Element with atomic number belongs to the period.........and the group ........

Options

A. `4^(th),12^(th)`
B. `4^(th),11^(th)`
C. `5^(th),12^(th)`
D. `5^(th),11^(th)`

Explanation

To determine the period and group of the element with atomic number 47, we can follow these steps:

### Step 1: Identify the Period
1. **Understanding the Periodic Table Structure**: The periodic table is arranged in periods (rows) and groups (columns). The period number corresponds to the highest principal quantum number (n) of the electrons in the element.

2. **Finding the Period for Atomic Number 47**: The atomic number 47 corresponds to the element Silver (Ag). To find its period, we look at the electron configuration.

3. **Electron Configuration**: The electron configuration for Silver (Ag) is:
\[ [Kr] 4d^{10} 5s^1 \]
Here, the highest principal quantum number (n) is 5, indicating that Silver is in the **5th period**.

### Step 2: Identify the Group
1. **Determining the Group**: The group number can be determined by looking at the electron configuration and the number of electrons in the outermost shell.

2. **Counting Valence Electrons**: From the electron configuration:
- The outermost shell (n=5) has 1 electron in the 5s subshell and 10 electrons in the 4d subshell.
- The total number of valence electrons is 1 (from 5s) + 10 (from 4d) = 11 electrons.

3. **Finding the Group Number**: In the periodic table, elements in the d-block (transition metals) have a group number that can be calculated as:
- For transition metals, the group number can be determined by the number of d electrons plus the s electrons in the outermost shell.
- Since Silver has 10 d electrons and 1 s electron, it belongs to **Group 11**.

### Final Answer
- The element with atomic number 47 (Silver, Ag) belongs to the **5th period** and **Group 11**.

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