ChemistryQ1–29 of 29 questions
Page 1 of 1 · English
1
ChemistryEasyMCQAIIMS · 1980
The spectrum of $He$ is expected to be similar to
A
$H$
B
$Li^{+}$
C
$Na$
D
$He^{+}$
Solution
(B) The spectrum of an atom or ion depends on the number of electrons present in it.
$He$ (Helium) has $2$ electrons.
$Li^{+}$ (Lithium ion) also has $3 - 1 = 2$ electrons.
Since both species are isoelectronic (contain the same number of electrons),their spectra are expected to be similar.
$He$ (Helium) has $2$ electrons.
$Li^{+}$ (Lithium ion) also has $3 - 1 = 2$ electrons.
Since both species are isoelectronic (contain the same number of electrons),their spectra are expected to be similar.
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2
ChemistryMediumMCQAIIMS · 1980
Which one is the correct outer electronic configuration of chromium $(Cr)$?
A
B
C
D
Solution
(C) The atomic number of chromium $(Cr)$ is $24$. The expected electronic configuration is $[Ar] \, 3d^4 \, 4s^2$. However,due to the extra stability associated with half-filled $d$-orbitals,one electron from the $4s$ orbital shifts to the $3d$ orbital. Thus,the actual ground state configuration is $[Ar] \, 3d^5 \, 4s^1$. This corresponds to $5$ unpaired electrons in the $3d$ subshell and $1$ unpaired electron in the $4s$ subshell,which is represented by option $(C)$.
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3
ChemistryMediumMCQAIIMS · 1980
Which molecule has a zero dipole moment?
A
$H_2O$
B
$CO_2$
C
$HF$
D
$HBr$
Solution
(B) $CO_2$ has a linear geometry with the two $C=O$ bond dipoles pointing in opposite directions,which cancel each other out. Therefore,the net dipole moment of $CO_2$ is $0 \ D$.
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4
ChemistryMediumMCQAIIMS · 1980
Which bond angle $\theta$ would result in the maximum dipole moment for the triatomic molecule $YXY$?
A
$\theta = 90^o$
B
$\theta = 120^o$
C
$\theta = 150^o$
D
$\theta = 180^o$
Solution
(A) The resultant dipole moment $\mu_R$ of two individual bond dipoles $\mu$ inclined at an angle $\theta$ is given by the formula: $\mu_R = \sqrt{\mu^2 + \mu^2 + 2 \mu^2 \cos \theta} = \sqrt{2 \mu^2 (1 + \cos \theta)}$.
As the bond angle $\theta$ increases from $0^o$ to $180^o$,the value of $\cos \theta$ decreases from $1$ to $-1$.
Consequently,the term $(1 + \cos \theta)$ decreases as $\theta$ increases.
Therefore,the dipole moment $\mu_R$ is maximum when $\theta$ is minimum,which is $90^o$ among the given options.
As the bond angle $\theta$ increases from $0^o$ to $180^o$,the value of $\cos \theta$ decreases from $1$ to $-1$.
Consequently,the term $(1 + \cos \theta)$ decreases as $\theta$ increases.
Therefore,the dipole moment $\mu_R$ is maximum when $\theta$ is minimum,which is $90^o$ among the given options.
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5
ChemistryEasyMCQAIIMS · 1980
Bond order is a concept in the molecular orbital theory. It depends on the number of electrons in the bonding and antibonding orbitals. Which of the following statements is true about it? The bond order
A
Can assume any positive or integral or fractional value including zero
B
Has always an integral value
C
Can have a negative quantity
D
Is a non zero quantity
Solution
(A) The formula for Bond Order $(B.O.)$ is given by:
$B.O. = \frac{N_b - N_a}{2}$
where $N_b$ is the number of electrons in bonding molecular orbitals and $N_a$ is the number of electrons in antibonding molecular orbitals.
Bond order can be a positive value,zero,or a fraction. For example,the bond order of $He_2$ is $0$,and the bond order of $O_2^+$ is $2.5$.
$B.O. = \frac{N_b - N_a}{2}$
where $N_b$ is the number of electrons in bonding molecular orbitals and $N_a$ is the number of electrons in antibonding molecular orbitals.
Bond order can be a positive value,zero,or a fraction. For example,the bond order of $He_2$ is $0$,and the bond order of $O_2^+$ is $2.5$.
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6
ChemistryMediumMCQAIIMS · 1980
Which concept best explains that $o-$nitrophenol is more volatile than $p-$nitrophenol?
A
Resonance
B
Hyperconjugation
C
Hydrogen bonding
D
Steric hindrance
Solution
(C) $o-$Nitrophenol exhibits intramolecular hydrogen bonding,which reduces its intermolecular attraction.
$p-$Nitrophenol exhibits intermolecular hydrogen bonding,leading to association of molecules and a higher boiling point.
Therefore,$o-$nitrophenol is more volatile than $p-$nitrophenol due to the difference in the type of hydrogen bonding.
$p-$Nitrophenol exhibits intermolecular hydrogen bonding,leading to association of molecules and a higher boiling point.
Therefore,$o-$nitrophenol is more volatile than $p-$nitrophenol due to the difference in the type of hydrogen bonding.
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7
ChemistryMediumMCQAIIMS · 1980
Which one is the highest melting halide?
A
$NaCl$
B
$NaBr$
C
$NaF$
D
$NaI$
Solution
(C) The melting point of ionic halides depends on the lattice energy of the crystal.
According to Fajan's rule,as the size of the anion increases $(F^- < Cl^- < Br^- < I^-)$,the covalent character increases and the ionic character decreases.
$NaF$ has the smallest anion,which results in the highest lattice energy and the strongest ionic character among the given options.
Therefore,$NaF$ has the highest melting point.
According to Fajan's rule,as the size of the anion increases $(F^- < Cl^- < Br^- < I^-)$,the covalent character increases and the ionic character decreases.
$NaF$ has the smallest anion,which results in the highest lattice energy and the strongest ionic character among the given options.
Therefore,$NaF$ has the highest melting point.
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8
ChemistryEasyMCQAIIMS · 1980
Which of the following is a buffer solution?
A
$CH_3COOH + CH_3COONa$
B
$CH_3COOH + CH_3COONH_4$
C
$CH_3COOH + NH_4Cl$
D
$NaOH + NaCl$
Solution
(A) buffer solution is typically a mixture of a weak acid and its conjugate base (salt of the weak acid with a strong base).
In option $A$,$CH_3COOH$ is a weak acid and $CH_3COONa$ is its salt with a strong base $(NaOH)$,which forms an acidic buffer.
Therefore,the correct option is $A$.
In option $A$,$CH_3COOH$ is a weak acid and $CH_3COONa$ is its salt with a strong base $(NaOH)$,which forms an acidic buffer.
Therefore,the correct option is $A$.
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9
ChemistryMCQAIIMS · 1980
Which equilibrium can be described as an acid-base reaction using the Lewis acid-base definition but not using the Bronsted-Lowry definition?
A
$2NH_3 + H_2SO_4 \rightleftharpoons 2NH_4^+ + SO_4^{2-}$
B
$NH_3 + CH_3COOH \rightleftharpoons NH_4^+ + CH_3COO^{-}$
C
$H_2O + CH_3COOH \rightleftharpoons H_3O^{+} + CH_3COO^{-}$
D
$[Cu(H_2O)_4]^{2+} + 4NH_3 \rightleftharpoons [Cu(NH_3)_4]^{2+} + 4H_2O$
Solution
(D) The $Bronsted-Lowry$ theory defines acids as proton $(H^+)$ donors and bases as proton acceptors.
In options $A$,$B$,and $C$,there is a clear transfer of a proton from the acid to the base.
In option $D$,the reaction involves the formation of a coordinate covalent bond where $NH_3$ acts as a Lewis base (electron pair donor) and $[Cu(H_2O)_4]^{2+}$ acts as a Lewis acid (electron pair acceptor).
Since there is no proton transfer involved in this reaction,it cannot be described by the $Bronsted-Lowry$ definition but fits the $Lewis$ acid-base definition perfectly.
Therefore,the correct option is $D$.
In options $A$,$B$,and $C$,there is a clear transfer of a proton from the acid to the base.
In option $D$,the reaction involves the formation of a coordinate covalent bond where $NH_3$ acts as a Lewis base (electron pair donor) and $[Cu(H_2O)_4]^{2+}$ acts as a Lewis acid (electron pair acceptor).
Since there is no proton transfer involved in this reaction,it cannot be described by the $Bronsted-Lowry$ definition but fits the $Lewis$ acid-base definition perfectly.
Therefore,the correct option is $D$.
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10
ChemistryEasyMCQAIIMS · 1980
Which of the following hydrogen halides has the highest boiling point?
A
$HF$
B
$HCl$
C
$HBr$
D
$HI$
Solution
(A) The boiling point of hydrogen halides depends on the strength of intermolecular forces.
In $HF$,the molecules are associated due to strong intermolecular hydrogen bonding.
This results in a significantly higher boiling point compared to $HCl$,$HBr$,and $HI$,where only weak van der Waals forces exist.
Therefore,$HF$ has the highest boiling point.
In $HF$,the molecules are associated due to strong intermolecular hydrogen bonding.
This results in a significantly higher boiling point compared to $HCl$,$HBr$,and $HI$,where only weak van der Waals forces exist.
Therefore,$HF$ has the highest boiling point.
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11
ChemistryDifficultMCQAIIMS · 1980
In the precipitation of the iron group in qualitative analysis,ammonium chloride is added before adding ammonium hydroxide to:
A
Decrease concentration of $OH^{-}$ ions
B
Prevent interference by phosphate ions
C
Increase concentration of $Cl^{-}$ ions
D
Increase concentration of $NH_4^+$ ions
Solution
(A) $NH_4Cl$ is a strong electrolyte that dissociates completely to provide $NH_4^+$ ions.
According to the common ion effect,the presence of excess $NH_4^+$ ions suppresses the ionization of the weak base $NH_4OH$ $(NH_4OH \rightleftharpoons NH_4^+ + OH^-)$.
This decreases the concentration of $OH^-$ ions in the solution.
This controlled concentration of $OH^-$ ions is sufficient to exceed the solubility product $(K_{sp})$ of $III$ group hydroxides (like $Fe(OH)_3$,$Al(OH)_3$,$Cr(OH)_3$) but not enough to precipitate hydroxides of $IV$,$V$,and $VI$ group radicals.
Therefore,the correct option is $(A)$.
According to the common ion effect,the presence of excess $NH_4^+$ ions suppresses the ionization of the weak base $NH_4OH$ $(NH_4OH \rightleftharpoons NH_4^+ + OH^-)$.
This decreases the concentration of $OH^-$ ions in the solution.
This controlled concentration of $OH^-$ ions is sufficient to exceed the solubility product $(K_{sp})$ of $III$ group hydroxides (like $Fe(OH)_3$,$Al(OH)_3$,$Cr(OH)_3$) but not enough to precipitate hydroxides of $IV$,$V$,and $VI$ group radicals.
Therefore,the correct option is $(A)$.
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12
ChemistryMediumMCQAIIMS · 1980
The solid $NaCl$ is a bad conductor of electricity since:
A
In solid $NaCl$ there are no ions.
B
Solid $NaCl$ is covalent.
C
In solid $NaCl$ there is no velocity of ions.
D
In solid $NaCl$ there are no electrons.
Solution
(C) Solid $NaCl$ is a bad conductor of electricity because the ions are held in fixed positions within the crystal lattice and are not free to move to conduct electricity.
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13
ChemistryMCQAIIMS · 1980
Which of the following is a buffer solution?
A
$CH_3COOH + CH_3COONa$
B
$CH_3COOH + CH_3COONH_4$
C
$CH_3COOH + NH_4Cl$
D
$NaOH + NaCl$
Solution
(A) buffer solution is typically formed by a mixture of a weak acid $(WA)$ and its salt with a strong base $(SB)$,or a weak base $(WB)$ and its salt with a strong acid $(SA)$.
In the given options,$CH_3COOH$ is a weak acid and $CH_3COONa$ is its salt with a strong base $(NaOH)$.
Therefore,the mixture of $CH_3COOH + CH_3COONa$ acts as an acidic buffer solution.
In the given options,$CH_3COOH$ is a weak acid and $CH_3COONa$ is its salt with a strong base $(NaOH)$.
Therefore,the mixture of $CH_3COOH + CH_3COONa$ acts as an acidic buffer solution.
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14
ChemistryMediumMCQAIIMS · 1980
Elements up to atomic number $103$ have been synthesized and studied. If a newly discovered element is found to have an atomic number $106,$ its electronic configuration will be
A
$[Rn] 5f^{14} 6d^4 7s^2$
B
$[Rn] 5f^{14} 6d^1 7s^2 7p^3$
C
$[Rn] 5f^{14} 6d^6 7s^0$
D
$[Rn] 5f^{14} 6d^4 7s^2$
Solution
(A) The atomic number of the element is $Z = 106.$
Radon $(Rn)$ has an atomic number of $86.$
The electronic configuration is filled according to the Aufbau principle: $86 (Rn) + 14 (5f) + 4 (6d) + 2 (7s) = 106.$
Thus,the configuration is $[Rn] 5f^{14} 6d^4 7s^2.$
Radon $(Rn)$ has an atomic number of $86.$
The electronic configuration is filled according to the Aufbau principle: $86 (Rn) + 14 (5f) + 4 (6d) + 2 (7s) = 106.$
Thus,the configuration is $[Rn] 5f^{14} 6d^4 7s^2.$
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15
ChemistryMediumMCQAIIMS · 1980
Which is the most volatile compound?
A
$HF$
B
$HCl$
C
$HBr$
D
$HI$
Solution
(B)
Volatility is inversely proportional to the boiling point. Since $HCl$ has the lowest boiling point $(189 \ K)$ among the given hydrogen halides,it is the most volatile compound.
| Hydride | $HF, HCl, HBr, HI$ |
| Boiling point (in $K$) | $293, 189, 206, 238$ |
Volatility is inversely proportional to the boiling point. Since $HCl$ has the lowest boiling point $(189 \ K)$ among the given hydrogen halides,it is the most volatile compound.
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16
ChemistryDifficultMCQAIIMS · 1980
In the group $III$ radicals,in place of $NH_4Cl$,which of the following can be used?
A
$NH_4NO_3$
B
$(NH_4)_2SO_4$
C
$(NH_4)_2CO_3$
D
$NaCl$
Solution
(A) The group $III$ radicals $(Fe^{3+}, Al^{3+}, Cr^{3+})$ are precipitated as hydroxides using $NH_4OH$ in the presence of $NH_4Cl$.
$NH_4Cl$ provides $NH_4^+$ ions,which suppress the ionization of $NH_4OH$ due to the common ion effect,ensuring that the concentration of $OH^-$ ions is low enough to prevent the precipitation of group $IV$ and $V$ radicals.
Any ammonium salt that provides $NH_4^+$ ions without introducing interfering anions can be used.
$NH_4NO_3$ is a suitable substitute because it provides $NH_4^+$ ions and does not interfere with the precipitation process.
$(NH_4)_2SO_4$ cannot be used because $SO_4^{2-}$ ions would precipitate $Ba^{2+}, Sr^{2+},$ and $Ca^{2+}$ as sulfates.
$NH_4Cl$ provides $NH_4^+$ ions,which suppress the ionization of $NH_4OH$ due to the common ion effect,ensuring that the concentration of $OH^-$ ions is low enough to prevent the precipitation of group $IV$ and $V$ radicals.
Any ammonium salt that provides $NH_4^+$ ions without introducing interfering anions can be used.
$NH_4NO_3$ is a suitable substitute because it provides $NH_4^+$ ions and does not interfere with the precipitation process.
$(NH_4)_2SO_4$ cannot be used because $SO_4^{2-}$ ions would precipitate $Ba^{2+}, Sr^{2+},$ and $Ca^{2+}$ as sulfates.
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17
ChemistryDifficultMCQAIIMS · 1980
Sodium carbonate cannot be used in place of ammonium carbonate for the precipitation of the fifth group radicals because
A
Sodium ions interfere with the detection of fifth group radicals
B
Concentration of carbonate ions is very low
C
Sodium will react with acidic radicals
D
Magnesium will be precipitated
Solution
(D) . The fifth group radicals $(Ba^{2+}, Sr^{2+}, Ca^{2+})$ are precipitated as carbonates using $(NH_4)_2CO_3$ in the presence of $NH_4Cl$ and $NH_4OH$.
When $(NH_4)_2CO_3$ is used,the concentration of $CO_3^{2-}$ ions is kept low due to the common ion effect of $NH_4^+$.
However,if $Na_2CO_3$ is used,the concentration of $CO_3^{2-}$ ions becomes very high.
This high concentration of $CO_3^{2-}$ causes $Mg^{2+}$ ions (which belong to the $VI^{th}$ group) to precipitate as $MgCO_3$ along with the $V^{th}$ group radicals,thus interfering with the analysis.
When $(NH_4)_2CO_3$ is used,the concentration of $CO_3^{2-}$ ions is kept low due to the common ion effect of $NH_4^+$.
However,if $Na_2CO_3$ is used,the concentration of $CO_3^{2-}$ ions becomes very high.
This high concentration of $CO_3^{2-}$ causes $Mg^{2+}$ ions (which belong to the $VI^{th}$ group) to precipitate as $MgCO_3$ along with the $V^{th}$ group radicals,thus interfering with the analysis.
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18
ChemistryMediumMCQAIIMS · 1980
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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19
ChemistryDifficultMCQAIIMS · 1980
$_{84}Po^{210} \xrightarrow{} _{82}Pb^{206} + _{2}He^{4}$. From the above equation,deduce the position of polonium in the periodic table (lead belongs to group $IV A$).
A
$II A$
B
$IV B$
C
$VI B$
D
$VI A$
Solution
(D) According to the Group Displacement Law,when an element emits an $\alpha$-particle $(_{2}He^{4})$,its atomic number decreases by $2$ and it moves two groups to the left in the periodic table.
Since $Pb$ (lead) belongs to group $IV A$,the parent element $Po$ (polonium) must be located two groups to the right of $Pb$.
Therefore,the group of $Po = IV A + 2 = VI A$.
Since $Pb$ (lead) belongs to group $IV A$,the parent element $Po$ (polonium) must be located two groups to the right of $Pb$.
Therefore,the group of $Po = IV A + 2 = VI A$.
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20
ChemistryMediumMCQAIIMS · 1980
When electric current is passed through a cell having an electrolyte,the positive ions move towards the cathode and the negative ions towards the anode. If the cathode is pulled out of the solution,what happens to the ions?
A
The positive and negative ions will move towards the anode.
B
The positive ions will start moving towards the anode,the negative ions will stop moving.
C
The negative ions will continue to move towards the anode and the positive ions will stop moving.
D
The positive and negative ions will start moving randomly.
Solution
(D) . When the cathode is removed,the external electric circuit is broken,and the electric field across the electrolyte vanishes. In the absence of an electric field,the ions in the solution no longer experience a directional force and move randomly due to thermal energy.
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21
ChemistryDifficultMCQAIIMS · 1980
‘Lapis-Lazuli’ is a blue-coloured precious stone. It is a mineral of the class:
A
Sodium-alumino silicate
B
Zinc cobaltate
C
Basic copper carbonate
D
Prussian blue
Solution
(A) Lapis lazuli is a deep-blue metamorphic rock used as a semi-precious stone.
Chemically,it is a complex mineral belonging to the class of $Sodium-alumino$ $silicates$ containing sulfur,specifically the mineral $Lazurite$ $(Na_8[Al_6Si_6O_{24}]S_n)$.
Therefore,the correct option is $(A)$.
Chemically,it is a complex mineral belonging to the class of $Sodium-alumino$ $silicates$ containing sulfur,specifically the mineral $Lazurite$ $(Na_8[Al_6Si_6O_{24}]S_n)$.
Therefore,the correct option is $(A)$.
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22
ChemistryEasyMCQAIIMS · 1980
Lucas reagent is
A
Anhydrous $ZnCl_2$ + conc. $HCl$
B
Hydrous $ZnCl_2$ + dil. $HCl$
C
Conc. $HNO_3$ + anhydrous $ZnCl_2$
D
Conc. $HNO_3$ + anhydrous $MgCl_2$
Solution
(A) Lucas reagent is a solution of anhydrous $ZnCl_2$ in concentrated $HCl$.
This reagent is used to classify alcohols of low molecular weight. The reaction involves the substitution of the hydroxyl group $(-OH)$ with a chloride ion $(Cl^-)$.
The reaction proceeds via an $SN1$ mechanism,where the rate of reaction depends on the stability of the carbocation intermediate. Tertiary alcohols react fastest,followed by secondary,while primary alcohols react very slowly or not at all at room temperature.
Therefore,the correct composition is anhydrous $ZnCl_2$ and concentrated $HCl$.
This reagent is used to classify alcohols of low molecular weight. The reaction involves the substitution of the hydroxyl group $(-OH)$ with a chloride ion $(Cl^-)$.
The reaction proceeds via an $SN1$ mechanism,where the rate of reaction depends on the stability of the carbocation intermediate. Tertiary alcohols react fastest,followed by secondary,while primary alcohols react very slowly or not at all at room temperature.
Therefore,the correct composition is anhydrous $ZnCl_2$ and concentrated $HCl$.
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23
ChemistryDifficultMCQAIIMS · 1980
What is the composition of Lucas reagent?
A
Concentrated $HCl +$ anhydrous $ZnCl_2$
B
Dilute $HCl +$ hydrated $ZnCl_2$
C
Concentrated $HNO_3 +$ anhydrous $ZnCl_2$
D
Concentrated $HCl +$ anhydrous $MgCl_2$
Solution
(A) Lucas reagent is a solution of anhydrous zinc chloride $(ZnCl_2)$ in concentrated hydrochloric acid $(HCl)$.
It is primarily used to distinguish between primary,secondary,and tertiary alcohols based on the rate of reaction to form alkyl chlorides.
It is primarily used to distinguish between primary,secondary,and tertiary alcohols based on the rate of reaction to form alkyl chlorides.
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24
ChemistryDifficultMCQAIIMS · 1980
$A$ sample of chloroform being used as an anaesthetic is tested by:
A
Fehling solution
B
$AgNO_3$ solution after boiling with alcoholic $KOH$ solution
C
$AgNO_3$ solution
D
Both $(b)$ and $(c)$
Solution
(C) The correct answer is $(C)$.
Chloroform $(CHCl_3)$ is tested for purity before use as an anaesthetic by adding aqueous silver nitrate $(AgNO_3)$ solution.
Pure chloroform does not react with aqueous $AgNO_3$ and thus does not form a white precipitate of silver chloride $(AgCl)$.
If the chloroform has been oxidized to phosgene $(COCl_2)$ or contains impurities like $HCl$,it will react with $AgNO_3$ to form a white precipitate of $AgCl$.
Chloroform $(CHCl_3)$ is tested for purity before use as an anaesthetic by adding aqueous silver nitrate $(AgNO_3)$ solution.
Pure chloroform does not react with aqueous $AgNO_3$ and thus does not form a white precipitate of silver chloride $(AgCl)$.
If the chloroform has been oxidized to phosgene $(COCl_2)$ or contains impurities like $HCl$,it will react with $AgNO_3$ to form a white precipitate of $AgCl$.
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25
ChemistryMediumMCQAIIMS · 1980
Which of the following is not a characteristic of alcohols?
A
They are lighter than water.
B
Their boiling points rise fairly uniformly with increasing molecular weight.
C
Lower members are insoluble in water and organic solvents,but solubility regularly increases with molecular weight.
D
Lower members have a pleasant smell and burning taste,while higher members are odourless and tasteless.
Solution
(C) The correct answer is $(C)$. Lower members of alcohols are soluble in water due to the formation of hydrogen bonds with water molecules. As the molecular mass increases,the hydrophobic alkyl group increases in size,which causes the solubility to decrease,not increase.
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26
ChemistryMediumMCQAIIMS · 1980
Glucose forms many derivatives. The derivative which will help to prove the furanose structure is
A
Acetyl
B
Benzoyl
C
Osazone
D
Isopropylidene
Solution
(D) The furanose structure of sugars (like fructose) is confirmed by the formation of isopropylidene derivatives.
When fructose reacts with acetone in the presence of an acid catalyst,it forms di-isopropylidene fructose,which confirms the presence of a five-membered furanose ring structure.
When fructose reacts with acetone in the presence of an acid catalyst,it forms di-isopropylidene fructose,which confirms the presence of a five-membered furanose ring structure.
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27
ChemistryMediumMCQAIIMS · 1980
$A$ wooden artifact and a freshly cut tree have carbon activity of $7.6$ and $15.2 \ min^{-1} g^{-1}$ respectively. Given the half-life $(t_{1/2})$ of carbon is $5760$ years,the age of the artifact is:
A
$5760$ years
B
$5760 \times \frac{15.2}{7.6}$ years
C
$5760 \times \frac{7.6}{15.2}$ years
D
$5760 \times (15.2 - 7.6)$ years
Solution
(A) The decay of carbon follows first-order kinetics.
Given: Initial activity $(r_o)$ = $15.2 \ min^{-1} g^{-1}$,Final activity $(r)$ = $7.6 \ min^{-1} g^{-1}$,and $t_{1/2} = 5760$ years.
Since $r = \frac{r_o}{2}$,the artifact has undergone exactly one half-life.
Therefore,the age of the artifact $(t)$ = $1 \times t_{1/2} = 5760$ years.
Alternatively,using the formula $t = \frac{t_{1/2}}{0.693} \ln(\frac{r_o}{r}) = \frac{5760}{0.693} \ln(\frac{15.2}{7.6}) = \frac{5760}{0.693} \ln(2) = 5760$ years.
Given: Initial activity $(r_o)$ = $15.2 \ min^{-1} g^{-1}$,Final activity $(r)$ = $7.6 \ min^{-1} g^{-1}$,and $t_{1/2} = 5760$ years.
Since $r = \frac{r_o}{2}$,the artifact has undergone exactly one half-life.
Therefore,the age of the artifact $(t)$ = $1 \times t_{1/2} = 5760$ years.
Alternatively,using the formula $t = \frac{t_{1/2}}{0.693} \ln(\frac{r_o}{r}) = \frac{5760}{0.693} \ln(\frac{15.2}{7.6}) = \frac{5760}{0.693} \ln(2) = 5760$ years.
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28
ChemistryMediumMCQAIIMS · 1980
What is the standard cell potential for the cell $Zn | Zn^{2+} (1M) || Cu^{2+} (1M) | Cu$? Given $E^o$ for $Zn^{2+} | Zn = -0.76 \ V$ and $E^o$ for $Cu^{2+} | Cu = +0.34 \ V$.
A
$-0.76 + (-0.34) = -0.42 \ V$
B
$-0.34 + 0.76 = +0.42 \ V$
C
$0.34 - (-0.76) = 1.10 \ V$
D
$-0.76 - (+0.34) = -1.10 \ V$
Solution
(C) The standard cell potential is calculated using the formula: $E^o_{cell} = E^o_{cathode} - E^o_{anode}$.
In the given cell $Zn | Zn^{2+} || Cu^{2+} | Cu$,the $Cu^{2+} | Cu$ electrode acts as the cathode and $Zn^{2+} | Zn$ acts as the anode.
Given $E^o_{cathode} = E^o_{Cu^{2+} | Cu} = +0.34 \ V$ and $E^o_{anode} = E^o_{Zn^{2+} | Zn} = -0.76 \ V$.
Substituting these values: $E^o_{cell} = 0.34 \ V - (-0.76 \ V) = 0.34 + 0.76 = 1.10 \ V$.
In the given cell $Zn | Zn^{2+} || Cu^{2+} | Cu$,the $Cu^{2+} | Cu$ electrode acts as the cathode and $Zn^{2+} | Zn$ acts as the anode.
Given $E^o_{cathode} = E^o_{Cu^{2+} | Cu} = +0.34 \ V$ and $E^o_{anode} = E^o_{Zn^{2+} | Zn} = -0.76 \ V$.
Substituting these values: $E^o_{cell} = 0.34 \ V - (-0.76 \ V) = 0.34 + 0.76 = 1.10 \ V$.
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29
ChemistryMediumMCQAIIMS · 1980
Which equilibrium can be described as an acid-base reaction using the Lewis acid-base definition but not using the Bronsted-Lowry definition?
A
$2NH_3 + H_2SO_4 \rightleftharpoons 2NH_4^+ + SO_4^{2-}$
B
$NH_3 + CH_3COOH \rightleftharpoons NH_4^+ + CH_3COO^{-}$
C
$H_2O + CH_3COOH \rightleftharpoons H_3O^{+} + CH_3COO^{-}$
D
$[Cu(H_2O)_4]^{2+} + 4NH_3 \rightleftharpoons [Cu(NH_3)_4]^{2+} + 4H_2O$
Solution
(D) The Bronsted-Lowry definition defines acids as proton $(H^+)$ donors and bases as proton acceptors.
In options $A$,$B$,and $C$,there is a clear transfer of a proton $(H^+)$ from the acid to the base.
In the reaction $[Cu(H_2O)_4]^{2+} + 4NH_3 \rightleftharpoons [Cu(NH_3)_4]^{2+} + 4H_2O$,there is no proton transfer.
Instead,$NH_3$ acts as a Lewis base by donating a lone pair of electrons to the $Cu^{2+}$ ion (Lewis acid) to form a coordinate covalent bond.
Therefore,this reaction is an acid-base reaction according to the Lewis definition but not the Bronsted-Lowry definition.
In options $A$,$B$,and $C$,there is a clear transfer of a proton $(H^+)$ from the acid to the base.
In the reaction $[Cu(H_2O)_4]^{2+} + 4NH_3 \rightleftharpoons [Cu(NH_3)_4]^{2+} + 4H_2O$,there is no proton transfer.
Instead,$NH_3$ acts as a Lewis base by donating a lone pair of electrons to the $Cu^{2+}$ ion (Lewis acid) to form a coordinate covalent bond.
Therefore,this reaction is an acid-base reaction according to the Lewis definition but not the Bronsted-Lowry definition.
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