1 D and F Block Elements
The highest magnetic moment is shown by the transition metal with the configuration:
1) 3d2 2) 3d6 3) 3d7 4) 3d9
The transition metal with most unpaired electrons shows the highest magnetic moment.
3d6 -> 4 unpaired electrons
3d2 -> 2 unpaired electrons
3d7 -> 3 unpaired electrons
3d9 -> 1 unpaired electrons
2 Polymers
Which of the following is a linear polymer ?
1) glycogen 2) amylopectin 3) amylose 4) starch.
Cellulose is a natural linear polymer, a long chain made by the linking of smaller glucose molecules.
Natural rubber is a linear polymer made up of Isoprene (2-Methyl -1, 3 – Butadiene) units.
3 Classification of Elements and Periodicity
Within a specified period, an increase in atomic number is usually accompanied by: 1. an increase in atomic radius and an increase in electronegativity. 2. a decrease in atomic radius and an increase in electronegativity. 3. an increase in atomic radius and a decrease in electronegativity. 4. a decrease in atomic radius and a decrease in electronegativity.
Answer:
Because of increasing affecting nuclear charge and electrostatic attraction, the atomic radius decreases. There are more protons and neutrons so electrons are needed to increase electronegativity.
4 Atomic Structure
How many electrons can exist in all the atomic orbitals that correspond to the principal quantum number 4? 1. 2 2. 8 3. 18 4. 32
The four energy level contains electrons in the s, p, d, and f orbitals. All you need to do is take the sum of the number of electrons and all four orbitals (s:2, p:6, d:10, f:14) to reach the correct answer of 32.
5 Chemical Equilibrium
Buffer solution can be obtained by the mixing aqueous solutions of:
1) CH3COONa and CH3COOH 2) CH3COOH and excess NaOH 3) CH33COONa and excess HCl 4) NaCl and HCl.
Solution
A mixture of acetic acid (a weak acid) and sodium acetate ( its salt with strong base sodium hydroxide) acts as acidic buffer.
6 Ionic Equilibrium
The solubility of CaF2 (Ksp=5.3×10−9 ) in 0.1M solution of NaF would be : (Assume no reaction of cation/anion)
1) 5.3×10−10M 2) 5.3×10−8M 3) 5.3×10−7M 4) 5.3×10−11M
Solution:
CaF2 ?Ca2+ + 2F−
Ksp=[Ca2+][F−]2=S(S +0.1)2 = S × 0.12 = 5.3×10−9
Note: S<<0.1so, S+0.1≈0.1
⇒S=5.3×10−7 M
7 Stoichiometry
Oxalic acid (H2C2O4) is present in many plants and vegetables. If 24.0 mL of 0.01M KMnO4 solution is needed to titrate 1.0 g of H2C2O4 to the equivalence-point, the mass percentage (as nearest integer) of H2C2O4 in the sample is : 1) 3 2) 2 3) 5 4) 6
$$2$$ moles of $$KMnO_4$$ $$\equiv$$ $$5$$ moles of $$H_2C_2O_4$$. Molar mass of oxalic acid = $$90$$ g/mol. Number of moles of $$KMnO_4= \dfrac {24.0} {1000} \times 0.01 = 0.00024$$ mol. Number of moles of $$H_2C_2O_4 = \dfrac {5}{2} \times 0.00024= 0.0006$$ mol. Mass of $$H_2C_2O_4 = 0.0006 \times 90=0.054$$ g. Mass percentage of $$H_2C_2O_4 = \dfrac {0.054}{1}\times 100 = 5.4\%$$. The nearest integer is 5.
8 Coordination Compounds
The hardness of water is estimated by
1) distillation method
2) titrimetric method
3) EDTA method
4) conductivity method
9 Hydrogen
In which of the following reactions, H2?O2? acts as reducing agent?
A) H2?O2?? + 2H+ + 2e− → 2H2??O B) H2??O2?? − 2e− → O2? ?+ 2H+ C) H2??O2? ?+ 2e− → 2OH− D) H2?O2? ??+ 2OH−2e− →O2? ?+ 2H2??O
1) A, C 2) B, D 3) A, B 4) C, D
SOLUTION
Reducing agents itself undergo oxidation (loss of electrons); hence in following two reactions, H2?O2?? acts as reducing agent.
B) H2??O2?? − 2e− → O2? ?+ 2H+ D) H2?O2? ??+ 2OH−2e− → O2? ?+ 2H2??O
10 Chemical Kinetics
The rate of a reaction quadruples when the temperature changes from 300 to 310K. What is the activation energy of this reaction? (Assume activation energy and preexponential factor are independent of temperature ; In 2= 0.693; R= 8.314 Jmol−1K−1) 1) 0.60 ×105Jmol−1 2) 2.46×105Jmol−1 3) .246×105Jmol−1 4) 4.00 ×105Jmol−1
$$\log \left(\cfrac {k_2}{k_1}\right)=\cfrac {E_A}{2.303R} \left[\cfrac {1}{T_1}-\cfrac {1}{T_2}\right]$$
Your Score:
