1 Solid State
Ferrous oxide has a cubic structure with an edge length of the unit cell equal to 5.0 Å. Assuming the density of ferrous oxide to be 3.84 g cm−3, the number of Fe2+ and O2− ions present in each unit cell will be respectively, (use NA=6×1023) 1) 4 and 4 2) 2 and 2 3) 1 and 1 4) 3 and 4
The expression for density $$(d)$$ is as follows: $$d=\dfrac{nM}{VN_A} \implies n=\dfrac{dVN_A}{M} =\dfrac{ 3.84\times (5\times 10^{-8})^3\times 6.023\times 10^{23}}{72} = 4$$
2 Classification of Elements and Periodicity
The first ionization potential of four consecutive elements in the second period of the periodic table are 8.3, 11.3, 14.5 and 13.6 eV respectively. Which one of the following is the first ionization potential of carbon? 1) 13.6 2) 11.3 3) 8.3 4) 14.5
SOLUTION
The first ionization potential of carbon is 11.3 V The four consecutive elements are B,C,N and O. We know that the first ionization potential of N is higher than the first ionization potential of O as in case of N, the electron is to be removed from stable half-filled 2p orbital.
3 Atomic Structure
the position of both, electron and helium atom is known with in 0.1nm. Further the momentum of electron is known with in 5.0×10-26 kg ms-1The minimum uncertainty in the measurement of momentum of helium atom is 1) 50 kgms-1 2) 5.0 ×10-26 kgms-1 3) 80 kg m s-1 4) 80.0 ×10-26 kgms-1
Δxe Δpe=ΔxHe ΔpHe As Δxe=ΔxHe Δpe=ΔpHe
4 P – Block Elements
Ammonia gas dissolves in water to form NH4OH. In this reaction water acts as :
1) A conjugate base 2) A non-polar solvent 3) An acid 4) A base.
Solution
NH3?+H2?O → NH4+? +OH−
Water acts as a proton donor in this reaction. Since, a proton donor is an acid, water acts a Bronsted-Lowry acid.
5 General Principles & Process of Isolation of Metals
Pure anhydrous magnesium chloride can be prepared from hydrated salt by
1) They exist in equilibrium
2) They possess same electronic arrangement but different atomic arrangements
3) They possess same molecular mass
4) They have different electronic as well as atomic arrangements
6 Redox Reactions
What is the oxidation number of H and Cl in HClO4 respectively?
1) +1,+7 2) +2,+6 3) +3,+7 4) +2,+7
The oxidation number of H=+1
Let the oxidation number of Cl be x..
(+1)×1+(+x)×1+(−2)×4=0
⇒x=8−1=+7
∴ Oxidation number of Cl is +7 .
7 Surface Chemistry
Calculate the surface area in (sq m) of a catalyst that adsorbs 103cm3 of nitrogen reduced to STP per gram in order to form the monolayer. The effective area (in m2) occupied by N2? molecule on the surface is: ( surface area of N2?=1.62×10−15cm2)
1) 4115 m2 2) 2355 m2 3) 4355 m2 4) 435.5 m2
According to the given data,
Number of N2? molecules = Volume(ml) × NA / Volume at STP(in ml)
= (103×6.023×1023) / 22400 = 2.69×1022
Total area covered by N2? = 2.69×1022 × 1.62×10−15 = 435×105 cm2 = 4355 m2.
8 Atomic Structure
A certain laser transition emits 6.37×1015 quanta per second per square metre. Calculate the power output in joule per square metre per second. Given λ = 632.8 nm. 1) $$3.14\times 10^{-19}Jm^{-2}$$ $$sec^{-1}$$ 2) $$4\times 10^{-3}Jm^{-2}$$ $$sec^{-1}$$ 3) $$2\times 10^{-3}Jm^{-2}$$ $$sec^{-1}$$ 4) $$4.03\times 10^{9}Jm^{-2}$$ $$sec^{-1}$$
Energy falling per square meter per second=No.of quanta falling per square meter per second\times Energy of one quanta $$=6.37\times 10^{15}\times hc/\lambda$$ $$=\displaystyle\dfrac{6.37\times 10^{15}\times 6.626\times 10^{-34} \times 10^8}{632.8\times 10^{-9}}$$ $$=2\times 10^{-3}Jm^{-2}$$ $$sec^{-1}$$
9 General Principles & Process of Isolation of Metals
Thermal decomposition method is used to purify
1) Ti
2) Ni
3) Zr
4) Cr
10 S – Block Elements
Alkaline earth metals have only positive oxidation state. This is due to their:
1) high melting points 2) high electronegativity 3) softness 4) low ionization potential
The alkaline earth metals have very low ionization energies. So, they easily lose electrons to form positively charged ions (cations).
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