Question 58

Match List - I with List - II.
List - I (Electronic configuration of neutral atom where n = 2) List - II (1st Ionization Energy in kJ mol$$^{-1}$$)
A. ns$$^2$$ I. 2080
B. ns$$^2$$np$$^1$$ II. 899
C. ns$$^2$$np$$^3$$ III. 800
D. ns$$^2$$np$$^6$$ IV. 1402
Choose the correct answer from the options given below :

Solution

As we move across a period from left to right in the periodic table, the first ionization energy generally increases due to an increase in effective nuclear charge and a decrease in atomic size.

Fully-filled subshells ($$ns^2$$): Beryllium ($$2s^2$$) has a stable, fully-filled subshell, making it harder to remove an electron compared to Boron ($$2s^2 2p^1$$). Therefore, $$IE_1(\text{Be}) > IE_1(\text{B})$$.

Half-filled subshells ($$ns^2 np^3$$): Nitrogen ($$2s^2 2p^3$$) has an extra stable, half-filled p-subshell, making its ionization energy higher than Oxygen ($$2s^2 2p^4$$).

Noble Gases ($$ns^2 np^6$$): Neon ($$2s^2 2p^6$$) has a completely filled valence shell (stable octet), giving it an exceptionally high ionization energy.

Element given in List - I:

A. $$2s^2 \rightarrow$$ Beryllium (Be)
B. $$2s^2 2p^1 \rightarrow$$ Boron (B)
C. $$2s^2 2p^3 \rightarrow$$ Nitrogen (N)
D. $$2s^2 2p^6 \rightarrow$$ Neon (Ne)

Using our periodic trends and stability rules, we can arrange these elements in increasing order of their first ionization energy:

$$\text{Boron (B)} < \text{Beryllium (Be)} < \text{Nitrogen (N)} < \text{Neon (Ne)}$$

Neon (D) must have the highest value $$\rightarrow$$ I ($$2080$$)

Nitrogen (C) must have the second-highest value $$\rightarrow$$ IV ($$1402$$)

Beryllium (A) is higher than Boron $$\rightarrow$$ II ($$899$$)

Boron (B) must have the lowest value $$\rightarrow$$ III ($$800$$)

A-II, B-III, C-IV, D-I (Option A)

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