Chemistry

Division of periodic table into blocks periodic table has been classified into 4 blocks S block    P block   d block    f block The division depends upon the type of atomic orbital .which receive the valence electron. for eg.  Mg the valence electron enter the s orbital .so Mg belongs to s block. S block : it is present in the L. H. S of the periodic table .in all these elements the valence electron  enter the s- orbital.  so they are called   s block  elements.  so s block consistint of 2 groups group (1) elements are called alkali metals                 for eg. Li, Na group (2) elements which soluble in water                           than formed hydroxide and form                            alkali.  for eg.  berillium, magnesium                   etc Electronic configurations of s block elements      (ns^1-2)  P block:  it is present in the R. H. S  of the  periodic table.  in all these elements the valence electron enter p orbital. so they are called p block el

Chapter 02  periodic table and atomic periodicity or properties: Modern periodic table:  the table in which elements are arrange in order of increasing atomic number. Modern periodic law: physical and chemical properties of elements are periodic function of their atomic number. periodic function: repelition of properties after certain interval of time. periodicity of elements: the repelition of properties after certain interval of time. the elements repeats their properties because their valence shell electron configuration repeats.  eg.   li     2   1          Na   2   8   1           k     2   8   8   1 114 elements are known today. for their systematic study they have been classified into group and period. Group :vertical column are called group there are 18 vertical column. so their are 18 groups. Periods:  horizontal rows are called periods there are 7 horizontal rows. so their are 7 periods  . A/c to periods no. of elements 1 period  2 elements 2 pe

Normal wave function:  if probability of finding electron in space is unity i.e 100℅ the wave function sie is said to be normalised  or Integration  of ꌏ^2 over the whole space must be equal to 1 ∫ꌏ^2 dxdydz = 1 if the wave function satisfied the above relation .it is said to be normalised. Orthogonal wave function : if ꌏ(1)  and ꌏ(2)  represent two different acceptable wave function.   the product of two wave functions integrated over the entire space must be equal to zero. ∫ꌏ(1)ꌏ(2) dxdydz = 0 the wave function which obey the above relation is said to be orthogonal. # if wave function -ve than called imaginary. # if wave function +ve than called real. wave function which are both normalised and orthogonal are called orthagonal normalised function.

Angular wave function:  Wave function depends upon quantum no. l, m and often referred to atomic orbital. the shape of these orbital depends upon angle 𝟅 and ⲫ.                                                    Mathematical expression for angular wave function for s & p orbital is given  orbital               Angular wave function  s                                     (1/4π) ^1/2                   pz                           (3/4π) ^1/2 cos   𝟅 px                     (3/4π) ^1/2 sin   𝟅 cos   𝟅 py                      (3/4π) ^1/2 sin   𝟅 sin ⲫ.  the expression for s orbital has no.  angle dependence ,hence s orbital are speherical symmetrical.  limitation of slater's rule                                            1.it is observed that slater's rule becomes less reliable for atoms of heavier elements.  2.  s & p orbital are grouped together for calculating effective nuclear charge.  3.while due to higher pene

Radial probability distribution for p and d orbital Radial and Angular wave function:  the radial part of wave function depends upon quantum no.  n and  l  and gives the distribution of electron w. r. t distance .it is governed mainly exponential term     e^-Zr/na°(a not)   here  e   based on natural log.  Z   Atomic number r     distance from nucleus  n    principal quantum no. or radial quantum           no.  a°    0.529A° for hydrogen  ( Bohr radii)  the exact mathematical expression for radial part of wave function for 1s or 2s and 2p orbitals. n= 1 ,l=0 s orbital    R(r) =2× (z/a°) ^3/2 ×(2-zr/a°) ×e^-zr/2a° n= 2 ,l=0 2s orbital    R(r) = (z/a°) ^3/2 ×(2-zr/a°) ×e^-zr/2a° n= 2 ,l=1 2p orbital    R(r) = 1√3×(z/2a°) ^3/2 ×(zr/a°) ×e^-zr/2a ° the radial wave function can be represented by plotting radial function R(r)  apart distance(r)

Radial probhjjhajajbability distribution curves:  The probability of finding the electron is given by the quantity  sie^2.By radial probability us probability of finding the electron within small Radial space around the nucleus. volume of spherical shell between radius r and r+dr =4πr^2 and probability of finding the electron will be 4πr^2dr sie^2. radial probability distribution curves are obtained by plotting radial probability at various distance from the nucleus . 1.Radial probability distribution curves for S orbital n=1, l= 0. distance from nucleus here A° is called angstrom. the probability of finding of electron in a shell is maximum at distance r=r° (r not)  which is 0.529 A° for H atom. diagram shows that probability plot for 2s has(two region of high probability)  or (two peaks)  separated by node. we conclude that  no.  of high probability region in S orbital = n no.  of node = ( n-1) 

Question answer 4.calculate zeff 4s electron of potassium?  ans Zn (30)     1s^2 2s^2 2p^6   3s^2 3p^6 3d^10   4s^2                     n-2                           n-1                    n S= 1×0.35 +18×0.85 + 10×1.00 S=  0.35+15.30+10.00 S= 25.65 zeff=A-S        = 30-25.65=4.35 5.calculate zeff 4s electron of iron?  ans Fe (26)        1s^2 2s^2 2p^6   3s^2 3p^6 3d^6  4s^2                     n-2                           n-1                    n S= 1×0.35 +14×0.85 + 10×1.00 S=  0.35+11.9+10.00 S= 22.25 Zeff= 26-22.25=3.75 6.calculate zeff 3d electron of zinc?  ans Zn (30)        1s^2 2s^2 2p^6 3s^2 3p^6   3d^10   4s^2                                                                     n         here 4s (not encluding in screening effect)  according to rule no. 6 S= 9×0.35  + 18×1.00 S= 21.15 Zeff= 30-21.15=8.85 7.calculate zeff 3d electron of iron?  ans fe (26)        1s^2 2s^2 2p^6 3s^2 3p^6   3d^6   4s^2                              

Question Answer 1.In an atom first shell has 2 electron  second shell has 8 electron third shell have 18 electron . explain this arrangement on the basis of quantum number? Answer K shell          no. of orbital   total no. of electrons n=1                                                      l=0                 1s                               2       m=0       (s=-1/2, +1/2) L shell n=2                                                        l=0                  2s                               2       m=0       (s=-1/2, +1/2)        l=1                2p                                   6       m=-1 (s=-1/2, +1/2)        m=0 (s=-1/2, +1/2)        m=1 (s=-1/2, +1/2) M shell n=3                                                        l=0                   3s                                  2       m=0       (s=-1/2, +1/2)        l=1                  3p                                  6       m=-1 (s=-1/2, +1/2)        m=0 (s=-1/2, +1/2)        m=1 (s

Screening effect (or shielding effect): The electron of valence shell (outer most shell)  are attracted towards the nucleus also the valence electron are repelled by the electron present in time shell. This decrease in force of attraction exerted by the nucleus ,on valence electron due to presence of electron in inner shell, is called screening effect or shielding effect. More are the no.  of electron in the inner shell more more will be screening effect.  slater's  rules :  Slater gave a set of emperical rules for calculating effective nuclear charge z(eff)  z* effective nuclear charge acting on a given electron is caluculated by substracting the screening constant from atomic number  (z)    z(eff)  = atomic nio.  -  screening constant    z(eff)  = Z- S Rules  1.are the purpose of estimitating screening constant ns and np electron considered together in a single group while nd and nf electron each from separate group.  2.write down electronic configuration of

Electronic configuration of ions of transition elements:  In first transition series atom loses 4s electron before they lose 3d electron. the reason is that when electron goes in 3d orbital the energy of 3d orbital became less than 4s orbital .In other words 3d orbital became more stable than 4s. once both the electron are lost from 4s orbital than electron are lost from 3d orbital. for eg.   fe (26)         [ Ar] 4s^2  3d^6 [fe^3+]         [At]  3d^5   4s^0 Question 1 Write down electronic configuration of elements with atomic no.  24,46,74,57.  Ans. electronic configuration of elements as: 1 .  cr(24)          [Ar] 4s^1 3d^5 2.   pd(46)         [kr]  5s^0 4d^10 3.   W(74)          [xe]  4f^14  6s^2  5d^4 4.    la(57)           [xe ]  6s^2  5d^1 Question   2 cu^2+ is more stable than cu^+1  ?  Ans  cu^2+ cupric  cu^+  cuprous cu^2+ compounds are more stable due to high lattice energy in solid state and high hydration energy in solution. and cu^2+siz

Electronic configuration of Lanthanoids; ce (58) to lu(71)   electronic configuration of lanthanoid cannot be predicted corrected by Aufbau principal. for eg. after filled 6s orbital in barium the next electron in lanthanum should enter 4f orbital in fact it does not happen and next electron goes in 5d orbital. so electronic configuration of  lanthanoid[ 57]   |xe|6s^2 5d^1. Electronic configuration of Actinoids; electronic configuration of actinoids cannot be predict correctly by aufbau principal for eg.  After filling 7s orbital in radium(88)  the next electron in actinium should 5f orbital. in fact it does not happen and next electron goes in 6d orbital .so electronic configuration of  actinium[89]  |Rn|7s^2 6d^1.  Bohr's bury's rule (n+l)  rule:  According to this rule the energy of an orbital depends upon (n+l)  value, the orbital with lower value of (n+l)  will be lower energy  and will filled first. for eg.                 3d                     4s  

Stability of half filled and completely filled orbital

Hund rule of maximum multiplicity & numericals: Hund rule of maximum multiplicity: pairing of electron in various orbital of a subshell connot take place until each orbital contain ane electron. Numericals : 1.what is the total capacity of electron of 4th principal energy shell ? sol. the no. of electron in a shell =2n^2 where n =4 than 2×(4) ^2= 32 2.how many electron confit in principal shell n=2? Sol. no.  of electron in a shell =2n^2 where n =2 than 2×(2) ^2= 8 3.how many electron can be accomodate in 4d subshell? Sol. d subshell have five orbital and each orbital can have 2 electron.  so 4d subshell can accomodate 2×5= 10 electrons

Pauli exclusive principal: It state that an orbital cannot accomodate more than two electron .they should be opposite spin . Question : Define pauli exclusive principal on the basis of quantum number ? ans. Two electron in an atom cannot be completly identically i. e no two electron in an atom can have same value all the four quantum no.            from picture for left side spin                      for right side spin n=4                                             n=4     l=0, 1,2,3                                    l=0, 1,2,3 for s than                                  for s than l=0                                               l=0            m=0                                             m=0 s=+1/2                                          s=-1/2 Energy is inversely proportional to Stability. which shown in picture.                                    

Physical significance of sie and sie^2  The wave function sie has no physical significance,  it simply represents amplitude of wave   while square of amplitude sie^2 represent intensity of electron.  i. e sie ^2 gives probability of finding the electron in space .p the space is called atomic orbital  A zero value of sie^2 means probability of finding the electron is zero and high value of sie^2 means greater chances of finding the electron .  the value of sie^2 lies between 0&1. if sie^2 =1   100℅  sie^2=0    0℅

Shape of p orbital : for p sub shell   l=1 m=-1, 0,+1   i. e P subshell has three possible orbital desigrated as named as These three orbital are of equal energy but differ in orbital or orientation. P orbital is domb-bell shaped and probability of finding the electron in both the lobes are same .  Shape of d orbital: for d sub shell l=2 m =-2, -1, 0,+1, +2 since there are five values of m. so there are five possible orientation named as dxy, dyz, dzx, dx^2y^2. The three orbital dxy ,dyz, dzx, are similar in shape each consisting of four lobes of high electron density lying b/w the axis.  dx^2y^2 orbital also have four lobes High electron density lying along the y axis & x-axis  dz^2 orbital is symmetrical about z-axis and  is dombbell shaped. in this orbital the part of orbital lying in the direction of magnetic field is enclosed and part of orbital lying perpendicular to it is contracted

Orbital : The space around the nucleus in which probability of finding the electron is maximum is called orbital . Shape is orbital For S subshell   l=0, m=0  There is only one possible orbital this means that probability of finding the electron is same in all direction so all s orbital are sperical in shape but size of s orbital increase as principal quantum number increase.  The gap in 2s or 3s is called nodal space This may be notice that inside the 2s and 3s orbital there is a finding the electron is zero that space is called nodal space.    Number of space in an s -orbital =(n-1)   where n is principal quantum number. 

Numericals on quantum no.  1.an electron is present in 3d orbital write down possible value of n, l, m, s sol.      (3d) n=3 l=2 m=-2, -1, 0,+1, +2 n= +1/2 or-1/2 for each value of m.  2 .an electron is present in the 4th orbital write down possible value of n, m, l, s sol.      (4f) n=4 l=3 m=-3, -2, -1, 0,+1, +2, +3 s=-1/2 or +1/2 of each value of m.  3.an electron is present in 2pz orbital write down n, l, m, s value Sol.   (2pz) n=2 l=1 m=0 s=-1/2 or +1/2 of m 4. an electron is present in 4dz^2 orbital write down n, l, m, s value Sol. (4dz^2) n=4 l=2 m=0 s=-1/2 or +1/2 of m

Spin quantum number:   the electron not only revolve around the nucleus but also spin around its own axis.  this no.  tell spinning of electron whether electron moves in clockwise direction or anticlockwise direction .it is denoted by (s) .  it can have value  clockwise +1/2 or anticlockwise -1/2. Name of orbital  (P)                   value of m         px                                         m=+1        py                                         m=-1        pz                                          m=0 Name of orbital  (d)                   value of m       dxy                                        m= +2        dyz                                        m=+1       dzx                                         m=-1       dx^2y^2                                m=-2                dz^2                                       m=0    

Azimuthal quantum number ( Angular quantum number) :  This quantum number tells of subshell in a shell.  it is denoted by (l)   .it value 0 to (n-1) .               value of l         name of sub shell if  n=1       l= 0                       s     n=2       l =0, 1                  sp     n=3       l=0, 1,2                spd     n=4       l=0, 1,2,3             spdf no.  of subshell in a shell is given by = n Magnitic   quantum number   : this no.  tells us no.  of orientations or orbitals in a subshell.  it is denoted by (m).   it can have value from -l to  +l and including zero.                       value of m                  no. of orbitals if l=0             m=0                                 1(s)      l= 1            m=-1, 0,1                         3(p)      l=2            m=-2, -1, 0,1,2                  5(d)      l=3              m=-3, -2, -1, 0,1,2,3        7(f)  no.  of orbitals in a shell = 2n-1

Quantum numbers A set of four numbers which gives complete idea about the energy and distance of electron in an atom. they are  1. Principal quantum number 2.Azimuthal quantum number 3. Magnetic quantum number 4. Spin quantum number Principal  quantum number : This quantum number tells us average distance and energy of electron  it is denoted by (n) . it can have value 1,2,3,4.......not including zero.  if n=1       K shell     n=2        L shell     n=3        M shell     n=4        N shell     n=5        O shell # Energy of an electron E(n) = -1312/n^2 KJ/mol # No.  of electron in a shell is given by formula 2n^2 . in both lines n is principal quantum number. 

Numericals 3 calculate the product of uncertainty in displacement and velocity for an electron with mass 9.1×10^-31kg . sol. mass of electron =9.1×10^-31kg h= 6.6×10^-34kgm^3/s  now  we know  ∆x.∆p= h/4π ∆x.m∆V=h/4π ∆x.∆V =h/(4πm)  putting all value there  ∆x.∆V= 5.77×10^-5 m^2/s       ans 4. On the basis of uncertainty principal so that electron cannot exist in the nucleus . sol.   radius of nucleus = 10^-14 m  h= 6.6×10^-34kgm^3/s mass of electron =9.1×10^-31kg we know  ∆x.∆p= h/4π ∆x.m∆V=h/4π ∆V =h/(4πm∆x)  putting all value  we get  ∆V= 5.77×10^9m/s since this value  5.77×10^9m/s is greater than  velocity 3×10^8m/s which is not possible. hence electron cannot be found in the nucleus of atom. 

Numerical 1. calculate uncertainty in velocity of an electron if uncertainty in position is  1Angstrom given mass of electron 9.1× 10^-34kgm^2s^-1. sol.  ∆x = 10^-10m  mass of electron = 9.1×10^-31kg h =6.6×10^-34kg m^2s^-1 ∆V=?   we know ∆x.∆p=h/4π ∆xm∆V =h/4π        because p=mv or ∆V= h/(4π×m×∆x)  putting all values we get  ∆V= 5.77×10^5m/s      ans 2. calculate uncertainty in position of electron if uncertainty in velocity is   0.0001℅ given  velocity of electron = 300m/s. sol. ∆x =?   ∆v= 0.0001℅×300       = (0.0001/100) ×300= 3×10^-4m/s mass of electron = 9.1×10^-31kg h =6.6×10^-34kg m^2s^-1 we know ∆x.∆p= h/4π               ∆x.m.∆V= h/4π               ∆x= h/(4π.m.∆V)  putting all value we get   ∆x= 1.92×10^ -3metre           ans

significance this principal has no significance in our daily life because we are to deal with macroscopic object.  evidently or clearly their position and velocity don't alter by strike with photon of light and this principal is applicable to microscopicparticle only. eg. if light falls on the electron that  its velocity be changed and if light falls on the big particle than its velocity not be changed so Hinesberg uncertainty principal gave significance for microscopic particle only.

Hinesberg uncertainty principal It stated that it is impossible to determine simultaneously both position and momentum of microscopic particle like electron. Hinesberg propose a mathematical relation for uncertainty principal as- ∆x.∆p~h/4π where ∆ denote for uncertainty and x denote position , so ∆x denote uncertainty position and p denote momentum so ∆p denote uncertainty momentum. If ∆x is very small that means position of particle know more exactly than ∆p would be large and vice-versa.

Numericals 1. Find out the no.  of wave by a bohr electron in one complete revolution in 3rd orbit? sol. A/c to bohr's postulates (m×V×r) =(n×h)/2π  n= 2πr×1/√   or √= h/mV here √ sign suppose lamda. or n= circumference/√ where n is number of waves since electron in 3rd orbit .so n= 3 so n=3  we can say that number of waves in one complete revolution = 3

Derivation of bohr's postulates of quantisation  of angular momentum from de -broglie eq.  proff...  we know circumference of circular orbit must be integral multiple of wavelength .     2πr= n×√    we also know that √ = h/ ( m×V)   putting value of √ in eq. we get     2πr= (n×h) /(m×V)  or mVr= (n×h)/2π which states that angular momentum should be integral multiple of h/2π  in other words angular momentum is quantized. it can have certain fixed value.

numericals 1 calculate De-broglie wavelength of electron moving at a speed of 1℅ of light given m=9.1×10^-31kg, h=6.63×10^-34kgm^-2s^-1. sol.  mass of the electron = 9.1× 10^-31kg  Planck's constant = 6.63×10^-34kgm^2s^-1 velocity of electron = 1℅ speed of light                                   = 1/100 × 3×10^8m/s=3×10^6 applying all value in relation      √= h/m×v  √= (6.63×10^-34) /  (9.1×10^-31 × 3×10^6)  √=2.43×10^-10metre   ans                       2 An electron moving with K. E of 2.275×10^-10J calculate its De-broglie wavelength.   sol.  mass of the electron = 9.1× 10^-31kg  Planck's constant = 6.63×10^-34kgm^2s^-1 K. E energy of electron= 2.275×10^-25J we know K. E= 1/2 m×V^2   so applying all value in relation we get  2.275 ×10^-25 = 1/2 m×V^2   2.275×10^-25=1/2×9.1×10^-31×V^2   V^2=0.5×10^6m^2/s^2     V=.707×10^3m/s   ans see next 

justification of dual nature of electron   Particle nature :  when an electron strikes a ZnS screen  it produced a spot called scintillation .which is localised and spread like a wave .this suggested that electron behave like a particle. wave nature : when electrons we're made to strike against nickel crystal,  concentrating ( having common centre)  were formed these rings are called diffraction rings. this explains wave nature of electron.   some important values: c = velocity of light =3×10^8m/s. h= Planck constant =6.626×10^-34kgm^2s^-2. √= wavelength (lamda) v= frequency =c/√ numerical related to de broglie principle in next page 

Unit 1 Atomic structure Idea of De-broglie matter waves :   Einstein in 1905 suggested that light has dual character .  In 1924 De-broglie proposed that matter also has dual character ( it means behave as waveas well as particle)  .   Principle :  The principal of wave and particle duality of matter is known as De- broglie principle.   If substance behavea particle it's energy is given by Einstein          E= mc^2     1eq .  if substance behave like a wave it's energy given by plancks quantum theory.          E= hv           2eq. (here v is called nue  )  from 1 & 2 eq.  we get hv=mc^2   & also know v= c/√   let is √ called lamda    now (h×c) /√ =mc^2 or √= h/m×c  Applying relation to an eletron moving with velocity(V).    so √= h/m×V &we know that p= m×V         or √= h/p This relation is called De-broglie wave eq .    Significance of de broglie Relationship  In our daily life we come a

                                              INORGANIC CHEMISTRY                                BSC SEMESTER -1                                                                                                                                                                                   section A                                     1.atomic structure              2.periodic properties                                                        section B              1.covalent bond              2.ionic bond