The elements in increasing order of ionization energy are

Options

A. Cu, Ag, Au
B. Ag, Cu, Au
C. Au, Cu, Ag
D. Ag, Au, Cu

Answer

To determine the increasing order of ionization energy for the given elements, we need to analyze the elements in the context of their electronic configurations and periodic trends. Here's a step-by-step solution:

### Step 1: Identify the Elements
From the video transcript, we identify the elements mentioned:
- D-block elements: Scandium (Sc), Titanium (Ti), Vanadium (V), Chromium (Cr), Manganese (Mn), Iron (Fe), Cobalt (Co), Nickel (Ni), Copper (Cu), Zinc (Zn).
- Additionally, we have elements from the last group: Silver (Ag) and Gold (Au).

### Step 2: Understand Ionization Energy Trends
Ionization energy generally increases across a period (left to right) and decreases down a group (top to bottom) in the periodic table. For transition metals:
- The ionization energy of 3d elements is generally higher than that of 4d and 5d elements due to increased shielding and electron-electron repulsion in the outer shells.

### Step 3: Compare the Elements
Now, we will compare the ionization energies of the mentioned elements:
- **3d Series**: Sc < Ti < V < Cr < Mn < Fe < Co < Ni < Cu
- **4d Series**: Zr < Nb < Mo < Tc < Ru < Rh < Pd < Ag < Cd
- **5d Series**: Hf < Ta < W < Re < Os < Ir < Pt < Au < Hg

### Step 4: Establish the Order
Based on the trends:
1. **Copper (Cu)** has a relatively high ionization energy due to its filled d-subshell.
2. **Silver (Ag)** has a lower ionization energy than Cu but is higher than Au.
3. **Gold (Au)** has the lowest ionization energy among these three due to its position in the periodic table.

### Step 5: Write the Increasing Order
Thus, the increasing order of ionization energy for the elements Au, Cu, and Ag is:
**Au < Ag < Cu**

### Final Answer
The increasing order of ionization energy is: **Au < Ag < Cu**

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