20 Most Important Questions from Periodic Properties | Periodic Table NEET Questions | NEET 2023

NEET 2023 is approaching, and it is time to do thorough revisions and give your best for the entrance exam. This comprehensive session consists of the 20 most important questions from Class 11 Chemistry Unit 3 Periodic Properties for NEET 2023. The Periodic Table is a list of elements arranged in ascending order of atomic number. Elements in the same group exhibit similar properties when atoms are grouped in this arrangement.

Question 1: Which of the following represents the correct order of electron affinity?

1. Cl > F > S > O
2. F > O > S > Cl
3. F > Cl > S > O
4. Cl > S > O > F

Answer: a) Cl > F > S > O

Explanation: The electron gain enthalpy (EA) of elements increases as the period progresses to the right. It also decreases down the group. The EA of Cl is higher than that of F, and EA of S is higher than that of O as F and O are very small in size. As a result, the correct EA order is Cl > F > S > O.

Question 2: Consider the isoelectronic ions, K⁺, S^2–, Cl^– and Ca²⁺. The radii of these ionic species follow the order:

1. Ca²⁺ > K⁺ > Cl^– > S^2–
2. Cl^– > S^2– > K⁺ > Ca²⁺
3. S^2– > Cl^– > K⁺ > Ca²⁺
4. K⁺ > Ca²⁺ > S^2– > Cl^–

Answer: c) S^2– > Cl^– > K⁺ > Ca²⁺

Explanation: As we progress along a period, size decreases, and as we proceed down, the group size increases, so here

Above are isoelectronic species, hence size ∝ 1/Z.

Thus, the radii of these ionic species follow the order S^2– > Cl^– > K⁺ > Ca²⁺.

Question 3: Which of the following has maximum ionisation energy?

1. N
2. O
3. O⁺
4. Na

Answer: c) O⁺

Explanation: The minimum energy needed to remove the most loosely bound electron from an isolated gaseous component is known as ionisation energy.

Along the period, ionisation energy tends to increase.

The effective nuclear charge determines the ionisation energy. The ionisation energy increases as the effective nuclear charge increases.

• ₇N – 1s² 2s² 2p³
• ₈O – 1s² 2s² 2p⁴
• ₈O+ – 1s² 2s² 2p³
• ₁₁Na – 1s² 2s² 2p² 3s¹

There is only one valence electron in Na. As a result, it can readily lose one electron. As a result, it has the least ionisation energy.

O⁺ has the maximum ionisation energy because its effective nuclear charge is greater than that of O and N.

Ionisation potential and ionisation energy are the same things.

Thus, option c is the correct response.

Question 4: Fluorine has the highest electronegativity among the ns²np⁵ group on the Pauling scale, but the electron affinity of fluorine is less than of chlorine because:

1. The atomic number of fluorine is less than that of chlorine.
2. Fluorine, being the first member of the family, behaves in an unusual manner.
3. Chlorine can accommodate an electron better than fluorine by utilising its vacant 3p-orbital.
4. Small size, high electron density and an increased electron repulsion make the addition of an electron to fluorine less favourable than that in the case of chlorine.

Answer: d) Small size, high electron density, and an increased electron repulsion make the addition of an electron to fluorine less favourable than that in the case of chlorine.

Explanation: We know that fluorine has a smaller atomic size than chlorine.

Flourine’s valence shell already has 7 electrons. As a result, if we add one more electron to the valence shell, the incoming electron will be repelled by electrons in the valence shell due to their smaller size and high electron density.

Chlorine has a larger dimension and a valence shell with seven electrons. As a result, it has a lesser electron density and less electron repulsions.

As a result of its small size, high electron density, and greater electron repulsion, adding an electron to fluorine is less advantageous than adding an electron to chlorine.

Thus, Cl > F [EA].

Question 5: The period number in the long form of the periodic table is equal to:

1. Magnetic quantum number of any element of the period
2. The atomic number of any element of the period
3. Maximum principal quantum number of any element of the period
4. Maximum azimuthal quantum number of any element of the period

Answer: c) Maximum principal quantum number of any element of the period

Explanation: The period number in the long form of the periodic table refers to the maximum principal quantum number of any element in the period because each period begins with the filling of electrons in a new principal quantum number.

Period number = maximum n of any element (where, n = principal quantum number).

Question 6: The screening effect of d-orbital electrons of a shell is:

1. Much less than s-electrons of the same shell
2. Much more than s-electrons of the same shell
3. Equal to s-electrons of the same shell
4. Equal to p-electrons of the same shell

Answer: a) Much less than s-electrons of the same shell

Explanation: In terms of screening, the ‘s’ has the maximum impact, followed by the letters p, d, and f (s > p > d > f).

Question 7: The periodic repetition of the properties of the elements is due to:

1. Recurrence of similar valence shell electronic configuration
2. Recurrence of same types of oxides formed
3. The same number of isotopic forms
4. Both ‘a’ and ‘b’

Answer: a) Recurrence of similar valence shell electronic configuration

Explanation: The repeating of similar electronic configurations of their atoms in the outermost energy shell after a regular interval is the reason for periodicity in element properties.

Question 8: Electrons enter 4s-subshell before 3d-subshell because of:

1. Hund’s rule
2. Pauli’s exclusion principle
3. Aufbau principle
4. None of the above

Answer: c) Aufbau principle

Explanation: The Aufbau principle governs how electrons are filled in an atom’s atomic orbitals when it is in its ground state. It claims that electrons fill atomic orbitals in the sequence of increasing orbital energy levels. The Aufbau principle states that the lowest energy atomic orbitals are occupied first, followed by the higher energy levels.

The orbitals with the smallest n + ℓ value will be filled first. In this case, the principal quantum number is n, and the azimuthal quantum number is ℓ. The following is an energy level diagram for orbitals:

Let us calculate the value of n + ℓ for 3d and 4s orbitals.

For 3d orbital, n = 3 and ℓ = 2

∴ n + ℓ = 5

For 4s orbital, n = 4 and ℓ = 0

∴ n + ℓ = 4

As a result, the electrons enter the 4s orbital before entering the 3d orbitals since the 4s orbital has less energy than the 3d orbital.

Question 9: Which of the following species has the highest ionisation potential?

1. Li⁺
2. Mg⁺
3. Al⁺
4. Ne

Answer: a) Li⁺

Explanation: Li⁺ has a noble gas electronic configuration. It has a 76.638 eV ionisation potential, while Ne has a 21.56 eV ionisation potential.

Question 10: The process that requires absorption of energy is:

1. N → N^–
2. F → F^–
3. Cl → Cl^–
4. H → H^–

Answer: a) N → N^–

Explanation: In fact, all elements, such as H, F, Cl, or N-atoms, can easily gain an electron to form ions.

However, all other atoms, with the exception of nitrogen, accept only one electron, releasing energy and causing the electron gain enthalpy to be negative.

Question 11: Which of the following is arranged in order of increasing radius?

1. K⁺(aq) < Na⁺(aq) < Li⁺(aq)
2. Na⁺(aq) < K⁺(aq) < Li⁺(aq)
3. K⁺(aq) < Li⁺(aq) < Na⁺(aq)
4. Li⁺(aq) < Na⁺(aq) < K⁺(aq)

Answer: d) Li⁺(aq) < Na⁺(aq) < K⁺(aq)

Explanation: The atomic and ionic radii increase as you move down the group. Li⁺, Na⁺, and K⁺ are all members of group 1. They occur in the order Li⁺, Na⁺, and K⁺ as you move from top to bottom. As a result, K⁺ has the highest ionic radius, followed by Na⁺ and Li⁺.

Question 12: When we move from left to right in a period, the electropositive character:

1. Increases
2. Decreases
3. No change
4. First increases, then decreases

Answer: b) Decreases

Explanation: As the tendency to lose electrons decreases as you move from left to right in a period, the electropositive character decreases.

Question 13: Which of the following order is wrong?

1. NH₃ < PH₃ < AsH₃ − Acidic nature
2. Li < Be < B < C − Ionisation energy
3. Al₂O₃ < MgO < Na₂O < K₂O – Basic nature
4. Li⁺ < Na⁺ < K⁺ < Cs⁺ − Ionic radius

Answer: b) Li < Be < B < C − Ionisation energy

Explanation: Li, Be, B, and C are all elements from the same period. Because the nuclear charge of the elements increases in the same direction, the value of the 1st ionisation potential increases as you move from left to right in a period. However, the ionisation potential of boron (B → 2s² p¹) is lower than that of beryllium (Be → 2s²) because, in the case of boron, 2p¹ electron must be removed to obtain B⁺ [B (2s² p¹) → B⁺ (2s²) + e⁻], whereas, in the case of Be, 2s² electron must be removed to obtain Be⁺ (2s¹). Because p electrons are easier to remove than s electrons, the energy required to remove an electron in boron is lower. Thus, the order will be Li < B < Be < C.

Question 14: Which of the following compounds will have the highest 2^nd I.E.?

1. Cr
2. Fe
3. V
4. Mn

Answer: a) Cr

Explanation: Cr+ :[Ar]3d⁵ has a stable half-filled configuration, so it has the highest second ionisation energy.

Question 15: The ionisation energy of nitrogen is more than that of oxygen because:

1. Of the extra stability of half-filled p-orbitals in nitrogen
2. Of the smaller size of nitrogen
3. The former contains less number of electrons
4. The former is less electronegative

Answer: a) Of the extra stability of half-filled p-orbitals in nitrogen

Explanation: The energy required to remove the outermost, or highest energy, an electron from a neutral atom in the gas phase is known as the first ionisation energy. Because of the half-filled electronic configuration of nitrogen, it is more stable than oxygen.

Question 16: Which of the following oxides is expected to react readily with NaOH?

1. Na₂O
2. CaO
3. NO
4. Cl₂O₇

Answer: d) Cl₂O₇

Explanation: When Cl₂O₇ is dipped in water, an acid called perchloric acid (HClO₄) forms.

When NaOH is dipped in water, it combines with the water to produce NaO, a basic compound. As a result, Cl₂O₇ is acidic in nature, whereas NaOH is basic in nature.

As we all know, compounds that are acidic in nature react strongly with substances that are basic.

As a result, Cl₂O₇ reacts strongly with NaOH.

Question 17: The element with the highest electron affinity will belong to:

1. Period 2, group 17
2. Period 3, group 17
3. Period 2, group 18
4. Period 2, group 1

Answer: b) Period 3, group 17

Explanation: Electron affinity refers to the amount of energy released when an electron is added to a neutral atom to form an anion.

Electron affinity tends to increase going left to right across a period

Electron affinity decreases going down the group.

Electron affinity = 1/Atomic size

Because of their small size, high effective nuclear charge, and nearly full outer shell of electrons, halogens have high electron affinities. When one electron is added to halogens with a high electron affinity, it releases a lot of energy.

Fluorine (F) is the more electronegative atom; however, when extra electrons are added, repulsion occurs due to its small size.

Because of the presence of unoccupied d-orbitals and the fact that electronegativity is not very low, electron repulsion is not observed in the case of chlorine as it is in the case of fluorine when extra electrons are added. As a result, chlorine has a greater electron affinity than fluorine.

Chlorine is a member of the 17th group, 3rd period of the periodic table.

Question 18: Pd has an exceptional electronic configuration of 4d¹⁰5s⁰. It belongs to:

1. 4^th period, group 11
2. 5^th period, group 10
3. 6^th period, group 9
4. 3^rd period, group 16

Answer: b) 5^th period, group 10

Explanation: According to the Aufbau principle, its electronic configuration should be [₃₆Kr] 4d⁸, 5s².

Thus its period = 5th as n = 5.

The element’s period is equal to the valence shell’s principal quantum number, and the d-block element’s group is equal to the number of electrons in an (n−1)d subshell + the number of electrons in the valence shell.

∴ Group = ns + (n − 1)d electrons = 2 + 8 = 10

Palladium is a chemical element with the symbol Pd and the atomic number 46. It’s a rare silvery-white metal that was discovered in 1803.

Question 19: Total number of elements present in the 5th period of the modern periodic table is:

1. 2
2. 8
3. 18
4. 32

Answer: c) 18

Explanation: The 5th period of the periodic table contains 18 elements.

Rubidium, the first element in the 5th period, has the atomic number 37. Rubidium, Strontium, Yttrium, Zirconium, Niobium, Molybdenum, Technetium, Ruthenium, Rhodium, Palladium, Silver, Cadmium, Indium, Tin, Antimony, Tellurium, Iodine and Xenon are the elements in the 5th period, in order of increasing atomic number.

Question 20: Which of the following is the correct order of metallic character for Si, Be, Mg, Na and P?

1. P < Si < Be < Na < Mg
2. P < Si < Be < Mg < Na
3. Na > Be > Mg > Be > P
4. Na > Si > Mg > Be > P

Answer: b) P < Si < Be < Mg < Na

Explanation: The reactivity of a metal is described by its metallic character. Metals have low ionisation energies, indicating that they lose electrons in chemical processes. Metal atoms have low electronegativities, which indicates that they have a low attraction for electrons within a compound. The trend in metallic character is the polar opposite of the ionisation energy trend. Metallic character tends to increase down a group and decreases across a period as we progress from left to right. As a result, the metallic character order is P < Si < Be < Mg < Na.

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