Periodic Table

To determine the correctness of the given statements, let's analyze each one based on our understanding of Chemistry, specifically the oxidation states of Group 16 elements.

1. Statement (I): Oxygen being the first member of group 16 exhibits only –2 oxidation state.
   • Oxygen is indeed the first member of Group 16 in the periodic table. It commonly exhibits an oxidation state of –2 in most of its compounds. However, it does exhibit other oxidation states under certain conditions. For example, in peroxides like , oxygen has an oxidation state of –1, and in compounds like , oxygen has an oxidation state of +2.
   • Therefore, Statement (I) is incorrect due to this variability in oxidation states.
2. Statement (II): Down the group 16 stability of +4 oxidation state decreases and +6 oxidation state increases.
   • In Group 16, which includes elements like Oxygen, Sulfur, Selenium, Tellurium, and Polonium, the stability of oxidation states varies with different elements.
   • As we move down the group from Oxygen to Polonium, the ability of elements to exhibit a +6 oxidation state generally decreases due to the inert pair effect, which is more pronounced in heavier elements. Conversely, in practice, the +4 state is more stable in heavier Group 16 elements like Selenium and Tellurium compared to the +6 state.
   • Therefore, Statement (II) is incorrect. The trend mentioned is reversed in the explanation.

Based on the analysis above, both Statement I and Statement II are incorrect, which makes the correct answer:

Both Statement I and Statement II are incorrect

.

The problem asks for the group number in the periodic table for an element given its IUPAC systematic name, "Unununnium".

Concept Used:

The IUPAC systematic name for an element is derived from its atomic number (Z). Each digit in the atomic number corresponds to a specific numerical root. The group number of an element can then be determined from its atomic number by either writing its electronic configuration or by locating its position within its period in the periodic table.

The numerical roots are:

• 0 = nil
• 1 = un
• 2 = bi
• 3 = tri
• 4 = quad
• 5 = pent
• 6 = hex
• 7 = sept
• 8 = oct
• 9 = enn

The name is formed by concatenating the roots and adding the suffix "-ium".

Step-by-Step Solution:

Step 1: Determine the atomic number (Z) from the IUPAC name.

The name "Unununnium" is broken down into its constituent roots:

• Un 1
• un 1
• un 1

Combining these digits gives the atomic number:

The element is Roentgenium (Rg).

Step 2: Determine the block and period of the element.

To find the position of the element, we can write its electronic configuration. The noble gas preceding the element with Z = 111 is Radon (Rn, Z = 86).

The electronic configuration of Rn is . After Rn, we start filling the 7th period.

The configuration for Z = 111 is:

(Note: Due to stability, the actual configuration is likely , similar to Gold in the same group, but either configuration will lead to the same group number).

Since the last electron enters the d-orbital (6d), the element belongs to the d-block. The highest principal quantum number is 7, so it is in the 7th period.

Step 3: Calculate the group number.

For a d-block element, the group number is the sum of the electrons in the outermost s-orbital and the penultimate d-orbital.

Using the configuration :

Alternatively, using the more stable configuration :

Both configurations place the element in Group 11.

Final Computation & Result:

The element with the IUPAC name "Unununnium" has an atomic number of 111 and belongs to Group 11 of the periodic table.

The problem states that the element belongs to the nth group, so .

The value of n is 11.

The question involves evaluating the relationship between the assertion and reason provided, both of which pertain to periodic trends in the covalent radius of elements down a group in the periodic table. To answer this, we need to examine the concepts involved.

1. Understanding the Assertions:
   • Assertion (A): "There is a considerable increase in covalent radius from N to P. However, from As to Bi, only a small increase in covalent radius is observed."
   • Reason (R): "Covalent and ionic radii in a specific oxidation state increase down the group."
2. Evaluating Assertion (A):
   • The covalent radius generally increases as we move down a group, due to additional electron shells being added, increasing the size of the atom.
   • From Nitrogen (N) to Phosphorus (P), a considerable increase is seen because of the addition of a new shell. However, from Arsenic (As) to Bismuth (Bi), filled d and f orbitals come into play, causing less effective nuclear charge which leads to only a smaller increase in size. Hence Assertion (A) is true.
3. Evaluating Reason (R):
   • The reason given is true. As we move down a group, the covalent and ionic radii generally increase due to the addition of electron shells.
4. Relationship Between Assertion (A) and Reason (R):
   • While both the assertion and the reason are true, (R) does not specifically explain the smaller increase from As to Bi due to poor shielding of f orbitals, which is crucial here.
5. Conclusion:
   • The most appropriate option is: "Both (A) and (R) are true but (R) is not the correct explanation of (A)." It correctly aligns with the given facts and trends.

Thus, the correct response is that while both statements are true, the reason does not adequately explain the assertion.

To solve this problem, we need to analyze the given statements and decide which option appropriately reflects the relationship between the assertion and the reason. Let's break it down:

1. Assertion (A): "The first ionisation enthalpy decreases across a period."
2. Reason (R): "The increasing nuclear charge outweighs the shielding across the period."

Explanation:

1. Across a period in the periodic table, as we go from left to right, elements have increasing atomic numbers. This means more protons are added to the nucleus, increasing the nuclear charge.
2. As the nuclear charge increases, electrons are pulled closer to the nucleus, making it more difficult to remove an electron. Thus, the first ionisation enthalpy generally increases across a period, not decreases. Therefore, Assertion A is false.
3. The reason provided, "The increasing nuclear charge outweighs the shielding across the period," is true. Although electrons are being added as we move across a period, they are added to the same shell. Therefore, the shielding effect does not increase significantly, and the increased nuclear charge effectively holds the electrons more strongly.

Based on the above explanation:

1. The Assertion A is incorrect because the first ionisation enthalpy actually increases across a period.
2. The Reason R is correct because it correctly describes why ionisation enthalpy changes as it does across a period.

Conclusion:

The correct choice is: A is false but R is true.

To determine which statements about the periodic table are NOT true, let's evaluate each statement one by one:

1. A. The properties of elements are a function of atomic weights.
   • This statement reflects the initial formulation of the periodic law by Dmitri Mendeleev, where elements were arranged by increasing atomic weights. However, this has since been revised to be based on atomic numbers, as certain anomalies in atomic weight-based ordering were resolved using atomic numbers.
   • Thus, this statement is NOT true according to the modern periodic law, which states that the properties of elements are a periodic function of their atomic numbers.
2. B. The properties of elements are a function of atomic numbers.
   • This statement is true and reflects the modern periodic law established by Henry Moseley, which arranges elements based on increasing atomic numbers, correcting anomalies from Mendeleev's version.
3. C. Elements having similar outer electronic configuration are arranged in the same period.
   • In a periodic table, elements with similar outer electronic configurations are arranged in the same group, not the same period.
   • Therefore, this statement is NOT true.
4. D. An element's location reflects the quantum numbers of the last filled orbital.
   • This statement is true. The position of an element in the periodic table, particularly the group and period, is determined by the quantum numbers, specifically the principal quantum number (n) and azimuthal quantum number (l) of the last filled orbital.
5. E. The number of elements in a period is the same as the number of atomic orbitals available in the energy level that is being filled.
   • The number of elements in a period corresponds to the number of electrons that can occupy the orbitals (subshells) in an energy level, not the number of atomic orbitals.
   • For example, the first period has 2 elements, corresponding to the filling of the 1s orbital, while the second period has 8 elements, corresponding to filling the 2s and 2p orbitals.
   • This statement is NOT true.

Given this analysis, the statements that are NOT true according to the modern understanding of the periodic table are A, C, and E.

To determine the most suitable spectral line of the hydrogen atom for heat treatment at a wavelength of about 900 nm, we need to analyze the series and transitions available within the hydrogen emission spectrum and how they correspond to this wavelength.

The transition wavelength can be calculated using the formula:

where is the wavelength, is the Rydberg constant, and and are the principal quantum numbers with .

We are given that , and the Rydberg constant .

Substituting the values into the formula:

Simplifying gives:

For the hydrogen atom, the spectral series have the following characteristics:

• Lyman Series: (UV region)
• Balmer Series: (Visible region)
• Paschen Series: (Infrared region)

The Paschen series transitions fit within the infrared region, typically including wavelengths near 900 nm. Hence, the transition in the Paschen series is likely to correspond to the mentioned wavelength of 900 nm.

For a spectral match near 900 nm, the most suitable transition is Paschen series, , as it aligns well with the infrared wavelength provided in the question.

To solve this matching problem, we need to align each group from LIST-I with its respective element from LIST-II. Let's break it down:

1. Pnicogen (group 15): The pnicogens are the elements in Group 15 of the periodic table. These traditionally include nitrogen, phosphorus, arsenic, antimony, and bismuth. The synthetic element Moscovium (Mc) also belongs to group 15.
2. Chalcogen (group 16): The chalcogens are in Group 16, including oxygen, sulfur, selenium, tellurium, and polonium. Livermorium (Lv) is a synthetic element that belongs to Group 16.
3. Halogen (group 17): The halogens include fluorine, chlorine, bromine, iodine, and astatine. Tennessine (Ts) is a synthetic element in Group 17.
4. Noble gas (group 18): This group contains helium, neon, argon, krypton, xenon, and radon. Oganesson (Og) displays properties similar to the noble gases, although it might not be a gas at room temperature.

Now, let's match them with LIST-II:

• A-IV: Pnicogen (group 15) matches with Mc (Moscovium).
• B-III: Chalcogen (group 16) matches with Lv (Livermorium).
• C-I: Halogen (group 17) matches with Ts (Tennessine).
• D-II: Noble gas (group 18) matches with Og (Oganesson).

Thus, the correct answer is: A-IV, B-III, C-I, D-II

The question asks to identify the element that does not belong to the same period as the others in the modern periodic table. Let's analyze the given elements one by one:

1. Palladium (Pd): Palladium is an element in the 5th period of the periodic table, and it is found in Group 10.
2. Iridium (Ir): Iridium belongs to the 6th period and is found in Group 9.
3. Osmium (Os): Osmium is located in the 6th period and belongs to Group 8.
4. Platinum (Pt): Platinum is also in the 6th period, found in Group 10.

From this analysis, we observe:

• Iridium, Osmium, and Platinum all belong to the 6th period of the periodic table.
• Palladium, on the other hand, belongs to the 5th period.

Therefore, the element only belonging to the 5th period and differing from the rest is Palladium (Pd).

Conclusion: The element that does not belong to the same period as the remaining elements is Palladium, which is the correct answer.