All questions of Classification of Elements & Periodicity in Properties for NEET Exam

Introduction:
In order to determine which element is highly electropositive among the elements with atomic numbers 9, 12, 16, and 36, we need to understand the concept of electronegativity and how it relates to electropositivity. Electronegativity is a measure of an atom's ability to attract electrons towards itself in a chemical bond. Electropositivity, on the other hand, is the tendency of an atom to lose electrons and form positive ions. Therefore, an element with low electronegativity and high electropositivity will readily lose electrons and form positive ions.

Explanation:
Let's analyze each element and its electropositivity based on their atomic numbers:

Element with atomic number 9 (Fluorine):
Fluorine is a halogen and has a high electronegativity. It readily gains an electron to achieve a stable electron configuration. Therefore, it is not highly electropositive.

Element with atomic number 12 (Magnesium):
Magnesium is an alkaline earth metal and has a low electronegativity. It readily loses its two valence electrons to achieve a stable electron configuration. Therefore, it is highly electropositive.

Element with atomic number 16 (Sulfur):
Sulfur is a nonmetal and has a moderate electronegativity. It can gain or lose electrons depending on the reaction conditions. However, it is not highly electropositive as it tends to gain electrons to achieve a stable electron configuration.

Element with atomic number 36 (Krypton):
Krypton is a noble gas and has a very high electronegativity. It has a stable electron configuration and tends not to lose or gain electrons. Therefore, it is not highly electropositive.

Conclusion:
Based on the analysis, the element with atomic number 12 (Magnesium) is highly electropositive among the elements with atomic numbers 9, 12, 16, and 36. It has a low electronegativity and readily loses its two valence electrons to form a stable 2+ ion.

In half filled and fully filled orbitals it's difficult to remove an electron from its outermost shell cause it's Highly stable electronic configuration so it's option B !!!

Incorrect Statement Explanation:

Decrease in electronegativities across a period is accompanied by an increase in non-metallic properties:
- This statement is incorrect because the decrease in electronegativities across a period is actually accompanied by an increase in metallic properties, not non-metallic properties.
- Electronegativity is the tendency of an atom to attract a shared pair of electrons towards itself in a chemical bond.
- As we move across a period from left to right, electronegativity generally increases. This means that elements on the left side of the periodic table (metals) have lower electronegativities compared to elements on the right side (non-metals).
- Metals tend to lose electrons and form cations, while non-metals tend to gain electrons and form anions. Therefore, a decrease in electronegativity across a period is associated with an increase in metallic properties, not non-metallic properties.

In conclusion, the statement that a decrease in electronegativities across a period is accompanied by an increase in non-metallic properties is incorrect.

The most non-metallic element among the options given is option 'B' with the electron configuration 1s22s22p5. Let's understand why this is the correct answer.

Understanding Metallic and Non-metallic Elements:
- Metals are elements that tend to lose electrons and form positive ions.
- Non-metals, on the other hand, tend to gain or share electrons and form negative ions or covalent bonds.

Electron Configuration:
The electron configuration of an element describes how its electrons are distributed into different energy levels and orbitals.

Analyzing the Options:
a) 1s22s22p63s1
- This electron configuration belongs to sodium (Na), which is a metal. Sodium readily loses its 3s1 electron to form a sodium ion (Na+). Therefore, it is a metallic element.

b) 1s22s22p5
- This electron configuration belongs to fluorine (F), which is a non-metal. Fluorine readily accepts an electron to complete its p orbital and form a fluoride ion (F-). Therefore, it is the most non-metallic element among the options.

c) 1s22s22p63s2
- This electron configuration belongs to magnesium (Mg), which is a metal. Magnesium readily loses its 3s2 electrons to form a magnesium ion (Mg2+). Therefore, it is a metallic element.

d) 1s22s22p3
- This electron configuration belongs to phosphorus (P), which is a non-metal. Phosphorus tends to gain three electrons to complete its p orbital and form a phosphide ion (P3-). Therefore, it is a non-metallic element.

Conclusion:
Among the given options, option 'B' with the electron configuration 1s22s22p5 represents fluorine (F), which is the most non-metallic element. Fluorine readily accepts an electron to achieve a stable electron configuration, making it highly non-metallic in nature.

The wrong statement on the basis of the periodic table is option C, which states that there are four transition series in the periodic table, each containing 10 elements.

Explanation:
The periodic table is a tabular arrangement of chemical elements, organized based on their atomic number, electron configuration, and recurring chemical properties. It is divided into several periods (rows) and groups (columns). Each period represents the energy level or shell occupied by the outermost electrons, while each group represents elements with similar properties due to their similar valence electron configurations.

The transition elements or transition metals are a group of elements that occupy a central block in the periodic table, specifically in groups 3 to 12. They are characterized by the presence of partially filled d orbitals in their electron configurations. The transition metals have similar chemical properties, including the ability to form complex ions, exhibit variable oxidation states, and act as good catalysts.

The correct properties and characteristics related to the periodic table are:

a) The most electronegative element in the periodic table is fluorine:
Fluorine, with an atomic number of 9, is located in group 17 (group VIIA) of the periodic table. It is the most electronegative element because it has the highest tendency to attract electrons towards itself when bonded to other elements. This high electronegativity is due to its small atomic size and high effective nuclear charge.

b) Scandium is the first transition element and belongs to the fourth period:
Scandium, with an atomic number of 21, is the first element in the d-block or transition metals. It belongs to the fourth period (row) of the periodic table as it has four electron shells. Scandium has the electron configuration [Ar] 3d1 4s2.

d) Along a period, halogens have the maximum negative electron gain enthalpy:
Halogens, which include elements like fluorine, chlorine, bromine, iodine, and astatine, belong to group 17 (group VIIA) of the periodic table. They have high electron affinity or electron gain enthalpy, which refers to the energy change when an atom gains an electron to form a negative ion. Halogens have the greatest negative electron gain enthalpy along a period due to their high effective nuclear charge and small atomic size.

c) There are four transition series in the periodic table, each containing 10 elements:
This statement is incorrect. There are actually five transition series in the periodic table, not four. These series are based on the filling of the 3d, 4d, 5d, 6d, and 7d orbitals. Each series contains 10 elements. The first transition series includes the elements from scandium (Sc) to zinc (Zn), the second series includes the elements from yttrium (Y) to cadmium (Cd), the third series includes the elements from lanthanum (La) to mercury (Hg), the fourth series includes the elements from actinium (Ac) to copernicium (Cn), and the fifth series includes the elements from rutherfordium (Rf) onwards.

Therefore, option C is the wrong statement because there are actually five transition series in the periodic table, each containing 10 elements.

Understanding Ionization Energies
Ionization energy refers to the energy required to remove an electron from an atom. A sudden large jump between the second and third ionization energies indicates that removing the third electron requires significantly more energy than removing the first two. This typically occurs when the electron being removed is from a more stable electron configuration.
Analysis of Electronic Configurations
Let’s analyze the provided electronic configurations:
- Option A: 1s2, 2s2, 2p6, 3s1, 3p2
- Removal of the first electron from 3p2 is relatively easy. Removal of a second electron from the same subshell remains manageable. The third ionization would not show a large jump since the remaining electrons are still in the same energy levels.
- Option B: 1s2, 2s2, 2p6, 3s2, 3p1
- Similar to option A, removing the first two electrons from 3s2 and 3p1 does not reach a stable noble gas configuration. Hence, the third ionization does not exhibit a significant jump.
- Option C: 1s2, 2s2, 2p6, 3s1
- Removing one electron from 3s1 still doesn’t lead to a substantial increase in ionization energy for subsequent removals, as there’s no stable electron configuration left behind.
- Option D: 1s2, 2s2, 2p6, 3s2
- Here, the removal of the first two electrons from 3s2 would occur easily, leading to a fully filled 2p subshell (a stable configuration). The third ionization would require removing an electron from a new, higher energy subshell (3p) which is much more stable, resulting in a significant jump in ionization energy.
Conclusion
Thus, option D represents an electronic configuration where the large jump between the second and third ionization energies can be attributed to the stability of the noble gas configuration achieved after the removal of two electrons from the 3s subshell.

Answer 'b' is correct....

Because all elements have same no of electron

The law of octaves is a concept proposed by John Newlands, an English chemist, in 1864. It was one of the earliest attempts to organize the elements based on their chemical properties. According to this law, when the elements are arranged in order of increasing atomic mass, the properties of every eighth element are similar to the first element.

Explanation:
1. Law of Octaves: The law of octaves states that the properties of elements are repeated after regular intervals of eight elements.
2. Arrangement of Elements: Newlands arranged the known elements in order of increasing atomic mass. He observed that when the elements were arranged in this manner, there was a periodic repetition of properties.
3. Similarity of Properties: Newlands found that every eighth element had properties that were similar to the first element. This means that the elements at the first and ninth position, second and tenth position, third and eleventh position, and so on, shared similar properties.
4. Example: For example, Newlands noticed that lithium (Li) and sodium (Na) had similar properties, so did beryllium (Be) and magnesium (Mg), boron (B) and aluminum (Al), and so on.
5. Limitations: Although the law of octaves showed periodicity in the properties of elements, it had limitations. It failed to accommodate all the known elements at that time, and it did not explain the underlying reason for the repetition of properties.
6. Recognition and Acceptance: The law of octaves was initially met with skepticism and criticism. However, it laid the foundation for the development of the modern periodic table.
7. Advancement: Dmitri Mendeleev, a Russian chemist, later expanded on Newlands' work and proposed the periodic law, which states that the properties of elements are periodic functions of their atomic numbers.
8. Modern Periodic Table: The modern periodic table is based on the periodic law and arranges the elements in order of increasing atomic number. It organizes the elements into groups and periods based on their similar properties and electronic configurations.

In conclusion, the correct answer is option 'A' - every eighth element has properties similar to the first element. This observation, known as the law of octaves, was proposed by John Newlands and served as a precursor to the development of the modern periodic table.

Understanding Variable Valency
Variable valency refers to the ability of an element to exhibit different oxidation states or valencies. This characteristic is most commonly associated with transition elements.
Why Transition Elements Exhibit Variable Valency
- Presence of d-orbitals: Transition metals have partially filled d-orbitals, which allows them to lose different numbers of electrons.
- Multiple oxidation states: They can exhibit multiple oxidation states due to their ability to lose different numbers of electrons from both s and d subshells.
- Formation of complex ions: Transition elements can form complex ions, which can further allow them to display variable valencies depending on the ligands attached.
Examples of Transition Elements with Variable Valency
- Iron (Fe): Exhibits +2 and +3 states.
- Copper (Cu): Shows +1 and +2 states.
- Manganese (Mn): Can have oxidation states ranging from +2 to +7.
Distinction from Other Elements
- Representative Elements: Typically have fixed valencies based on their group number and do not exhibit the same level of variability.
- Noble Gases: Generally do not form compounds and have a complete valence shell, hence no variable valency.
- Non-metals: While some non-metals can have multiple oxidation states (like sulfur), they do not exhibit the extensive variability seen in transition metals.
In conclusion, the transition elements are unique in their ability to showcase variable valency due to their electronic configuration and involvement in complex formation. This property is essential in various chemical reactions and applications, making them a significant focus in chemistry.

Explanation:

In the periodic table, the atomic radius refers to the size of an atom. It is defined as the distance between the nucleus and the outermost electron shell of an atom. The atomic radius varies across the periodic table due to the distribution of electrons and the effective nuclear charge experienced by the outermost electrons.

Statement (a): In a group, there is continuous increase in size with increase in atomic number
This statement is true. In a group, elements have the same number of valence electrons and similar electron configurations. As we move down a group, the number of electron shells increases, resulting in an increase in atomic size. This is due to the shielding effect of inner electron shells, which reduces the attraction between the nucleus and the outermost electrons.

Statement (b): In 4f-series, there is a continuous decrease in size with increase in atomic number
This statement is true. The 4f-series consists of the lanthanides, which are located in the f-block of the periodic table. As we move across the lanthanide series, the atomic number increases, resulting in a decrease in atomic size. This is because the increasing number of protons in the nucleus exerts a stronger attraction on the electrons, leading to a contraction in atomic radius.

Statement (c): The size of inert gases is larger than halogens
This statement is true. Inert gases, also known as noble gases, have completely filled electron shells. They are located in Group 18 of the periodic table. The halogens, on the other hand, are located in Group 17. Inert gases have larger atomic radii compared to halogens because inert gases have more electron shells and experience less effective nuclear charge due to greater electron-electron repulsion.

Statement (d): In 3rd period, the size of atoms increases with increase in atomic number
This statement is not true. In the 3rd period, the atomic size does not strictly increase with atomic number. The trend is not continuous due to the difference in electron configurations. While the atomic size generally increases from left to right across a period, there are exceptions due to variations in the effective nuclear charge and electron-electron repulsion. For example, the atomic radius of sodium (Na) is larger than that of magnesium (Mg) in the 3rd period.

Therefore, the correct answer is option D - In the 3rd period, the size of atoms increases with increase in atomic number.

A

The order of increasing metallic character is as follows:
P

Explanation:
- The electronic states X and Y of an atom are given as follows:

X: 1s2 2s2 2p6 3s1
Y: 1s2 2s2 2p6 3s2 3p6 4s1

a) X represents an alkali metal:
- Alkali metals are the elements in Group 1 of the periodic table.
- The electronic configuration of the alkali metals in the ground state is ns1, where n is the principal quantum number.
- The electronic state X has the electronic configuration 1s2 2s2 2p6 3s1, which does not match the electronic configuration of alkali metals.
- Therefore, statement a) is not correct.

b) Energy is required to change X into Y:
- The electronic state X has only one electron in the 3s orbital, while the electronic state Y has one electron in the 4s orbital.
- To change from X to Y, energy is required to move the electron from the 3s orbital to the higher energy 4s orbital.
- Therefore, statement b) is correct.

c) Y represents the ground state of the element:
- The ground state of an atom is the electronic state with the lowest energy.
- The electronic state Y has the full electronic configuration 1s2 2s2 2p6 3s2 3p6 4s1, which is not the ground state configuration.
- Therefore, statement c) is not correct.

d) Less energy is required to remove an electron from X than from Y:
- The energy required to remove an electron from an atom is known as ionization energy.
- In general, ionization energy increases as you move across a period and decreases as you move down a group in the periodic table.
- The electronic state X has an electron in the 3s orbital, while the electronic state Y has an electron in the 4s orbital.
- Since the 4s orbital is farther from the nucleus than the 3s orbital, it is easier to remove an electron from the 4s orbital.
- Therefore, less energy is required to remove an electron from Y than from X.
- Hence, statement d) is not correct.

Conclusion:
- The correct statement among the given options is d) Less energy is required to remove an electron from X than from Y.