delta E$
Solution
(C) The relationship between enthalpy change $(\Delta H)$ and internal energy change $(\Delta E)$ is given by the equation: $\Delta H = \Delta E + \Delta n_g RT$.
For the reaction $CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)}$,the change in the number of moles of gaseous species is calculated as: $\Delta n_g = n_{p(g)} - n_{r(g)} = 1 - (1 + 0.5) = 1 - 1.5 = -0.5$.
Since $\Delta n_g = -0.5$ (which is negative),the term $\Delta n_g RT$ is negative.
Therefore,$\Delta H = \Delta E - 0.5 RT$,which implies that $\Delta H < \Delta E$.
For the reaction $CO_{(g)} + \frac{1}{2} O_{2(g)} \to CO_{2(g)}$,the change in the number of moles of gaseous species is calculated as: $\Delta n_g = n_{p(g)} - n_{r(g)} = 1 - (1 + 0.5) = 1 - 1.5 = -0.5$.
Since $\Delta n_g = -0.5$ (which is negative),the term $\Delta n_g RT$ is negative.
Therefore,$\Delta H = \Delta E - 0.5 RT$,which implies that $\Delta H < \Delta E$.
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9
ChemistryEasyMCQAIIMS · 1983
The oxygen molecule $(O_2)$ exhibits which of the following magnetic properties?
A
Paramagnetism
B
Diamagnetism
C
Ferromagnetism
D
Ferrimagnetism
Solution
(A) According to Molecular Orbital Theory $(MOT)$, the electronic configuration of $O_2$ is: $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$.
Since the oxygen molecule contains two unpaired electrons in the antibonding molecular orbitals ($\pi^* 2p_x$ and $\pi^* 2p_y$), it exhibits paramagnetism.
Since the oxygen molecule contains two unpaired electrons in the antibonding molecular orbitals ($\pi^* 2p_x$ and $\pi^* 2p_y$), it exhibits paramagnetism.
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10
ChemistryMediumMCQAIIMS · 1983
The normality of $1 \ M$ solution of $H_3PO_4$ will be.....$N$
A
$1$
B
$0.5$
C
$2$
D
$3$
Solution
(D) The relationship between normality and molarity is given by the formula: $\text{Normality} = \text{Molarity} \times \text{Basicity}$.
For phosphoric acid $(H_3PO_4)$,the basicity is $3$ because it can donate $3 \ H^+$ ions per molecule.
Given that the molarity is $1 \ M$,the normality is calculated as: $1 \ M \times 3 = 3 \ N$.
For phosphoric acid $(H_3PO_4)$,the basicity is $3$ because it can donate $3 \ H^+$ ions per molecule.
Given that the molarity is $1 \ M$,the normality is calculated as: $1 \ M \times 3 = 3 \ N$.
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11
ChemistryMediumMCQAIIMS · 1983
$CH_3-CH=CH_2 + HBr \xrightarrow{\ \ } .........$,the product formed is
A
$CH_3-CH_2-CH_2-Br$
B
$CH_3-CH(Br)-CH_3$
C
$BrCH_2-CH=CH_2$
D
$CH_2=C=CH_2$
Solution
(B) The reaction of propene $(CH_3-CH=CH_2)$ with hydrogen bromide $(HBr)$ follows Markovnikov's rule.
According to Markovnikov's rule,the negative part of the addendum $(Br^-)$ attaches to the carbon atom of the double bond that has the lesser number of hydrogen atoms.
Therefore,the reaction is: $CH_3-CH=CH_2 + HBr \rightarrow CH_3-CH(Br)-CH_3$.
The product formed is $2-$bromopropane.
According to Markovnikov's rule,the negative part of the addendum $(Br^-)$ attaches to the carbon atom of the double bond that has the lesser number of hydrogen atoms.
Therefore,the reaction is: $CH_3-CH=CH_2 + HBr \rightarrow CH_3-CH(Br)-CH_3$.
The product formed is $2-$bromopropane.
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12
ChemistryMCQAIIMS · 1983
In order to determine whether variations of a character in a population were genetically controlled,the most appropriate procedure will be to
A
Count chromosomes and find out the variations in number in the population
B
Examine $DNA$ and see if the population shows any variation
C
Measure the variations and see if they are continuous or discontinuous
D
Cross individuals of both extremes and see if the offsprings and parents show the range of variations
Solution
(D) The most appropriate procedure to determine if variations are genetically controlled is to perform breeding experiments. By crossing individuals representing the extreme phenotypes,one can observe if the traits are inherited by the offspring. If the variations are genetic,they will follow Mendelian patterns of inheritance and appear in the subsequent generations,confirming that the character is controlled by genes.
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13
ChemistryMCQAIIMS · 1983
What is the decreasing order of reactivity of $HX$ in the given reaction: $ROH + HX \rightarrow RX + H_2O$?
A
$HI > HBr > HCl > HF$
B
$HBr > HCl > HI > HF$
C
$HCl > HBr > HI > HF$
D
$HF > HBr > HCl > HI$
Solution
(A) The reaction $ROH + HX \rightarrow RX + H_2O$ involves the cleavage of the $C-O$ bond in the alcohol.
This reaction is facilitated by the nucleophilic attack of the halide ion $(X^-)$ on the protonated alcohol.
The reactivity of $HX$ depends on the strength of the $H-X$ bond.
As the size of the halogen atom increases from $F$ to $I$,the $H-X$ bond strength decreases ($HI < HBr < HCl < HF$ in terms of bond energy).
Consequently,the ease of breaking the $H-X$ bond increases in the order $HF < HCl < HBr < HI$.
Therefore,the decreasing order of reactivity of $HX$ is $HI > HBr > HCl > HF$.
This reaction is facilitated by the nucleophilic attack of the halide ion $(X^-)$ on the protonated alcohol.
The reactivity of $HX$ depends on the strength of the $H-X$ bond.
As the size of the halogen atom increases from $F$ to $I$,the $H-X$ bond strength decreases ($HI < HBr < HCl < HF$ in terms of bond energy).
Consequently,the ease of breaking the $H-X$ bond increases in the order $HF < HCl < HBr < HI$.
Therefore,the decreasing order of reactivity of $HX$ is $HI > HBr > HCl > HF$.
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14
ChemistryDifficultMCQAIIMS · 1983
Identify $Z$ in the following series: $C_2H_5I$ $\xrightarrow{Alco. KOH} X$ $\xrightarrow{Br_2} Y$ $\xrightarrow{KCN} Z$
A
$CH_3-CH_2-CN$
B
$NC-CH_2-CH_2-CN$
C
$Br-CH_2-CH_2-CN$
D
$Br-CH=CH-CN$
Solution
(B) The reaction sequence is as follows:
$1.$ $C_2H_5I \xrightarrow{Alco. KOH} CH_2=CH_2$ $(X)$
$2.$ $CH_2=CH_2 \xrightarrow{Br_2} Br-CH_2-CH_2-Br$ $(Y)$
$3.$ $Br-CH_2-CH_2-Br \xrightarrow{KCN} NC-CH_2-CH_2-CN$ $(Z)$
Therefore,$Z$ is $NC-CH_2-CH_2-CN$.
$1.$ $C_2H_5I \xrightarrow{Alco. KOH} CH_2=CH_2$ $(X)$
$2.$ $CH_2=CH_2 \xrightarrow{Br_2} Br-CH_2-CH_2-Br$ $(Y)$
$3.$ $Br-CH_2-CH_2-Br \xrightarrow{KCN} NC-CH_2-CH_2-CN$ $(Z)$
Therefore,$Z$ is $NC-CH_2-CH_2-CN$.
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15
ChemistryMediumMCQAIIMS · 1983
Which one of the following combines with $Fe(II)$ ions to form a brown complex?
A
$N_2O$
B
$NO$
C
$N_2O_3$
D
$N_2O_5$
Solution
(B) The brown ring test is used to detect the presence of nitrate ions $(NO_3^-)$ in a solution.
In this test,$Fe(II)$ ions react with $NO$ (nitric oxide) to form a brown-colored coordination complex.
The reaction is: $[Fe(H_2O)_6]^{2+} + NO \to [Fe(H_2O)_5(NO)]^{2+} + H_2O$.
This complex,$[Fe(H_2O)_5(NO)]^{2+}$,is responsible for the brown ring observed at the junction of the two layers.
In this test,$Fe(II)$ ions react with $NO$ (nitric oxide) to form a brown-colored coordination complex.
The reaction is: $[Fe(H_2O)_6]^{2+} + NO \to [Fe(H_2O)_5(NO)]^{2+} + H_2O$.
This complex,$[Fe(H_2O)_5(NO)]^{2+}$,is responsible for the brown ring observed at the junction of the two layers.
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16
ChemistryMediumMCQAIIMS · 1983
The highest catenation ability is shown by
A
Oxygen
B
Sulphur
C
Selenium
D
Tellurium
Solution
(B) The tendency towards catenation depends on the strength of the element-element bond.
As the atomic size increases down the group from $S$ to $Te$,the bond dissociation energy decreases.
Although $O$ is smaller than $S$,the $O-O$ bond energy is lower than the $S-S$ bond energy due to significant inter-electronic repulsions between the lone pairs of the small oxygen atoms.
Therefore,$S$ exhibits the highest tendency towards catenation among the group $16$ elements.
As the atomic size increases down the group from $S$ to $Te$,the bond dissociation energy decreases.
Although $O$ is smaller than $S$,the $O-O$ bond energy is lower than the $S-S$ bond energy due to significant inter-electronic repulsions between the lone pairs of the small oxygen atoms.
Therefore,$S$ exhibits the highest tendency towards catenation among the group $16$ elements.
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17
ChemistryMediumMCQAIIMS · 1983
Which one is the anhydride of $HClO_4$?
A
$Cl_2O$
B
$ClO_2$
C
$Cl_2O_6$
D
$Cl_2O_7$
Solution
(D) An acid anhydride is an oxide that forms an acid when reacted with water.
To find the anhydride of $HClO_4$,we remove water $(H_2O)$ from two molecules of the acid:
$2HClO_4 \to H_2O + Cl_2O_7$.
Thus,$Cl_2O_7$ is the anhydride of perchloric acid $(HClO_4)$.
To find the anhydride of $HClO_4$,we remove water $(H_2O)$ from two molecules of the acid:
$2HClO_4 \to H_2O + Cl_2O_7$.
Thus,$Cl_2O_7$ is the anhydride of perchloric acid $(HClO_4)$.
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18
ChemistryEasyMCQAIIMS · 1983
Which one of the following noble gases is the least polarizable?
A
$Xe$
B
$Ar$
C
$Ne$
D
$He$
Solution
(D) Polarizability depends on the size of the electron cloud. Larger atoms have more loosely held electrons,making them more polarizable.
Since $He$ has the smallest atomic size among the given noble gases,its electrons are most tightly held by the nucleus.
Therefore,$He$ is the least polarizable.
Since $He$ has the smallest atomic size among the given noble gases,its electrons are most tightly held by the nucleus.
Therefore,$He$ is the least polarizable.
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19
ChemistryMediumMCQAIIMS · 1983
In the following reaction: $CH_3-CH_2-CH(OH)-CH_3 \xrightarrow[475 \ K]{H_2SO_4}$
A
$CH_3-CH=CH-CH_3$ predominates
B
$CH_2=CH-CH_2-CH_3$ predominates
C
Both are formed in equal amounts
D
The amount of production depends on the nature of catalyst
Solution
(A) According to $Saytzeff's$ rule,in the dehydration of alcohols,the more stable (more substituted) alkene is the major product.
$CH_3-CH=CH-CH_3$ $(but-2-ene)$ is more substituted and stable than $CH_2=CH-CH_2-CH_3$ $(but-1-ene)$.
Therefore,$CH_3-CH=CH-CH_3$ is the major product.
$CH_3-CH=CH-CH_3$ $(but-2-ene)$ is more substituted and stable than $CH_2=CH-CH_2-CH_3$ $(but-1-ene)$.
Therefore,$CH_3-CH=CH-CH_3$ is the major product.
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20
ChemistryMediumMCQAIIMS · 1983
Identify $Z$ in the following series:
$CH_2=CH_2$ $\xrightarrow{HBr} X$ $\xrightarrow{Hydrolysis} Y$ $\xrightarrow[I_2 \ \text{excess}]{Na_2CO_3} Z$
$CH_2=CH_2$ $\xrightarrow{HBr} X$ $\xrightarrow{Hydrolysis} Y$ $\xrightarrow[I_2 \ \text{excess}]{Na_2CO_3} Z$
A
$C_2H_5I$
B
$C_2H_5OH$
C
$CHI_3$
D
$CH_3CHO$
Solution
(C) $CH_2=CH_2 \xrightarrow{HBr} CH_3-CH_2Br (X)$
$CH_3-CH_2Br \xrightarrow{Hydrolysis} CH_3-CH_2OH (Y)$
$CH_3-CH_2OH \xrightarrow[I_2 \ \text{excess}]{Na_2CO_3} CHI_3 (Z) + HCOONa + NaI + H_2O$
This is the iodoform test,where ethanol reacts with iodine in the presence of base to form a yellow precipitate of iodoform $(CHI_3)$.
$CH_3-CH_2Br \xrightarrow{Hydrolysis} CH_3-CH_2OH (Y)$
$CH_3-CH_2OH \xrightarrow[I_2 \ \text{excess}]{Na_2CO_3} CHI_3 (Z) + HCOONa + NaI + H_2O$
This is the iodoform test,where ethanol reacts with iodine in the presence of base to form a yellow precipitate of iodoform $(CHI_3)$.
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21
ChemistryMediumMCQAIIMS · 1983
Which one of the following nuclear transformations is of the $(n, p)$ type?
A
$_{33}As^{75} + _{2}He^{4} \rightarrow _{35}Br^{78} + _{0}n^{1}$
B
$_{83}Bi^{209} + _{1}H^{2} \rightarrow _{84}Po^{210} + _{0}n^{1}$
C
$_{3}Li^{7} + _{1}H^{1} \rightarrow _{4}Be^{7} + _{0}n^{1}$
D
$_{21}Sc^{45} + _{0}n^{1} \rightarrow _{20}Ca^{45} + _{1}H^{1}$
Solution
(D) In an $(n, p)$ nuclear reaction,a neutron is captured by the target nucleus and a proton is emitted.
This is represented as $Target(n, p)Product$.
In option $D$,the reaction is $_{21}Sc^{45} + _{0}n^{1} \rightarrow _{20}Ca^{45} + _{1}H^{1}$.
Here,a neutron $(_{0}n^{1})$ is absorbed and a proton ($_{1}H^{1}$ or $p$) is released,which corresponds to the $(n, p)$ type transformation.
This is represented as $Target(n, p)Product$.
In option $D$,the reaction is $_{21}Sc^{45} + _{0}n^{1} \rightarrow _{20}Ca^{45} + _{1}H^{1}$.
Here,a neutron $(_{0}n^{1})$ is absorbed and a proton ($_{1}H^{1}$ or $p$) is released,which corresponds to the $(n, p)$ type transformation.
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22
ChemistryDifficultMCQAIIMS · 1983
What is $X$ in the following nuclear reaction?
$_{7}N^{14} + _{1}H^{1} \xrightarrow{} _{8}O^{15} + X$
$_{7}N^{14} + _{1}H^{1} \xrightarrow{} _{8}O^{15} + X$
A
$_{+1}e^{0}$
B
$_{0}n^{1}$
C
$\gamma$
D
$_{-1}e^{0}$
Solution
(C) In a nuclear reaction,the sum of atomic numbers and the sum of mass numbers must be conserved on both sides.
For the given reaction: $_{7}N^{14} + _{1}H^{1} \to _{8}O^{15} + X$
Sum of atomic numbers on the left side: $7 + 1 = 8$.
Sum of atomic numbers on the right side: $8 + Z = 8$,so $Z = 0$.
Sum of mass numbers on the left side: $14 + 1 = 15$.
Sum of mass numbers on the right side: $15 + A = 15$,so $A = 0$.
$A$ particle with atomic number $0$ and mass number $0$ is a gamma photon,denoted as $\gamma$.
Therefore,the correct option is $C$.
For the given reaction: $_{7}N^{14} + _{1}H^{1} \to _{8}O^{15} + X$
Sum of atomic numbers on the left side: $7 + 1 = 8$.
Sum of atomic numbers on the right side: $8 + Z = 8$,so $Z = 0$.
Sum of mass numbers on the left side: $14 + 1 = 15$.
Sum of mass numbers on the right side: $15 + A = 15$,so $A = 0$.
$A$ particle with atomic number $0$ and mass number $0$ is a gamma photon,denoted as $\gamma$.
Therefore,the correct option is $C$.
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23
ChemistryMediumMCQAIIMS · 1983
Which of the following rate laws has an overall order of $0.5$ for a reaction involving substances $x$,$y$,and $z$?
A
Rate $= K[x][y][z]$
B
Rate $= K[x]^{0.5}[y]^{0.5}[z]^{0.5}$
C
Rate $= K[x]^{1.5}[y]^{-1}[z]^0$
D
Rate $= K[x][z]^n/[y]^2$
Solution
(C) The overall order of a reaction is the sum of the exponents of the concentration terms in the rate law expression.
For option $C$,the rate law is $\text{Rate} = K[x]^{1.5}[y]^{-1}[z]^0$.
The overall order $= 1.5 + (-1) + 0 = 0.5$.
Therefore,option $C$ is correct.
For option $C$,the rate law is $\text{Rate} = K[x]^{1.5}[y]^{-1}[z]^0$.
The overall order $= 1.5 + (-1) + 0 = 0.5$.
Therefore,option $C$ is correct.
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24
ChemistryMediumMCQAIIMS · 1983
Which method of purification is represented by the equation
$Ti + 2I_2$ (Impure) $\xrightarrow{500 \ K} TiI_4$ $\xrightarrow{1675 \ K} Ti + 2I_2$ (Pure)
$Ti + 2I_2$ (Impure) $\xrightarrow{500 \ K} TiI_4$ $\xrightarrow{1675 \ K} Ti + 2I_2$ (Pure)
A
Cupellation
B
Poling
C
Van Arkel
D
Zone refining
Solution
(C) The given reaction represents the $Van \ Arkel$ method.
This method is specifically used for the refining of metals like $Ti$ and $Zr$.
In this process,the crude metal is converted into a volatile compound,which is then decomposed to obtain the pure metal.
This method is specifically used for the refining of metals like $Ti$ and $Zr$.
In this process,the crude metal is converted into a volatile compound,which is then decomposed to obtain the pure metal.
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25
ChemistryMediumMCQAIIMS · 1983
Which ion has the maximum magnetic moment?
A
$V^{3+}$
B
$Mn^{3+}$
C
$Fe^{3+}$
D
$Cu^{2+}$
Solution
(C) The magnetic moment $(\mu)$ is calculated using the formula $\mu = \sqrt{n(n+2)} \text{ B.M.}$,where $n$ is the number of unpaired electrons.
Comparing the values,$Fe^{3+}$ has the highest number of unpaired electrons $(n=5)$,resulting in the maximum magnetic moment of $5.92 \text{ B.M.}$ Therefore,the correct option is $(C)$.
| Ion | Outer Configuration | Unpaired Electrons $(n)$ | Magnetic Moment ($B$.$M$.) |
|---|---|---|---|
| $V^{3+}$ | $3d^2$ | $2$ | $\sqrt{2(2+2)} = 2.83$ |
| $Mn^{3+}$ | $3d^4$ | $4$ | $\sqrt{4(4+2)} = 4.90$ |
| $Fe^{3+}$ | $3d^5$ | $5$ | $\sqrt{5(5+2)} = 5.92$ |
| $Cu^{2+}$ | $3d^9$ | $1$ | $\sqrt{1(1+2)} = 1.73$ |
Comparing the values,$Fe^{3+}$ has the highest number of unpaired electrons $(n=5)$,resulting in the maximum magnetic moment of $5.92 \text{ B.M.}$ Therefore,the correct option is $(C)$.
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26
ChemistryEasyMCQAIIMS · 1983
Which one of the following statements is true for transition elements?
A
They exhibit diamagnetism
B
They exhibit inert pair effect
C
They do not form alloys
D
They show variable oxidation states
Solution
(D) Transition elements show variable oxidation states due to the participation of $(n-1)d$ orbital electrons along with $ns$ orbital electrons in bond formation.
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27
ChemistryEasyMCQAIIMS · 1983
What is the decreasing order of reactivity of $HX$ in the reaction $ROH + HX \to RX + H_2O$?
A
$HI > HBr > HCl > HF$
B
$HBr > HCl > HI > HF$
C
$HCl > HBr > HI > HF$
D
$HF > HBr > HCl > HI$
Solution
(A) The reactivity of hydrogen halides $(HX)$ towards alcohols depends on the strength of the $H-X$ bond.
As the size of the halogen atom increases from $F$ to $I$,the bond dissociation energy of the $H-X$ bond decreases.
Therefore,the ease of cleavage of the $H-X$ bond increases in the order $HF < HCl < HBr < HI$.
Thus,the decreasing order of reactivity is $HI > HBr > HCl > HF$.
As the size of the halogen atom increases from $F$ to $I$,the bond dissociation energy of the $H-X$ bond decreases.
Therefore,the ease of cleavage of the $H-X$ bond increases in the order $HF < HCl < HBr < HI$.
Thus,the decreasing order of reactivity is $HI > HBr > HCl > HF$.
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28
ChemistryDifficultMCQAIIMS · 1983
Identify the product $Z$ in the series
$CH_3CN$ $\xrightarrow{Na + C_2H_5OH} X$ $\xrightarrow{HNO_2} Y$ $\xrightarrow[H_2SO_4]{K_2Cr_2O_7} Z$
$CH_3CN$ $\xrightarrow{Na + C_2H_5OH} X$ $\xrightarrow{HNO_2} Y$ $\xrightarrow[H_2SO_4]{K_2Cr_2O_7} Z$
A
$CH_3CHO$
B
$CH_3CONH_2$
C
$CH_3COOH$
D
$CH_3CH_2NHOH$
Solution
(C) $CH_3CN \xrightarrow{Na + C_2H_5OH} CH_3CH_2NH_2$ ($X$ is ethylamine).
$CH_3CH_2NH_2 \xrightarrow{HNO_2} CH_3CH_2OH$ ($Y$ is ethanol).
$CH_3CH_2OH \xrightarrow[H_2SO_4]{K_2Cr_2O_7} CH_3COOH$ ($Z$ is acetic acid).
$CH_3CH_2NH_2 \xrightarrow{HNO_2} CH_3CH_2OH$ ($Y$ is ethanol).
$CH_3CH_2OH \xrightarrow[H_2SO_4]{K_2Cr_2O_7} CH_3COOH$ ($Z$ is acetic acid).
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29
ChemistryDifficultMCQAIIMS · 1983
Which one of the following is not a property of hydrophilic sols?
A
High concentrations of dispersed phase can be easily attained
B
Coagulation is reversible
C
Viscosity and surface tension are about the same as for water
D
The charge of the particle depends on the $pH$ values of the medium; it may be positive,negative or even zero
Solution
(C) The correct answer is $(C)$.
Hydrophilic sols (lyophilic colloids) have a strong affinity between the dispersed phase and the dispersion medium.
Due to this interaction,the viscosity of these sols is significantly higher than that of the dispersion medium (water),and their surface tension is generally lower than that of water.
Therefore,the statement that viscosity and surface tension are about the same as for water is incorrect.
Hydrophilic sols (lyophilic colloids) have a strong affinity between the dispersed phase and the dispersion medium.
Due to this interaction,the viscosity of these sols is significantly higher than that of the dispersion medium (water),and their surface tension is generally lower than that of water.
Therefore,the statement that viscosity and surface tension are about the same as for water is incorrect.
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