ChemistryQ1–11 of 11 questions
Page 1 of 1 · English
1
ChemistryEasyMCQAIIMS · 1987
In $HCHO$,the carbon atom has which hybridization?
A
$sp$
B
$sp^2$
C
$sp^3$
D
All of the above
Solution
(B) In $HCHO$ (formaldehyde),the carbon atom is bonded to two hydrogen atoms by single bonds and one oxygen atom by a double bond.
The steric number of the carbon atom is calculated as: $\text{Steric Number} = \text{Number of sigma bonds} + \text{Number of lone pairs} = 3 + 0 = 3$.
$A$ steric number of $3$ corresponds to $sp^2$ hybridization.
Therefore,the correct option is $(B)$.
The steric number of the carbon atom is calculated as: $\text{Steric Number} = \text{Number of sigma bonds} + \text{Number of lone pairs} = 3 + 0 = 3$.
$A$ steric number of $3$ corresponds to $sp^2$ hybridization.
Therefore,the correct option is $(B)$.
0 likesView Solution
2
ChemistryMediumMCQAIIMS · 1987
The elements of group $IA$ provide a colour to the flame of a Bunsen burner due to
A
Low ionization potential
B
Low melting point
C
Softness
D
Presence of one electron in the outermost orbit
Solution
(A) The alkali metals $(Group \ IA)$ exhibit a characteristic flame color because they possess low ionization potential.
When heated in a Bunsen burner,the valence electrons absorb energy and are excited from a lower energy level to a higher energy level.
As these excited electrons return to their ground state,they emit the absorbed energy in the form of visible light,which corresponds to a specific color.
When heated in a Bunsen burner,the valence electrons absorb energy and are excited from a lower energy level to a higher energy level.
As these excited electrons return to their ground state,they emit the absorbed energy in the form of visible light,which corresponds to a specific color.
0 likesView Solution
3
ChemistryEasyMCQAIIMS · 1987
Solvay's process is used for the preparation of
A
Ammonia
B
Sodium bicarbonate
C
Sodium carbonate
D
Calcium carbonate
Solution
(C) The Solvay process is an industrial method used for the large-scale production of $Na_2CO_3$ (Sodium carbonate).
In this process,$CO_2$ and $NH_3$ are passed through a concentrated solution of $NaCl$ (brine) to form $NaHCO_3$,which is then heated to yield $Na_2CO_3$.
In this process,$CO_2$ and $NH_3$ are passed through a concentrated solution of $NaCl$ (brine) to form $NaHCO_3$,which is then heated to yield $Na_2CO_3$.
0 likesView Solution
4
ChemistryMediumMCQAIIMS · 1987
Which of the following is not formed by the reaction of $Cl_2$ on $CH_4$ in sunlight?
A
$CHCl_3$
B
$CH_3Cl$
C
$CH_3CH_3$
D
$CH_3CH_2CH_3$
Solution
(D) The reaction of $CH_4$ with $Cl_2$ in the presence of sunlight follows a free radical substitution mechanism.
The successive chlorination products are $CH_3Cl$, $CH_2Cl_2$, $CHCl_3$, and $CCl_4$.
Additionally, the coupling of methyl free radicals $(\cdot CH_3 + \cdot CH_3 \to CH_3-CH_3)$ leads to the formation of ethane $(CH_3CH_3)$.
Propane $(CH_3CH_2CH_3)$ is not formed in this reaction because it requires a higher carbon chain length than what is available from the reactants.
The successive chlorination products are $CH_3Cl$, $CH_2Cl_2$, $CHCl_3$, and $CCl_4$.
Additionally, the coupling of methyl free radicals $(\cdot CH_3 + \cdot CH_3 \to CH_3-CH_3)$ leads to the formation of ethane $(CH_3CH_3)$.
Propane $(CH_3CH_2CH_3)$ is not formed in this reaction because it requires a higher carbon chain length than what is available from the reactants.
0 likesView Solution
5
ChemistryMediumMCQAIIMS · 1987
$A$ compound '$X$' on ozonolysis forms two molecules of $HCHO$. Compound '$X$' is
A
$C_2H_4$
B
$C_2H_2$
C
$C_2H_6$
D
$C_6H_6$
Solution
(A) Ozonolysis of an alkene involves the cleavage of the $C=C$ double bond to form carbonyl compounds.
For a compound '$X$' to yield two molecules of formaldehyde $(HCHO)$ upon ozonolysis,the structure must be ethene $(CH_2=CH_2)$.
The reaction is: $CH_2=CH_2 \xrightarrow[(2) Zn/H_2O]{(1) O_3} 2HCHO$.
Therefore,the compound '$X$' is $C_2H_4$.
For a compound '$X$' to yield two molecules of formaldehyde $(HCHO)$ upon ozonolysis,the structure must be ethene $(CH_2=CH_2)$.
The reaction is: $CH_2=CH_2 \xrightarrow[(2) Zn/H_2O]{(1) O_3} 2HCHO$.
Therefore,the compound '$X$' is $C_2H_4$.
0 likesView Solution
6
ChemistryMediumMCQAIIMS · 1987
Which of the following bonds makes a hydrocarbon most reactive?
A
$C \equiv C$
B
$C = C$
C
$C - C$
D
All of these
Solution
(A) The correct answer is $A$. The $C \equiv C$ bond (alkyne) is generally considered more reactive towards electrophilic addition reactions compared to $C = C$ (alkene) and $C - C$ (alkane) due to the presence of two $\pi$-bonds and the nature of $sp$-hybridized carbon atoms.
0 likesView Solution
7
ChemistryMediumMCQAIIMS · 1987
What happens when a mixture of acetylene and hydrogen is passed over heated Lindlar's catalyst?
A
$Ethane$ and water are formed
B
$Ethylene$ is formed
C
$Acetylene$ and $ethane$ are formed
D
None of these
Solution
(B) The reaction of acetylene $(CH \equiv CH)$ with hydrogen $(H_2)$ in the presence of Lindlar's catalyst $(Pd/BaSO_4)$ is a partial hydrogenation reaction.
$CH \equiv CH + H_2 \xrightarrow[Lindlar \ Catalyst]{Pd/BaSO_4} CH_2 = CH_2$
Thus,acetylene is reduced to ethylene $(CH_2 = CH_2)$.
$CH \equiv CH + H_2 \xrightarrow[Lindlar \ Catalyst]{Pd/BaSO_4} CH_2 = CH_2$
Thus,acetylene is reduced to ethylene $(CH_2 = CH_2)$.
0 likesView Solution
8
ChemistryMediumMCQAIIMS · 1987
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.
0 likesView Solution
9
ChemistryEasyMCQAIIMS · 1987
$Fluorine$ does not form positive oxidation states because
A
It is the most electronegative element
B
It forms only anions in ionic compounds
C
It cannot form multiple bonding
D
It shows non-bonded electron pair repulsion due to small size
Solution
(A) $Fluorine$ is the most electronegative element in the periodic table.
Due to its very high electronegativity,it cannot lose electrons to form positive oxidation states and always exhibits a $-1$ oxidation state in its compounds.
Due to its very high electronegativity,it cannot lose electrons to form positive oxidation states and always exhibits a $-1$ oxidation state in its compounds.
0 likesView Solution
10
ChemistryMediumMCQAIIMS · 1987
Which 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.
When a freshly prepared ferrous sulfate solution is added to the nitrate solution followed by the careful addition of concentrated sulfuric acid along the sides of the test tube,a brown ring is formed at the junction of the two layers.
The reaction involves the reduction of nitrate to nitric oxide $(NO)$,which then reacts with the $Fe(II)$ ions to form the brown complex $[Fe(H_2O)_5(NO)]SO_4$.
When a freshly prepared ferrous sulfate solution is added to the nitrate solution followed by the careful addition of concentrated sulfuric acid along the sides of the test tube,a brown ring is formed at the junction of the two layers.
The reaction involves the reduction of nitrate to nitric oxide $(NO)$,which then reacts with the $Fe(II)$ ions to form the brown complex $[Fe(H_2O)_5(NO)]SO_4$.
0 likesView Solution
11
ChemistryMediumMCQAIIMS · 1987
Which reagent gives an orange-coloured precipitate with acetaldehyde?
A
$2,4-DNP$
B
$NH_2OH$
C
Iodine
D
$NaHSO_3$
Solution
(A) $2,4-Dinitrophenylhydrazine$ $(2,4-DNP)$,also known as Brady's reagent,reacts with aldehydes and ketones to form $2,4-dinitrophenylhydrazones$.
These derivatives typically appear as orange,yellow,or red crystalline precipitates.
Acetaldehyde $(CH_3CHO)$ reacts with $2,4-DNP$ to form an orange crystalline solid,which is acetaldehyde $2,4-dinitrophenylhydrazone$.
These derivatives typically appear as orange,yellow,or red crystalline precipitates.
Acetaldehyde $(CH_3CHO)$ reacts with $2,4-DNP$ to form an orange crystalline solid,which is acetaldehyde $2,4-dinitrophenylhydrazone$.
0 likesView Solution
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real AIIMS style covering Chemistry with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D Chemistry papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Run live AIIMS mock exams with unlimited students, 360° analytics & white-label branding.
See DemoFrequently Asked Questions
How many Chemistry questions are in AIIMS 1987?
There are 11 Chemistry questions from the AIIMS 1987 paper on Vedclass, each with a detailed step-by-step solution in English.
Are AIIMS 1987 Chemistry solutions available in English?
Yes. All solutions on this page are in English. You can also switch to English or Hindi using the language buttons above the questions.
Can I practice AIIMS 1987 Chemistry as a timed test?
Yes. Use the Vedclass Test Series to attempt a full AIIMS mock test covering Chemistry with time limits and instant score analysis.
Can teachers create Chemistry papers from AIIMS previous year questions?
Yes. The Vedclass Exam Paper Generator lets teachers mix AIIMS Chemistry questions and generate Set A/B/C/D papers in minutes.
For Teachers & Institutes
Build a Custom Chemistry Paper
Pick AIIMS 1987 Chemistry questions, set difficulty, and generate Set A/B/C/D in 2 minutes.