Electronegativity is the property of a bonded atom. The tendency of an atom to attract the shared pair of electrons towards itself is called electronegativity. Different scales of electronegativity have been proposed by different scientists, some of them being Pauling's scale, Mulliken's scale and Allred Rochow scale.
The ionization potential of two atoms A and B are 14 eV/atom and 10.8 eV/atom respectively and their electron affinities are 8.4 eV/atom and 6 eV/atom respectively.
With respect to the above given information answer the following question,
If A is an atom of the periodic table, then the value of (n+1) for the unpaired electron in A is 

Options

A. 4
B. 2
C. 1
D. 3

Answer

To solve the question regarding the value of (n+1) for the unpaired electron in atom A, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Information**:
- Ionization potential of atom A = 14 eV/atom
- Ionization potential of atom B = 10.8 eV/atom
- Electron affinity of atom A = 8.4 eV/atom
- Electron affinity of atom B = 6 eV/atom

2. **Use the Ionization Energy Formula**:
The formula for ionization energy (IE) in terms of atomic number (Z) and principal quantum number (n) is given by:
\[ \text{IE} = \frac{13.6 Z^2}{n^2} \]
For atom A, we can set up the equation:
\[ 14 = \frac{13.6 Z^2}{n^2} \]

3. **Rearranging the Equation**:
Rearranging the equation to find \( n^2 \):
\[ n^2 = \frac{13.6 Z^2}{14} \]

4. **Assuming Z**:
Since we do not know Z directly, we can assume a value for Z. A reasonable assumption is to start with \( Z = 1 \) (for hydrogen) and increase it to find a suitable integer that satisfies the equation.

5. **Testing Values for Z**:
- If \( Z = 2 \):
\[ n^2 = \frac{13.6 \times 2^2}{14} = \frac{54.4}{14} \approx 3.886 \quad \Rightarrow \quad n \approx 1.97 \quad (\text{not an integer}) \]
- If \( Z = 3 \):
\[ n^2 = \frac{13.6 \times 3^2}{14} = \frac{122.88}{14} \approx 8.75 \quad \Rightarrow \quad n \approx 2.96 \quad (\text{not an integer}) \]
- If \( Z = 4 \):
\[ n^2 = \frac{13.6 \times 4^2}{14} = \frac{217.6}{14} \approx 15.4 \quad \Rightarrow \quad n \approx 3.93 \quad (\text{not an integer}) \]

6. **Finding the Integer Value of n**:
After testing various values, we find that for \( Z = 2 \), \( n \) is approximately 2. Therefore, we can conclude \( n = 2 \).

7. **Calculating (n+1)**:
Now, we can find the value of \( n + 1 \):
\[ n + 1 = 2 + 1 = 3 \]

### Final Answer:
Thus, the value of \( (n+1) \) for the unpaired electron in atom A is **3**.

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