The electronic configuration of few elements is given below. Mark the statement which is not correct about these elements.
(i) `1s^(2)2s^(2)2p^(6)3s^(1)`
(ii) `1s^(2)2s^(2)2p^(5)`
(iii) `1s^(2)2s^(2)2p^(6)`
(iv) `1s^(2)2s^(2)2p^(3)`

Options

A. (i) is an alkali metal
B. (iii) is a noble metal
C. (i) and (ii) form ionic compound
D. (iv) has high ionisation enthalpy than accepted

Answer

To solve the question, we need to analyze the electronic configurations provided and determine which statement about the elements is not correct.

1. **Identify the Elements:**
- (i) `1s^(2)2s^(2)2p^(6)3s^(1)` corresponds to Sodium (Na), atomic number 11.
- (ii) `1s^(2)2s^(2)2p^(5)` corresponds to Fluorine (F), atomic number 9.
- (iii) `1s^(2)2s^(2)2p^(6)` corresponds to Neon (Ne), atomic number 10.
- (iv) `1s^(2)2s^(2)2p^(3)` corresponds to Nitrogen (N), atomic number 7.

2. **Analyze Each Statement:**
- **Statement 1:** The first element (Sodium) is an alkali metal.
- This statement is **correct** because Sodium is indeed an alkali metal (Group 1).

- **Statement 2:** The third element (Neon) is a noble gas.
- This statement is **correct** because Neon has a complete outer shell and is classified as a noble gas (Group 18).

- **Statement 3:** The first and second elements (Sodium and Fluorine) form ionic compounds.
- This statement is **correct** because Sodium (Na) and Fluorine (F) combine to form Sodium Fluoride (NaF), which is an ionic compound.

- **Statement 4:** The fourth element (Nitrogen) has higher ionization enthalpy than expected.
- This statement is **correct** because Nitrogen has a half-filled p subshell (2p³), which is more stable and requires more energy to remove an electron, leading to higher ionization enthalpy.

3. **Conclusion:**
- All the statements provided are correct regarding the elements' properties based on their electronic configurations. Therefore, there is no incorrect statement among the options given.

**Final Answer:** All statements are correct; none is incorrect.

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