NEET Topicwise PYQs Previous Year Questions & Solutions | Periodic Table & Classification of Elements

Anand Classes presents NEET Topicwise PYQ Previous Year Questions & Solutions on Periodic Table & Classification of Elements. These detailed, chapter-wise solutions are designed to help NEET aspirants strengthen their concepts in the periodic table, modern periodic law, classification of elements, electronic configuration, periodic trends, and properties of elements. By practicing previous year NEET questions with step-by-step solutions, students can build strong exam strategies and boost their problem-solving skills. Click the print button to download study material and notes.

NEET Topicwise PYQs Solutions (Class 11 Chemistry) of Periodic Table and Classification of Elements

Iso-electronic Ion Pairs Question [NEET 2021]

Q.1 : From the following pairs of ions, which one is not an iso-electronic pair? [NEET 2021]
(a) $O^{2-}, F^-$
(b) $Na^+, Mg^{2+}$
(c) $Mn^{2+}, Fe^{3+}$
(d) $Fe^{2+}, Mn^{2+}$

Answer. (d)

Number of Electrons in Each Ion

Explanation

• $O^{2-}$ and $F^-$ are iso-electronic pair.
• $Na^+$ and $Mg^{2+}$ are iso-electronic pair.
• $Mn^{2+}$ and $Fe^{3+}$ are iso-electronic pair.
• $Fe^{2+}$ and $Mn^{2+}$ are not an iso-electronic pair. ✅

Incorrect Match Question [NEET (Sep.) 2020]

Q.2 : Identify the incorrect match. [NEET (Sep.) 2020]

Options:
(a) (B), (ii)
(b) (C), (iii)
(c) (D), (iv)
(d) (A), (i)

Answer. (c)

Given Data

Explanation

• In the given question, (D) Unununnium ($Z = 111$) is matched with (iv) Darmstadtium (Ds).
• But Darmstadtium has $Z = 110$.
• The correct name for $Z = 111$ is Roentgenium (Rg).

✅ So, D-(iv) is the incorrect match.

Question on Element Z = 114 [NEET 2017]

Q.3 : The element $Z = 114$ has been discovered recently. It will belong to which of the following family/group and electronic configuration? [NEET 2017]
(a) Halogen family, [Rn] $5f^{14} \, 6d^{10} \, 7s^2 \, 7p^5$
(b) Carbon family, [Rn] $5f^{14} \, 6d^{10} \, 7s^2 \, 7p^2$
(c) Oxygen family, [Rn] $5f^{14} \, 6d^{10} \, 7s^2 \, 7p^4$
(d) Nitrogen family, [Rn] $5f^{14} \, 6d^{10} \, 7s^2 \, 7p^6$

Answer. (b)

Explanation

• The element with atomic number $Z = 114$ is Flerovium (Fl).
• It is a super-heavy artificial chemical element.
• In the periodic table, it is a transactinide element in the p-block.
• It belongs to the 7th period and is the heaviest known member of the carbon family (Group 14).

Electronic Configuration of Flerovium ($Z = 114$)

$$
\text{[Rn]} \, 5f^{14} \, 6d^{10} \, 7s^2 \, 7p^2
$$

Question on Oxidation States [CBSE AIPMT 2009]

Q.4 : Which one of the elements with the following outer orbital configurations may exhibit the largest number of oxidation states? [CBSE AIPMT 2009]
(a) $3d^3 \, 4s^2$
(b) $3d^5 \, 4s^1$
(c) $3d^5 \, 4s^2$
(d) $3d^2 \, 4s^2$

Answer. (c)

Concept

• The sum of the number of unpaired electrons in the $d$-orbitals and the number of electrons in the $s$-orbital gives the number of possible oxidation states exhibited by a $d$-block element.

Calculation

(a) $3d^3 \, 4s^2 \;\;\Rightarrow\;\; \text{OS} = 3 + 2 = 5$

(b) $3d^5 \, 4s^1 \;\;\Rightarrow\;\; \text{OS} = 5 + 1 = 6$

(c) $3d^5 \, 4s^2 \;\;\Rightarrow\;\; \text{OS} = 5 + 2 = 7$

(d) $3d^2 \, 4s^2 \;\;\Rightarrow\;\; \text{OS} = 2 + 2 = 4$

Conclusion

Hence, the element with configuration $3d^5 \, 4s^2$ exhibits the largest number of oxidation states. ✅

Question on Electronic Configuration [CBSE AIPMT 2002]

Q.5 : An atom has electronic configuration:
$1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^3 \, 4s^2$
You will place it in:
(a) Fifth group
(b) Fifteenth group
(c) Second group
(d) Third group

Answer. (a)

Explanation

• Total electrons = 23 → Atomic number $Z = 23$.
• The element is Vanadium (V).
• Outer configuration: $3d^3 \, 4s^2$.
• Group number = $3 + 2 = 5$.
• Period number = 4 (highest $n = 4$).

✅ Therefore, the element belongs to the 5th group, 4th period of the periodic table.

Question on Atomic Number 118 [CBSE AIPMT 1996]

Q.6 : The element with the atomic number 118 will be:
(a) Alkali
(b) Noble gas
(c) Lanthanide
(d) Transition element [CBSE AIPMT 1996]

Answer. (b)

Explanation

• Atomic number $Z = 118$ corresponds to the element Oganesson (Og).
• It is the heaviest known noble gas, placed in Group 18 of the periodic table.

Complete Electronic Configuration of Oganesson ($Z = 118$) :

$
1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^{10} \, 4s^2 \, 4p^6 \, 4d^{10} \, 4f^{14} \, 5s^2 \, 5p^6 \, 5d^{10} \, 5f^{14} \, 6s^2 \, 6p^6 \, 6d^{10} \, 7s^2 \, 7p^6
$

• Condensed form:

$$
[\text{Rn}] \, 5f^{14} \, 6d^{10} \, 7s^2 \, 7p^6
$$

Special Note on Oganesson

• Oganesson (Og) is a synthetic, radioactive element.
• Discovered in 2002 by a joint team of Russian and American scientists.
• Only a few atoms have been produced in particle accelerators.
• Unlike lighter noble gases, relativistic effects may make Oganesson more reactive than expected.

Periodic Position Summary (Oganesson)

• Group: 18 (Noble gases)
• Period: 7
• Block: p-block
• Category: Noble gas (synthetic, radioactive)

✅ Therefore, the element with $Z = 118$ is a noble gas.

Question on Successive Elements in Periodic Table [CBSE AIPMT 1995]

Q.7 : The electronic configuration of an element is:
$1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^3$
What is the atomic number of the element which is present just below the above element in the periodic table?
(a) 33
(b) 34
(c) 36
(d) 49 [CBSE AIPMT 1995]

Answer. (a)

Explanation

• Total electrons = 15 → $Z = 15$ → element is Phosphorus (P).
• Phosphorus belongs to Group 15, Period 3.
• To find the element just below it, we look at the vertical group trend in the periodic table:

Group 15 Elements and Their Electronic Configurations

Observation:

• All Group 15 elements have the general outer electronic configuration:
  $$ns^2 \, np^3$$
• The element just below P in the periodic table (same group) is Arsenic (As).
• Atomic number of Arsenic = 33.
• This is why they exhibit similar valency and chemical properties.

Electronic configuration of Arsenic ($Z = 33$):

$$
1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^{10} \, 4s^2 \, 4p^3
$$

✅ Therefore, the element just below Phosphorus (P, $Z=15$) is Arsenic (As, $Z=33$).

Question on Atomic Number 33 [CBSE AIPMT 1993]

Q.8 : If the atomic number of an element is 33, it will be placed in the periodic table in the:
(a) First group
(b) Third group
(c) Fifth group
(d) Seventh group [CBSE AIPMT 1993]

Answer. (c)

Explanation

• The element with atomic number $Z = 33$ is Arsenic (As).
• Its electronic configuration is:

$$
1s^2 \, 2s^2 \, 2p^6 \, 3s^2 \, 3p^6 \, 3d^{10} \, 4s^2 \, 4p^3
$$

• In the valence shell ($n=4$), we have:

$$
4s^2 \, 4p^3
$$

• Total valence electrons $= 2 + 3 = 5$.
• Therefore, it belongs to Group 15 (VA or pnictogens), also called the fifth group in older notation.

✅ Hence, the correct answer is fifth group (Group 15).

Mnemonic for Group 15 Elements

Group 15 elements: N, P, As, Sb, Bi

👉 Mnemonic:
“Never Put Arsenic in Small Bottles”

• N → Nitrogen
• P → Phosphorus
• As → Arsenic
• Sb → Antimony
• Bi → Bismuth

Question on Families in Periodic Table [CBSE AIPMT 1989]

Q.9 : The electronic configuration of four elements are given below. Which element does not belong to the same family as others?
(a) [Xe] $4f^{14} \, 5d^{10} \, 6s^2$
(b) [Kr] $4d^{10} \, 5s^2$
(c) [Ne] $3s^2 \, 3p^5$
(d) [Ar] $3d^{10} \, 4s^2$

Answer. (c)

Explanation

• In a family (group), all elements have the same outermost electronic configuration.
• Let’s check each:

1. (a) [Xe] $4f^{14} \, 5d^{10} \, 6s^2$ → Mercury (Hg), $Z=80$

• Belongs to Group 12 (transition metals).

1. (b) [Kr] $4d^{10} \, 5s^2$ → Cadmium (Cd), $Z=48$

• Belongs to Group 12.

1. (c) [Ne] $3s^2 \, 3p^5$ → Chlorine (Cl), $Z=17$

• Belongs to Group 17 (Halogens).

1. (d) [Ar] $3d^{10} \, 4s^2$ → Zinc (Zn), $Z=30$

• Belongs to Group 12.

Conclusion

• Elements (a), (b), and (d) belong to Group 12 (Zn, Cd, Hg family).
• Element (c) has outer configuration $ns^2 \, np^5$, i.e., halogen family (Group 17).

✅ Therefore, (c) is the odd one out.

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