Chemistry Pre-u Chemistry Sem 1 Chap 2

PRE-U STPM CHEMISTRY SEMESTER 1 CHAPTER 2 ELECTRONIC STRUCTURE OF ATOMS

Past Year Questions Analysis

2.1 2.2 2.3 2.4

CHAPTER 2 : ELECTRONIC STRUCTURE OF ATOMS Electronic Energy Levels of Atomic Hydrogen Atomic orbitals Electronic configuration Classification of elements into s, p, d and f blocks in the Periodic Table

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2.1

Electronic Energy Levels of Atomic Hydrogen

According to Maxwell’s theory, an electromagnetic wave has an electric field component and a magnetic field component. Electromagnetic radiation is the emission and transmission of energy in the form of electromagnetic waves, which travel 3.00 x 108 meters per second (This constant is more well-known as speed of light, c)., Electromagnetic radiation is characterized by a frequency (f), and wavelength (λ) where

speed of light (3.00 × 10 8 m s −1 ) c Frequency = ; f= wavelength ( m ) λ

White light consists of continuous distribution of all possible wavelengths spanning the entire visible region of the electromagnetic spectrum. When a narrow beam of white light is passed through a glass prism, different wavelengths travel through the glass at different rates. As a result, the white light dispersed into component colours, ranging from red at the longwavelength end of the spectrum (700 nm) to violet at the shortwavelength end (400 nm)

When radiation from a particular source is passed through a spectrometer, it will be separated into its components of different frequencies, producing a spectrum. • i. A continuous spectrum is one where light is emitted over a broad range of wavelengths (or frequencies); showing emission of a wide range of energies. The spectrum is smooth and continuous. • ii. A line spectrum is one where exact frequencies or wavelengths appear as lines (indicating that only certain amounts of energy are emitted and none in between those energies). A line spectrum is normally produced by atoms that have been excited and is also called atomic emission spectrum

• When a sample of hydrogen gas (H2) is subjected to an electrical discharge, the hydrogen molecules dissociate forming hydrogen atoms. Equation : ½ H2 (g)
H (g) • The hydrogen atoms formed, then absorb different amounts of energy and the electron in each of the atoms will be raised to higher energy level. Spectroscopists studying the spectrum of atomic hydrogen had identified several series of spectral lines in different regions of electromagnetic spectrum

a) These spectra line formed from spectroscopy are specific and can be quantisised according to the radiation source by using Rydberg’s Equation

 1 1  = R H  2 − 2  λ n2   n1 1

λ = wavelength (in m) n1 = ground state energy level n2 = energy level where electron fall from compared to ground state RH = Rydberg constant = 1.097 x 107 m-1

b) Hydrogen spectrum produced under different sources produce different series with different characteristics. Table 2.2 below compared the hydrogen spectrum produced under ultraviolet and visible ray

Ultraviolet rays

Visible rays

♠ Produced Lyman series

♥ Produced Balmer series

♠ Usually use to calculate ionisation energy of hydrogen gaseous atom

♥ Can be used to determine wavelength produced by each spectra given by dispersion of light

♠ Under emission spectrum, ♥ Under emission spectrum, electrons from higher electrons from higher energy energy level settled at n = 1 level settled at n = 2 ♠ Series of convergence lines ♥ Series of convergence lines produced have higher produced have lower frequency frequency

2.1.1 The Bohr Model of the Hydrogen Atom 1. Neil Bohr suggested a model for the hydrogen atom that predicted the existence of line spectra. He then outline 3 postulates about hydrogen atom where : • The H atom has only certain allowable energy levels, which Bohr called stationary states. Each of these states is associated with a fixed circular orbit of the electron around the nucleus. • The atom does not radiate energy while in one of its stationary states (atom does not change energy while the electron moves within an orbit). • The atom changes to another stationary state (the electron moves to another orbit) only by absorbing or emitting a photon whose energy equals the difference in energy between the two states: Estate A – Estate B @ ΔE =hf or ΔE =hc / λ • (where the energy of state A is higher than that of state B, and h is the constant proposed by Planck’s theory where h = 6.63 x 10-34 J s.)

• A spectral line results when a photon of specific energy (and thus specific frequency) is emitted as the electron moves from a higher energy state to a lower one. • Therefore, Bohr’s model explains that an atomic spectrum is not continuous because the atom’s energy has only certain discrete levels, or states

2. The quantum number, n (1, 2, 3, . . .) is associated with the “shell” of an electron orbit, which is directly related to the electron’s energy: the lower the n value, the closer the orbit to the nucleus, and the lower the energy level. – When the electron is in the first orbit (n = 1), the orbit closest to the nucleus, the hydrogen atom is in its lowest energy level, called the ground state. – If the hydrogen atom absorbs a photon whose energy equals the difference between the first and second energy levels, the electron moves to the second orbit (n = 2), the next orbit further from the nucleus. When the electron is in the second or any higher orbit (energy level),the atom is said to be in an excited state. The process where electron moved from ground state to higher energy level is called as absorption

• If the H atom in the first excited state emits a photon of that same energy, it returns to the ground state. The process where electron dropped from higher energy level (excited state) back to ground state is called as emission

• When a sample of gaseous Hydrogen atoms is excited, different atoms absorb different quantities of energy. Each atom has one electron, but so many atoms are present that all the energy levels (orbits) are populated by electrons. When the electrons drop from outer orbits to the n = 3 orbit (second excited state), the emitted photons create the infrared series of lines. The visible series arises when electrons drop to the n = 2 orbit (first excited state).

Ultraviolet Lyman Series

Infrared Passchen Series

Visible light Balmer Series

• When an electron drops from an outer orbit to an inner one, the atom emits a photon of specific energy that gives rise to a spectral line. In a given series, each electron drop, and thus each emission, has the same inner orbit, that is, the same value of n1 in the Rydberg equation, where the orbit radius is proportional to n2 value. • An energy diagram shows how the ultraviolet series arises. Within each series the greater the difference in orbit radii, the greater the difference in energy levels, and the higher the energy of the photon emitted. For example, in the ultraviolet series, in which n = 1, a drop from n2 = 5 to n1 = 1 emits a photon with more energy (shorter wavelength, higher frequency) than a drop from n2 = 2 to n1 = 1. [The axis shows negative values because n = ∞ is defined as the atom with zero energy.] • Since Bohr’s model is a one-electron model. It works beautifully for the H atom and for other one-electron species, such as He+ (Z = 2), Li2+ (Z = 3) and Be3+ (Z = 4),

• One of the usefulness of Bohr’s theory, applied when calculating the energy levels of an atom, which he derived from the classical principles of electrostatic attraction and circular motion, where the equation is describe as

E = − 2.18 × 10 −18

Z2  J ×  2  n 

• For hydrogen atom, since the atomic number, Z = 1. Therefore :

E = − 2.18 × 10

−18

 1  J × 2  n 

• If the ground level (under Lyman series) n = 1, the energy at ground state is

E = − 2.18 × 10

−18

1 J ×  2  ; E = − 2.18 × 10 −18 J 1 

a. Note that even though the energy value is negative, however, as mentioned above, under zero energy where E = 0 kJ when n = ∞. In terms of magnitude, more energy will be released when electron fall from n = ∞ to n = 1. If the ground state energy level is higher, lesser energy will be released. b. Derivation from equation above allowed us to find the energy difference between two energy level, where

∆E = En2 − En1 @   1    1  −18 −18 ∆E = − 2.18×10 J ×  2  − − 2.18×10 J ×  2   n2    n1   1 1 ∆E = 2.18×10 J ×  2 − 2   n1 n2  −18

c. Further derivation from equation (b) also allowed us to find the wavelength produce in absorption / emission process. Using Planck’s equation, where ΔE = hf or ΔE = hc / λ

 1 1  hc 2.18 × 10 J ×  2 − 2  = ; λ  n1 n 2  rearrange and substitute −18

 1 1 2.18 × 10 −18 J 1  = ×  2 − 2  − 34 8 λ (6.63 × 10 )(3.00 × 10 )  n1 n 2   1 1 1  7 = 1.10 × 10 ×  2 − 2  λ  n1 n 2 

• Ionisation energy of one mole of electron in hydrogen atom can also be calculated using Bohr’s equation. H (g)
H+ (g) + eΔH = + x kJ mol-1 (Ionisation energy) • In order to remove an electron from hydrogen atom, electron must at least reached the convergence limit of the energy level n = ∞. Consider the ground state energy level of Lyman series, n = 1. 1 1 ∆E = 2.18 × 10 J ×  2 − 2  ;  n1 n 2  1 1  −18 ∆E = 2.18 × 10 J ×  2 − 2  ; ∆E = 2.18 × 10−18 J 1 ∞    1 kJ For 1 mol e − ; ∆H = ∆E × N A × 3 ; 10 J −18

∆H = 2.18 × 10−18 J × (6.02 × 1023 ) ×

1 kJ −1 ; ∆ H = 1310 kJ mol 103 J

• To remove 1 mol of electron from ground state, n1 = 1 to convergence limit, n2= ∞, a total of 1310 kJ is required.

Example 1 : Calculate the wavelength of the first line of Lyman series

Example 2 : Calculate the wavelength of the third line of Balmer series

In Lyman series, n1 = 1 First line in series, so n2 = 1 + 1 = 2  1 1 1  = R H  2 − 2  λ n2   n1

In Balmer series, n1 = 2 First line in series, so n2 = 2 + 3 = 5

 1 1  7 1 = 1 .097 × 10 − 2 2  λ 2  1 λ = 122 nm

 1 1  = RH  2 − 2  λ n2   n1 1

 1 1  7 1 = 1 .097 × 10 − 2  22 5  λ 

λ = 434 nm

Example 3 : Calculate the wavelength of the forth line of Paschen series

Example 4 : Calculate the wavelength of the last line of Balmer series

In Passchen series, n1 = 3 First line in series, so n2 = 3 + 4 = 7  1 1 1  = RH  2 − 2  λ n2   n1

In Balmer series, n1 = 2 Last line in series, so n2 = 2 + ∞ = ∞

 1 1  7 1 = 1 .097 × 10 − 2 2  λ 7  3 λ = 1005 nm

 1 1   = RH  2 − 2  λ n2   n1 1

 1 1  7 1 = 1 .097 × 10 − 2 2  λ ∞  2

λ = 365 nm

Example 5 : Calculate the frequency of the second line of Paschen series

Example 6 : Calculate the frequency of the last line of Lyman series

In Passchen series, n1 = 3 Second line in series, n2 = 3 + 2 = 5

In Lyman series, n1 = 1 last line in series, so n2 = 1 + ∞ = ∞

 1 1 1   = R H  2 − 2  λ  n1 n 2 

 1 1  = RH  2 − 2  λ n2   n1 1

 1 1  7 1 = 1 .097 ×10 − 2 2   λ 3 5  

 1 1  7 1 = 1 .097 × 10 − 2 2   λ 1 ∞  

1 / λ = 7.80 x 105 m-1 f = c x (1 / λ) = (3.0 x 108)(7.80 x 105) f = 2.34 x 1014 s-1

1 / λ = 1.097 x 107 m-1 f =c x (1 / λ) =(3.0 x 108)(1.097 x 107) f = 3.29 x 1015 s-1

c) the third line of Lyman series In Lyman series, n1 = 1 First line in series, so n2 = 3 + 1 = 4

d) the fifth line of Balmer series In Balmer series, n1 = 2 Fifth line in series, so n2 = 2 + 5 = 7

 1 1 1   = R H  2 − 2  λ  n1 n 2 

 1 1  = RH  2 − 2  λ n2   n1

 1 1  7 1 = 1 .097 ×10 − 2 2   λ 1 4  

 1 1  7 1 = 1 .097 ×10 − 2 2   λ 2 7  

1 / λ= 1.028 x 107 m-1 f = c x (1 / λ) = (3.0 x 108)( 1.028 x 107) f = 3.09 x 1015 s-1

1

1 / λ = 2.519 x 107 m-1 f = c x (1 / λ) = (3.0 x 108)( 2.519 x 107) f = 7.56 x 1014 s-1

Example 3 : Using Bohr’s Equation, calculate the energy required to cause the emission of spectral line below b) the first line of Balmer series a) the fifth line of Lyman series In Lyman series, n1= 1 In Balmer series, n1 = 2 First line in series, so n2 = 1 + 5 = 6 First line in series, so n2 = 2 + 1 = 3  1 1 1 1 −18 ∆E = 2.18 × 10 −18 J ×  2 − 2  ∆E = 2.18 × 10 J ×  2 − 2   n1 n2   n1 n2   1 1 1  1  −18 −18    ∆E = 2.18 × 10 J × 2 − 2 ∆E = 2.18 × 10 J × 2 − 2 2  1 6  3     -19 ΔE = 3.03 x 10 J / e ΔE = 2.12 x 10-18 J / ec) the third line of Passchen series In Passchen series, n1 = 3 First line in series, so n2 = 3 + 3 = 6 ∆E = 2.18 × 10

−18

 1 1  J ×  2 − 2   n1 n2 

1 1   ∆E = 2.18 × 10 J × 2 − 2 3 6   ΔE = 1.82 x 10-19 J / e−18

d) the last line of Passchen series In Balmer series, n1 = 3 Last line in series, so n2 = 3 + ∞ = ∞

 1 1 ∆E = 2.18 × 10−18 J ×  2 − 2   n1 n2  1 1   ∆E = 2.18 × 10 J × 2 − 2 3 ∞   ΔE = 2.42 x 10-19 J / e−18

2.2

Atomic Orbital

• The position of electrons cannot be specified as electron behaves like wave as it extended it space. Werner Karl Heisenberg, then formulated what is now known as the Heisenberg uncertainty principle: it is impossible to know simultaneously both the momentum and the position of a particle with certainty. – However, Bohr’s theory had made a significant contribution to our understanding of atoms, and his suggestion that the energy of an electron in an atom is quantized. This concept is the perfected by an Austrian physicist, Erwin Schrödinger, through his well-known equation – Schrödinger’s equation, where the energy of atom can be calculate. – Even though Schrödinger equation specifies the possible energy states the electron can occupy in a hydrogen atom, however, it cannot pinpoint the location of electron in an atom. Therefore, to counter this problem, we replaced with the term orbital, a region with high probability to find an electron.

• An atomic orbital is specified by three quantum numbers. One is related to the orbital’s size, another to its shape, and the third to its orientation in space. The quantum numbers have a hierarchical relationship: the size-related number limits the shape-related number, which limits the orientation-related number

– The principal quantum number (n) (better known as shell) is a positive integer (1, 2, 3, and so forth). It indicates the relative size of the orbital and therefore the relative distance from the nucleus of an atom. – The angular momentum quantum number(l) is an integer from 0 to n – 1. It is related to the shape of the orbital and is sometimes called the orbital shape (or azimuthal) quantum number. Note that the principal quantum number sets a limit on the values for the angular momentum quantum number; that is, n limits l. • For an orbital (shell) with n = 1, l can have a value of only 0. • For orbitals (shell) with n = 2, l can have a value of 0 or 1 • For orbitals (shell) with n = 3, l can be 0, 1, or 2; and so forth. Note that the number of possible l values equals the value of n

• The magnetic quantum number (ml) is an integer from [–l] through 0 to [+l]. It prescribes the orientation of the orbital in the space around the nucleus (or simple, number of orbitals presence in l). The possible values of an orbital’s magnetic quantum number are set by its angular momentum quantum number. • For (l = 0), magnetic quantum number, (ml) = 0 [therefore 1 orbital] • For (l = 1), magnetic quantum number, (ml) = –1, 0, +1 [therefore 3 orbitals] • For (l = 2), magnetic quantum number, (ml) = –2, –1, 0, +1, +2 [therefore 5 orbitals] • The electron spin quantum number (ms) ~ represents the assumption of electrons act like tiny magnets. According to electromagnetic theory, a spinning charge generates a magnetic field, and it is this motion that causes an electron to behave like a magnet. Therefore, in each ml, two oppositely spin quantum is filed accordingly and has a value of +½ and –½, and are usually denote as ↑ (for +½) and ↓ (for –½)

• The energy states and orbitals of the atom are described with specific terms and associated with one or more quantum numbers – Level. The atom’s energy levels, or shells, are given by the n value: the smaller the n value, the lower the energy level and the greater the probability of the electron being closer to the nucleus. [n = 1 is the closest to nucleus, followed by n = 2, 3 and so forth] – Sublevel. The atom’s levels contain sublevels, or subshells, which designate the orbital shape. Each sublevel has a letter designation: l Name of sublevel (orbital)

0

1

2

3

4

5

s

p

d

f

g

h

• Orbital. Each allowed combination of n, l, and ml values specifies one of the atom’s orbitals. Thus, the three quantum numbers that describe an orbital express its size (energy), shape, and spatial orientation. You can easily give the quantum numbers of the orbitals in any sublevel if you know the sublevel letter designation and the quantum number hierarchy Energy Sublevel, n level, l

orbital, ml

No of orbital

Atomic Orbital Designation

0

0

1

1s

2

0 1

0 –1, 0 +1

1 3

2s 2px , 2py , 2pz,

3

0 1 2

0 –1, 0 +1 –2, –1, 0 , +1 , +2

1 3 5

3s 3px , 3py , 3pz, 3dxy , 3dyz , 3dxz , 3dx2-y2 , 3dz2

1

2. Shape of each orbitals • The s Orbital– An orbital with l = 0 has a spherical shape with the nucleus at its center and is called an s orbital

• The p orbitals – An orbital with l = 1, called a p orbital, has two regions (lobes)of high probability, one on either side of the nucleus. The nucleus lies at the nodal plane of this dumbbell-shaped orbital as described in diagram below

• The d Orbital – An orbital with l = 2 is called a d orbital. There are five possible ml values for the l = 2 value: –2, –1, 0 , +1 , +2. Thus, a d orbital can have any one of five different orientations, as describe in diagram below

• Orbitals with Higher l values– Orbitals with l = 3 are f orbitals and must have a principal quantum number of at least n = 4. There are seven f orbitals (2 l + 1 = 7), each with a complex, multi-lobed shape

• Special case of hydrogen atom – The energy state of the H atom depends only on the principal quantum number, n = 1. When an electron occupies an orbital with a higher n value, it occurs from the nucleus, so the atom is higher in energy. But the H atom is a special case because it has only one electron. The energy states of all other atoms depend on both the n and l values of the occupied orbitals because of additional nucleus – electron attractions and electron – electron repulsions. In other words, for the H atom only, all four orbitals in n = 2 (one 2s and three 2p) have the same energy, while all nine orbitals in n = 3 (one 3s, three 3p, and five 3d) have the same energy and so forth.

2.3

Electronic configuration

• Electron configuration of the atom shows how the electrons are distributed among the various atomic orbitals, in order to understand electronic behaviour of that atom. Using the principle of n, l, ml and ms learned earlier, it allows us to understand how the arrangement of electrons occurs in many-electrons atom. • The arrangement of electrons in its orbitals are guided under 3 basic rule and principles, which are Aufbau's Principle, Pauli Exclusion's Principle and Hund's Rule.

• Aufbau's principle stated that electrons are filled up in orbitals from the lowest energy orbital available. This will results in groundstate electron configurations to build up eventually

• Pauli Exclusion's Principle ~ an atomic orbital can hold a maximum of two electrons with opposing quantum spins. From the quantum spin number, we understand that electrons behaviour resemble to that of a magnet when spinning charge generates a magnetic field. In general, we represent a positive spin quantum, ms = +1/2 as ↑ (sometimes ) ; while a negative spin quantum, ms = -1/2 as ↓ (sometimes ). Correct

2 electrons occupied an orbital with opposite spin

Incorrect

2 electrons occupied an orbital with same spin

2 electrons occupied an orbital with same spin

3 electrons occupied an orbital with different spin

• Hund's Rule ~ when orbitals of equal energy are available, the electron configuration of lowest energy has the maximum number of unpaired electrons with parallel spins. In order to fulfil Hund's rule, sub-shell must have at least 2 or more orbitals. Therefore, porbitals, d-orbitals and f-orbitals are filled according to Hund's rule. For example, in filling 2 and 3 electrons in p-orbitals and filling 5 and 7 electrons in d-orbitals Filling in p - orbitals

Filling in d - orbitals

Filling in 2 electrons in p-orbitals

Filling in 5 electrons in d-orbitals (more stable)

Filling in 3 electrons in p-orbitals (more stable)

Filling in 7 electrons in d-orbitals (After positive spin is first filled, negative spin is then filled to each orbital)

Element

No of e-

Electronic configuration

Hydrogen H

1

____ 1s

1s1

Helium He

2

____ 1s

1s2

Lithium Li

3

____ ____ 1s 2s

1s22s1

Beryllium Be

4

____ ____ 1s 2s

1s22s2

Boron B

5

____ ____ ____ ____ ____ 1s 2s 2p

1s22s22p1

Carbon C

6

____ ____ ____ ____ ____ 1s 2s 2p

1s22s22p2

Nitrogen N

7

____ ____ ____ ____ ____ 1s 2s 2p

1s22s22p3

Orbital diagram

Element

No of e-

Oxygen O

8

____ ____ ____ ____ ____ 1s 2s 2p

1s22s22p4

Fluorine F

9

____ ____ ____ ____ ____ 1s 2s 2p

1s22s22p5

Neon Ne

10

____ ____ ____ ____ ____ 1s 2s 2p

1s22s22p6

Sodium Na

11

____ ____ ____ ____ ____ ____ 1s 2s 2p 3s

1s22s22p63s1

12

____ ____ ____ ____ ____ ____ 1s 2s 2p 3s

1s22s22p63s2

Al

13

____ ____ ____ ____ ____ ____ 1s 2s 2p 3s

___ ___ ___ 3p

1s22s22p63s23p1

Silicon Si

14

____ ____ ____ ____ ____ ____ ____ ____ ____ 1s 2s 2p 3s 3p

1s22s22p63s23p2

Magnesium

Mg Aluminium

Electronic configuration

Orbital diagram

Element

No of e-

Phosphorous, P

15

____ ____ ____ ____ ____ ____ ____ ____ ____ 1s 2s 2p 3s 3p

1s22s22p63s23p3

Sulphur S

16

____ ____ ____ ____ ____ ____ ____ ____ ____ 1s 2s 2p 3s 3p

1s22s22p63s23p4

Chlorine Cl

17

____ ____ ____ ____ ____ ____ ____ ____ ____ 1s 2s 2p 3s 3p

1s22s22p63s23p5

Argon Ar

18

____ ____ ____ ____ ____ ____ ____ ____ ____ 1s 2s 2p 3s 3p

1s22s22p63s23p6

Potassium K

19

____ ____ ____ ____ ____ ____ ____ ____ ____ ___ 1s 2s 2p 3s 3p 4s

1s22s22p63s23p64s1

Calcium Ca

20

____ ____ ____ ____ ____ ____ ____ ____ ____ ___ 1s 2s 2p 3s 3p 4s

1s22s22p63s23p64s2

Orbital diagram

Electronic configuration

1s22s22p63s23p63d14s2

1s22s22p63s23p63d24s2

1s22s22p63s23p63d34s2

1s22s22p63s23p63d54s1

1s22s22p63s23p63d54s2

1s22s22p63s23p63d64s2

1s22s22p63s23p63d74s2

1s22s22p63s23p63d84s2

1s22s22p63s23p63d104s1

1s22s22p63s23p63d104s2

a. Note that from Scandium to Vanadium, each electron is filled according to Hund's rule, with a single positive spin electron is filled in each 3d-subshells. b. When expressing the electronic configuration for Chromium, 24Cr, the valence electron of Cr is filled as 3d5 4s1 instead of 3d4 4s2.

c. This is due to, according to Hund's rule, half-filled 3d orbitals have extra stability, compared to a partial-filled 3d orbital.

• Another anomaly of filling the electronic configuration occur on the element copper, Cu. Supposedly, After nickel, 28Ni is filled as 1s22s22p63s23p63d84s2, Cu should be filled : 1s22s22p63s23p63d94s2. However, due to full-filled 3d orbitals have extra stability compared to a partial-filled 3d orbitals, henceforth valence electrons of Cu is filled as 3d104s1.

2.3.1 Electronic Configuration of Ions 1. Ions are formed when an atom or molecule donate / received electron(s). Ions can be positively charged or negatively charged. A positively charged ion is also known as cation, while a negatively charged ion is also known as anion. Table below compared the properties of the formation for both cation and anion Ions

Cation (Positively charged ion)

Anion (Negatively charged ion)

Occur when

Electron(s) are donated

Electron(s) are received

Formation of Na → Na+ + +1 and -1 1s22s22p63s1 1s22s22p6

e-

Formation of Mg → Mg2+ + 2e+2 and -2 1s22s22p63s2 1s22s22p6

F + 1s22s22p5

e- →

F1s22s22p6

O + 2e- → O21s22s22p4 1s22s22p6

2. From the example above, it is shown that, when electron(s) are donated, electron(s) are first removed from higher energy level, and conversely when electron(s) are received, electron(s) are filled from the lower possible energy level. Most of the main group elements donate and received electron(s) to achieve a stable valence electronic configuration of ns2 np6 (also known as octet configuration) Al3+ : 1s22s22p6 [Since electronic configuration of aluminium is 1s22s22p63s23p1, when 3 electrons are removed, it shall be removed from 3p, then 3s]

P3- 1s22s22p63s23p6

V3+ 1s22s22p63s23p63d2 [Since electronic configuration of vanadium is 1s22s22p63s23p63d34s2 when 3 electrons are removed, it shall be removed from 4s, then 3d]

N3- 1s22s22p6 [Since electronic configuration of nitrogen is 1s22s22p3, when 3 electrons are added, it shall be added to 3p orbital as its not yet complete]

[Since electronic configuration of aluminium is 1s22s22p63s23p3, when 3 electrons are added, it shall be added to 3p orbital as its not yet complete]

Fe2+ 1s22s22p63s23p63d6 [Since electronic configuration of iron is

S4-

22s22p63s23p63d64s2 1s …………………………….…when 2

22s22p63s23p4 1s is …………….,,,,,,,,,,,,, when 4 electrons

electrons are removed, it shall be

are added, it shall be added to

4s removed from …………….]

3p 4s orbitals] …………. and ………..

1s22s22p63s23p64s2

[Since electronic configuration of sulphur

Ga4+ 1s22s22p63s23p63d9 Br- 1s22s22p63s23p63d104s24p6 [Since electronic configuration of gallium [Since electronic configuration of Br is 22s22p63s23p63d104s24p5 22s22p63s23p63d104s24p1 1s 1s is …………………………….……….. when ………………………………………,

4 electrons are removed, it shall be

when 1 electrons are added, it shall be

4p , 4s , then 3d removed from ………….…………….]

4p orbitals] added to ………….

Mn4+

1s22s22p63s23p63d3

Cl- 1s22s22p63s23p6

Ca+

1s22s22p63s23p64s1

O3-

1s22s22p63s1

Co3+

1s22s22p63s23p63d6

F-

1s22s22p6

2.4 Classification of elements into s, p, d and f blocks in the Periodic Table