Periodic Table & Periodicity — NEET UG practice

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Sample questions with solutions

Q1 · 2026

Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R.

Assertion A: The first ionization enthalpy of O is lower than that of N and F.

Reason R: The loss of an electron from O leads to stable half-filled $p$ orbital

In light of the above statements, choose the most appropriate answer from the options given below:

A.
A is not correct but R is correct
B.
Both A and R are correct and R is the correct explanation of A
C.
Both A and R are correct and R is NOT the correct explanation of A
D.
A is correct but R is not correct.

Answer: B

Write the ground state electron configurations of nitrogen, oxygen and fluorine.

N: 1s^2 2s^2 2p^3, \quad O: 1s^2 2s^2 2p^4, \quad F: 1s^2 2s^2 2p^5

First ionization enthalpy is the energy needed to remove the most loosely bound electron from a neutral gaseous atom.
Nitrogen has a half-filled $2p^3$ configuration, which is extra stable due to symmetrical distribution and exchange energy, so removing an electron from N requires more energy than expected.
The measured first ionization enthalpies confirm this:

\Delta_iH(N) = 1402\ \text{kJ mol}^{-1}, \quad \Delta_iH(O) = 1314\ \text{kJ mol}^{-1}, \quad \Delta_iH(F) = 1681\ \text{kJ mol}^{-1}

So oxygen's first ionization enthalpy is lower than both nitrogen's and fluorine's, confirming Assertion A is correct.
When oxygen loses one electron, its configuration changes from $2s^2 2p^4$ to $2s^2 2p^3$ , which is a stable half-filled p subshell:

O(g) \rightarrow O^+(g) + e^-

This extra stability gained upon losing an electron makes the removal easier, which correctly explains why oxygen's ionization enthalpy is lower than nitrogen's — so Reason R is correct and is the right explanation of A.

Hence, the answer is option B: both A and R are correct, and R correctly explains A.

Q2 · 2026

The correct order of increasing metallic character of $Na, Be, P, Mg$ and Si is

A.
$P < Si < Be < Mg < Na$
B.
$P < Si < Na < Mg < Be$
C.
$P < Mg < Be < Si < Na$
D.
$Be < Si < P < Mg < Na$

Answer: A

Metallic character depends on how easily an atom loses electrons; it decreases across a period (left to right) because effective nuclear charge increases, pulling the outer electrons in more tightly.
Metallic character increases down a group because atomic size increases and the outermost electron is held less tightly.
Placing the given elements according to periodic position: Na and Mg lie in Period 3 groups 1 and 2 (strongly metallic), Be lies in Period 2 group 2, while Si and P are on the right side of Period 3, closer to non-metallic behaviour.
Since electronegativity is inversely related to metallic character,

\text{Electronegativity} \propto \frac{1}{\text{Metallic character}}

comparing the Pauling electronegativity values helps confirm the ranking: $Na(0.9) < Mg(1.2) < Be(1.5) < Si(1.8) < P(2.1)$ .

A lower electronegativity means higher metallic character, so reversing the ranking of metallic character gives:

P < Si < Be < Mg < Na

Hence, the correct increasing order of metallic character is $P < Si < Be < Mg < Na$ , matching option A.

Q3 · 2026

Identify the incorrect statement from the following :

A.
The largest and the smallest species among $Mg, Mg^{2+}, Al$ and $Al^{3+}$ are Al and $Mg^{2+}$ respectively.
B.
The IUPAC name of the element with atomic number 107 is Unnilseptium.
C.
The similarity in behaviour of Li with Mg is referred to as 'diagonal relationship'
D.
The oxidation state and covalency of Al in $[AlCl(H_2O)_5]^{2+}$ are 3 and 6, respectively.

Answer: A

Compare the sizes of $Mg, Mg^{2+}, Al$ and $Al^{3+}$ : removing electrons from a metal atom to form a cation always decreases its size, because the same nuclear charge now pulls fewer electrons closer.
Since $Mg$ has fewer protons than $Al$ , but both lose electrons to form much smaller cations, the actual size order works out to be:

Mg > Al > Mg^{2+} > Al^{3+}

So the largest species is Mg (not Al) and the smallest is $Al^{3+}$ (not $Mg^{2+}$ ) — this means statement A is incorrect.
The element with atomic number 107 is indeed named Unnilseptium under IUPAC's temporary systematic naming rules, so statement B is correct.
Lithium and magnesium, though in different groups, show similar chemical behaviour — this is the well-known diagonal relationship, so statement C is correct.
In the complex $[AlCl(H_2O)_5]^{2+}$ , aluminium retains its usual +3 oxidation state; since it is bonded to one $Cl^-$ ion and five $H_2O$ molecules, its covalency (coordination number) is 6, so statement D is correct.

Hence, the incorrect statement is option A.

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