Formula pack-Structure of Atom-IITJEE-AIPMT-JEE(Main)

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Chemistry Coaching for Class-8, 9, 10, 11, 12-IITJEE-AIPMT- JEE (MAIN) IN DURGAPUR 0343 2546991, 9434008713 STRUCTURE OF ATOM Sr . No . Particle Symbol Natu re Charge (in esu) × 10 –10 Mass (in amu) Discovered by 1. Electron e 4.8029 0.0005486 J.J. Thomson 2. Proton p + + + 4.8029 1.0072 Goldstein 3. Neutron N 0 1.0090 Chadwick 4. Positron e + , 1e 0 , + + + 4.8029 0.0005486 Anderson (1932) (a) Charge to Mass ratio of electron = 1.758820 X 10 11 C kg -1 (b) Mass on electron = 9.1094 X 10 -31 kg (c) Rutherford’s Nuclear Model of Atom: Radius of the atom = 10 -10 m & Radius of the nucleus = 10 -15 m (d) Atomic number (Z) = Number of protons in the nucleus of an atom = Number of electrons in a neutral atom (e) Mass number (A) = Number of protons (Z) + Number of neutrons (n) (f) Isotopes = Identical (Z) & different (A); Isobars = different (Z) & Identical (A) of different elements 4. (i) mv 2 r = Ze 2 r 2 (ii) mvr= nh 2 π (iii) v= 2 πZe 2 nh = 2.188 × 10 8 Z/n cm/sec (iv) r= n 2 h 2 4 π 2 mZe 2 =0.529× ( n 2 Z ) Å (v) E T = 2 π 2 mZ 2 e 4 n 2 h 2 =−KE = PE 2 (vi) E T = Ze 2 2 r = K. E . + P. E . (vii) E T = 2 π 2 mK 2 Z 2 e 4 n 2 h 2

Transcript of Formula pack-Structure of Atom-IITJEE-AIPMT-JEE(Main)

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STRUCTURE OF ATOM

Sr. No.

Particle Symbol NatureCharge

(in esu) × 10–10

Mass

(in amu)Discovered by

1. Electron e– – 4.8029 0.0005486 J.J. Thomson

2. Proton p+ + + 4.8029 1.0072 Goldstein

3. Neutron N 0 1.0090 Chadwick

4. Positron e+, 1e0, + + + 4.8029 0.0005486 Anderson (1932)

(a) Charge to Mass ratio of electron = 1.758820 X 1011 C kg-1

(b) Mass on electron = 9.1094 X 10-31 kg

(c) Rutherford’s Nuclear Model of Atom: Radius of the atom = 10-10 m & Radius of the nucleus = 10-15 m

(d) Atomic number (Z) = Number of protons in the nucleus of an atom = Number of electrons in a neutral atom

(e) Mass number (A) = Number of protons (Z) + Number of neutrons (n)

(f) Isotopes = Identical (Z) & different (A); Isobars = different (Z) & Identical (A) of different elements

4. (i)

mv2

r= Ze

2

r 2

(ii)mvr= nh

(iii)v=2π Ze2

nh= 2. 188 × 108 Z/n cm/sec

(iv)

r= n2h2

4 π2mZe2=0 . 529×( n2

Z ) Å

(v)

ET=−2π2mZ2 e4

n2 h2=−KE=PE

2

(vi)

ET=−Ze2

2r= K. E .+ P. E .

(vii)

ET=−2π2mK 2Z2 e4

n2h2

(viii) ET = – 21.8 × 10–19 Z2/n2 J/atom

= – 13.6 Z2/n2 eV/atom

= 21.8 × 10–12

Z2

n2

erg

= – 1312 kJ/mol

(x) E1 < E2< E3 < ...... < E (E = 0)

(xi) (E2– E1) > (E3–E2) > (E4–E3) > .......,

(xii)For H-atom : E1 = –13.6 eV

E2 = – 3.4 eV

E3 = –1.5 eV

E4 = –0.85 eV

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0343 2546991, 9434008713E5 = –0.54 eV

(xiii) I.E. = 13.6 Z2 eV/atom

(xiv) Number of revolution per sec by an e–

=

0 .657×1016 z2

n3

(xv) Time taken for one revolution

=

1. 52×10−16 n3

z2

(i) Various Series of spectrum lines :

Series

of LinesTransition

Spectrum

Zone

Wave

Length

1. Lyman n2 = 2,3,4,

... to n1 = 1

Ultraviolet < 3800 Å

2. Balmer n2 = 3,4,5

... to n1 = 2

Visible 3800-7800 Å

3. Paschen n2 = 4,5,6

... to n1 = 3

Infrared > 7800 Å

4. Bracket n2 = 5,6, 7

... to n1 = 4" "

5. Pfund n2 = 6, 7, 8

... to n1= 5" "

6. Humphury n2 = 7, 8, 9

... to n1 = 6" "

(ii) Wave number ( ν ) and wavelength () of spectral lines :

υ=1λ=RH Z

2 [ 1

n12−

1

n22 ]

Here, RH =

2π2me4

ch3 (RH = Rydberg constant ;

= 109678 cm–1)

(iii) Total no. of spectrum line =

n(n−1)2

(i)λ= hmv

= hp

(ii)

12

mv 2 = eV (for electron)

(iii)λ= h

√2emV

(iv)λ=12 .25

√V (Here is in Å, V is in volt) (for electron)

(v)λ= h

√2mK .E .

(vi)λ=12 .25

√K . E . (for electron)(Here is in Å and K.E. is in eV)

(x) (p)

h4 π & (x) (v)

h4 πm

E. t

h4 π (For energy and time)

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0343 2546991, 9434008713

(a) Schrodinger wave equation :

(i) 2ψ +

8π 2mh2

(E–V)ψ = 0….(i) or

(ii) E = H Ψ …..(ii)

(i) Radial nodes or spherical nodes = n – – 1

(ii) Angular nodes =

(iii) Total nodes = n – 1

Sub

Shell

Value

of l

Value

of m

No. of

orbitals

Max.

no. of e–

S 0 0 1 2

P 1 0, ±1 3 6

D 2 0, ±1, ±2 5 10

F 3 0, ±1, ±2, ±3 7 14

(i) Boundary surface diagram for 1s orbital.

(ii) Boundary surface diagrams of the three 2p orbitals.

(iii) Boundary surface diagrams of the five 3d orbitals.

(a) No. of e–s in any subshell = 2(2l +1)(b) No. of orbitals in any subshell = (2l +1)(c) Orbital angular momentum of e–

= [l(l+1)] h/2(d) Spin angular momentum of e– =

[s(s+1)] h/2 (e) No. of Max. e– in any shell =2n2

(f) Max. number of orbitals in any shell=n2

(g) Max. number of subshell in any shell = n(h) Spin multiplicity = 2S + 1

(Here S = n/2 and n = total no. of unpaired electrons)

Magnetic Moment = n (n+2) B.M,

1 B.M. =

eh4 πme , = 9.27 × 10–24 JT–1 ;

n = total no. of unpaired e–

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h= w +

12 mv2 ………………(i)

w = h0 (work function)………(ii)

(i) Aufbau Principle :According to this principle, "In the ground state, the atomic orbitals are filled in order of increasing energies". i.e. in the ground state the electrons occupy the lowest orbitals available to them.

Order of filling of e– 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, ……… , 5f, 6d, 7p.

(ii) Pauli's Exclusion Principle :According to this principle,"No two electrons in an atom can have all the four quantum numbers n, l, m and s identical”.

(iii) Hund's Rule of Maximum Multiplicity :According to this rule "Electron pairing will not take place in orbitals of same energy until all the available orbital of a given sub shell contain one electron each with parallel spin".

Stability of Completely Filled and Half Filled sub-shells

(i) Symmetrical distribution of electrons:Completely filled and half filled subshells have symmetrical distribution of electrons, electrons are in the same sub-shells have equal energy and less screening effect, more strongly attracted by the nucleus.

(ii) Exchange energy:These electrons with same spin change their positions in degenerate orbitals for half filled subshells. Exchange energy may be released for symmetrical distribution of electrons, eg, 5 electrons in 3d orbitals gives smaller repulsion, smaller screening effect and large exchange energy.

Key points related to Quantum Mechanical Model of atom:

(a) Wave Function Ψ :

All the information about the electron in an atom is stored in its orbital wave function Ψ.

(b) Probability density Ψ2:

From the value of Ψ2, it is possible to predict the region around the nucleus where electron will most probably be found.

(c) Hamiltonian operator H˄:

H˄Ψ = EΨ, it is Schrodinger Equation that indicates total energy of the system.

SUPRATIM DAS, CITY CENTRE, DURGAPUR, WB [email protected] # 9434008713

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