Electronic Configurations Of Elements And The Periodic Table

Since there is a single hydrogen atom, so only
5 → 4, 4 → 3, 3 → 2, 2 → 1 lines are obtained.

The given electronic configuration is . This configuration can be deciphered as follows:

1. refers to the noble gas Radon, which is used here as a core reference for electronic configuration.
2. indicates the f-block is completely filled in the 5th energy level.
3. shows that there is one electron in the d sub-shell in the 6th energy level.
4. means there are two electrons in the s sub-shell in the 7th energy level.

To find the element's identity, we calculate its atomic number. The atomic number of Radon is 86, and the additions from the configuration are:

• adds 14 electrons.
• adds 1 electron.
• adds 2 electrons.

Thus, the total atomic number is:

The element with atomic number 103 is Lawrencium (Lr). In the IUPAC naming system for elements beyond 100, each digit of the atomic number is translated into a syllable:

• '1' is translated to "Un".
• '0' is translated to "nil".
• '3' is translated to "trium".

Combining these syllables, the systematic IUPAC name for element 103 is "Unniltrium". Therefore, the correct answer is:

Unniltrium

To match the tetrahedral complexes in List-I with their corresponding electronic configurations in List-II, we need to consider the oxidation states of the metal ions and their electronic configurations:

1. Titanium Tetrachloride, TiCl₄:
   • Titanium is in the +4 oxidation state in TiCl₄, resulting in an electron configuration of d⁰.
   • Electronic configuration: e⁰, t₂² (Configuration III matches d⁰ state with unpaired electrons being first placed in e).
2. Ferrate Ion, [FeO₄]^2-:
   • Iron in [FeO₄]^2- typically occurs in the +6 oxidation state, which removes all 3d electrons, giving a vacant orbital.
   • Electronic configuration: e², t₂⁰ (Configuration I is a match considering vacant t₂ orbitals filled e).
3. Tetrachloroferrate(III), [FeCl₄]^-:
   • Iron is in the +3 oxidation state in [FeCl₄]^-, giving a d⁵ electronic configuration.
   • Electronic configuration: e², t₂³ (Configuration IV aligns with high spin d⁵ configuration).
4. Tetrachlorocobaltate(II), [CoCl₄]^2-:
   • Cobalt in the +2 oxidation state in this complex has a d⁷ configuration.
   • Electronic configuration: e⁴, t₂³ (Configuration II matches with high spin d⁷ configuration, distinguishing between paired and unpaired).

Consequently, the correct matching of List-I with List-II is:

• (A) TiCl₄ - (III) e⁰, t₂²
• (B) [FeO₄]^2- - (I) e², t₂⁰
• (C) [FeCl₄]^- - (IV) e², t₂³
• (D) [CoCl₄]^2- - (II) e⁴, t₂³

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

To solve this question, we need to match the complex ions from List I with their correct electronic configurations in List II.

We'll examine each complex ion and determine its electronic configuration based on the oxidation state of the central metal and the coordination chemistry.

1. A.
   • The oxidation state of Cr is +3.
   • Electronic configuration of Cr:
   • Cr³⁺ loses 3 electrons:
   • In octahedral complexes, the electron configuration is .
   • So, this matches with II.
2. B.
   • The oxidation state of Fe is +3.
   • Electronic configuration of Fe:
   • Fe³⁺ loses 3 electrons:
   • In octahedral complexes, the electron configuration is .
   • So, this matches with III.
3. C.
   • The oxidation state of Ni is +2.
   • Electronic configuration of Ni:
   • Ni²⁺ loses 2 electrons:
   • In octahedral complexes, the electron configuration is .
   • So, this matches with IV.
4. D.
   • The oxidation state of V is +3.
   • Electronic configuration of V:
   • V³⁺ loses 3 electrons:
   • In octahedral complexes, the electron configuration is .
   • So, this matches with I.

Based on the analysis, the correct matching is:

A-II, B-III, C-IV, D-I

Thus, the correct answer is:

A-II, B-III, C-IV, D-I