ChemistryQ1–4 of 4 questions
Page 1 of 1 · English
1
ChemistryEasyMCQAIIMS · 1984
If the electron pair forming a bond between two atoms $A$ and $B$ is not in the centre,then the bond is
A
Single bond
B
Polar bond
C
Non-polar bond
D
$\pi$ bond
Solution
(B) When there is a difference in electronegativity between two bonded atoms,the shared electron pair is attracted more towards the more electronegative atom.
As a result,the electron pair does not remain in the centre,leading to the formation of a polar bond.
As a result,the electron pair does not remain in the centre,leading to the formation of a polar bond.
0 likesView Solution
2
ChemistryMediumMCQAIIMS · 1984
The decomposition of $N_2O_4$ to $NO_2$ is carried out at $280 \ K$ in chloroform. When equilibrium has been established,$0.2 \ mol$ of $N_2O_4$ and $2 \times 10^{-3} \ mol$ of $NO_2$ are present in $2 \ L$ solution. The equilibrium constant for the reaction $N_2O_4 \rightleftharpoons 2NO_2$ is:
A
$1 \times 10^{-2}$
B
$2 \times 10^{-3}$
C
$1 \times 10^{-5}$
D
$2 \times 10^{-5}$
Solution
(C) The reaction is $N_2O_4 \rightleftharpoons 2NO_2$.
First,calculate the molar concentrations of the species at equilibrium:
$[N_2O_4] = \frac{0.2 \ mol}{2 \ L} = 0.1 \ M$.
$[NO_2] = \frac{2 \times 10^{-3} \ mol}{2 \ L} = 10^{-3} \ M$.
The equilibrium constant expression is $K_c = \frac{[NO_2]^2}{[N_2O_4]}$.
Substituting the values: $K_c = \frac{(10^{-3})^2}{0.1} = \frac{10^{-6}}{10^{-1}} = 10^{-5}$.
First,calculate the molar concentrations of the species at equilibrium:
$[N_2O_4] = \frac{0.2 \ mol}{2 \ L} = 0.1 \ M$.
$[NO_2] = \frac{2 \times 10^{-3} \ mol}{2 \ L} = 10^{-3} \ M$.
The equilibrium constant expression is $K_c = \frac{[NO_2]^2}{[N_2O_4]}$.
Substituting the values: $K_c = \frac{(10^{-3})^2}{0.1} = \frac{10^{-6}}{10^{-1}} = 10^{-5}$.
0 likesView Solution
3
ChemistryEasyMCQAIIMS · 1984
What is the oxidation number of nickel in $Ni(CO)_4$?
A
$0$
B
$+4$
C
$-4$
D
$+2$
Solution
(A) In the coordination compound $Ni(CO)_4$,the ligand is carbonyl $(CO)$,which is a neutral ligand with an oxidation state of $0$.
Let the oxidation state of $Ni$ be $x$.
The sum of oxidation states of all atoms in a neutral complex is $0$.
$x + 4 \times (0) = 0$
$x = 0$.
Therefore,the oxidation number of nickel in $Ni(CO)_4$ is $0$.
Let the oxidation state of $Ni$ be $x$.
The sum of oxidation states of all atoms in a neutral complex is $0$.
$x + 4 \times (0) = 0$
$x = 0$.
Therefore,the oxidation number of nickel in $Ni(CO)_4$ is $0$.
0 likesView Solution
4
ChemistryDifficultMCQAIIMS · 1984
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.
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 1984?
There are 4 Chemistry questions from the AIIMS 1984 paper on Vedclass, each with a detailed step-by-step solution in English.
Are AIIMS 1984 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 1984 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 1984 Chemistry questions, set difficulty, and generate Set A/B/C/D in 2 minutes.