1 Chemical Bonding and Molecular Structure
According to MO theory which of the list ranks the oxygen species in terms of increasing bond order $$O_2^{2 - },\,O_2^ + ,O_2^ - \,,\,O_2^{2 - }$$ 1) $$O_2^{2 - },\,O_2^ - ,O_2^ + \,,\,{O_2}$$ 2) $$O_2^ + ,\,{O_2},O_2^ - \,,\,O_2^{2 - }$$ 3) $${O_2},\,{O_2},O_2^ + \,,\,O_2^{2 - }$$ 4) $$O_2^{2 - },\,O_2^ - ,{O_2}\,,\,O_2^ + $$
SOLUTION
According to MOT bond order is given by the formula:
where NA?, NB? are the number of electrons in antibonding and bonding
Bond order of the following is:
$$O_2^{2 - },\,O_2^ - ,{O_2}\,,\,O_2^ + $$
2 States of Matter
When the temperature of an ideal gas is increased from 270C to 9270C, the kineticenergy will be:
1) same 2) eight times 3) four times 4) twice
T1=273+27=300K T2=273+927=1200K Vrms∝√T (T) has become four times. Therefore, Vrms will become two times.
3 Chemical Bonding and Molecular Structure
The percentage s-character of the hybrid orbitals in methane, ethene and ethyne respectively 1. 10, 20, 40 2. 50, 75, 100 3. 25, 50, 75 4. 25, 33, 50
Solution
% of s-character in
$$\displaystyle{u}{n}{d}{e}{r}{s}{e}{t}{\left({\left({s}{p}^{{{3}}}\right)}\right)}{\left({C}{H}_{{{4}}}\right)}=\frac{100}{{4}}-{25},$$
$$\displaystyle{u}{n}{d}{e}{r}{s}{e}{t}{\left({\left({s}{p}^{{{2}}}\right)}\right)}{\left({C}_{{{2}}}{H}_{{{4}}}\right)}=\frac{100}{{3}}={33}$$
$$\displaystyle{u}{n}{d}{e}{r}{s}{e}{t}{\left({\left({s}{p}\right)}\right)}{\left({C}_{{{2}}}{H}_{{{2}}}\right)}=\frac{100}{{2}}={50}$$
4 Classification of Elements and Periodicity
The correct order of ionisation energy of C,N,O & F is: 1) F<N<C<O 2) C<N<O<F 3) C<O<N<F 4) F<O<N<C
The correct order of ionization energy of C,N,O & F is C<O<N<F. In a period on moving from left to right, the ionization energy increases. This is due to increase in the effective nuclear charge. However the ionization energy of O is lower than the ionization energy of N. This is because in case of N an electron is to be removed from stable half filled 2p sub-shell which requires large energy.
5 Chemical Equilibrium
Starting with 100 mm (Hg) pressure of SO2Cl2 in a closed container, the following equilibrium is set up at some temperature: SO2Cl2(g) ? SO2(g) + Cl2(g). If mole fraction of SO2Cl2 at equilibrium is 1/3, Kp for the reaction is: 1) 60 mm 2) 50 mm 3) 120 mm 4) 25 mm
SO2Cl2(g) ? SO2(s) + Cl2(g) Total pressure = 100 - x + x = 100 + x mm
given, $$\cfrac{100-x}{100+x}\,=\,\cfrac{1}{3}$$ , x = 50 mm
Kp = $$\cfrac{{{P}_{S{{O}_{2}}}}\times {{P}_{C{{l}_{2}}}}}{{{P}_{S{{O}_{2}}C{{l}_{2}}}}}\,=\,\cfrac{x\times x}{\left( 100-x \right)}$$
= 50 mm
6 Chemical Kinetics
The rate constant of a reaction depends on :
1) the temperature of the system 2) the time of reaction 3) the extent of reaction 4) the initial concentration of the reactants
7 Haloalkanes and Haloarenes
The reaction of 1-hexene with HBr in presence of peroxide yields 1-bromo cyclohexane. The mechanism of the reaction involves : 1 attack on alkene by a Br+ ion 2) attack on alkene by a bromide ion 3) attack on alkene by a bromine free radical 4) None of these
The reaction of 1-Hexene with HBr in presence of peroxide yields 1-bromo cyclohexane.
The mechanism of the reaction involves an attack on alkene by a bromine atom. The reaction proceeds through a free radical mechanism.
In the presence of peroxide, bromine free radical is generated. This bromine free radical then attacks alkene.
8 Hydrocarbons
The ease of nitration of the following three hydrocarbons follows the order.
1) II =III =I 2) III > II > I 3) II > III > I 4) I = III > II
−CH3 is a ring activator and it increases electron density to ortho and para positions where nitration can happen.
In case of the given figure i, both the −CH3 groups activate same position. As electron density at the indicated positions, is much more than the other 2 given compounds and so nitration will happen most easily..
This is followed by the given figure ii and iii.
9 Classification of Elements and Periodicity
The correct order of decreasing ionic radius among the following anions would be: 1) Se2−,I−,Br−,F−,O2− 2) F−.Br−,O2−,Se2−,I− 3) I−,Se2−,Br−,O2−,F− 4) F−,O2−,Br−,Se2−,I−
The correct order of decreasing ionic radius among the following anions would be I−,Se2−,Br−,O2−,F−
On moving down the group, the ionic radii increases as new shells of electrons are added. The radius of the isoelectronic species decreases with increasing charge.
10 D and F Block Elements
In which of the following compounds transition metal has zero oxidation state?
1) CrO5 2) NH2⋅NH2 3) NOClO4 4) [Fe(CO)5]
Solution:In [Fe(CO)5] carbon monoxide is a natural ligand, therefore no involvement of Fe electrons in complex, Fehas zero oxidation state.
Your Score:
where NA?, NB? are the number of electrons in antibonding and bonding 