MHT CET · Chemistry · Classification of Elements and Periodicity in Properties

Which from following is a CORRECT decreasing order of ionisation enthalpy for different elements?

A \(\mathrm{Ar}>\mathrm{Ne}>\mathrm{S}>\mathrm{Cl}\)
B \(\mathrm{Ne}>\mathrm{Ar}>\mathrm{Cl}>\mathrm{S}\)
C \(\mathrm{Ne}>\mathrm{S}>\mathrm{Cl}>\mathrm{Ar}\)
D \(\mathrm{Cl}>\mathrm{S}>\mathrm{Ne}>\mathrm{Ar}\)

Verified Solution

Answer & Solution

Correct Answer

(B) \(\mathrm{Ne}>\mathrm{Ar}>\mathrm{Cl}>\mathrm{S}\)

Step-by-step Solution

Detailed explanation

The ionisation enthalpy decreases down the group due to increase in the atomic size.
Therefore, ionisation enthalpy of \(\mathrm{Ne}\) is more than that of Ar.
Across a period ionisation enthalpy increases with increase of atomic number.
Therefore, ionisation enthalpy of \(\mathrm{S}, \mathrm{Cl}\) and \(\mathrm{Ar}\) is in the order: \(\mathrm{S} < \mathrm{Cl} < \mathrm{Ar}\).
Hence, the correct decreasing order of ionisation enthalpy for the given elements is:
\(\mathrm{Ne}>\mathrm{Ar}>\mathrm{Cl}>\mathrm{S}\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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