MCQEasyJEE 2026Trends in Atomic Radius & Ionisation Energy

JEE Chemistry 2026 Question with Solution

The correct trend in the first ionization enthalpies of the elements in the 3rd3^{\text{rd}} period of periodic table is :

  • A

    AlAl

  • B

    SiSi

  • C

    SS

  • D

    AlAl

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The trend in first ionization enthalpies of selected elements of the 3rd3^{\text{rd}} period is to be identified.

Find: The correct increasing order among AlAl, SiSi, SS, PP and ClCl.

Ionization enthalpy generally increases across a period. However, there are important exceptions due to the extra stability of filled or half-filled subshells and the penetration difference between ss- and pp-electrons.

From the solution text, the relevant trend across the 3rd3^{\text{rd}} period is:

Na<Mg>Al<Si<P>S<Cl<ArNa<Mg>Al<Si<P>S<Cl<Ar

This indicates two exceptions:

  1. Mg>AlMg > Al because 3s23s^2 is more stable than 3p13p^1.
  2. P>SP > S because P(3p3)P(3p^3) has a half-filled stable configuration, whereas S(3p4)S(3p^4) has one paired electron.

Therefore, among the asked elements:

AlAl

So the correct option is A.

Half-filled Subshell Shortcut

Given: Ionization enthalpy usually increases from left to right across a period.

Find: Which option correctly accounts for the exception in the 3rd3^{\text{rd}} period.

A quick check is to compare the nearby elements and remember the exception: half-filled configuration wins. Thus P(3p3)P(3p^3) has higher ionization enthalpy than S(3p4)S(3p^4).

Now move left to right overall: AlAl is lower than SiSi, and ClCl is highest among the given elements. Therefore the increasing order becomes:

AlAl

Hence, the correct option is A.

Common mistakes

  • Assuming ionization enthalpy always increases strictly across a period is incorrect because exceptions occur for AlAl and SS. Always check subshell stability before writing the final order.

  • Placing SS above PP is a common error. This is wrong because P(3p3)P(3p^3) has a half-filled pp-subshell, which is more stable and therefore harder to ionize than S(3p4)S(3p^4).

  • Ignoring the role of orbital type in comparing MgMg and AlAl leads to confusion in the general trend. A 3p3p electron in AlAl is easier to remove than a 3s3s electron in a filled subshell.

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