Explore topic-wise InterviewSolutions in Current Affairs.

Physical properties of metals: 

1. Lustrous: When we rub the surface of metals with sand paper they will shine. This property is called lustrous, 

e.g.: Iron, zinc etc. 

2. Sonority: When we hit the metal surface they give a ringing sound is called sonority, 

e.g.: Iron, copper etc. 

3. Malleability: Metals can be flattened into thin sheets. The property of flattening metals into thin sheets is called malleability, 

e.g.: Silver, Iron, copper.

4. Ductility: Metals can be drawn into wires. The property of drawing a metal to make fine wire is called ductility, 

e.g.: Silver, gold. 

5. Electric conductivity: Electricity can be easily pass through metals. So they are called good conductors of electricity. 

e.g.: Silver, copper, iron. 

6. Conductivity of heat: Metal absorbs heat quite easily. They are good conductors of heat, 

e.g.: Copper, iron.

Aluminium has far greater affinity for oxygen as compared to carbon because of which carbon cannot separate oxygen from aluminium oxide. Hence Al metal cannot be obtained by the reduction of Al₂O₃ with coke.

The metal has corroded because of being exposed to moist air. Green compound is basic copper carbonate [CuCO₃. Cu(OH)₂]

The metal is copper.

Two ways to prevent this process :

Painting, Greasing, Oiling, Galvanizing

The surface of some metals acquires a dull appearance when exposed to air for a long time because metals form a thin layer of oxides, carbonates or sulphide on their surface by the slow action of various gases present in air.

(a) Sodium metal: Soft, low melting point Carbon non-metal: graphite conducts electricity; diamond has a very high melting point.

(b) Gallium and cesium.

(i) The reaction is as follows: 

3Fe + 4H₂O → Fe₃O₄ + 4H₂ 

(ii) The reaction is as follows: 

Ca + 2H₂O → Ca(OH)₂ + H₂ 

(iii) The reaction is as follows: 

2K + 2H₂O → 2KOH + H₂

(a) Metals react with the things like moisture, oxygen, carbon dioxide to form a layer of substance like rust in iron which hides their shiny surface. This is what we call corrosion. 

(b) Nitric acid is a strong oxidising agent and Zinc is not a very reactive metal according to the reactivity series. Hence the nitric acid oxidises the hydrogen gas which is produced may be in the first step if reaction between acid or metal had taken place. So the hydrogen gas does not evolve. 

(c) Metal sulphides are insoluble in water and halides are soluble so they get washed away in spite of getting deposited over the crust.

(c) transition elements

The answer is (c) Fe₃O₄

3Fe + 4H₂O → Fe₃O₄ + 4H₂

Cu (Z = 29) Electronic configuration is [Ar] 3d¹⁰4s¹

Cu : Electronic configuration is [Ar] 3d⁹ . In Cu²⁺ , promotion of electrons take place in outer dorbital by the absorption of light form visible region involves d-d transition. Due to this Cu²⁺ compounds are coloured. Where in Zn²⁺ electronic configuration is [Ar]3d¹⁰. It has completely filled d-orbital. 

So there is no chance of d – d transition. So Zn²⁺ compounds are colourless.

Metals lose their shine or brightness on keeping in air for a long time and acquire a dull appearence due to the formation of a thin layer of oxide, carbonate or sulphide on their surface by the slow action of various gases present in air.

Brightness of metals can be restored by rubbing the dull surface of the metal object with a sand paper, then the outer corroded layer is removed and the metal object becomes shiny and bright once again.

The answer is (c) Dullness

Dulctility is one of the properties of metals which enable metals to be drawn into thin wires.

The answer is (d) (iii) and (iv)

Calcium reacts vigorously with water and forms hydrogen which will make calcium to float.

(d) Transition elements

W  is used in light bulb filaments. 

1. As we move from one element to another in 4f series (Ce to Lu) the nuclear charge increases by one unit and an additional electron is added into the same inner 4f sub-shell. 

2. 4f sub-shell have a diffused shapes and therefore the shielding effect of 4f electrons are relatively poor. Hence, with increase of nuclear charge, the valence shell is pulled slightly towards nucleus. 

3. As a result, the effective nuclear charge experienced by the 4f elelctoms increases and the size of Ln³⁺ ions decreases.

(a) Ionic compounds are made up of oppositely charged ions which are held together by strong electrostatic force of attraction. Due to this reason, they are hard solids. When large pressure is applied, they tend to break into pieces. 

(b) Conduction of electricity is achieved by the movement of charged particles. In the molten state or in their aqueous solution form, ionic compounds contain ions which can move throughout the solution/molten solid, which helps in conducting electricity.

The answer is (c) HNO₃

Nitric acid is a powerful oxidizing agent. It reacts with metal to form water.

Given : 

Mass of urea = 5.6 g 

To find : 

The number of moles and molecules of urea 

Formulae : 

i. Number of moles = \(\frac{Mass\,of\,a\,substance}{Molar\,mass\,of\,a\,substance}\) 

ii. Number of molecules = Number of moles × Avogadro’s constant

Mass of urea = 5.6 g

Molecular mass of urea, 

NH₂CONH₂

= (2 × Average atomic mass of N) + (4 × Average atomic mass of H) + (1 × Average atomic mass of C) + (1 × average atomic mass of O)

= (2 × 14 u) + (4 × 1 u) + (1 × 12 u) + (1 × 16 u) 

= 60 u

∴ Molar mass of urea = 60 g mol^-1

∴ Number of moles = \(\frac{Mass\,of\,a\,substance}{Molar\,mass\,of\,a\,substance}\)

= \(\frac{5.6g}{60g\,mol^{-1}}\)

= 0.09333 mol

[Calculation using log table : \(\frac{5.6}{60}\) 

= Antilog₁₀[log₁₀(5.6) – log₁₀(60)]

= Antilog₁₀ [0.7482 – 1.7782]

= Antilog₁₀ \(\overline{2} .9700\)

= 0.09333]

Now, 

Number of molecules of urea 

= Number of moles × Avogadro’s constant

= 0.09333 mol × 6.022 × 10²³ molecules/mol

= 0.5616 × 10²³ molecules (by using log table)

= 5.616 × 10²² molecules

∴ Number of moles of urea = 0.0933 mol 

Number of molecules of urea = 5.616 × 10²² molecules

[Calculation using log table : 

0.09333 × 6.022

= Antilog₁₀[log₁₀(0.09333) + log₁₀(6.022)]

= Antilog₁₀[ \(\overline{2} .9698\) + 0.7797]

= Antilog₁₀ \(\overline{1} .7495\)

= 0.5616]

(c) Fe and Cu

As we move across 4f series, the atomic and ionic radii of lanthanoids show gradual decrease with increase in atomic number. This decrease in ionic size is called lanthanoid contraction. 

Effects (or) Consequences of lanthanoid contraction: 

1. Basicity differences: As we move from Ce³⁺ to Lu³⁺ , the basic character of Ln³⁺ ions decrease. Due to the decrease in the size of  Ln³⁺ ions, the ionic character of Ln – OH bond decreases (covalent character increases) which results in the decrease in the basicity. 

2. Similarities among lanthanoids – In the complete fseries only 10 pm decrease in atomic radii and 20 pm decrease in ionic radii is observed. Because of this very small change in radii of lanthanoids, their chemical properties are quite similar. 

The elements of second and third transition series resemble each other more closely than the elements of first and second transition series due to lanthanoid contraction. For example, 

• 4d series – Zr – Atomic radius 145 pm 
• 5d series – Hf – Atomic radius 144 pm

i. Formula mass of KCl

= Average atomic mass of K + Average atomic mass of Cl 

= 39 u + 35.5 u 

= 74.5 u

ii. Formula mass of AgCl 

= Average atomic mass of Ag + Average atomic mass of Cl 

= 108 + 35.5 

= 143.5 u

∴ i. Formula mass of KCl = 74.5 u

ii. Formula mass of AgCl = 143.5 u

In Ionic compounds, ions are held tightly together by strong electrostatic forces of attraction making them quite rigid and hence even when a small force is applied to the compound, the force causes the ions to get displaced from the lattice and they suffer repulsion from similar charged ions in the lattice causing the ionic solids to shatter.

The answer is (c) Conc.HCl : Conc.HNO₃ is 3 : 1

(d) Pt 

Reason: Pt belongs to 5d series whereas others are 4d series.

(a) 2, 3, 4, 1

(b) 2, 3, 4, 1

Scandium has electronic configuration,

₂₁Sc : Is², 2s², 2p⁶, 3s², 3p⁶, 3d¹, 4s² Sc shows only two oxidation states namely + 2 and + 3.

• Due to the loss of two electrons from the 4s-orbital, Sc acquires + 2 oxidation state Sc²⁺ : Is² 2s² 2p⁶ 3s² 3p⁶ 3d¹
• Due to the loss of one more electron from the 3d-orbital, it acquires + 3 oxidation state with the extra stability of an inert element  ₁₈Ar. Sc⁺³ : Is² 2s² 2p⁶ 3s² 3p⁶.
• Since Sc³⁺ acquires extra stability of inert element [Ar]¹⁸, it does not form higher oxidation state. 

The answer is (b) (ii) and (iii)

HCl and CCl₄ are covalent compound hence they cannot be ionic.

(a) 5.92 BM

Mn²⁺ ⇒ 3d contains 5 unpaired electrons 

n = 5

\( = \sqrt n(n+2)BM\)

= \(\sqrt 5(5+2)=\sqrt 35\) = 5.92 BM

(a) 3, 4, 1, 2

1. Ru and Os have +8 as the highest oxidation state. 

2. The highest oxidation state of 4d and 5d elements are found in their compounds with the higher electronegative elements like O, F and Cl. For example: RuO₄ , OsO₄

1. An interstitial compound or alloy is a compound that is formed when small atoms like hydrogen, boron, carbon or nitrogen are trapped in the interstitial holes in a metal lattice.

2. They are usually non-stoichiometric compounds. 

3. Transition metals form a number of interstitial compounds such as TiC, ZrH_1.92 , Mn₄N etc.

4. The elements that occupy the metal lattice provide them new properties. 

• They are hard and show electrical and thermal conductivity 
• They have high melting points higher than those of pure metals 
• Transition metal hydrides are used as powerful reducing agents 
• Metallic carbides are chemically inert.

1. The element of second and third transition series resemble each other more closely than the elements of first and second transition series due to lanthanoid contraction. 

2. e.g., Zr – 4d series -Atomic radius 145 pm Hf – 5d series – Atomic radius 144 pm 

3. The radii are very similar even though the number of electrons increases.

4. Zr and Hf have very similar chemical behaviour, having closely similar radii and electronic configuration. 

5. Radius dependent properties such as lattice energy, solvation energy are similar. 

6. Thus lanthanides contraction leads to formation of pair of elements and those known as chemical twins, e.g., Zr – Hf

(c) [Rn] 5f^0-7 6d^0-1 7s² 

(a) 3, 1, 4, 2

Gold (Au) and Silver (Ag) are also known as Noble metals as they are less reactive and exist in their native state in nature.

The answer is (b) Electrical conductivity in solid state

In ionic compound free ions are not available in solid state hence solid ionic compounds cannot conduct electricity

The answer is (d) (iii) and (iv)

Sodium and potassium are at the top in reactivity series hence they can be refined by electrolytic refining.

1. Mn²⁺ , Fe³⁺ configuration is d⁵ .

2. µ = \(\sqrt{5(5+2)}\) = \(\sqrt{35}\) = 5.916 µ_B

Among 3d series, Mn²⁺ , Fe³⁺ have high magnetic moment as 5.916 µ_B .

Silver has the highest electrical conductivity at room temperature.

(c) 2, 4, 1, 3

1. Mn³⁺ + e^- → Mn²⁺

E° = + 1.51V 

Fe³⁺ + e^- → Fe²⁺

E° = + 0.77V

2. The high reduction potential of Mn³⁺ / Mn²⁺ indicates Mn²⁺ is more stable than Mn³⁺ . For Fe³⁺ / Fe²⁺ the reduction potential is 0.77V, this low value indicates that both Fe³⁺ and Fe²⁺ can exist under normal conditions. 

3. The drop from Mn to Fe is due to the electronic structure of the ions concerned. Mn³⁺ has 3d⁴ configuration while that of Mn²⁺ is 3d⁵ . The extra stability associated with a half filled d sub-shell makes the reduction of Mn³⁺ very feasible.

An interstitial compound or alloy is a compound that is formed when small atoms like hydrogen, boron, carbon or nitrogen are trapped in the interstitial holes in a metal lattice. 

They are usually non-stoichiometric compounds, 

e.g., TiC, ZrH_1.92 , Mn₄N. 

1. They are hard and show electrical and thermal conductivity. 

2. They have high melting points. 

3. Transition metal hydrides are used as powerful reducing agents. 

4. Metallic carbides are chemically inert.

(c) [Ar] 3d⁵ 4s¹

The answer is (c) (ii) and (iv)

Gold and silver are non-reactive metals because of they are non-reactive they exist in native state in nature.

(a)  (i) only

1. In transition series, as we move down from Ti to Zn, the standard reduction potential E⁰ _M²⁺/M³ value is approaching towards less negative value and copper has a positive reduction potential, i.e. elemental copper is more stable than Cu²⁺ . 

2. E⁰ _M²⁺/M​ value for manganese and zinc are more negative than regular trend. It is due to extra stability arises due to the half filled d⁵ configuration in Mn²⁺ and completely filled d¹⁰ configuration in Zn²⁺ . 

3. The standard electrode potential for the M³⁺ / M²⁺ half cell gives the relative stability between M³⁺and M²⁺. 

4. The high reduction potential of M³⁺/ M²⁺ indicates M²⁺ is more stable than M³⁺ . 

5. For Fe³⁺ / Fe²⁺ the reduction potential is 0.77 V, and this low value indicates that both Fe³⁺ and Fe²⁺can exist under normal condition. 

6. Mn³⁺ has a 3d² configuration while that of Mn²⁺ is 3d⁵ . The extra stability associated with a half filled d sub-shell makes the reduction of Mn³⁺ very feasible [E° = +1.51 V]

1. In general, atomic radius decreases along a period. But for the 3d transition elements, the expected decrease in atomic radius is observed from Sc to V , thereafter upto Cu the atomic radius nearly remains the same.

2. As we move from Sc to Zn in 3d series, the extra electrons are added to the 3d orbitals, the added 3d electrons only partially shield the increased nuclear charge and hence the effective nuclear charge increases slightly.

3. However, the extra electrons added to the 3d sub shell strongly repel the 4s electrons and these two forces are operated in opposite direction and as they tend to balance each other, it leads to constancy in atomic radii.

1. An alloy is formed by blending a metal with one or more other elements. The elements may be metals or non-metals or both.

2. The bulk metal is named as solvent, and the other elements in smaller portion is called solute. 

3. According to Hume – Rothery rule to form an alloy, the difference between the atomic radii of the solvent and solute is less than 15%. Both the solvent and solute must have the same crystal structure and valence and their electro negativity difference must be close to zero.

4. Since their atomic sizes are similar and one metal atom can be easily replaced by another metal atom from its crystal lattice to form an alloy. The alloys are hard and have high melting points. Examples – Gold – copper alloy.