Classification Of Elements And Periodicity In Properties

In inert gases, all the orbitals are completely filled, therefore they have no tendency to accept an additional electron and hence their electron gain enthalpies are positive

Answer (b) More negative across a period and less negative down a group

Ionization enthalpy is the energy required to remove an electron from an atom in the gaseous state. The trend in ionization enthalpy follows the atomic structure and the number of protons in the nucleus.
- B < C: Boron (B) has a lower ionization enthalpy than Carbon (C) because it has a single electron in the 2p orbital, which is easier to remove compared to the 2p orbital in carbon.
- C < O: Oxygen (O) has a higher ionization enthalpy than carbon due to increased effective nuclear charge and a smaller atomic radius, which makes it harder to remove an electron.
- O < N: Nitrogen (N) has a higher ionization enthalpy than oxygen due to its stable half-filled 2p orbitals (which provides extra stability).
Thus, the correct order of ionization enthalpy is (D) B < C < O < N.

The atomic radius generally decreases across a period due to the increase in nuclear charge, which pulls the electrons closer to the nucleus. Going down a group, the atomic radius increases due to the addition of electron shells. - Fluorine (F) has the smallest atomic radius in this group because it has the highest nuclear charge and the smallest number of electron shells. - Oxygen (O) has a slightly larger atomic radius than fluorine, as it has one less proton and one more electron shell. - Nitrogen (N) has a larger atomic radius compared to oxygen because of its lower nuclear charge and electron shielding. - Carbon (C) has the largest atomic radius in this sequence due to its even lower nuclear charge and electron shielding effects. Thus, the correct increasing order of atomic radii is (E) F < O < N < C.

Step 1: Understanding the Concept:
The International Union of Pure and Applied Chemistry (IUPAC) assigns official, permanent names and symbols to new elements once their discovery is confirmed. Before confirmation, temporary systematic names based on Latin and Greek roots for the digits are used (e.g., Ununnilium for 110).
Step 2: Key Formula or Approach:
This question requires direct factual recall of the periodic table, specifically the modern nomenclature for the superheavy transactinide elements (atomic numbers ).
Step 3: Detailed Explanation:
Let's review the official IUPAC names for the elements listed in the options:
- Element 109: Meitnerium (Mt), named after Lise Meitner.
- Element 110: Darmstadtium (Ds), named after the city of Darmstadt in Germany, where it was discovered.
- Element 111: Roentgenium (Rg), named after Wilhelm Conrad Röntgen.
- Element 112: Copernicium (Cn), named after Nicolaus Copernicus.
- Element 113: Nihonium (Nh), named after Japan (Nihon).
The question asks for the element with atomic number 110. Based on the list, this is Darmstadtium.
Step 4: Final Answer:
The official IUPAC name is Darmstadtium.

Step 1: Understanding the Concept:
Generally, atomic radius increases as we move down a group in the periodic table due to the addition of new electron shells. However, in the transition elements (d-block), a phenomenon called lanthanoid contraction drastically alters this expected trend for the elements of the 5d series.
Step 2: Key Formula or Approach:
The approach involves identifying which of the given pairs consists of a 4d transition metal and the corresponding 5d transition metal immediately below it in the same group. Lanthanoid contraction causes these pairs to have nearly identical atomic radii.
Step 3: Detailed Explanation:
The lanthanoid contraction is caused by the poor shielding effect of the 14 electrons present in the subshell. As we move across the lanthanoid series (Ce to Lu), the increasing nuclear charge draws the electrons closer, causing a steady contraction in size.
By the time we reach the 5d transition elements (Hf, Ta, W, etc.), this significant contraction perfectly offsets the expected increase in size that should occur from adding a new principal quantum shell. Consequently, the elements of the 3rd transition series (5d) have atomic radii very similar to the elements of the 2nd transition series (4d) lying directly above them.
Let's analyze the pairs:
- A) Mo (Molybdenum) & W (Tungsten): Mo is in the 4d series (Group 6), and W is immediately below it in the 5d series (Group 6). Due to lanthanoid contraction, their radii are almost identical ( ). This is a classic textbook example.
- B) Ti & La: Ti is a 3d element, La is a 5d element. They are not in the same group and differ significantly in size.
- C) Ag & Ni: Ag is 4d (Group 11), Ni is 3d (Group 10). Different groups and periods; radii are not the same.
- D) Mn & S: Mn is a transition metal, S is a p-block non-metal. Vastly different.
- E) U & W: Uranium is an actinide; Tungsten is a transition metal.
Step 4: Final Answer:
Mo and W exhibit nearly the same atomic radius due to lanthanoid contraction.

Step 1: Understanding the Concept:
Isoelectronic species are those that possess the exact same number of electrons. The size of an ion in an isoelectronic series is inversely proportional to its effective nuclear charge (which corresponds to the number of protons or the atomic number, ).
Step 2: Key Formula or Approach:
For a given isoelectronic series, the theoretical approach is based on the relation:

Step 3: Detailed Explanation:
First, let's verify that all species are isoelectronic by calculating their total number of electrons:
- : Atomic number ( ) = 13. Electrons =
- : Atomic number ( ) = 11. Electrons =
- : Atomic number ( ) = 9. Electrons =
- : Atomic number ( ) = 8. Electrons =
All species have 10 electrons.
Now, compare their nuclear charge ( ):

A higher nuclear charge exerts a greater pull on the same number of electrons, shrinking the electron cloud. Therefore, the ion with the highest positive charge will be the smallest, and the ion with the highest negative charge will be the largest.
Thus, the increasing order of ionic radius is:

Step 4: Final Answer:
The correct option is i) .