Explore topic-wise InterviewSolutions in Current Affairs.

Ramesh is a grateful and thankful person.

Ramesh’s friends called him ‘the Couple ’Man Friday’ because he used to run errands for them and do odd jobs like going to the post office, buying fruit and vegetables, watering the plants and pumping the water out of the tank.

I could help old people by satisfying their needs. Going to shop, post office.

for a single day and treated him like his own son. If Ramesh was late in paying his college fees, they would spare some money from their pension and help him out.

A) Galvanizing

(D) Heating copper wire in presence of air at high temperature.

The correct option is (a) It is a decomposition reaction and endothermic in nature

It is a decomposition reaction. Since heat energy is supplied, the reaction is of endothermic nature

The lesson ‘The greatefultenant’ is written by SUDHA MURTHY. She describes a grateful tenant, Ramesh who had named his house ‘ShyamkamaP, which is the combination of the names of two people Shyama Rao and Kamala. when he was staying in Dharwad as tenant their. They did not treat him like a tenant but as their own child. Ramesh with them for six years. The old couple used to take care of him when he was studying at college. They shared their meals with him. And even they would even lend some money from their meager pension and help him pay his college fees. Their love and generosity changed the life of Ramesh. To show his gratitude, he named his house ‘SHYAMKAMAL’. People like Ramesh reaffirm our faith in humanity.

(D) Helium or nitrogen

To prevent rancidification of foods containing fats and oils, the packed food is surrounded by unreactive gas (like helium, nitrogen, etc.). The inert atmosphere thus created prevents oxidation of fats and oils.

The correct option is (d) Helium or nitrogen

• Helium, as well as nitrogen, can be used for storage of a fresh sample of an oil for a long time because they abstain the oil from rancidity. 
• The contact of air and oil is prevented by using nitrogen or helium as blanketing gas.

Correct option is B) 2, 3

• Elements present in the first period i.e., H and He contain only one shell which is also their valence shell and can accommodate maximum two electrons.
• As first shell can accommodate only two electrons, first period ends at He which has a complete duplet. Hence, first period in the modem periodic table contains only two elements.

Correct option is C) 2 or 3

• The periodic trends are explained in the terms of two fundamental factors, namely, attraction of extranuclear electrons towards the nucleus and repulsion between electrons belonging to the same atom.
• These attractive and repulsive forces operate simultaneously in an atom.
• This results in two interrelated phenomena called effective nuclear charge and screening effect in an atom.

i. In a multi-electron atom, the positively charged nucleus attracts the negatively charged electrons around it, and there is mutual repulsion between the negatively charged extranuclear electrons.

ii. The repulsion by inner shell electrons results in pushing the outer shell electrons further away from the nucleus. 

Thus, the outer shell electrons are held less tightly by the nucleus.

iii. As a result, the attraction of the nucleus for an outer electron is partially cancelled and hence, an outer shell electron does not experience the actual positive charge present on the nucleus. This effect of the inner electrons on the outer electrons is called screening effect or shielding effect.

iv. The net nuclear charge actually experienced by an electron is called the effective nuclear charge (Z_eff).

The effective nuclear charge is lower than the actual nuclear charge (Z).

v. Effective nuclear charge (Z_eff) = Z – electron shielding

= Z – σ

Here σ (sigma) is called shielding constant or screening constant and the value of σ depends upon type of the orbital that the electron occupies.

• The fourth period corresponds to the filling of fourth shell, n = 4.
• Therefore, it begins with filling of 4s subshell. The first two elements of the fourth period are K (Z = 19) : [Ar] 4s¹ and Ca (Z = 20) : [Ar] 4s².
• According to the aufbau principle, the next higher energy subshell is 3d, which can accommodate up to 10 electrons. Thus, filling of the 3d subshell results in the next 10 elements of the fourth period i.e., from Sc (Z = 21) : [Ar] 4s²3d¹ to Zn (Z = 30): [Ar] 4s²3d¹⁰.
• After filling of 3d subshell, the electrons enter the 4p subshell which can accommodate maximum 6 electrons. Hence, filling of 4p subshell results in the next 6 elements i.e., from Ga (Z = 31): [Ar] 4s²3d¹⁰ 4p1 to Kr (Z = 36): [Ar] 4s² 3d¹⁰ 4p⁶.
• Thus, the elements in fourth period are: 2 elements (with 4s subshell), 10 elements (with 3d subshell) and 6 elements (with 4p subshell).
• Hence, there are 18 elements in the fourth period of the modem periodic table.

When zinc granules are heated with NaOH solution, sodium zincate is formed with the evolution of hydrogen gas.

2NaOH(aq) + Zn(s) +heat  Na₂ZnO₂(aq) + H₂(g)

The main product formed in this reaction is H, gas

The group, period and the block of the element can be determined on the basis of its electronic configuration.

i. Period: The principal quantum number of the valence shell corresponds to the period of the element.

e. g. The principal quantum number (n) of the valence shell (3s¹) of Na (1s² 2s² 2p⁶ 3s¹) is 3.

This corresponds to third period.

ii. Block: The subshell in which the last electron enters, corresponds to the block of the elements (with exception being He). 

e. g. The subshell 3d (in which the last electron enters) for Sc (1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹ 4s²) corresponds to d block.

iii. Group: The group of the element is determined on the basis of number of electrons present in the outermost or penultimate [next to outermost, i.e. (n-1)] 

shell:

• For s-block elements, group number = number of valence electrons.
• For p-block elements, group number = 18 – number of electrons required to complete octet.
• For d-block elements, group number = 2 + number of (n-1)d electrons.

i. Filling of electrons in the second period:

• In the second period, electrons are filled in the second shell i.e., n = 2.
• This shell can accommodate a maximum of eight electrons and gets filled as the atomic number increases along the second period.
• The second period begins with Li (Z = 3): 1s² 2s¹ and ends up with Ne (Z = 10): 1s² 2s² 2p⁶.
• Neon has complete octet with 8 electrons in its valence shell. Therefore, the second period contains eight elements.

ii. Filling of electrons in the third period:

• The third period corresponds to the filling of the third shell i.e. n = 3.
• The third period also contains eight elements.
• It begins with the filling of electrons in the first element Na (Z = 11) : [Ne] 3s¹ and ends with the last element Ar (Z = 18) = [Ne] 3s² 3p⁶.
• The condensed electronic configurations for the elements of third period is [Ne] 3s^1-2 3p^1-6.

i. Across a period:

• As we move across a period, atomic number increases by one and thus, actual nuclear charge (Z) increases by +1 at a time.
• However, the valence shell remains the same and the newly added electron gets accommodated in the same shell. There is no addition of electrons to the core i.e., inner shells. Thus, shielding due to core electrons remains the same even though the actual nuclear charge increases.
• As a result, the effective nuclear charge (Z_eff) goes on increasing across a period.

ii. Down a group:

• As we move down a group, a new larger valence shell is added. As a result, there is an additional shell in the core.
• The shielding effect of the increased number of core electrons outweighs the effect of the increased nuclear charge. Thus, the effective nuclear charge experienced by the outer electrons decreases largely down a group.
• Hence, the effective nuclear charge (Z_eff) decreases down a group.

Option : c. principal quantum number

a. i. Across a period, the screening effect is the same while the effective nuclear charge increases.

ii. As a result, the outer electron is held more tightly and hence, the ionization enthalpy increases across a period.

iii. Since the alkali metals are present in the group 1 of the modem periodic table, they have low ionization energies.

b. i. Across a period, the screening effect is the same and the effective nuclear charge increases.

ii. As a result, the outer electron is held more tightly and hence, the ionization enthalpy increases across a period.

iii. Inert gases are present on the extreme right of the periodic table i.e., in group 18. Also, inert gases have stable electronic configurations i.e., complete octet or duplet. 

Due to this, they are extremely stable and it is very difficult to remove electrons from their valence shell.

Hence,

Inert gases have exceptionally high ionization potential.

c. The less electron affinity of fluorine is due to its smaller size. 

Adding an electron to the 2p orbital in fluorine leads to a greater repulsion than adding an electron to the larger 3p orbital of chlorine.

Hence,

Fluorine has less electron affinity than chlorine.

d. i. Noble gases have completely filled valence shell i.e., complete octet (except He with complete duplet).

ii. Since their valence shell contains eight electrons, they experience greater electronic repulsion and this results in increased atomic size (atomic radii) of the noble gas elements. 

Hence,

Noble gases possess.

i. Across a period:

a. As we move across a period from left to right in the periodic table, the effective nuclear charge increases steadily.

b. Hence, due to the increase in effective nuclear charge, the tendency to attract shared electron pair in a covalent bond increases i.e., electronegativity increases from left to right across a period.

e. g. Li < Be < B < C < N < O

ii. Down a group:

a. As we move down the group from top to bottom in the periodic table, the size of the valence shell goes on increasing.

b. However, the effective nuclear charge decreases as the shielding effect of the core electrons increases due to the increase in the size of the atoms.

c. Thus, the tendency to attract shared electron pair in a covalent bond decreases, decreasing the electronegativity down the group.

e.g. F > Cl > Br > I > At.

The general outer electronic configuration of the d-block elements is ns^0-2 (n-1)d^1-10 while the general outer electronic configuration of the f-block elements is ns² (n-1)d^0-1 (n-2)f^1-14.

Option : d. bottom

The atoms or ions which have the same number of electrons are called isoelectronic species.

i. Variation in atomic size down the group : 

a. As we move down the group from top to bottom in the periodic table, the atomic size increases with the increase in atomic number. 

b. This is because, as the atomic number increases, nuclear charge increases but simultaneously the number of shells in the atoms also increases.

c. Asa result, the effective nuclear charge decreases due to increase in the size of the atom and shielding effect increases down the group. 

Thus,

The valence electrons experience less attractive force from nucleus and are held less tightly.

d. Hence, the atomic size increases in a group from top to bottom.

e. g.

• In group 1, as we move from top to bottom i.e., from Li to Cs, a new shell gets added in the atom of the elements and the electrons are added in this new shell.
• As a result of this, the effective nuclear charge goes on decreasing and screening effect goes on increasing down a group.
• Therefore, The atomic size is the largest for Cs and is the smallest for Li in group 1.

[Note : Atomic radii of Li and Cs are 152 pm and 262 pm respectively.]

ii. Variation in atomic size across a period : 

a. As we move across a period from left to right in the periodic table, the atomic size of an element decreases with the increase in atomic number.

b. This is because, as the atomic number increases, nuclear charge increases gradually but addition of electrons takes place in the same shell.

c. Therefore, as we move across a period, the effective nuclear charge increases but screening effect caused by the core electrons remains the same.

d. As a result of this, attraction between the nucleus and the valence electrons increases. 

Therefore,

Valence electrons are more tightly bound and hence, the atomic radius goes on decreasing along a period resulting in decrease in atomic size.

e. g.

• In the second period, as we move from left towards right i.e., from Li to F, the electrons are added in the second shell of all the elements in second period (except noble gas Ne).
• As a result of this, the effective nuclear charge goes on increasing from Li to F, however, screening effect remains the same.
• Therefore, the atomic size is the largest for Li (alkali metal) and is the smallest for F (halogen).

[Note : Atomic radii of Li and F are 152 pm and 64 pm respectively.]

i. The radii of isoelectronic species vary according to actual nuclear charge. Larger nuclear charge exerts greater attraction on the electrons and thus, the radius of that isoelectronic species becomes smaller.

ii. For example, F^- and Na⁺ both have 10 electrons but the nuclear charge on F^- is +9 which is smaller than that of Na+ which has the nuclear charge +11.

Hence, F has larger ionic radii (133 pm) than Na⁺ (98 pm).

A cation (or positively charged ion) is formed by the removal of one or more electrons from the atom of an element whereas an anion (or negatively charged ion) is formed when the atom of an element gains one or more electrons.

i. Li⁺ has smaller ionic radius than K⁺ 

ii. F^- has smaller ionic radius than N^3-.

Option : b. halogens

a. Fe²⁺ has a larger size than Fe³⁺. 

b. Ca²⁺ has a larger size than Mg²⁺.

• Atomic radius decreases across the period. Hence, the atomic radius of Mg is larger than that of Al.
• Parent atoms have larger radius than their corresponding cations. Hence, the radius of Mg is larger than that of Mg²⁺ and the radius of Al is larger than that of Al³⁺.
• Mg²⁺ and Al³⁺ are isoelectronic. Among isoelectronic species, the one with larger nuclear charge will have smaller radius. Al³⁺ (Z = 13) has a larger nuclear charge than that of Mg²⁺ (Z = 12). Hence, the ionic radius of Al³⁺ is smaller than Mg²⁺.
• Therefore, the decreasing order of radius is Mg > Al > Mg²⁺ > Al³⁺.

Hence, species with the largest size is Mg and with the smallest size is Al³⁺.

• A cation is formed by the loss of one or more electrons, therefore, it contains fewer electrons that the parent atom but has the same nuclear charge.
• As a result, the shielding effect is less and effective nuclear charge is larger within a cation. Thus, radius of a cation is smaller than the parent atom.
• However, an anion is formed by the gain of one or more electrons and therefore, it contains a greater number of electrons than the parent atom.
• Due to these additional electrons, anion experiences increased electronic repulsion and decreased effective nuclear charge. As a result, an anion has larger radius than its parent atom.

Hence, radius of a cation is smaller and that of an anion are larger as compared to that of their parent atoms.

(D) I^- > I > I⁺

a. Ionic radius : 

Ionic radius is defined as the distance of valence shell of electrons from the centre of the nucleus in an ion.

b. Electronegativity : 

The ability of a covalently bonded atom to attract the shared electrons toward itself is called electronegativity (EN).

Option : b. Mg²⁺ and Ne

i. Decomposition reaction

ii. Displacement reaction

iii. Reduction

iv. Displacement reaction

v. Combination reaction

vi. White.

Correct option is D) H₂

a. 1s² 

Here,

n = 1. 

Therefore, 

The element belongs to the 1st period.

The outer electronic configuration 1s² corresponds to the maximum capacity of 1s, the complete duplet. 

Therefore,

The element is placed at the end of the 1st period in the group 18 of inert gases in the modem periodic table,

b. 1s²2s²2p⁶

Here,

n = 2. 

Therefore, 

The element belongs to the 2nd period.

The outer electronic configuration 2s² 2p⁶ corresponds to complete octet. 

Therefore, 

The element is placed in the 2nd period of group 18 in the modem periodic table.

Correct option is (C) Fe

Correct option is A) O₂ , H₂ 

Option : b. 3, 12 

Correct option is: (D) electronic configuration

According to quantum mechanical model of the atom, the properties of elements can be corelated to their electronic configuration

(D) electronic configuration

Option : a. 13, 31

i. Metals (like alkali metals) react vigorously with oxygen to form oxides which reacts with water to form strong bases. 

e. g. Sodium (Na) reacts with oxygen to form Na₂O which produces NaOH on reaction with water.

ii. Nonmetals (like halogens) react with oxygen to form oxides which on reaction with water form strong acids.

e.g. Chlorine reacts with oxygen to form Cl₂O₇ which produces HClO₄ on reaction with water.

i. The nuclear charge is the same in Na and Na⁺. But Na⁺ has a smaller number of electrons and a smaller number of occupied shells (two shells in Na⁺, while three shells in Na). 

Therefore, radius of Na is larger.

ii. Na⁺ and Mg²⁺ are isoelectronic species.

Mg²⁺ has a larger nuclear charge than that of N^a+.

Therefore, N^a+ has larger radius.

Correct option is C) 2, 2, 3

• The d-block in the modem periodic table is formed as a result of filling the last electron in d orbital.
• As there are five orbitals in a d subshell, 10 electrons can successively be accommodated.

Hence, there are total 10 groups in the d-block of the modem periodic table i.e., group 3 to 12.

The principal quantum number ‘n’ represents the outermost or valence shell of an element (which corresponds to period number) while azimuthal quantum number ‘l’ constitutes a subshell belonging to the shell for the given ‘n’.