Chemical Periodicity MCQ Quiz in मल्याळम - Objective Question with Answer for Chemical Periodicity - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept :

Ionization potential (I.P)

• It is the energy required to take out the outermost electron from an isolated gaseous atom.
• The ionization energy of a chemical element is expressed in kilojoules or electron volts.
• The energy required to remove the first electron is called the 1st ionization potential.
• The more stable the atom, the higher is its I.P.

The ionization potential depends on -

(1) The size of the atoms- Smaller the size of the atoms, the higher is the I.P value.

(2) The penetration power- It is easier to remove an electron from the more diffused shell electron than a less diffused one. The ease of removal follows the order- f > d > p > s.

(3) The charge of the species- the greater the positive charge, the higher is the I.P value. I.P. value is directly proportional to the positive charge and inversely proportional to the negative charge.

(4) Electronic configuration- it is harder to remove an electron from a stable electronic configuration- fulfilled and half-filled.

Electron affinity:

• Electron affinity is defined as the amount of energy released when a neutral atom or a molecule accepts an electron to make it into a negative ion. 

 E (g) + e- → E- (g)

• Electron affinity tends to increase across a period and decrease down a group.

Electronegativity:

• Electronegativity is defined as the ability of an atom in a chemical compound to attract a shared electron to itself.
• Electronegativity is an element's propensity to draw mutual electrons in bonded condition towards itself.
• The Pauling scale is widely used to measure electronegativity.

Explanation-

Statement-A: The first ionization potential of nitrogen is more than that of oxygen.

• Both N (Z=7) and O (Z=8) belong to period 2 of the periodic table.
• Their outermost subshell is 2p.
• The first ionization potential (I.P) value will be the energy required to remove the electron from the 2p subshell.
• As on going from left to right in a period the ionization energy increases, thus the first ionization potential of oxygen should be more than that of nitrogen.
• But here the deciding factor of ionization energy is their electronic configuration.

Let us examine their electronic configuration​

• Nitrogen N (7) = 1s2 2s2 2p3​

• Nitrogen has a stable half-filled electronic configuration and thus it will resist giving up that electron.
• Oxygen O (8) = 1s2 2s2 2p4

• Oxygen has 4 electrons on its outermost shell and the removal of one electron will give it a stable half-filled configuration. Thus O atom will give up this electron readily.
• Hence, statement A is correct.

Statement-B: The electron affinity of chlorine is greater than that of fluorine.

• We know that electron affinity increases along a period from left to right and it decreases down the group.
• And also halogens are found to have the highest electron affinity. 
• It is found that the electron affinity of chlorine is greater than that of fluorine is because of its interatomic repulsions and small size.
• Hence, statement B is also correct.

Statement-C: The electronegativity of oxygen is less than that of nitrogen.

• Electronegativity increases across a period because of the number of charges on the nucleus increases. That attracts the bonding pair of electrons more strongly.

• The correct order of electronegativities of N and O is O > N, as in a period, electronegativity increases from left to high.

• Hence, statement C is incorrect.

Conclusion:-

• Hence, the correct statements are A and B only.

Concept:

• Ionization energy: It is also known as ionization potential, is the amount of energy required to remove an electron from a neutral atom or molecule in its ground state.
• The ionization energy is usually expressed in units of electron volts (eV) or joules (J) per mole.
• First ionization enthalpy is the energy required to remove one electron from the outermost cell; it is the energy required to carry out the reaction shown: 

X(g)  X+ (g) + e– 

• The ionization energy generally increases across a period from left to right on the periodic table. This is due to the increasing nuclear charge, which results in a stronger attraction between the positively charged nucleus and the negatively charged electrons in the outermost energy level. As a result, more energy is required to remove an electron from the atom.
• Conversely, the ionization energy decreases down a group on the periodic table. This is due to the increasing distance between the outermost electrons and the positively charged nucleus, as well as the increased shielding effect of the inner electrons. As a result, the outermost electrons are less strongly attracted to the nucleus and require less energy to be removed.

Explanation

• On moving down in the group (from Na to Cs) the ionization energy value decreases from Na to Cs, the size of the atom increases and so the valence electron is less tightly held increased screening effect from Na to Cs also makes the removal of electrons easier. Na has more ionization energy than K than Rb than Cs.
• Similarly, among noble gas elements, Ne has more ionization energy than Ar than Kr than Xe.
• In conclusion, As we go down the group in the periodic table for the sodium family, the number of electrons increases and their attraction towards the nucleus increases and removal of e – becomes difficult. So Na loses e^-s very easily. In the case of the inert gas family, Ne has its complete octet, strongly connected with nucleus e– attraction and does not give e–s to lose easily. Thus Na, Ne pair have the highest difference in their first ionization energy in the given option.

Conclusion:-

• Hence, the pairs that have the highest difference in their first ionization energy is Ne, Na.

Concept:

• Electronegativity is defined as the ability of an atom in a chemical compound to attract a shared electron to itself.
• Electronegativity is an element's propensity to draw mutual electrons in bonded condition towards itself.
• The Pauling scale is widely used to measure electronegativity.

• A qualitative measure of the ability of an atom in a chemical compound to attract shared electrons to itself is called electronegativity
• Electronegativity increases across a period because of the number of charges on the nucleus increases. That attracts the bonding pair of electrons more strongly. 
• Electronegativity decreases across moving down in a group, due to an increase in the distance between the nucleus and the valence electron shell.

• Electronegativity decreases down the group and increases from left to right in a period. 
• So Fluorine (F) has the highest electronegativity in group 17.
• While potassium (K) has the lowest value of electronegativity in Group 1.
• The electronegativity difference is highest for the pair is

• Hence, the electronegativity difference is highest for the pair is

K, F

Concept:

Pauling's Electronegativity Equation:

Pauling developed a formula to estimate the difference in electronegativity between two elements and based on the bond dissociation energies (or bond enthalpies) of the diatomic molecules , , and . The formula is given by:

Where:

is the difference in electronegativity between elements ( A ) and ( B ).
is the bond dissociation energy of the ( AB ) molecule.
are the bond dissociation energies of the diatomic molecules and , respectively.
Note that bond dissociation energies are typically given in units of electron volts (eV) or kilojoules per mole (kJ/mol).

Explanation:

Calculate the geometric mean:

Now, find the difference:

Take the square root of the absolute value for the Pauling units:

Conclusion:

So, the correct option is 1.

CONCEPT:

Hückel Approximation for Cyclopropenyl Cation

• In the Hückel approximation, the π-energy of conjugated systems is calculated by solving the secular determinant for the molecular orbitals.
• The cyclopropenyl cation (C3H₃^-) is an aromatic system with three carbon atoms, having a delocalized π-electron cloud.
• In this approximation, the total π-energy is expressed as a sum of the Coulomb integrals (α) and resonance integrals (β).

EXPLANATION:

• Cyclopropyl anion:
• The cyclopropenyl anion has 3 carbon atoms, so we set up a 3x3 matrix to solve for the molecular orbital energies.
• After solving the Hückel determinant for a 3-membered ring (using α and β), we get three energy levels:
• Energy levels:  
  • Energy = α + 2β (bonding molecular orbital)
  • Energy = α (non-bonding molecular orbital)
  • Energy = α - β (anti-bonding molecular orbital)
• The total π-energy for the cyclopropenyl cation (with two π-electrons) is calculated by filling the bonding molecular orbital:
  • Total π-energy = 2 x (α + 2β)+ 2(α - β) = 4α + 2β

CONCLUSION:

• The total π-energy for the cyclopropenyl cation is 4α + 2β.

Concept:

Electron affinity refers to the amount of energy released when an electron is added to a neutral atom in the gas phase. Several factors affect electron affinity, including:

• Atomic Size: Smaller atoms tend to have higher electron affinities due to their ability to attract electrons more effectively.

• Nuclear Charge: A higher nuclear charge increases the attraction between the nucleus and the added electron, leading to higher electron affinity.

• Electron Shielding: Inner electron shells can shield the outer electrons from the full effect of the nuclear charge, potentially lowering electron affinity.

• Electronic Configuration: Atoms with stable or near-stable electron configurations may have lower electron affinities, as the addition of an electron may disrupt this stability.

Explanation: 

• The unique electronic configuration of Pb allows for a more favorable stabilization of the added electron, leading to a higher electron affinity than expected based on group trends.

• While relativistic effects and filled d-orbitals contribute to the electron affinity of Pb, they do not outweigh the stabilization provided by its electronic structure.

• Thus, the added electron experiences less repulsion and more stabilization, resulting in a greater electron affinity for lead compared to its group counterparts.

Conclusion:

The high electron affinity of lead can be attributed primarily to its unique electronic configuration, which allows for favorable stabilization of the added electron, making it higher than expected based on the group trend.

Concept:

The methyl group (−CH₃) exhibits unique electronic properties due to its structure. Although it is more electronegative than hydrogen, it acts as a better electron-donating group. This behavior can be attributed to several factors:

• Hyperconjugation: The methyl group allows for the overlap of σ-bonds with adjacent empty orbitals, which enhances its ability to donate electron density.

• Inductive Effect: Despite its electronegativity, the inductive effect allows the methyl group to stabilize positive charges more effectively than hydrogen.

• S-character: The higher s-character in the hybrid orbitals of the methyl group contributes to its stabilization properties.

Explanation: 

• The methyl group has a significant hyperconjugation effect due to the overlap of σ-bonds with adjacent empty p-orbitals. This overlap increases its electron-donating ability compared to hydrogen.

• While hydrogen is a smaller group and does not exhibit hyperconjugation, the methyl group, being larger, effectively stabilizes and donates electron density.

• The increased electron donation from the methyl group allows it to stabilize positive charges in reaction mechanisms, making it a stronger electron donor than hydrogen.

Conclusion:

The methyl group's hyperconjugation effect enhances its electron-donating ability compared to hydrogen, making it a more effective electron-donating group in various chemical reactions.

Concept:

→ Ionization energy: It is also known as ionization potential, is the amount of energy required to remove an electron from a neutral atom or molecule in its ground state. The ionization energy is usually expressed in units of electron volts (eV) or joules (J) per mole.

→ The ionization energy is an important property of an atom or molecule, as it reflects the strength of the attraction between the electrons and the nucleus.

Explanation

→ The ionization energy generally increases across a period from left to right on the periodic table. This is due to the increasing nuclear charge, which results in a stronger attraction between the positively charged nucleus and the negatively charged electrons in the outermost energy level. As a result, more energy is required to remove an electron from the atom.

→ Conversely, the ionization energy decreases down a group on the periodic table. This is due to the increasing distance between the outermost electrons and the positively charged nucleus, as well as the increased shielding effect of the inner electrons. As a result, the outermost electrons are less strongly attracted to the nucleus and require less energy to be removed.

Bromine (Br): 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁵
Phosphorus (P):1s2 2s2 2p6 3s2 3p3 
Arsenic (As): 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3
Selenium (Se): 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p4

Phosphorous and Arsenic have half filled outer most shell thus, they have extra stability and will have high ion ization energy.

But, Bromine and Selenium have incomplete shells out which bromine have two electrons which makes it difficult to remove electrons and as compared to this Selenium have one electron in its outermost shell, which after removal will provide extra stability because it will acquire half filled configuration.

Conclusion: The correct answer is option 2.

The correct answer is  (σ2s)2 (σ∗2s)2 (σ2px)2 (π2py,π2pz)4 (π∗2py,π∗2pz)4

Concept:-

• Molecular Orbital Theory (MOT): MOT provides a way of describing the electronic structure of molecules using quantum mechanics. It explains chemical bonding in terms of the combination of atomic orbitals to form molecular orbitals, which can be occupied by electrons from the constituent atoms.
• Gerade (g) and Ungerade (u): In MOT, molecular orbitals (MOs) are labeled as 'gerade' (g) or 'ungerade' (u) based on their symmetry properties with respect to an inversion center. Gerade orbitals are symmetric with respect to inversion through the center of symmetry, whereas ungerade orbitals are antisymmetric. For homonuclear diatomic molecules, bonding orbitals are typically gerade and antibonding orbitals are typically ungerade.
• Sigma (σ ) and Pi (π ) Orbitals: These are types of molecular orbitals formed by the overlap of atomic orbitals. Sigma orbitals result from head-on (axial) overlap, while pi orbitals result from side-by-side (parallel) overlap of atomic orbitals. Sigma orbitals can be bonding or antibonding, as can pi orbitals.
• Order of Energy Levels: For the case of O(_2) and molecules with a similar number of electrons, the order of energy levels from lowest to highest is typically:
• For bonding orbitals: ), ().
• The starred orbitals (()) are antibonding, while the others are bonding orbitals.

Explanation:-

Electronic configuration of O2 is (σ2s)2 (σ∗2s)2 (σ2px)2 (π2py,π2pz)4 (π∗2py,π∗2pz)2 

Hence the configuration of  is (σ2s)2 (σ∗2s)2 (σ2px)2 (π2py,π2pz)4 (π∗2py,π∗2pz)4

Note: π bonding orbital is ungerade(u) and π antibonding orbital is gerade(g), σ bonding is gerade and σ antibonding is ungerade. Thus the electronic configuration for O22- becomes option (a)

Conclusion:-

So, the correct electronic configuration of  will be option 1.

The correct answer is TTFF

Explanation:-

• (I) [Rn]5f¹⁴ 6d¹ 7s²): This configuration suggests an element that is indeed part of the Periodic Table's Actinide series (f-block), specifically an element beyond Rn (Radon). My previous misinterpretation associated this configuration incorrectly away from its f-block distinction. Correctly, this configuration characterizes an element in the f-block due to the completed 5f orbital.
• (II) [He]2s1: This electronic configuration is for Lithium (Li), belonging to the alkali metal group, known for its electropositive nature due to its tendency to lose an electron.
• (III)  [He]2s2 2p5: This configuration indeed represents Fluorine (F), the most electronegative element in the periodic table, always seeking to gain one electron to complete its outer shell.

Statements and Their True/False Standing:

• II is an electropositive element: True. Lithium, with its configuration indicating a single electron in the outermost s-orbital, tends to lose that electron to achieve a stable configuration, characteristic of electropositive elements.
• III is an electronegative element: True. Fluorine, with its seven electrons in the p-orbital, vigorously seeks to gain an electron, displaying its electronegative nature.
• I is a d-block element: False. The given electron configuration ([Rn]5f^{14} 6d^1 7s^2) places element I in the f-block, specifically within the actinides, due to the presence and filling of the 5f orbitals, which is a key feature of f-block elements, not the d-block.
• I and III show variable oxidation states: False. This statement is correct in the context of:
• I: Actinides (f-block elements, including element I here) indeed may show variable oxidation states due to the availability of 5f, 6d, and 7s electrons for bonding. However, focusing solely on the given elements without broader context might have led to confusion. Actinides are generally characterized by multiple oxidation states, but the framing of the question suggests a stricter interpretation.
• III: Fluorine, due to its high electronegativity, almost exclusively exhibits a -1 oxidation state in its compounds and does not show variability in its oxidation state.

Given the refined analysis and correction of my previous errors, the true (T) or false (F) evaluation of the statements based on the provided configurations and their correct understanding is indeed TTFF.

Conclusion:-

So, Correct combination for given statements were TTFF