Presentation is loading. Please wait. # Chemistry. Atomic structure Session Objectives Session objectives Schrodinger wave equation Shapes of orbitals Nodal Plane Quantum Numbers Rules for.

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Chemistry

Atomic structure

Session Objectives

Session objectives Schrodinger wave equation Shapes of orbitals Nodal Plane Quantum Numbers Rules for electronic configuration of an atom

Schrodinger wave equation Describes the probability of finding an electron in a given volume element.

Quantum mechanical model of atom The energy of electrons in atoms is quantised. The number of possible energy levels for electrons in atoms of different elements is a direct consequence of wave-like properties of electrons. The position and momentum of an electron cannot be determined simultaneously. Electrons of different energies are likely to be found in different regions. The region in which an electron with a specific energy will most probably be located is called an atomic orbital

Orbit and Orbital Orbit is a fixed circular path around the nucleus in which electron moves(proposed by Bohr) whereas orbital is the quantum mechanical concept and refers to the wave function.

Nodal Plane The plane where the probability of finding the electron is almost zero. Total nodes in a shell = (n -1) Angular nodes = l Spherical nodes= (n –l -1)

Quantum Numbers Principal AzimuthalMagnetic Spin They specify the address of each electron in an atom. These are four types.

Principal Quantum Number (n) Average distance of the electron from the nucleus Energy Level of electron Possible values (n=1,2,3…..) Maximum number of electrons in any shell is 2n 2. In 3d orbital,principal quantum number n is 3. For example:

Azimuthal or Angular Momentum Quantum number:(l) It describes Energy sub-level Shape of the orbital p d s f

Quantum Numbers Orbital angular momentum of an electron is given by Subshell Values of l s p df 0123

Illustrative example The orbital angular momentum of an electron is 4s orbital is (b) zero AIEEE 2003 Solution Orbital angular momentum = For s electrons, l = 0 For 4s electrons, orbital angular momentum is zero. Hence, answer is (b)

Magnetic quantum number(m) Magnetic Quantum Number (m l ) Orientations of an orbital in space. Explains the Stark and Zeeman effect. Takes (2l+1) values,m=- l to + l. Specifies the exact orbital within each sublevel Orbitals combine to form a spherical shape: 2s 2p z 2p y 2p x

Spin Quantum Number(s) Describes the direction of spin of an electron Clockwise or anticlockwise (+1/2,-1/2) Spin angular momentum= Magnetic moment = n= no. of unpaired electrons m s = +½ m s = -½

Relation between quantum numbers For every value of n, l = 0 to (n-1) For every value of l, m = -l to +l For every value of m, s =

Quantum Numbers New Delhi n New Delhi (shell) Area l Okhla PhaseI (sub shell) Street m B Block (orbital) House number s 52 (spin)

Illustrative example Oxygen (Z=8) having configuration i.e., 8 O 16 = 1s 2 2s 2 2p 4 hence, for the last electron n=2, l=1, m=-1, s=-1/2 Write down all the four quantum numbers for the last electron of oxygen. Solution:

Illustrative example The electrons, identified by quantum numbers n and 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 lowest to the highest as (a) (iv) < (ii) < (iii) < (i)(b) (ii) < (iv) < (i) < (iii) (c) (i) < (iii) < (ii) < (iv) (d) (iii) < (i) < (iv) < (ii) Solution (i)n = 4, l = 1(4p) (ii)n = 4, l = 0(4s) (iii)n = 3, l = 2(3d)(iv)n = 3, l = 1(3p) According to Aufbau ’ s rule order of increasing energy of subshells is 3p < 4s < 3d < 4p or (iv) < (ii) < (iii) < (i) Hence, answer is (a)

Illustrative Example A compound of vanadium has a magnetic moment of 1.73 B.M. Find out the oxidation state of vanadium in the compound. Magnetic moment = The compound contains only one unpaired electron, electronic configuration of V will be [Ar] 3d 1 4s 0 n = 1 Electronic configuration of vanadium (23) is [Ar] 3d 3 4s 2 (five unpaired electrons) V is present as V 4+- Solution

Aufbau’s principle “Fill up” electrons in lowest energy orbitals Order of orbitals (filling) in multi-electron atom 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s

Aufbau’s principle For example: Consider 3d and 4s orbitals, the electron will first enter the orbital having minimum value of (n+l). Electron will therefore enter 4s orbital (4+0=4) before entering 3d orbital (3+2=5).

Aufbau’s principle Incase (n+l) values are same!!! Incase of 3d orbital (3+2=5) and 4p orbital (4+1=5), the (n+l) values are same. In such a case,electron enters the orbital for which n is minimum. The electron will thus enter 3d orbital before entering 4p orbital.

Pauli’s exclusion principle For example: Incase of 1s 2,there are two electrons in the 1s orbital. The quantum numbers of the two electrons are: n=1, l=0, m=0, s=+1/2 n=1, l=0, m=0, s=-1/2 It is impossible for two electrons in a given atom to have same set of four quantum numbers.

Hund’s rule The most stable arrangement of electrons in sub shells is the one with the greatest number of parallel spins. Electron pairing starts only after all the degenerate orbitals are filled with electrons having same direction of spin. For example: Nitrogen (Atomic number=7) Electronic configuration 1s 2 2s 2 2p 3 Degenerate refers to orbitals having same energy.

Illustrative example If the nitrogen atom had electronic configuration 1s 7, it would have energy lower than that of the normal ground state configuration 1s 2 2s 2 2p 3, because the electrons would be closer to the nucleus. Yet 1s 7 is not observed because it violates (a) Heisenberg’s uncertainty principle (b) Hund’s rule (c) Pauli exclusion principle (d) Bohr postulate of stationary orbits Solution According to Pauli’s exclusion principle, an orbital cannot have more than two electrons and these two with opposite spin. Hence, answer is (c)

Exceptional electronic configuration Orbitals in the same sub shell tend to become completely filled or half filled since such orbitals are more stable. Such as electronic configuration of Cr(24): [Ar]3d 4 4s 2 But actually it is [Ar]3d 5 4s 1

Illustrative example The electronic configuration of an element is 1s 2 2s 2 2p 6 3s 2 3p 6 3d 5 4s 1. This represents its (a) excited state(b) ground state (c) cationic form(d) anionic form The above electronic configuration is for ground state of chromium (3d 5 4s 1 ) Solution Hence, answer is (b)

Class exercise

Class exercise-1 An electron is in one of 4d orbitals. Which of the following quantum number is not possible? (a) n = 4(b) l = 1 (c) m = 1(d) m = 2 Solution bIn 4d orbital, n = 4 = 2 m = –2, –1, 0, 1, 2 Therefore, the quantum number value which is not possible is (b). Hence correct option is (b)

Class exercise-2 Solution An electron in an isolated atom may be described by four quantum numbers: n, l, m and s. The value of m for the most easily removed electron from a gaseous atom of an alkaline earth metal is (a) same as the maximum value of n for the element (b) any number from –(n – 1) to +(n – 1) (c) any positive number from 1 to ( n – 1) (d) zero dAlkaline earth metal belongs to s block. The value of n for s block elements is 1. l = (n – 1) = 0 m = 0 Hence correct option is (d)

Class exercise-3 Solution The set of quantum numbers that represents the electron of highest energy is (a) n=1, l=0, m=0, s=+1/2(b) n=2, l = 0, m=0, s=+1/2 (c) n=4, l=0, m=0, s=1/2 (d) n=3, l=2, m=0, s=+1/2 dThe set of quantum number which represents an electron of the highest energy has the maximum value of (n + l). Hence correct option is (d)

Class exercise-4 Solution The number of electrons that can be accommodated by p- orbitals are (a) two electrons with parallel spins(b) six electrons (c) four electrons(d) eight electrons bp orbital can accomodate a total of six electrons (two each in p x, p y, p z ). Hence correct option is (b)

Class exercise-5 Solution Which among the following set of quantum numbers is not possible? (a) n = 3, l = 0, m = 0(b) n = 3, l = 1, m = 1 (c) n = 2, l = 1, m = 0(d) n = 2, l = 0, m = –1 dThe set of quantum number which is not possible is n = 2, l = 0, m = –. In this case, m = 0. Hence correct option is (d)

Class exercise-6 Solution Which orbital has equal probability of finding an electron in all directions? The s orbital has equal probability of finding an electron in all directions as it is spherical in shape, which is symmetrical around the nucleus.

Class exercise-7 Solution What is the angular momentum of 4s orbital? Orbital angular momentum = In case of 4s orbital, = 0  Orbital angular momentum is zero.

Class exercise-8 Solution Why do some atoms exhibit exceptional electronic configuration? Some atoms exhibit exceptional electronic configuration because half-filled orbitals and fully filled orbitals are more stable than partially filled orbitals.

Class exercise-9 Solution Write down the electronic configuration for Si (atomic number 14), V (atomic number 23), Zn (atomic number 30), Kr (atomic number 36). Si = 1s 2 2s 2 sp 6 3s 2 3p 2 V = 1s 2 2s 2 2p 6 3s 2 3p 6 3d 3 4s 2 Zn = 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 Kr = 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 6

Class exercise-10 Solution An electron is in a 4d orbital. What possible values of the quantum numbers can it have? n = 4 l = 2 m = ±2, ±1, 0 s =

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