Chapter 01: Periodic Table and Periodic Properties

The amount of energy which is required to remove one valence electron from one mole of isolated gaseous atoms to form mono-positive gaseous ions is called 1st ionization energy. Example: refilm. Na (g) → + Na(g) ΔH =+ 496 kJ/mol

Sulfur has four electrons in its 3p sub- shell while phosphorus has three electrons in its 3p sub-shell which makes phosphorus more stable than sulfur due to half filled stability rule, as shown in given example. 1s P=[ Ne] 3s 3p, 3p, 3p. 16S = [Ne] 3s 3p? 3p, 3p. So, it is easy to remove one electron from sulfur but difficult to ionize phosphorus.

The elements in group 13 to 17 are called p-block elements because their valence electrons are ended up in p-sub shell. According to valence electrons the elements are classified as s, p, d and f-block elements.

See Q11. of theory. 2.4 Explain with the help of equations acidic and basic behaviour of oxides and chlorides. co,+H2O->H,CO3

Factors affecting electronegativity are: Atomic size: (i): Greater the atomic size, lesser will be the electronegativity. 1 E.N oc. Atomic Size (ii) Effective nuclear charge (Z-effect) Greater the value of effective nuclear charge, larger will be the electronegativity. E.N oc Effective nuclear charge

responsible for increasing the reactivity of alkali metals are "increase in atomic size" and "low ionization energy" down the group. As we know alkali metals have only one valence electron in their shells, a low remove ionization energy is required to valence electron and hence their reactivity increases. Increase in atomic size down the group also makes them reactive.

See Q8. of theory.

Some of the elements can show variable oxidation numbers because they can expand their octet by exciting their electrons to the empty orbitals if they have: If the. elements do not have empty orbitals, then they cannot show variable oxidation number. Examples: Oxidation number of P is +3 in PCl3 and +5 in PCls, while Na, Mg, Ca cannot show variable oxidation states. 07. Identity the element which is in period 5 and group 15? Ans. X is the element present in 5"h period. So, its valence shell configuration ended up 'in 5t* shell having principal quantum number 5. And its electron is showing its group number "15" X= 1s 2s 2p'3s 3p°45 3d°4p"554d"5p? n = 5, 5"h period I5th group 55 4d' 5p 2 + 10 + 3 = 15 X = Sb=51

See Q7. of theory.

Na and Mg are present at the left side of the periodic table and they are metals as well. So, that's why their oxides are more ionic in nature. While moving to the right side of the periodic table ionic character is decreased and the elements are non-metals as well. So, their ionic character decreased like in nitrogen and phosphorus oxides. Give reasons for the different

Both Na and oxygen. Na gives golden yellow flame and forms peroxide while in limiting amount of oxygen Na can form normal oxide. 2Nas) + 02(g) Na,02(s) (white solid) (excess) (peroxide) 4Na s) + 02(g) →2Na,O(s) (white solid) (limited) (normal oxide) Mg gives intense white flame with oxygen 2Mg(s) + 02ig) →2MgO (white solid) (normal oxide) Both Na and Mg give chlorides with chlorine. 2Na(s) + Cl2 2(g) 2NaCl (s) (white solid) Mg(s) +Cl 2(g) MgCl 2(s) (white to colourless solid)

to lose electron easily. While helium is a noble gas and has filled valence shell, so its ionization energy is very high as compared to lithium despite the fact that helium has +2 charge while lithium has +3 nuclear charge.

electronic configuration of Be which has filled 2s sub-shell and unstable configuration of B which has one electron in its 2p sub- shell. 4Be=15 2s sB=15 2s 2p}

The common thing in Nat, Mgt, Al+ t, Ne° and F is that all these are iso- electronic species and have 10 electrons in them. Their increasing sizes order is Al't

Neutral chlorides: These chlorides in which chloride ion and metal ions are surrounded by water (hydrates) are called neutral chlorides. In these chlorides, they simply ionizes and do not react with water. Their pH near to 7. (pH = 7) NaCl (s) → Na (ag) + Cl (ag) MCl 2(5) - → M8 (ag) + 2C (ag) (pH = 6.5) Acidic chlorides These chlorides which can react with water and produced H+ ion to give acidic solutions. Chlorides of Al to S from IIIA to VIA give acidic chlorides. This process is called hydrolysis: of the middle element is the average of atomic' masses of first and third elements. In this way he noticed some periodicity in the elements. Example Li, ' Na, "K 7+39 Na = - - = 23 2

By 1700 A.D., only 12 elements were recognized Gold (Au). Silver (Ag), Copper (Cu), Iron (Fe), Lead(Pb), Tin(Sn), Mercury (Hg), Phosphorus (P), Sulfur (S), Carbon (C), Zinc (Zn) and Arsenic (As).

In 1829, Dobereiner grouped three elements in such a way that the atomic mass

After every eight elements the next element show periodicity with first element if they are arranged on the basis of their atomic masses. It is called Newland's Octaves.

periodic law, it is stated as: Properties of the elements are the periodic functions of their atomic masses.

In 1869, Lother Meyer developed the famous curves between atomic weight and atomic volume of the elements, These curves also showed periodicity. Periodic Moseley's

Properties of the elements are the periodic functions of their atomic numbers. This Moseley's law also called modern periodic law.

There are 7 periods and 18 groups, vertical columns are groups while periods are horizontal row. 18 groups are further divided into 8-A and 10-B group

Elements which have the tendency to lose electron easily are called metals. e.g: Li, K, Cs etc Elements which have the tendency to accept electrons easily are called non-metals. e.g: N, O, Fetc.

Elements which have the properties some of the metals and some of the non metals are called metalloids: They are also referred as semi metals. Position: They are present in the P-block in "Stair-step line" arrangement including B from IIIA and to the Po of VIA

Elements can be classified as s, p, d and f-block elements, because of their valence electrons are ended up in respective sub shell. If the valence electron are ended up in s-sub shell, then they are called s-block elements. And if their valence electrons are present in p-sub shell then they are called p- block elements. and

The series of 14 elements after the Lanthanum -57 element • called are Lanthanides and the 14 elements after the element Actinium -89 are called Actinides. Both these series are also called f-block element

Some of the families in the periodic table are: • Alkali metals (Li, Na, K, Rb, Cs, Fr) • Alkaline earth metals (Be, Mg, Ca, Sr, Ba, Ra) • Transition metals (Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu, Zn) • Chalcogens (O, S, Se, Te, Po, Lv) • Halogens (F, Cl, Br, I; At, Ts) • Noble gases (He, Ne, Ar, Kr, Xe, Rn, Og)

Chalogens word derived from two Greek words "Chalcos" means "copper" and "gen" means "form". So, group 16 elements are called Chalcogens are also called copper giver. Elements of this group are O, S, Se, Te, Po, Lv.

The word "halogen" means "salt- former" because these elements easily react with alkali metals and alkaline earth metals to form stable halide salts. Elements include in this group are F, CI, Br, I, At, Ts.

Due to their stable electronic configuration (complete outer most shell), they are almost entirely unreactive under normal conditions and rarely form compounds. Elements in this group are He, Ne, Ar, Kr, Xe, Rn, Og. Anions When an atom carries negative charge, it forms anion. It forms by accepting electron. It's size is greater than its parental size. eg: F, CI IA IVA B Be Si Al Mg AH, = +1403kJ/ mol Na(g) → Nag) tè (8) AH, = +1365kJ / mol → (g) tè

Cations When an atom carries positive charge, it forms cation. It forms by losing electron. It's size is smaller than its parental size. eg: Lit, Nat

Elements diagonally positioned in the periodic table show some similarities in "diagonal relationship" despite the fact that they are present in different groups. For example lithium and magnesium, sodium and calcium have some common physical and chemical properties like sizes of their atoms and ions, charge densities, polarizing powers etc. 1A Li Period 2 Na Period 3 ionization Energy, Electron Affinity

It is the amount of energy which is an electron from required to remove outermost shell of an isolated gaseous atom to form positive ion. Na(g) → Nag) té Д, = +496 kJ/ mol

Those electrons which are present in same orbital experience some repulsion. It is easy to remove electron that is present in the paired form as compared to those electrons which are present in unpaired form due to spinning of electrons. It is called "Spin- paired repulsion".

He has 4 electrons as 1s2 -, 2s which makes it stable configuration of filled s-sub as shell while B has five electron 25 and it's configuration 2p! shows that it is relatively unstable due to one electron in its p-sub shell. That's why its value is 801kJ/mol while that of Be is 899kJ/mol even that it has smaller size than Be.

Electron affinity value increases in a period as we move from left to right, because of smaller size and increase in effective nuclear charge (z-effect or proton effect). Electronegativity electronegativity. Give

It is the power of an atom to attract the shared pair of electrons towards itself in a molecule trend of electronegativity in the 2" period is increased from left to right. 2nd F Li Be B C NO Period 4 1 E.N 1.5 2 2.5 3 3.5 Metallic Character

tendency to lose electron. It increases down the group and decreases in a period. 5J- Oxide

in air to give peroxide while in limited amount of oxygen it forms normal oxide. (Peroxide) 2Na+Oz→Na,0z (Normal oxide) 4Na+Oz→2Na,O

Non metallic oxides are acidic in nature because when they react with water, they form acids. DESCRIPTIVE QUESTIONS (EXERCISE)

Metallic oxides are basic in nature because when they react with water, they form base. Na, O+ H2O →→2NaOH

These oxides which act as an acid and as well as a base are called amphoteric oxides. Example: ZnO, Al203 and BeO Al,O, + 6HCe-2ACCC,+3H2O Al,O,+2NaOH-2NaACO, +H2O Oxidation State

4(O.N of P) +6(O.N of O) = 0 4(O.N of P) +6(-2) =0 4(O.N of P)-12 = 0 4(O.N of P) = 12 12 P=→= +3 P.O10: 4(ON of P) + 10(ON of 0) = 0 4(O.N of P) +10(-2) =0 4(O.N of P)-20=0 4P-20=0 20 P = - 4 P= +5 Ans. See Q12. and Q13. of theory.