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Q3.30(ii):
Assign the position of the element having outer electronic configuration (ii) ((n-1)d^{2}ns^{2}) for (n=4)

Solution :

Step 1: Substitution of Principal Quantum Number

Given the general outer electronic configuration for the element as $(n-1)d^{2}ns^{2}$ and the specified value of $n=4$, we perform the substitution:

  • For $n = 4$:
  • $(n-1)d = (4-1)d = 3d$
  • $ns = 4s$
[Per the Aufbau Principle and the definition of valence shell configuration]. The configuration becomes $3d^{2}4s^{2}$.

Step 2: Determination of Atomic Number and Identification

To identify the element, we determine the full electronic configuration by filling the preceding inner orbitals ($1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6}$): $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{2}$ Summing the electrons: $2 + 2 + 6 + 2 + 6 + 2 + 2 = 22$. The element with atomic number $Z = 22$ is Titanium (Ti).

Step 3: Determination of Group and Period

We classify the element based on its orbital occupancy:

  • Period: The highest principal quantum number ($n$) in the configuration is 4. Therefore, the element belongs to Period 4. [Rule: The principal quantum number of the valence shell corresponds to the period number].
  • Group: For $d$-block elements, the group number is calculated as the sum of electrons in $(n-1)d$ and $ns$ orbitals. Group Number = (Number of $d$-electrons) + (Number of $s$-electrons) Group Number = $2 + 2 = 4$. Therefore, the element belongs to Group 4. [Rule: For transition elements, Group = $(n-1)d + ns$ electrons].

Step 4: Verification via Visualization

Configuration: 3d² 4s² Element: Titanium (Z=22) Period: 4 Group: 4

Summary of Classification

Attribute Value
Configuration $3d^{2}4s^{2}$
Element Name Titanium (Ti)
Period 4
Group 4
Block $d$-block

Final Solution: The element is Titanium (Ti), situated in Period 4, Group 4 of the periodic table.


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