Some Basic Concepts of Chemistry 1

$IBQUFS
4PNF
#BTJD
$PODFQUT
! PG
$IFNJTUSZ
1. Suppose the elements X and Y combine to (a) 8 : 16 : 1 (b) 16 : 8 : 1
form two compounds XY 2 and X 3 Y 2 . When (c) 16 : 1 : 2 (d) 8 : 1 : 2 (2014)
0.1 mole of XY2 weighs 10 g and 0.05 mole of 7. When 22.4 litres of H2(g) is mixed with 11.2
X3Y2 weighs 9 g, the atomic weights of X and
litres of Cl2(g), each at S.T.P, the moles of HCl(g)
Y are
(a) 40, 30 (b) 60, 40 formed is equal to
(c) 20, 30 (d) 30, 20 (a) 1 mol of HCl(g) (b) 2 mol of HCl(g)
(NEET-II 2016) (c) 0.5 mol of HCl(g) (d) 1.5 mol of HCl(g).
2. What is the mass of the precipitate formed (2014)
when 50 mL of 16.9% solution of AgNO3 is 8. 1.0 g of magnesium is burnt with 0.56 g O2 in
mixed with 50 mL of 5.8% NaCl solution? a closed vessel. Which reactant is left in excess
(Ag = 107.8, N = 14, O = 16, Na = 23, and how much? (At. wt. Mg = 24, O = 16)
Cl = 35.5) (a) Mg, 0.16 g (b) O2, 0.16 g
(a) 3.5 g (b) 7 g (c) 14 g (d) 28 g (c) Mg, 0.44 g (d) O2, 0.28 g
(2015) (2014)
3. If Avogadro number N A , is changed from 9. 6.02 × 1020 molecules of urea are present in
6.022 × 1023 mol–1 to 6.022 × 10 20 mol–1, this 100 mL of its solution. The concentration of
would change solution is
(a) the mass of one mole of carbon (a) 0.001 M (b) 0.1 M
(b) the ratio of chemical species to each other (c) 0.02 M (d) 0.01 M
in a balanced equation (NEET 2013)
(c) the ratio of elements to each other in a
compound 10. In an experiment it showed that 10 mL of 0.05 M
(d) the definition of mass in units of grams. solution of chloride required 10 mL of 0.1 M
(2015) solution of AgNO 3, which of the following
will be the formula of the chloride (X stands
4. The number of water molecules is maximum in
for the symbol of the element other than
(a) 1.8 gram of water
(b) 18 gram of water chlorine)
(c) 18 moles of water (a) X2Cl2 (b) XCl2 (c) XCl4 (d) X2Cl
(d) 18 molecules of water. (2015) (Karnataka NEET 2013)

5. A mixture of gases contains H2 and O2 gases 11. Which has the maximum number of molecules
in the ratio of 1 : 4 (w/w). What is the molar among the following?
ratio of the two gases in the mixture? (a) 44 g CO2 (b) 48 g O3
(a) 16 : 1 (b) 2 : 1 (c) 1 : 4 (d) 4 : 1 (c) 8 g H2 (d) 64 g SO2
(2015, Cancelled) (Mains 2011)
6. Equal masses of H 2, O 2 and methane have 12. The number of atoms in 0.1 mol of a triatomic
been taken in a container of volume V at gas is (NA = 6.02 × 10 23 mol–1)
temperature 27 °C in identical conditions. The (a) 6.026 × 1022 (b) 1.806 × 1023
23
ratio of the volumes of gases H2 : O2 : methane (c) 3.600 × 10 (d) 1.800 × 1022
would be (2010)
2

13. 25.3 g of sodium carbonate, Na 2 CO 3 is 21. Percentage of Se in peroxidase anhydrous

dissolved in enough water to make 250 mL of enzyme is 0.5% by weight (at. wt. = 78.4) then
solution. If sodium carbonate dissociates minimum molecular weight of peroxidase
completely, molar concentration of sodium ion, anhydrous enzyme is
Na+ and carbonate ions, CO 32– are respectively
(a) 1.568 × 104 (b) 1.568 × 103
(Molar mass of Na2CO3 = 106 g mol–1)
(c) 15.68 (d) 2.136 × 104
(a) 0.955 M and 1.910 M
(2001)
(b) 1.910 M and 0.955 M
22. Molarity of liquid HCl, if density of solution is
(c) 1.90 M and 1.910 M
1.17 g/cc is
(d) 0.477 M and 0.477 M (2010)
(a) 36.5 (b) 18.25
14. 10 g of hydrogen and 64 g of oxygen were
(c) 32.05 (d) 42.10 (2001)
filled in a steel vessel and exploded. Amount
of water produced in this reaction will be 23. Specific volume of cylindrical virus particle is
(a) 3 mol (b) 4 mol 6.02 × 10–2 cc/g whose radius and length are
(c) 1 mol (d) 2 mol (2009) 7 Å and 10 Å respectively. If NA = 6.02 × 1023,
15. What volume of oxygen gas (O 2) measured find molecular weight of virus.
at 0°C and 1 atm, is needed to burn completely (a) 15.4 kg/mol (b) 1.54 × 104 kg/mol
1 L of propane gas (C 3H 8) measured under (c) 3.08 × 10 kg/mol (d) 3.08 × 103 kg/mol
4

the same conditions? (2001)

(a) 5 L (b) 10 L (c) 7 L (d) 6 L 24. In quantitative analysis of second group in
(2008) laboratory, H2S gas is passed in acidic medium
16. How many moles of lead (II) chloride will be for precipitation. When Cu2+ and Cd2+ react
formed from a reaction between 6.5 g of PbO with KCN, then for product, true statement is
and 3.2 g HCl? (a) K2[Cu(CN)4] more soluble
(a) 0.011 (b) 0.029 (c) 0.044 (d) 0.333 (b) K2[Cd(CN)4] less stable
(2008) (c) K3[Cu(CN)2] less stable
17. An organic compound contains carbon, (d) K2[Cd(CN)3] more stable. (2000)
hydrogen and oxygen. Its elemental analysis 25. Volume of CO 2 obtained by the complete
gave C, 38.71% and H, 9.67%. The empirical decomposition of 9.85 g of BaCO3 is
formula of the compound would be (a) 2.24 L (b) 1.12 L
(a) CHO (b) CH4O (c) 0.84 L (d) 0.56 L (2000)
(c) CH3O (d) CH2O (2008)
26. Oxidation numbers of A, B, C are +2, +5 and
18. An element, X has the following isotopic –2 respectively. Possible formula of compound is
composition:
(a) A2(BC2) 2 (b) A3(BC 4) 2
200 199 202
X : 90% X : 8.0% X : 2.0% (c) A2(BC3) 2 (d) A3(B 2C) 2
The weighted average atomic mass of the (2000)
naturally occurring element X is closest to
27. The number of atoms in 4.25 g of NH 3 is
(a) 201 amu (b) 202 amu approximately
(c) 199 amu (d) 200 amu (2007) (a) 4 × 1023 (b) 2 × 2023
19. The maximum number of molecules is present in (c) 1 × 10 23
(d) 6 × 1023 (1999)
(a) 15 L of H2 gas at STP
28. Given the numbers: 161 cm, 0.161 cm, 0.0161 cm.
(b) 5 L of N2 gas at STP
The number of significant figures for the three
(c) 0.5 g of H2 gas numbers is
(d) 10 g of O2 gas. (2004) (a) 3, 3 and 4 respectively
20. Which has maximum molecules? (b) 3, 4 and 4 respectively
(a) 7 g N2 (b) 2 g H2 (c) 3, 4 and 5 respectively
(c) 16 g NO2 (d) 16 g O2 (2002) (d) 3, 3 and 3 respectively. (1998)
Some Basic Concepts of Chemistry 3

29. Haemoglobin contains 0.334% of iron by 37. A 5 molar solution of H2SO4 is diluted from 1
weight. The molecular weight of haemoglobin litre to a volume of 10 litres, the normality of
is approximately 67200. The number of iron the solution will be
atoms (Atomic weight of Fe is 56) present in (a) 1 N (b) 0.1 N
one molecule of haemoglobin is (c) 5 N (d) 0.5 N (1991)
(a) 4 (b) 6 (c) 3 (d) 2
38. The number of gram molecules of oxygen in
(1998)
6.02 × 1024 CO molecules is
30. In the reaction,
(a) 10 g molecules (b) 5 g molecules
4NH3(g) + 5O2(g) ® 4NO(g) + 6H2O(l)
(c) 1 g molecules (d) 0.5 g molecules.
when 1 mole of ammonia and 1 mole of O2 are
(1990)
made to react to completion :
(a) All the oxygen will be consumed. 39. Boron has two stable isotopes, 10B(19%) and
11
(b) 1.0 mole of NO will be produced. B(81%). Calculate average at. wt. of boron in
(c) 1.0 mole of H2O is produced. the periodic table
(d) All the ammonia will be consumed. (a) 10.8 (b) 10.2
(1998) (c) 11.2 (d) 10.0 (1990)
31. Among the following which one is not 40. The molecular weight of O2 and SO2 are 32 and
paramagnetic? [Atomic numbers; Be = 4, 64 respectively. At 15°C and 150 mmHg
Ne = 10, As = 33, Cl = 17] pressure, one litre of O2 contains ‘N’ molecules.
(a) Ne2+ (b) Be+ (c) Cl– (d) As + The number of molecules in two litres of SO2
(1998)
under the same conditions of temperature and
32. 0.24 g of a volatile gas, upon vaporisation, gives pressure will be
45 mL vapour at NTP. What will be the vapour (a) N/2 (b) N
density of the substance? (Density of H2 = 0.089) (c) 2 N (d) 4 N (1990)
(a) 95.93 (b) 59.93 (c) 95.39 (d) 5.993
41. A metal oxide has the formula Z2O3. It can be
(1996)
reduced by hydrogen to give free metal and
33. The amount of zinc required to produce 224 mL water. 0.1596 g of the metal oxide requires 6 mg
of H2 at STP on treatment with dilute H2SO4 of hydrogen for complete reduction. The atomic
will be weight of the metal is
(a) 65 g (b) 0.065 g (c) 0.65 g (d) 6.5 g
(a) 27.9 (b) 159.6
(1996)
(c) 79.8 (d) 55.8 (1989)
34. The dimensions of pressure are the same as
that of 42. Ratio of Cp and CV of a gas ‘X’ is 1.4. The
(a) force per unit volume number of atoms of the gas ‘X’ present in 11.2
(b) energy per unit volume litres of it at NTP will be
(c) force (a) 6.02 × 1023 (b) 1.2 × 1023
23
(d) energy. (1995) (c) 3.01 × 10 (d) 2.01 × 1023
(1989)
35. The number of moles of oxygen in one litre of
air containing 21% oxygen by volume, under 43. What is the weight of oxygen required for the
standard conditions, is complete combustion of 2.8 kg of ethylene?
(a) 0.0093 mol (b) 2.10 mol (a) 2.8 kg (b) 6.4 kg (c) 9.6 kg (d) 96 kg
(c) 0.186 mol (d) 0.21 mol. (1989)
(1995)
44. The number of oxygen atoms in 4.4 g of CO2
36. The total number of valence electrons in 4.2 g is
of N3– ion is (NA is the Avogadro’s number) (a) 1.2 × 1023 (b) 6 × 1022
(a) 2.1 NA (b) 4.2 NA (c) 6 × 10 23
(d) 12 × 1023
(c) 1.6 NA (d) 3.2 NA (1994) (1989)
4

45. At S.T.P. the density of CCl 4 vapour in g/L 47. 1 cc N 2O at NTP contains
will be nearest to
1.8
(a) 6.87 (b) 3.42 (c) 10.26 (d) 4.57 (a) u 1022 atoms
(1988) 224
6.02
46. One litre hard water contains 12.00 mg Mg2+. (b) u 1023 molecules
Milli-equivalents of washing soda required to 22400
remove its hardness is 1.32
(a) 1 (b) 12.16 (c) u 1023 electrons
224
(c) 1 × 10–3 (d) 12.16 × 10–3
(1988) (d) All the above. (1988)

Answer Key

1. (a) 2. (b) 3. (a) 4. (c) 5. (d) 6. (c) 7. (a) 8. (a) 9. (d) 10. (b)
11. (c) 12. (b) 13. (b) 14. (b) 15. (a) 16. (b) 17. (c) 18. (d) 19. (a) 20. (b)
21. (a) 22. (c) 23. (a) 24. (c) 25. (b) 26. (b) 27. (d) 28. (d) 29. (a) 30. (a)
31. (c) 32. (b) 33. (c) 34. (b) 35. (a) 36. (c) 37. (a) 38. (b) 39. (a) 40. (c)
41. (d) 42. (a) 43. (c) 44. (a) 45. (a) 46. (a) 47. (d)
Some Basic Concepts of Chemistry 5

1. (a) : Let atomic weight of element X is x and Mass

that of element Y is y. So, no. of moles =
Mol. mass
O X w w w
For XY2,
.PM XU nH 2 ; n ; n
2 O 2 32 CH 4 16
  w w w
 ⇒ x + 2y =  ...(i) So, the ratio is : : or 16 :1: 2.
Y  Z  2 32 16

O X 7. (a) : 1 mole ≡ 22.4 litres at S.T.P.

For X3Y2,
.PM XU nH 2
22.4
1 mol ; nCl 2
11.2
0.5 mol
  22.4 22.4
 ⇒ Y   Z  ...(ii)
Y   Z  Reaction is as,
On solving equations (i) and (ii), we get y = 30 H2(g) + Cl2(g) 2HCl(g)
x + 2(30) = 100 ⇒ x = 100 – 60 = 40 Initial 1 mol 0.5 mol 0
Final (1 – 0.5) (0.5 – 0.5) 2 × 0.5
2. (b) : 16.9% solution of AgNO3 means 16.9 g of = 0.5 mol = 0 mol 1 mol
AgNO3 in 100 mL of solution. Here, Cl2 is limiting reagent. So, 1 mole of HCl(g) is
16.9 g of AgNO3 in 100 mL solution ≡ 8.45 g of formed.
AgNO3 in 50 mL solution.
1
Similarly, 5.8% of NaCl in 100 mL solution 8. (a) : nMg = 0.0416 moles
º 2.9 g of NaCl in 50 mL solution. 24
The reaction can be represented as : 0.56
nO 2 = 0.0175 moles
AgNO3 + NaCl AgCl + NaNO3 32
Initial 8.45/170 2.9/58.5 0 0 The balanced equation is
mole = 0.049 = 0.049 1O
Final moles 0 0 0.049 0.049
Mg + 2 2
MgO
\ Mass of AgCl precipitated = 0.049 × 143.3 Initial 0.0416 moles 0.0175 moles 0
= 7.02 » 7 g Final (0.0416 – 2 × 0.0175) 0 2 × 0.0175
= 0.0066 moles (O 2 is limiting reagent.)
3. (a) : Mass of 1 mol (6.022 × 1023 atoms) of carbon
= 12 g ∴ Mass of Mg left in excess = 0.0066 × 24 = 0.16 g
If Avogadro number is changed to 6.022 × 1020 atoms ×
then mass of 1 mol of carbon 9. (d) : Moles of urea = = 0.001
×
12 u 6.022 u 10 20
12 u 10 3 g Concentration of solution = × 1000 = 0.01 M
6.022 u 10 23
10. (b) : Millimoles of solution of chloride
6.023 u 10 23 = 0.05 × 10 = 0.5
4. (c) : 1.8 gram of water = u 1.8
18 Millimoles of AgNO3 solution = 10 × 0.1 = 1
= 6.023 × 1022 molecules So, the millimoles of AgNO3 are double than the
18 gram of water = 6.023 × 1023 molecules chloride solution.
18 moles of water = 18 × 6.023 × 1023 molecules ∴ XCl2 + 2AgNO3 → 2AgCl + X(NO3)2
11. (c) : 8 g H2 has 4 moles while the others has
1
5. (d) : Number of moles of H2 = 1 mole each.
2
4 12. (b) : No. of atoms = NA × No. of moles × 3
Number of moles of O2 = = 6.023 × 1023 × 0.1 × 3 = 1.806 × 1023
32
1 4 13. (b) : Given that molar mass of Na2CO3 = 106 g
Hence, molar ratio = : =4:1 25.3 × 1000
2 32 ∴ Molarity of solution =
106 × 250
6. (c) : According to Avogadro’s hypothesis,
= 0.9547 M = 0.955 M
ratio of the volumes of gases will be equal to the
ratio of their no. of moles. Na2CO3 → 2Na+ + CO2– 3
6

[Na+] = 2[Na2CO3] = 2 × 0.955 = 1.910 M 2 g H2 = 6.023 × 1023

[CO2–
3 ] = [Na2CO3] = 0.955 M
23
0.5 g H2 = 6.023 × 10 × 0.5 = 1.505 × 10 23
14. (b) : H2 + 1/2O2 → H2O 2
2g 16 g 18 g 32 g O2 = 6.023 × 1023
1 mol 0.5 1 mol
10 g of H2 = 5 mol and 64 g of O2 = 2 mol 6.023 × 10 23 × 10
10 g of O2 = = 1.882 × 10 23
∴ In this reaction, oxygen is the limiting reagent so 32
amount of H2O produced depends on that of O2. 20. (b) : 1 mole of any element contain 6.023 × 1023
Since 0.5 mol of O2 gives 1 mol H2O number of molecules.
∴ 2 mol of O2 will give 4 mol H2O 1 g mole of O2 = 32 g O2
15. (a) : ⇒ 16 g of O2 = 0.5 g mole O2
1 g mole of N2 = 28 g N2
⇒ 7 g N2 = 0.25 g mole N2
According to the above equation
1 g mole of H2 = 2 g H2
1 vol. or 1 litre of propane requires to 5 vol.
⇒ 2 g H2 = 1.0 g mole H2
or 5 litre of O2 to burn completely. 1 g mole NO2 = 14 + 16 × 2 = 46
16. (b) : PbO + 2HCl → PbCl2 + H2O ⇒ 16 g of NO2 = 0.35 mole NO2
n mol 2n mol n mol 2 g H2 (1 g mole H2) contain maximum molecules.
6.5 3.2 21. (a) : In peroxidase anhydrous enzyme 0.5% Se
mol mol
224 36.5 is present means, 0.5 g Se is present in 100 g of
0.029 mol 0.087 mol enzyme. In a molecule of enzyme one Se atom must
Formation of moles of lead (II) chloride depends be present. Hence 78.4 g Se will be present in
upon the no. of moles of PbO which acts as a limiting 100
factor here. So, no. of moles of PbCl2 formed will be × 78.4 = 1.568 × 104
0.5
equal to the no. of moles of PbO i.e. 0.029.
22. (c) : Density = 1.17 g/cc.
17. (c) : ⇒ 1 cc. solution contains 1.17 g of HCl
Element % Atomic mole simple 1.17 × 1000
∴ Molarity = = 32.05
mass ratio ratio 36.5 × 1
38.71 3.22 23. (a) : Specific volume (vol. of 1 g) cylindrical
C 38.71 12 = 3.22 =1 virus particle = 6.02 × 10–2 cc/g
12 3.22 Radius of virus, r = 7 Å = 7 × 10–8 cm
9.67 9.67 Volume of virus = πr2l
H 9.67 1 = 9.67 =3 22
1 3.22 = × (7 × 10 −8 ) 2 × 10 × 10 −8 = 154 × 10–23 cc
7
51.62 3.22 Volume
O 51.62 16 = 3.22 =1 wt. of one virus particle = Specific volume
16 3.22
Hence empirical formula of the compound would be 154 × 10 −23
⇒ g
CH3O. 6 .0 2 × 1 0 − 2
∴ Molecular wt. of virus = wt. of NA particle
18. (d) : Average isotopic mass of X
154 × 10 −23
200 × 90 + 199 × 8 + 202 × 2 = −2
× 6.02 × 10 −23 g/mol.
= 6.02 × 10
90 + 8 + 2 = 15400 g/mol = 15.4 kg/mol
18000 + 1592 + 404 24. (c) : K3[Cu(CN)2] = 3(+1) + x + 2(–1) = 0
= = 199.96 a.m.u. ≈ 200 a.m.u. ⇒ x = –1
100
As the oxidation no. of ‘Cu’ is –1 (–ve), so this
19. (a) : At STP, 22.4 L H2 = 6.023 × 1023 molecules complex is unstable and is not formed.
6.023 × 10 23 × 15 25. (b) : BaCO3 → BaO + CO2
15 L H2 = = 4.033 × 10 23
22.4 197 ⋅ 34 g → 22 ⋅ 4 L at N.T.P.
6.023 × 10 23 × 5 22 ⋅ 4
5 L N2 = = 1.344 × 10 23 9 ⋅ 85 g → × 9 ⋅ 85 = 1 ⋅ 118 L
22.4 197 ⋅ 34
Some Basic Concepts of Chemistry 7

⇒ 9 ⋅ 85 g BaCO3 will produce 1.118 L CO2 at N.T.P. 65

= × 224 = 0.65 g
on the complete decomposition. 22400
Force
26. (b) : In A3(BC4)2, (+2) × 3 + 2[+5 + 4(-2)] 34. (b) : Pressure =
⇒ + 6 + 10 − 16 = 0 Area
MLT −2
Hence in the compound A3(BC4)2, the oxidation no. Therefore dimensions of pressure =
of ‘A’, ‘B’ and ‘C’ are +2, +5 and −2 respectively. L2
= ML−1T–2
27. (d) : No. of molecules in 4.25 g NH 3 and dimensions of energy per unit volume
4.25 Energy ML2T −2
= × 6.023 × 1023 = 2.5 × 6.023 × 1022 = = = ML−1T−2
17 Volume L3
Number of atoms in 4.25 g NH3
35. (a) : Volume of oxygen in one litre of air
= 4 × 2.5 × 6.023 × 10 22 = 6.023 × 1023 21
= × 1000 = 210 mL
28. (d) : Zeros placed left to the number are never 100
significant, therefore the no. of significant figures 210
Therefore no. of mol = = 0.0093 mol
for the numbers. 22400
161 cm = 0.161 cm and 0.0161 cm are same, i.e. 3 36. (c) : Each nitrogen atom has 5 valence
29. (a) : Quantity of iron in one molecule electrons, therefore total number of electrons in N3−
67200 ion is 16. Since the molecular mass of N3 is 42,
= × 0.334 = 224.45 amu therefore total number of electrons in 4.2 g of N3− ion
100
No. of iron atoms in one molecule of haemoglobin 4.2
224.45 = × 16 × N A = 1.6 NA
= 56 = 4 42
37. (a) : 5M H2SO4 = 10N H2SO4
30. (a) : 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l) N1V1 = N2V2 Þ 10 × 1 = N2 × 10 Þ N2 = 1N
4 mole + 5 mole → 4 mole + 6 mole 38. (b) : Avogadro’s No., NA = 6.02 × 1023 molecules.
5 \ 6.02 × 1024 CO molecules = 10 moles CO
⇒ 1 mole of NH3 requires = = 1.25 mole of oxygen
4 = 10 g atoms of O = 5 g molecules of O2
4
while 1 mole of O2 requires = = 0.8 mole of NH3. 39. (a) : Average atomic mass
5
As there is 1 mole of NH3 and 1 mole of O2, so all 19 u 10  81 u 11
oxygen will be consumed. 10.81
100
31. (c) : Ne2+(8) ⇒ 1s22s2 2 p x2 2 p1y 2 p1z 40. (c) : If 1L of one gas contains N molecules,
2 L of any gas under the same conditions will contain
Be+(3) ⇒ 1s22s1
2 N molecules.
Cl–(18) ⇒ 1s22s22p63s23p6
As + (32) ⇒ 1s 2 2 s 2 2 p 6 3s 2 3 p 6 4 s 2 3d 10 4 p1x 4 p1y 41. (d) : Z2O3 + 3H2 ® 2Z + 3H2O
Cl– is not paramagnetic , as it has no unpaired electron. Valency of metal in Z2O3 = 3
0.1596 g of Z2O3 react with 6 mg of H2.
32. (b) : Weight of gas = 0.24 g, Volume of gas = [1 mg = 0.001 g = 10–3g]
45 mL = 0.045 litre and density of H2 = 0.089.
We know that weight of 45 mL of H2 = 0.1596
Density × Volume = 0.089 × 0.045 = 4.005 × 10−3 g \ 1 g of H2 react with = 26.6 g of Z2O3
0.006
Therefore vapour density \ Eq. wt. of Z2O3 = 26.6
Weight of certain volume of substance Now, Eq. wt. of Z + Eq. wt. of O = Eq. wt. of Z + 8 = 26.6
=
Weight of same volume of hydrogen Þ Eq. wt. of Z = 26.6 – 8 = 18.6
0.24 \ At. wt. of Z = 18.6 × 3 = 55.8
= = 59.93
4.005 × 10−3 ª Atomic wt. º
33. (c) : Zn + H2SO4 → ZnSO4 + H2 « Eq. wt = Valency of metal »
¬ ¼
(65 g) (22400 mL)
Since 65 g of zinc reacts to liberate 22400 mL of H2 42. (a) : Here, Cp/CV = 1.4, which shows that the
at STP, therefore amount of zinc needed to produce gas is diatomic.
224 mL of H2 at STP 22.4 L at NTP = 6.02 × 1023 molecules
8

\ 11.2 L at NTP = 3.01 × 1023 molecules 46. (a) : Mg2+ + Na2CO3 → MgCO3 + 2Na+
Since gas is diatomic. 1g eq. 1g eq.
\ 11.2 L at NTP = 6.02 × 1023 atom 1g eq. of Mg2+ = 12g of Mg2+ = 12000 mg
43. (c) : C2H4 + 3O2 ® 2CO2 + 2H2O Now, 1000 millieq. of Na2CO3 = 12000 mg of Mg2+
28 g 96 g \ 1 millieq. of Na2CO3 = 12 mg of Mg2+
47. (d) : As we know,
96 g
2.8 kg C2H4 = 28 g u 2.8 kg 22400 cc of N2O contain 6.02 × 1023 molecules
6.02 u 1023
96 \ 1 cc of N2O contain molecules
u 2.8 u 103 g = 9.6 × 103 g = 9.6 kg 22400
28
Since in N2O molecule there are 3 atoms
44. (a) : 1 mol of CO2 = 44 g of CO2
3 u 6.02 u 1023
\ 4·4 g CO2 = 0.1 mol CO2 = 6 × 1022 molecules \ 1 cc N2O atoms
22400
[Since, 1 mole CO2 = 6 × 1023 molecules] 1.8 u 10 22
atoms
= 2 × 6 × 1022 atoms of O = 1.2 × 1023 atoms of O 224
No. of electrons in a molecule of N2O = 7 + 7 + 8 = 22
45. (a) : 1 mol CCl4 vapour = 12 + 4 × 35.5
= 154 g = 22.4 L 6.02 u 1023
Hence, no. of electrons u 22 electrons
154 22400
\ Density of CCl4 vapour g L–1
22.4 1.32
–1 = u 1023 electrons
= 6.875 g L 224
Structure of Atom 9

$IBQUFS
" 4USVDUVSF
PG
"UPN
1. Which one is the wrong statement? 6. The number of d-electrons in Fe2+ (Z = 26) is
(a) The uncertainty principle is not equal to the number of electrons in which
I one of the following?
Δ& × ΔU ≥
π (a) d-electrons in Fe (Z = 26)
(b) Half filled and fully filled orbitals have (b) p-electrons in Ne (Z = 10)
greater stability due to greater exchange (c) s-electrons in Mg (Z = 12)
energ y, greater symmetry and more (d) p-electrons in Cl (Z = 17)
balanced arrangement. (2015, Cancelled)
(c) The energy of 2s-orbital is less than the
energy of 2p-orbital in case of hydrogen 7. The angular momentum of electron in ‘d’ orbital
like atoms. is equal to
(d) de-Broglie’s wavelength is given by (a) 2 3= (b) 0 =
I
λ= , where m = mass of the particle, (c) 6= (d) 2=
NW
v = group velocity of the particle. (2015, Cancelled)
(NEET 2017) 8. What is the maximum number of orbitals that
2. How many electrons can fit in the orbital for can be identified with the following quantum
which n = 3 and l = 1? numbers?
(a) 2 (b) 6 (c) 10 (d) 14 n = 3, l = 1, ml = 0
(NEET-II 2016) (a) 1 (b) 2 (c) 3 (d) 4
(2014)
3. Which of the following pairs of d-orbitals will
have electron density along the axes? 9. Calculate the energy in joule corresponding
(a) dz2, dxz (b) dxz, dyz to light of wavelength 45 nm. (Planck’s
(c) dz2, dx2 – y2 (d) dxy, dx2 – y2 constant, h = 6.63 × 10–34 J s, speed of light,
(NEET-II 2016) c = 3 × 108 m s–1)
(a) 6.67 × 1015 (b) 6.67 × 1011
4. Two electrons occupying the same orbital are –15
(c) 4.42 × 10 (d) 4.42 × 10–18
distinguished by
(2014)
(a) azimuthal quantum number
(b) spin quantum number 10. Be2+ is isoelectronic with which of the following
(c) principal quantum number ions?
(d) magnetic quantum number. (a) H+ (b) Li+ (c) Na+ (d) Mg2+
(NEET-I 2016) (2014)
5. Which is the correct order of increasing energy 11. What is the maximum numbers of electrons
of the listed orbitals in the atom of titanium? that can be associated with the following set
(At. no. Z = 22) of quantum numbers?
(a) 4s 3s 3p 3d (b) 3s 3p 3d 4s n = 3, l = 1 and m = –1
(c) 3s 3p 4s 3d (d) 3s 4s 3p 3d (a) 4 (b) 2 (c) 10 (d) 6
(2015) (NEET 2013)
10

18. The orbital angular momentum of a p-electron

⎛Z ⎞
12. Based on equation E = – 2.178 × 10–18 J ⎜ ⎟ , is given as
⎝n ⎠
h h h h
certain conclusions are written. Which of them (a)
is not correct? S (b) S
(c)
S
(d)
S
(a) Equation can be used to calculate the (Mains 2012)
change in energy when the electron
changes orbit. 19. The total number of atomic orbitals in fourth
(b) For n = 1, the electron has a more negative energy level of an atom is
energy than it does for n = 6 which means (a) 8 (b) 16 (c) 32 (d) 4
that the electron is more loosely bound in (2011)
the smallest allowed orbit.
(c) The negative sign in equation simply 20. The energies E1 and E2 of two radiations are
means that the energy of electron bound 25 eV and 50 eV respectively. The relation
to the nucleus is lower than it would be if between their wavelengths i.e., λ1 and λ2 will
the electrons were at the infinite distance be
from the nucleus. (a) λ1 = λ2 (b) λ1 = 2λ2
(d) Larger the value of n, the larger is the
orbit radius. (NEET 2013) (c) λ1 = 4λ2 (d) O O (2011)
–34
13. The value of Planck’s constant is 6.63 × 10 J s.
The speed of light is 3 × 1017 nm s–1. Which 21. If n = 6, the correct sequence for filling of
value is closest to the wavelength in nanometer electrons will be
of a quantum of light with frequency of (a) ns → (n – 2)f → (n – 1)d → np
6 × 1015 s–1? (b) ns → (n – 1)d → (n – 2)f → np
(a) 50 (b) 75 (c) 10 (d) 25 (c) ns → (n – 2)f → np → (n – 1)d
(NEET 2013) (d) ns → np(n – 1)d → (n – 2)f (2011)
14. The outer electronic configuration of Gd 22. According to the Bohr theory, which of the
(At. No. 64) is following transitions in the hydrogen atom
(a) 4f 5 5d4 6s1 (b) 4f 7 5d1 6s2 will give rise to the least energetic photon?
3 5 2
(c) 4f 5d 6s (d) 4f 4 5d5 6s1 (a) n = 6 to n = 1 (b) n = 5 to n = 4
(Karnataka NEET 2013) (c) n = 6 to n = 5 (d) n = 5 to n = 3
15. According to law of photochemical equivalence (Mains 2011)
the energy absorbed (in ergs/mole) is given 23. A 0.66 kg ball is moving with a speed of 100 m/s.
as (h = 6.62 × 10–27 ergs, c = 3 × 1010 cm s–1, The associated wavelength will be
NA = 6.02 × 10–23 mol–1) (h = 6.6 × 10–34 J s)
u u (a) 6.6 × 10–32 m (b) 6.6 × 10–34 m
(a) (b) –35
(c) 1.0 × 10 m (d) 1.0 × 10–32 m
O O
(Mains 2010)
u u
(c) (d) 24. Maximum number of electrons in a subshell
O O
(Karnataka NEET 2013) of an atom is determined by the following
(a) 2l + 1 (b) 4l – 2
16. Maximum number of electrons in a subshell (c) 2n2 (d) 4l + 2 (2009)
with l = 3 and n = 4 is
(a) 14 (b) 16 (c) 10 (d) 12 25. Which of the following is not permissible
(2012) arrangement of electrons in an atom?
(a) n = 5, l = 3, m = 0, s = +1/2
17. The correct set of four quantum numbers for (b) n = 3, l = 2, m = –3, s = –1/2
the valence electron of rubidium atom (c) n = 3, l = 2, m = –2, s = –1/2
(Z = 37) is (d) n = 4, l = 0, m = 0, s = –1/2 (2009)
(a) 5, 1, 1, +1/2 (b) 6, 0, 0, +1/2
(c) 5, 0, 0, +1/2 (d) 5, 1, 0, +1/2 26. If uncertainty in position and momentum are
(2012) equal, then uncertainty in velocity is
Structure of Atom 11

1 h h 32. The frequency of radiation emitted when the

(a) (b) electron falls from n = 4 to n = 1 in a hydrogen
m π π
atom will be (Given ionization energy of
1 h h H = 2.18 × 10–18 J atom–1 and h = 6.625 × 10–34 J s)
(c) (d) (2008)
2m π 2π (a) 1.54 × 1015 s–1 (b) 1.03 × 1015 s–1
15 –1
27. The measurement of the electron position is (c) 3.08 × 10 s (d) 2.00 × 1015 s–1
associated with an uncertainty in momentum, (2004)
which is equal to 1 × 10 –18 g cm s –1 . The 33. The value of Planck’s constant is 6.63 × 10–34 J s.
uncertainty in electron velocity is (mass of The velocity of light is 3.0 × 108 m s–1. Which
an electron is 9 × 10–28 g) value is closest to the wavelength in
(a) 1 × 105 cm s–1 (b) 1 × 1011 cm s–1 nanometers of a quantum of light with
9 –1
(c) 1 × 10 cm s (d) 1 × 106 cm s–1 frequency of 8 × 1015 s –1 ?
(Prelims 2008) (a) 2 × 10–25 (b) 5 × 10–18
1
28. Consider the following sets of quantum (c) 4 × 10 (d) 3 × 107 (2003)
numbers: 34. In hydrogen atom, energy of first excited state
n l m s is –3.4 eV. Then find out K.E. of same orbit
(i) 3 0 0 +1/2 of hydrogen atom
(ii) 2 2 1 +1/2 (a) +3.4 eV (b) +6.8 eV
(iii) 4 3 –2 –1/2 (c) –13.6 eV (d) +13.6 eV (2002)
(iv) 1 0 –1 –1/2
(v) 3 2 3 +1/2 35. Main axis of a diatomic molecule is z, molecular
orbital px and p y overlap to form which of the
Which of the following sets of quantum number following orbitals.
is not possible? (a) π molecular orbital
(a) (i), (ii), (iii) and (iv) (b) σ molecular orbital
(b) (ii), (iv) and (v) (c) δ molecular orbital
(c) (i) and (iii) (d) No bond will form. (2001)
(d) (ii), (iii) and (iv). (2007)
36. The following quantum numbers are possible
29. The orientation of an atomic orbital is governed for how many orbitals : n = 3, l = 2, m = +2 ?
by (a) 1 (b) 2 (c) 3 (d) 4
(a) principal quantum number (2001)
(b) azimuthal quantum number 37. For given energy, E = 3.03 × 10 –19 Joules
(c) spin quantum number corresponding wavelength is
(d) magnetic quantum number. (2006) (h = 6.626 × 10 –34 J sec, c = 3 × 10 8 m/sec)
30. Given : The mass of electron is 9.11 × 10–31 kg, (a) 65.6 nm (b) 6.56 nm
Planck constant is 6.626 × 10 –34 J s, the (c) 3.4 nm (d) 656 nm (2000)
uncertainty involved in the measurement of 38. Isoelectronic species are
velocity within a distance of 0.1 Å is (a) CO, CN–, NO+, C22–
(a) 5.79 × 105 m s–1 (b) 5.79 × 106 m s–1 (b) CO–, CN, NO, C2–
(c) 5.79 × 107 m s–1 (d) 5.79 × 108 m s–1 (c) CO+, CN+, NO–, C2
(2006) (d) CO, CN, NO, C2 (2000)
31. The energy of second Bohr orbit of the 39. The uncertainty in momentum of an electron
hydrogen atom is –328 kJ mol–1; hence the is 1 × 10 –5 kg m/s. The uncertainty in its
energy of fourth Bohr orbit would be position will be (h = 6.62 × 10 –34 kg m2/s)
(a) – 41 kJ mol–1 (b) –82 kJ mol–1 (a) 5.27 × 10–30 m (b) 1.05 × 10–26 m
–1
(c) –164 kJ mol (d) –1312 kJ mol–1 –28
(c) 1.05 × 10 m (d) 5.25 × 10–28 m
(2005) (1999)
12

40. Who modified Bohr’s theory by introducing 48. The radius of hydrogen atom in the ground
elliptical orbits for electron path? state is 0.53 Å. The radius of Li2+ ion (atomic
(a) Rutherford (b) Thomson number = 3) in a similar state is
(c) Hund (d) Sommerfield (a) 0.53 Å (b) 1.06 Å
(1999) (c) 0.17 Å (d) 0.265 Å (1995)
41. The de Broglie wavelength of a particle with 49. For which of the following sets of four
mass 1 g and velocity 100 m/s is quantum numbers, an electron will have the
(a) 6.63 × 10–35 m (b) 6.63 × 10–34 m highest energy?
–33
(c) 6.63 × 10 m (d) 6.65 × 10–35 m n l m s
(1999) (a) 3 2 1 +1/2
42. The Bohr orbit radius for the hydrogen atom (b) 4 2 – 1 +1/2
(n = 1) is approximately 0.530 Å. The radius (c) 4 1 0 –1/2
for the first excited state (n = 2) orbit is (in Å) (d) 5 0 0 –1/2 (1994)
(a) 4.77 (b) 1.06 (c) 0.13 (d) 2.12 50. Which one of the following is not isoelectronic
(1998) with O2–?
43. The position of both, an electron and a helium (a) Tl+ (b) Na+ (c) N3– (d) F–
atom is known within 1.0 nm. Further the (1994)
momentum of the electron is known within
51. Electronic configuration of calcium atom can
5.0 × 10–26 kg m s–1. The minimum uncertainty
be written as
in the measurement of the momentum of the
(a) [Ne] 4p2 (b) [Ar] 4s2
helium atom is 2
(c) [Ne] 4s (d) [Kr] 4p2 (1992)
(a) 8.0 × 10–26 kg m s–1
(b) 80 kg m s–1 52. The energy of an electron in the nth Bohr orbit
(c) 50 kg m s–1 of hydrogen atom is
(d) 5.0 × 10–26 kg m s–1 (1998) 13.6 13.6
(a) 4
eV (b) eV
44. The ion that is isoelectronic with CO is n n3
(a) CN– (b) N2+ (c) O2– (d) N–2 13.6 13.6
(1997) (c) eV (d) eV (1992)
n2 n
45. What will be the longest wavelength line in
53. In a given atom no two electrons can have
Balmer series of spectrum?
the same values for all the four quantum
(a) 546 nm (b) 656 nm
numbers. This is called
(c) 566 nm (d) 556 nm (1996)
(a) Hund’s Rule
46. In a Bohr’s model of an atom, when an electron (b) Aufbau principle
jumps from n = 1 to n = 3, how much energy (c) Uncertainty principle
will be emitted or absorbed? (d) Pauli’s Exclusion principle. (1991)
(a) 2.389 × 10–12 ergs
(b) 0.239 × 10–10 ergs 54. For azimuthal quantum number l = 3, the
(c) 2.15 × 10–11 ergs maximum number of electrons will be
(d) 0.1936 × 10–10 ergs (1996) (a) 2 (b) 6 (c) 0 (d) 14
(1991)
47. Uncertainty in position of an electron (mass
= 9.1 × 10–28 g) moving with a velocity of 55. The order of filling of electrons in the orbitals
3 × 104 cm/s accurate upto 0.001% will be of an atom will be
(Use h/(4π) in uncertainty expression where (a) 3d, 4s, 4p, 4d, 5s
h = 6.626 × 10–27 erg second) (b) 4s, 3d, 4p, 5s, 4d
(a) 5.76 cm (b) 7.68 cm (c) 5s, 4p, 3d, 4d, 5s
(c) 1.93 cm (d) 3.84 cm (1995) (d) 3d, 4p, 4s, 4d, 5s (1991)