bcpldtpBC Note 201306 ACC J · BANSAL CLASSES Private Ltd. ‘Gaurav Tower’, A-10, Road No.-1, I.P.I.A., Kota-5 ACC_CHEMISTRY_General Chemistry 3 (2) Atomicmass: Massof anatomrelativeto

7 inch

2.5 inch

2.5 inch2.5 inchMEDICAL ENTRANCE

GENERAL CHEMISTRY

I N D E X

Topic Page No.

PHYSICAL CHEMISTRY

GENERAL CHEMISTRY

01. Basic Definitions (Mole concept) 02

02. Commonly used units and their conversions 04

03. Basic Definitions (Atomic structure) 05

04. Quantum Number 06

05. Electronic Configuration 08

06. Rules forElectronic Configuration 08

07. Extra stabilityof Half-filled and Fully-filled orbitals 09

08. Summary 10

09. Exercise - 1 12

Exercise - 2 15

10. Answer Key 16

11. Hints/Solution 17

MEDICAL ENTRANCE

GENERAL CHEMISTRY

BANSAL CLASSES Private Ltd. ‘Gaurav Tower’, A-10, Road No.-1, I.P.I.A., Kota-5

ACC_CHEMISTRY_General Chemistry 1

Syllabus :

• Quantum Numbers,Aufbau Principle, Pauli Exclusion Principles and Hund's rule

Contents :

• Basic Definitions (Mole concept)

• Commonly used units and their conversions

• Basic Definitions (Atomic structure)

• Quantum Number

• Electronic Configuration

• Rules forElectronic Configuration

• Extra stabilityof Half-filled and Fully-filled orbitals

GENERAL CHEMISTRY



GENERAL CHEMISTRY

Chemistry is the branch of science which deals with the study of matter, its physical & chemical proper-ties, its chemical composition, the physical and chemical changes which it undergoes and the energychanges that accompany there process.

BASIC DEFINITIONS (MOLE CONCEPT)(1) Mole : One mole is a collection of that many entities as there are number of atoms exactly in 12

gm of C-12 isotope.or 1 mole = collection of 6.02 × 1023 species

6.02 × 1023 = NA = Avogadro's No.

1 gm -atom is same as 1 mole of an atom & hence will have wt. equal to atomic

weight expressed in gms.

1 gm- molecule is same as 1 mole of the molecule & hence will have weight equalto molecular weight expressed in gms.

1 gm - Ion is same as 1 mole of an ion.

Special Highlights

BRAIN TEASERS: GENERAL MISTAKE:

TEACHER’S ADVICE:



(2) Atomic mass : Mass of an atom relative to th12

1of mass of an atom of C–12

Atomic weight =amu1

amuinatomanofMass

12Ccarbonofatomanofmass12

1

elementtheofatomanofMass

(1 amu = 1.67 ×10–24 gm)

(3) GramAtomic mass : Mass of 1 mol atoms of an element. It is also called molar mass ofelement.

(4) Average atomic weight =% of isotope × molar mass of isotope.The % used in above expression is by number (i.e. its a mole%)

(5) Average molecular weight =

i

ii

n

Mn

where ni = no. of moles of any compound and Mi = molecular mass of any compound.

Make yourselves clear in the difference between mole% and mass% in question

related to above.

Shortcut for % determination if average atomic weight is given for X having

isotopes XA & XB.

% of XA = BA

B

X&Xofweightatomicindifference

XofweightatomicweightatomicAverage × 100

Try working out of such a shortcut for XA, XB, XC

Methods of Calculations of mole :

(a) If no. of some species is given, then number of moles =AN

.noGiven

(b) If weight of a given species is given, then number of moles =.wtAtomic

)gm.(wtGiven(for atoms),

or =.wtMolecular

)gram(.wtGiven(for molecules)

For ionic compounds word " Molecular mass" is replaced by "formula mass";

because ionic compound is not exist as a molecule but it exist as a formula unit.



(c) If volume of a gas is given along with its temperature (T) and pressure (P)

use n =RT

PV

where R = 0.0821 lit-atm/mol-K (when P is in atmosphere and V is in litre.)

1 mole of any gas at STP (1 bar & 273 K) occupies 22.7 litre.

Gases do not have volume. What is meant by "Volume of gas"?

Do not use this expression (PV = nRT) for solids/liquids.

How would I calculate moles if volume of a solid is given?

(i) The numerical value of universal gas constant (R) changes with change in

units of P, V and T.

(ii) To determine electrons, protons, neutrons or total atoms in given quantityof substance, first of all determine number of molecules.

(iii) To determine neutrons, always subtract protons from mass number.

COMMONLY USED UNITSAND THEIR CONVERSIONS

Units of pressure1 atm = 76 cm of Hg = 760 torr = 1.01325 bar = 1.01325 ×105 Pascal

Units of temperatureK = °C + 273.15

Can you guess the formula for converting temperatures in different

scales?

Units of RR = 0.0821 lit-atm K–1 mol–1

= 8.314 JK–1 mol–1

= 1.987 Cal K–1 mol–1



Units of Volume1 dm3 = 103 cm3 = 1 litre = 10–3 m3 = 103 mL

BASIC DEFINITIONS (ATOMIC STRUCTURE)

Atom : An atom is the smallest particle of an element (made up of still smaller particles like electrons,protons, neutrons, etc.) which can take part in a chemical reaction. It may or may not exist free in nature.

* Discovery details of electron, proton and neutron should not discussed

Proton's have positive charge but they remain inside the nucleus together.

why?

Representation of atom : ZXA

AMass numberdetailsgive

ZAtomic number

Isotope: Atoms of a given element which have same atomic number but different mass numberare called isotopes. e.g. 1H

1, 1H2, 1H

3 etc.

Isobar: Atoms of different element with the same mass number but different atomic number.e.g. 18Ar40, 19K

40 and 20Ca40.



Iso-electronic species : Species (atoms, molecules or ions) having same number of electronsare called iso-electronic.e.g. H–,He, Li+ and Be2+ have 2 electron each.

Orbital :An orbital is defined as that zone in space where electron is most likely to be found. Theorbitals are characterized by a set of 3 quantum numbers (n,l,m).

QUANTUM NUMBERPrincipal Quantumnumber (n) :(i) Permissible value of n 1 –.(ii) It represents shell number / energy level.(iii) The energy states corresponding to different principal quantum numbers are denoted by letters

K,L, M, N etc.n : 1 2 3 4 5 6Designation of shell : K L M N O P(iv) It indicates the distance of an electron from the nucleus.(v) It also determines the energy of the electron. In general higher the value of ‘n’, higher is the

energy of an electron.

In H & hydrogen like species energy of electron increases with increases in 'n'

but energy difference between successive energy level decreases.

(vi) It give an idea of total numberof orbitals & electrons (which may) present in a shell & that equalsto n2 & 2n2 respectively.

Azimuthal Quantum number (l) :

(i) The values of l depends upon the value of ‘n’ and possible values are ‘0’ to (n–1).(ii) It gives the name of subshells associated with the energy level and number of subshells within an

energylevel.(iii) The different value of ‘l’ indicates the shape of orbitals and designated as follows:

Value Notation Name Shapel = 0 s Sharp Sphericall = 1 p Principal Dumbelll = 2 d Diffused DoubleDumbelll = 3 f Fundamental Complex

Shape of 2zd orbital is different from other d-orbitals in a d-subshell.



(iv) It also determines the energy of orbital along with n.For a particular energy level/ shell, energy of subshell is in the following:Order s < p < d < f

(v) It gives the total number of orbitals in a subshell & that equals to 2 l + 1 and number of electronin a subshell = 2 (2l + 1).

How many electrons are there in a orbital of f-subshell?

Maximum number of orbitals in a shell = n2 or

1n

0

)12(l

l

l .

Magnetic Quantum number (m):(i) The value of m depends upon the value of l and it may have integral value –l to +l including zero(ii) Its gives the number of orbitals in a given subshell and orientation of different orbitals in space.

e.g. for n = 4, l = 0 to 3

syllabusinNotd,d,d

d,dp,p,psOrbitals

7531nsOrientatio

Possible3,2,1,0,1,2,32,1,0,1,21,0,10m

3210

xzyzxy

yxzzyx

222

l

(iii) The orbital havingsame value of n and l but different value of m, have same energy in absence ofexternal electric & magnetic field. These orbitals having same energy of a particular subshell isknown as Degenerate orbitals.

An additional quantum number is used to characterize an electron present

in the orbital.

Spin Quantum number (s) :

(i) While moving around the nucleus, the electron always spins about its own axis either clockwiseor anticlockwise. The spin quantum number represents the direction of electron spin(rotation)around its own axis (clockwise or anticlockwise)

(ii) There are two possible values of ‘s’ are +2

1& –

2

1and represented by the two arrows (spin

up) and (spin down).



ELECTRONIC CONFIGURATION :Electronic configuration of atoms :The distribution of electrons in various shells, subshells and orbitals, in an atom of an element, is called itselectronicconfiguration.

3Li ; 1s2 , 2s1 or [He] 2s1 or1s 2s

or He

Electronic Configuratione.g. Nitrogen :

7N : 1s2, 2s2 2p3 [Orbital notation method]1s2, 2s2, 2px1 2py1 2pz1

or

1s 2s 2p[Orbital diagram method]

or[He] 2s2 2p3 [Condensed form]

RULES FOR ELECTRONIC CONFIGURATION :-AUFBAU PRINCIPLE (Means Building up)The electrons are added progressively to the various orbitals in the order of increasing energies startingwith the orbital of the lowest energy.

1s2s 2p

3s 3p 3d

4s 4p 4d 4f

5s 5p 5d 5f

6s 6p 6d 6f

7s 7p ..............

1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s < 5f < 6d < 7p.Alternatively, the order of increase of energy of orbitals can be calculated from (n + l) rule.

(i) The lower the value of (n + l) for an orbital, the lower is its energy.(ii) If two orbitals have the same (n + l) value, the orbital with lower value of n has the lower energy.

For H and hydrogen like species (n + l) rule is not applicable.



HUND’S RULE OF MAXIMUM MULTIPLICITYThis rule deals with the filling of electrons into the orbitals belonging to the same subshell i.e. orbitals ofequal energy, called degenerate orbitals.“Electrons are distributed among the orbitals of a subshell in such a way as to give the maximum numberof unpaired electrons with parallel spins”.

PAULI’S EXCLUSION PRINCIPLE‘No two electrons in an atom can have same values of all the four quantum numbers.’An orbital accommodates two electrons with opposite spin. These two electrons have same values ofprincipal, azimuthal and magnetic quantum number but the fourth, i.e. spin quantum number will bedifferent.

Can you guess maximum how many quantum numbers may be same for two

electrons in an atom if they are present in different orbitals.

EXTRASTABILITY OFHALF-FILLEDAND FULLY-FILLED ORBITALS.The electronic configuration of most of the atoms follows theAufbau’s rule. However, in certain elementssuch as Cr, Cu etc. where the two subshells (4s and 3d) differ slightly in their energies (4s < 3d), anelectron shifts from a subshell of lower energy (4s) to a subshell of higher energy (3d), provided such ashift results in all orbitals of the subshell of higherenergy gettingeither completely filled orhalf-filled.

24Cr [Ar] 3d5, 4s1 and not [Ar] 3d4 4s

29Cu [Ar] 3d10, 4s1 and not [Ar] 3d9 4s2

Ithasbeenfoundthat there isextrastabilityassociatedwith theseelectronicconfigurations.Thisstabilizationis due to the following two factors.

(i) Symmetrical distribution of electron : It is well known that symmetry leads to stability. Thecompletely filled orhalf-filled subshell have symmetrical distribution of electrons in themand aretherefore more stable. This effect is more dominant in d and f-orbitals. This means three or sixelectrons in p-subshell, 5 or 10 electrons in d-subshell and 7 or 14 subshell forms a stablearrangement.

(ii) Exchange energy : This stabilizing effect arises whenever two or more electrons with the samespin are present in the degenerate orbitals of a subshell. These electrons tend to exchange theirpositions and the energy released due to this exchange is called exchange energy.The number ofexchanges that can take place is maximum when thesubshell is eitherhalf-filled or fully filled.Asa result the exchange energy is maximum and so is the stability.


ACC_CHEMISTRY_General Chemistry 1 0

SUMMARY

Atoms are the building blocks of elements. They are the smallest parts of an element that chemically

react.The first atomic theory, proposed by John Dalton in 1808, regarded atom as the ultimate indivisible

particle of matter. Towards the end of the nineteenth century, it was proved experimentally that atoms

are divisible and consist of three fundamental particles: electrons, protons and neutrons. The discovery

of sub-atomic particles led to the proposal of various atomic models to explain the structure of atom.

Thomson in 1898 proposed that an atom consists of uniform sphere of positive electricity with

electrons embedded into it. This model in which mass of the atom is considered to be evenly spread over

the atom was proved wrong by Rutherford's famous alpha-particle scattering experiment in 1909.

Rutherford concluded that atom is made of a tiny positively charged nucleus, at its centre with electrons

revolving around it in circular orbits. Rutherford model, which resembles the solar system, was no

doubt an improvement over Thomson model but it could not account for the stability of the atom i.e.,

why the electron does not fall into the nucleus. Further, it was also silent about the electronic structure of

atoms i.e., about the distribution and relative energies of electrons around the nucleus. The difficulties of

the Rutherford model were overcome by Niels Bohr in 1913 in his model of the hydrogen atom. Bohr

postulated that electron moves around the nucleus in circular orbits. Only certain orbits can exist and

each orbit corresponds to a specific energy. Bohr calculated the energy of electron in various orbits and

for each orbit predicted the distance between the electron and nucleus. Bohr model, though offering a

satisfactory model for explaining the spectra of the hydrogen atom, could not explain the spectra of

multi-electron atoms. The reason for this was soon discovered. In Bohr model, an electron is regarded

as a charged particle moving in a well defined circular orbit about the nucleus. The wave character of the

electron is ignored in Bohr's theory.An orbit is a clearly defined path and this path can completely be

defined only if both the exact position and the exact velocity of the electron at the same time are known.

This is not possible according to the Heisenberguncertainty principle. Bohr model of the hydrogen atom,

therefore, not only ignores the dual behaviour of electron but also contradicts Heisenberg uncertainty

principle.

Erwin Schrodinger, in 1926, proposed an equation called Schrodinger equation to describe the

electron distributions in space and the allowed energy levels in atoms. This equation incorporates de

Broglie's concept of wave-particle duality and is consistent with Heisenberg uncertainty principle.

When Schrodinger equation is solved for the electron in a hydrogen atom, the solution gives the possible

energy states the electron can occupy [and the corresponding wave function(s) () (which in fact are the

mathematical functions) of the electron associated with each energy state]. These quantized energy

states and corresponding wave functions which are characterized by a set of three quantum numbers

(principal quantumnumber n, azimuthal quantum number l and magnetic quantumnumber ml)

arise as a natural consequence in the solution of the Schrodinger equation. The restrictions on the values

of these three quantum numbers also come naturally from this solution. The quantum mechanical model

of the hydrogen atom successfully predicts all aspects of the hydrogen atom spectrum including some

phenomena that could not be explained by the Bohr model.



According to the quantum mechanical model of the atom, the electron distribution of an atom

containing a number of electrons is divided into shells. The shells, in turn, are thought to consist of one

or more subshells and subshells are assumed to be composed of one or more orbitals, which the

electrons occupy. While for hydrogen and hydrogen like systems (such as He+, Li2+ etc.) all the orbitals

within a given shell have same energy, the energy of the orbitals in a multi-electron atom depends upon

the values of n and l : The lower the value of (n + l ) for an orbital, the lower is its energy. If two orbitals

have the same (n + l ) value, the orbital with lower value of n has the lower energy. In an atom many such

orbitals are possible and electrons are filled in those orbitals in order of increasing energy in accordance

with Pauli exclusion principle (no two electrons in an atom can have the same set of four quantum

numbers) and Hund’s rule of maximum multiplicity (pairing of electrons in the orbitals belonging to

the same subshell does not take place until each orbital belonging to that subshell has got one electron

each, i.e., is singly occupied). This forms the basis of the electronic structure of atoms.



EXERCISE-1

[SINGLE CORRECT CHOICE TYPE]

Q.1 The number of g-atoms of nitrogen in its 7 gm is equal to number of g-atoms in(1) 6 gm Mg (2) 28 gm Fe (3) 30 gm Ca (4) 20 gm Hg

Q.2 One atomic mass unit in kilogram is(1) 1/NA (2) 12 / NA (3) 1/1000 NA (4) 1000 / NA

Q.3 From 2 mg calcium, 1.2 × 1019 atoms are removed. The number of g-atoms of calcium left is(1) 5 × 10–5 (2) 2 × 10–5 (3) 3 × 10–5 (4) 5 × 10–6

Q.4 2 isotopes of an element are present in 1 : 2 ratio of number, having mass number M and (M + 0.5)respectively. The mean mass number of element will be

(1) 3M + 1 (2) 1.5 M + 0.5 (3) 0.5 M + 0.5 (4) M +3

1

Q.5 The number of neutrons in 0.45 g water, assuming that all the hydrogen atoms are H1atoms and all theoxygen atoms are O16 atoms, is(1) 8 (2) 0.2 (3) 1.2 × 1023 (4) 4.8 × 1024

Q.6 A gaseous mixture contains CO2 (g) and N2O (g) in a 2 : 5 ratio by mass. The ratio of the number ofmolecules of CO2 (g) and N2O (g) is(1) 5 : 2 (2) 2 : 5 (3) 1 : 2 (4) 5 : 4

Q.7 1.61 gm of Na2SO4.10H2O contains same number of oxygen atoms as present in(1) 0.98 gm H2SO4 (2) 0.08 gm SO3(3) 1.78 gm of H2S2O7 (4) 0.05 gm CaCO3

Q.8 Which of the following contain largest number of carbon atoms?(1) 15 gm ethane, C2H6 (2) 40.2 gm sodium oxalate, Na2C2O4(3) 72 gm glucose, C6H12O6 (4) 35 gm pentene, C5H10

Q.9 If molecular weight of glucose-1-phosphate is 260 and its density is 1.5 g/ml. What is the averagevolume occupied by 1 molecule of this compound?(1) 43 × 10–23 ml (2) 0.67 ml (3) 0.17 × 1023 ml (4) 29 × 10–23 ml

Q.10 The number of hydrogen atoms in 0.9 gm glucose, (C6H12O6), is same as(1) 0.048 gm hydrazine, N2H4 (2) 0.17 gm ammonia, NH3(3) 0.30 gm ethane, C2H6 (4) 0.03 gm hydrogen, H2

Q.11 The number of g-molecules of oxygen in 6.02 × 1024 CO molecules is(1) 1 g-molecule (2) 0.5 g-molecule (3) 5 g-molecule (4) 10 g-molecule



Q.12 What time, it would take to spendAvogadro’s number of rupees at the rate of 10 lac rupees per second?(1) 6.023 × 1017 sec (2) 1.909 × 1010 year(3) 1.673 × 1014 hour (4) all of these

Q.13 Ethanol, C2H5OH, is the substance commonly called alcohol. The density of liquid alcohol is 0.7893 g/ml at 293 K. If 1.2 mole of ethanol are needed for a particular experiment, what volume of ethanolshould be measured out?(1) 55 ml (2) 58 ml (3) 70 ml (4) 79 ml

Q.14 Which of the followingwill occupy greatervolume under the similar conditions of pressure and tempera-ture?(1) 6 gm oxygen (2) 0.98 gm hydrogen(3) 5.25 gm nitrogen (4) 1.32 gm helium

Q.15 How many mole of electron weight one kilogram?

(1)023.6108.9

1

× 108 (2)

108.9

1×1031

(3)108.9

1× 1054 (4) 6.023 × 1023

Q.16 Correct set of four quantum numbers for valence electron of rubidium( Z = 37) is

(1) 5, 0, 0, +2

1(2) 5, 1, 0, +

2

1(3) 5, 1, 1, +

2

1(4) 6, 0, 0, +

2

1

Q.17 The correct set of quantum numbers for the unpaired electron of chlorine atom isn l m n l m

(1) 2 1 0 (2) 2 1 1(3) 3 1 1 (4) 3 0 0

Q.18 The total number of neutrons in dipositive zinc ions with mass number 70 is(1) 34 (2) 40 (3) 36 (4) 38

Q.19 Principal quantum number of an atom represents(1) Size of the orbit (2) Spin angular momentum(3)Orbital angular momentum (4) Space orientation of the orbital

Q.20 Which of the following sets of quantum numbers represent an impossible arrangementn l m ms n l m ms

(1) 3 2 –22

1(2) 4 0 0

2

1

(3) 3 2 –32

1(4) 5 3 0

2

1

Q.21 The explanation for the presence of three unpaired electrons in the nitrogen atom can be given by(1) Pauli’s exclusion principle (2) Hund’s rule(3)Aufbau’s principle (4)Uncertainty principle



Q.22 The maximum number of electrons that can be accommodated in the Mth shell is(1) 2 (2) 8 (3) 18 (4) 32

Q.23 Which quantum number will determine the shape of the subshell(1)Principal quantum number (2)Azimuthalquantumnumber(3) Magnetic quantum number (4) Spin quantum number

Q.24 Which of the following has maximum number of unpaired electron (atomic number of Fe 26)(1) Fe (2) Fe (II) (3) Fe (III) (4) Fe (IV)

Q.25 It is known that atom contain protons, neutrons and electrons. If the mass of neutron is assumed to halfof its original value where as that of proton is assumed to be twice of its original value then the atomic

mass of C146 will be

(1) same (2) 25% more (3) 14.28 % more (4) 28.5% less

Q.26 The quantum numbers of four electrons (e1 to e4) are given belown l m s n l m s

e1 3 0 0 +1/2 e2 4 0 0 1/2e3 3 2 2 –1/2 e4 3 1 –1 1/2The correct order of decreasing energy of these electrons is:(1) e4 > e3 > e2 > e1 (2) e2 > e3 > e4 > e1 (3) e3 > e2 > e4 > e1 (4) none

Q.27 If the nitrogen atom had electronic configuration 1s7, it would have energy lower that of normal groundstate configuration 1s2 2s2 2p3 , because the electrons would be closer to the nucleus. Yet 1s7 is notobserved because it violates :–(1)Heisenberg uncertainity principle (2) Hunds rule(3) Pauli’s exclusion principle (4) Bohr postulate of stationary orbits

Q.28 The electrons, identified by n & l ; (i) n = 4 , l = 1 (ii) n = 4 , l = 0(iii) n = 3 , l = 2 (iv) n = 3 , l = 1 can be placed in order of increasing energy, from the lowestto highest as :(1) (iv) < (ii) < (iii) < (i) (2) (ii) < (iv) < (i)(3) (i) < (iii) < (ii) < (iv) (4) (iii) < (i) < (iv) < (ii)



EXERCISE-2

[ASSERTION & REASON TYPE QUESTIONS]

Each of the questions given below consist of Assertion and Reason. Use the following Key to choose

the appropriate answer.

(A) If both Assertion and Reason are correct, and Reason is the correct explanation of Assertion.

(B) If both Assertion and Reason are correct, and Reason is not the correct explanation of Assertion.

(C) If Assertion is correct and Reason is incorrect.

(D) IfAssertion is incorrect but Reason is correct.

(E) If bothAssertion and Reason are incorrect.

Q.1 Assertion : The ground state configuration of Cr is [Ar] 3d54s1

Reason : The energy of atom is lesser in 3d5 4s1 configuration compared to 3d4 4s2 configuration.(1)A (2) B (3) C (4) D (5) E

Q.2 Assertion : Minimum principal quantum number of an orbital belonging to 'g' sub-shell is 5.Reason : For a given value of principal quantum number (n), l may have values 0 to (n–1) only.(1)A (2) B (3) C (4) D (5) E



EXERCISE-1

Q.1 2 Q.2 3 Q.3 3 Q.4 4Q.5 3 Q.6 2 Q.7 3 Q.8 4Q.9 4 Q.10 3 Q.11 3 Q.12 4Q.13 3 Q.14 2 Q.15 1 Q.16 1Q.17 3 Q.18 2 Q.19 1 Q.20 3Q.21 2 Q.22 3 Q.23 2 Q.24 3Q.25 3 Q.26 3 Q.27 3 Q.28 1

EXERCISE-2

Q.1 1 Q.2 1



EXERCISE-1

1. g–atoms moles of atomsg–molecules Moles of Moleculesg–ions Moles of ions.

g–atoms of Nitrogen = Moles of N =14

7= 0.5

g–atom of Fe =56

28= 0.5

(At mass of Fe = 56)

2. 1 g = NA a.m.u.

1 a.m.u. =AN

1g =

AN1000

1Kg

3. g–atoms of Ca taken = Moles of Ca taken

= MassMolar

ginMass

=40

102 3= 5 × 10–5

g–atoms of Ca removed = 23

19

106

102.1

= 2 × 10–5

g–atoms of Ca left = 5 × 10–5 – 2 × 10–5 = 3 × 10–5

4. Average or mean mass number =21

2211

XX

XMXM

X1 & X2 Ratio or percentage of isotopeM1 & M2 Mass numbers of isotope

Mean Mass number =

3

25.0M1M

=3

1M3 = M +

3

1

5. In 1H1 Mass number = 1

(no of protons + no of neutrons)Atomic number = 1(number of proton) Number of neutron = 0



In one molecule of H2Onumber of protons = 8 + 2 = 10number of neutrons = 8 + 0 = 8

Moles of H2O =18

45.0= 2.5 × 10–2

Molecules of H2O = 2.5 × 10–2 × NA= 2.5 × 10–2 × 6 × 1023 = 15 × 1021

number of Neutrons = 15 × 1021 × 8 = 120 × 1021 = 1.2 × 1023

6. For compoundsRatio of Molecules = Ratio of Moles( 1 mole of any compound contain NA molecules)

CO2 N2OMass Ratio 2 : 5

Mole Ratio44

2:

44

5 2 : 5

Molecules Ratio 2 : 5

7. Moles of Na2SO4 . 10H2O =1018643246

61.1

=

322

61.1=

2

1× 10–2

Moles of O =2

1× 10–2 × 14 = 7 × 10–2

In 0.98 g of H2SO4

Moles of H2SO4 =98

98.0= 10–2

Moles of O = 4 × 10–2

In 0.08 g of SO3

Moles of SO3 =80

08.0= 10–2

Moles of O = 3 × 10–2

In 1.78 g of H2S2O7

Moles of H2S2O7 =112642

78.1

=

178

78.1= 10–2

Moles of O = 7 × 10–2

In 0.05 g CaCO3

Moles of CaCO3 =100

05.0= 5 × 10–4

Moles of O = 5 × 10–4 × 3= 1.5 × 10–3

8. For largest number of C–atom, moles of C must be largestIn 15 g of C2H6

Moles of C2H6 =30

15= 0.5



Moles of C = 0.5 × 2 = 1In 40.2 g Na2C2O4

Moles of Na2C2O4 =642446

2.40

=

134

2.40= 0.3

Moles of C = 0.3 × 2 = 0.6In 72 g of C6H12O6

Moles C6H12O6 =180

72= 0.4

Moles of C = 6 × 0.4 = 2.4In 35 g C5H10

Moles of C5H10 =70

35= 0.5

Moles of C = 0.5 × 5 = 2.5 (Maximum)

9. Mass of 1 Molecule of glucose–1–phosphate = 260 a.m.u. =AN

260g

density =Volume

Mass

Volume =density

Mass

Volume of 1 Molecule =5.1

N

260

A = 29 × 10–23 ml

10. Moles of glucose =180

9.0= 5 × 10–3

Moles of H atom = 5 × 10–3 × 12 = 6 × 10–2

for atoms to be same, moles of atoms should also be sameIn 0.048 g N2H4

Moles of N2H4 =32

048.0= 1.5 × 10–3

Moles of H = 1.5 × 10–3 × 4 = 6 × 10–3

In 0.17 g NH3

Moles of NH3 =17

17.0= 0.01

Moles of H = 3 × 0.01 = 0.03In 0.3 g C2H6

Moles C2H6 =30

3.0= 0.01

Moles of H = 6 × 0.01 = 0.06 = 6 × 10–2

In 0.03 g H2



Moles of H2 =2

03.0= 0.015

Moles of H2 = 0.015 × 2 = 0.03

11. Moles of CO = 23

24

1002.6

1002.6

= 10

Moles of O = 10

Moles of O2 =2

10= 5 = g–molecules of oxygen

12. 10 lac/sec = 106 sec–1

Time taken to spendAvogadro's number of rupees

= 6

23

10

10023.6 = 6.023 × 1017 sec

=6060

10023.6 17

hour = 1.673 × 1014 hour

=36524

10673.1 14

year = 1.909 × 1010 year

13. Mass of liquid alcohol = 1.2 × 46 = 55.2 g

Volume =density

Mass=

7893.0

2.55= 70 ml

14. For ideal gasPV = nRTat const P & TV n for volume to be maximum, moles of gas should be maximum.

Moles of oxygen =32

6

Moles of hydrogen =2

98.0= 0.49 (Maximum)

Moles of Nitrogen =28

25.5

Moles of He =4

32.1= 0.33

15. Mass of one electron = 9.108 × 10–31 kg

Number of electrons in 1 kg = 3110108.9

1

Number of Mole of electrons =A

31 N10108.9

1

=023.6108.9

1

× 108



16. Configuration Rb (Z = 37) is[Kr] 5s1

for valence electronn = 5 = 0m = 0

s = +2

1or –

2

1

17. 17Cl = 1s2 2s2 2p6 3s2 3p5

unpaired electron of Cl is present in 3p n = 3

= 1m = 0 or –1 or +1

18. Zn+2 Mass number = 70 (number of protons + number of neutrons)Atomic number = 30 (number of protons)

number of neutrons = 70 – 30 = 40In formation of ion of any element only electrons are gained or last, So nochange in number of protons or number of neutrons.

19. Principal quantum number (n) represents size of orbit. (higher the value of n, bigger will be its size).

20. varies from 0 to (n – 1)m varies from – to 0 to + If = 2, m value can't be –3.

21. Hund's rule states that in degenerate orbitals pairing of electrons in an orbital can not take place until allthe orbitals of that subshell are singly occupied by electrons of parallel spin.

22. Maximum number of electrons in nth shell = 2n2

Maximum number of electrons in Mth shell = 2 × 32 = 18(n = 3)

23. Azimuthal quantum number (l) determines subshell & shape of subshell.

24. In Fe configuration is [Ar] 3d6 4s2

32 4s

number of unpaired electrons = 4In Fe (II) (i.e. Fe+2) configuration is [Ar] 3d6

number of unpaired electrons = 4In Fe (III) (i.e. Fe+3) configuration is [Ar] 3d5



number of unpaired electrons = 5 (maximum)In Fe (IV) (i.e. Fe+4) configuration is [Ar] 3d4

number of unpaired electrons = 4

25. In 6C14

number of proton = 6number of neutrons = 14 – 6 = 8Atomic Mass = 14 u ( Mass of neutron Mass of proton 1 u)

If mass of neutron is assumed to half of its original value

Mass of neutrons = 8 ×2

1= 4 u

If mass of proton is assumed to be twice of its original valueMass of protons = 6 × 2 = 12 u

Now atomic mass = 12 + 4 = 16 uIncrease inAtomic mass of 6C

14 16 – 14 = 2

percentage increase =14

2× 100 = 14.28 %

100

massoriginal

massinIncrease

26. Higher the value of (n + ) higher will be the energy. If two electrons are having same value of (n + ) thenelectron having value of n will have higher energy. energy order is

e3 > e2 > e4 > e1(n+ = 5) (n+ = 4) (n+ = 4) (n+ = 3)

(n = 4, = 0) (n = 3, = 1)

27. Pauli's exclusion principal states that number two electrons in an atom can have the same set of all thefour quantum numbers or in an orbital there can be maximum two electrons & those must be withopposite spin.

28. Energy order (iv) < (ii) < (iii) < (i)(higher the value of (n + ) higher will be energy if two electrons have same value of (n + ) then electronhaving lower value of n will have lower energy)

EXERCISE-2

1. Energy of 3d5 4s1 configuration is lesser as compared to 3d4 4s2 as in this there is more symmetrical

distribution of electrons also exchange energy released is higher.

2. For g–subshell = 4 n value should be greater than or equal to 5.{as varies from 0 to (n – 1)}

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