Question

Identify general electronic configuration exhibited by \(2^{\text{nd}}\) series of transition elements.

• A. \([\text{Ar}] 4d^{1-10} 4s^2\)
• B. \([\text{Kr}] 4d^{1-10} 5s^{0-2}\)
• C. \([\text{Xe}] 4d^{1-10} 4s^2\)
• D. \([\text{Rn}] 4d^{1-10} 6s^2\)

Hint:

The second series fills 4d with 5s outer electrons, after krypton.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The second transition series (Y to Cd) fills the \(4d\) subshell. The core before it is krypton, \([\text{Kr}]\).

Step 2: Write the Configuration:
The 5s orbital holds 0 to 2 electrons, because some elements such as Pd (\(4d^{10}5s^0\)) and Ag (\(4d^{10}5s^1\)) are exceptions.
\[ [\text{Kr}]\,4d^{1-10}\,5s^{0-2} \]

Step 3: Check the Other Options:
Options (A) and (C) place \(4s^2\) after the 4d, which is wrong because the 4s subshell is already filled in the core. (C) also uses [Xe] as the core, which belongs to the third series, and (D) with [Rn] and 6s belongs to the 6th period. Only (B) is the right pattern.

Final Answer:
The 4d series has the configuration \([\text{Kr}]\,4d^{1-10}5s^{0-2}\), option (B). \[ \boxed{\text{(B) } [\text{Kr}]\,4d^{1-10}\,5s^{0-2}} \]