Explore topic-wise InterviewSolutions in Current Affairs.

Correct option is A) conductors

Correct option is D) Detergent solution

(a) Zinc sulphate.

(b) Iron (II) sulphate.

(c) Copper (II) sulphate.

(d) Sodium sulphate.

Acids : 

1. Lemon juice 

2. Tamarind juice 

Bases :

1. Soap water 

2. Surf water 

3. Lime water 

Salts : 

1. Salt water

The crystals of washing soda undergo efflorescence because it is hydrated salt containing 10 molecules of water of crystallization. When the crystals of washing soda exposed to air loose water of crystallization and form white powder of sodium carbonate monohydrates.

Na ₂CO₃ .10 H₂ O ------air---- --> Na₂ CO₂. H₂ O + 9H₂O

The loss of water of crystallization a substance when exposed to air is called efflorescence.

When crystals of washing soda (Na₂SO₄.10H₂O) are kept open in the air for long time then the water of crystallization will lose and the crystals will turn to white. This process of loosing water of crystallization when kept in air is called efflorescence. 

Na₂SO₄.10H₂O → Na₂SO₄ + 10H₂O

1. CaO + 2HCl  →  CaCl₂ + H₂O

2. Zn + S  →  ZnS

3. Fe + 2HCl  →  FeCl₂ + H₂

4. PbO + 2HNO₃  →  Pb(NO₃)₂ + H₂O

5. CuO + H₂SO₄   →  CuSO₄+ H₂O

1.  CuSO₄ (aq) + Na₂CO₃ (uq) → Na₂SO₄ (aq) + CuCO₃ (s)

2. 2NH₄OH(aq) + H₂SO₄ (aq) → (NH₄)₂SO₄ + 2H₂O

Three different classes of substances are a metal, a base and a metallic carbonate, which react with an acid to form salts.

(i) Zinc, a metal reacts with dilute sulphuric acid to form zinc sulphate and hydrogen gas is liberated.

Zn + H₂SO₄ → ZnSO₄ + H₂

(ii) Sodium hydroxide, a base reacts with dilute nitric acid to form sodium nitrate and water.

NaOH + HNO₃ → NaNO₃ + H₂O

(iii) Magnesium carbonate a metallic carbonate reacts with dilute hydrochloric acid to form magnesium chloride, water and carbon dioxide gas is liberated. 

When exposed to air washing soda crystals lose their water of crystallisation and become amorphous.

∴ The phenomenon is called efflorescence

1. A deliquescent salt  = MgCl₂

2. An insoluble chloride = AgCl

(a) The name given to the water in the compound copper sulphate-5-water is water of crystallization.

(b) The anhydrous copper sulphate is white in colour.

(c) By adding dehydrating substances such as conc. sulphuric acid as they remove the water of crystallisation.

(d) anhydrous calcium chloride

(i) → (B) Neutralisation

(ii) → (E) Direct synthesis

(iii) → (D) Double decomposition

(iv) → (A) Simple displacement

(v) → (C) Decomposition by acid.

(a) Deliquescent

(b) Deliquescent

(c) None of these

(d) None of these

(e) None of these

(f) Efflorescent

(g) None of these

(h) Highly deliquescent. 

(i) Due to high ionization enthalpy and small size, Be does not react with hydrogen by direct heating.

(b) A - 3, B - 4, C - 1, D - 2

(a) Baking soda.

(b) Washing soda.

Na₂CO₃.10H₂O - Sodium carbonate decahydrate.

Na₂CO₃ - Anhydrous sodium carbonate

10H₂O - Water of crystallization which impacts shape and colour to the crystals.

Preparation:

NaCl + H₂O + CO₂ + NH₃ → NH₄CI +NaHCO₃  

2NaHCO₃ +Heat→ Na₂CO₃ + H₂O + CO₂  

Na₂CO₃ + 10H₂O → Na₂CO₃.10H₂O

Uses:

(i) Used in glass, soap and paper industries.

(ii) Used in the manufacture of borax.

(iii) Used as a cleansing agent for domestic Purpose.

(i) Water of crystallization.

(ii) (a) White. (b) By adding cone. H₂SO₄ .

(iii) Compounds that take up enough water from the air to dissolve in the water, are called deliquescent, e.g. Calcium chloride (CaCl₂) and sodium hydroxide are deliquescent.

(iv) Calcium chloride.

(i) Sodium carbonate. 10-water.

(ii) Copper sulphate. 5H₂O.

(iii) Anhydrous calcium chloride.

(iv) A white precipitate of calcium carbonate is seen.

(v)  CaCl₂ (aq.) + Na₂CO₃ (aq.) →  CaCO₃ (White ppt) + 2NaCl (aq.)

(i) CuSO₄ . 5H₂O is blue in colour.

(ii) Na₂CO₃ .10H₂O is efflorescent in nature. 

(iii) CaO is hydroscopic in nature.

(iv) Anhydrous calcium chloride is deliquescent in nature.

(v) A white ppt. of calcium carbonate appears when the solutions of CaCl₂ and Na₂CO₃ are mixed. The white ppt. gradually settles at the base of test tube. 

The formula of soda ash is (iv) Na₂CO₃

Correct option is b. 5.74 g

Correct answer is W = 4.87 g

Isoelectronic species: Species having same number of electrons.

F^- ……..Ne       Ar……..K⁺

(ii) Assertion and reason both are correct statements and reason is correct explanation of assertion.

(i) Na < Mg > Al < Si

(b) Reason is correct explanation for assertion.

The electronic configurations of Na and Mg are: Na:1s² 2s² 2p⁶ 3s¹ And Mg: 1s² 2s² 2p⁶ 3s² thus, the first electron in both the cases has to be removed from the 3s orbital, but the nuclear charge of Na is lower than that of Mg, therefore, the first ionization energy of Na is lower than that of Mg. But after the removal of the first electron, Na acquire the nearest noble gas configuration (Ne) which makes Na highly stable, but an electron is still left in the p orbital of Mg. Hence second ionization energy of Na is higher than that of Mg.

Electronic configuration of ₁₃Al = 2,8, 3

= 1s² , 2s² , 2p⁶ , 3s² , 3p¹

Electronic configuration of ₃₁Ga = 2, 8, 18, 3

= 1s² , 2s² , 2p⁶ , 3s² , 3p⁶ , 3d¹⁰ , 4s² , 4p¹

In Ga, 10 electrons present in 3d- sub shell do not shield the outer electrons from the nucleus effectively. As a result, effective nuclear charge in Ga increases. So, the ionization enthalpy of Ga is slightly higher than AL.

(i), (iv)

(i) Electronegativity

(iv) Metallic character

(a) They all have 10 electrons, i.e they are isoelectronic species. 

(b) Al³⁺< Mg²⁺< Na+< F^-< O^2-< N^3-

(b) Correct explanation: in Al, 3p electron is to be removed while in Mg, a 3s electron is to be removed .

(ii), (iii)

(ii) B < C < N < O (increasing first ionisation enthalpy)

(iii) I < Br < Cl < F (increasing electron gain enthalpy)

Number of electrons in Na⁺ = 11 – 1 = 10

Number of another element having 10 electrons = Ne

\(\therefore\) Ne is iso-electronic with Na⁺ .

(ii), (iii)

(ii) K⁺, Ca²⁺, Sc³⁺ , Cl^–

(iii) P3^–, S^2–, Cl^–, K⁺

The electronic configurations of the elements are:- 

Z(9) : 1s² 2s² 2p⁵ 

Z(12) : 1s² 2s² 2p⁶ 3s² 

Z(36) : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰4s² 4p⁶ 

(a). The element with Z =9 (F), is highly electronegative because it requires only one electron to have the configuration of nearest noble gas element and has also very small size. It is infact, the most electronegative (4.0) in the period table. 

(b). The element with Z= 36 (Kr) is an inert gas element as it has ns² p⁶ electronic configuration. 

(c). The element with Z =12 (Mg) is highly electropositive in nature since by using two valence electrons, it will have the configuration of the noble gas Neon. 

As we move down from top to bottom in a group, electro negativity decreases due to increased atomic radius. Fluorine has the highest value of clectro negativity among all the elements.

The elements of a group show the following important similar characteristics: 

(i) Electronic configuration- All elements in a particular group have similar outermost electronic configuration e.g., all elements of group 1, i.e., alkali metals have ns¹ configuration in their valency shell. Similarly group 2 elements (alkaline Earths) have ns² outermost electronic configuration and halogens (group 17) have ns² np⁵ configuration (where n is the outermost shell). 

(ii) Valency-- The valency of an element depends upon the number of electrons in the outermost shell. So elements of a group show the same valency, e.g., elements of group 1 show +1 valency and group 2 show +2 valency i.e., NaCl, MgCl2 etc. 

(iii) Chemical properties-The chemical properties of the elements are related to the number of electrons in the outermost shell of their atoms. Hence, all elements belonging to the same group show similar chemical properties. But the degree of reactivity varies gradually from top to bottom in a group. For example, in group 1, all the elements are highly reactive metals but the degree of reactivity increases from Li to Cs. Similarly elements of group 17, i.e., halogens: F, CI, Br, I are all non-metals and their reactivity goes on decreasing from top to bottom.

Modern Periodic Law: This law states that, the physical and chemical properties of the elements are the periodic function of their atomic numbers’.

Main features of the long form of the Periodic Table are given as below:

(i) This periodic table has 18 vertical columns called the groups, which are numbered from 1 to 18.

(ii) Elements of group 1, 2 and 13 to 17 are called main group or typical or representative elements. Elements of group 1 and 2 are also known alkali metals and alkaline earth metals respectively.

(iii) Elements of group 3 to 12 are called transition elements.

(iv) Lanthanoids (elements with atomic number 58 to 71) and actinoids (elements with atomic number 90 to 103) are placed in two separate rows below the periodic table in order to avoid the unnecessary sidewise expansion of the periodic table.

(v) This periodic table contains seven horizontal rows, known as periods.

The ionic radius of the cation is always smaller than the parent atom because the loss of one or more electrons increases the effective nuclear charge. As a result, the force of attraction of nucleus for the electrons increases and hence, the ionic radii decreases. In contrast, the ionic radius of an anion is always larger than its parent atom because the addition of one or more electrons decreases the effective nuclear charge. As a result, the force of attraction of the nucleus for the electrons decreases and hence the ionic radii increases

(i), (iii), (iv)

(i) Helium has the highest first ionisation enthalpy in the periodic table.

(iii) Mercury and bromine are liquids at room temperature.
(iv) In any period, atomic radius of alkali metal is the highest.

Highly charged cations such as 3⁺ and 4⁺ ions are formed by the removal of 3 and 4 erespectively from the neutral atoms. For this purpose, a large amount of energy( E₁+E₂+E₃ OR E₁+E₂+E₃+E₄) is required.

They have fully filled s-orbitals and the additional electron cannot be placed in the much higher energy p-orbitals of valence shell.

Electro negativity decreases down the group due to increase in atomic size. It increases along the period from left to right due to decrease in atomic size. Fluorine (F) has highest electro negativity.