31 Years NEET Previous Year Questions: Classification of Elements & Periodicity

Statement A: "Unlike Ga that has a very high melting point, Cs has a very low melting point."

• This statement is False. Gallium (Ga) has a relatively high melting point, while Cesium (Cs) has an almost equal melting point.
  

Statement B: "On Pauling scale, the electronegativity values of N and Cl are same."

• This statement is incorrect. Nitrogen (N) has an electronegativity of 3.04, and Chlorine (Cl) has an electronegativity of 3.16 on the Pauling scale, so they have different electronegativities.

Statement C: "K⁺, Cl^-, Ca²⁺ and S^2- are all isoelectronic species."

• This statement is true. All these ions (K⁺, Cl^-, Ca²⁺, and S^2-) have the same number of electrons (18 electrons), making them isoelectronic species.

Statement D: "The correct order of the first ionization enthalpies of Na, Mg, Al, and Si is Si > Al > Mg > Na."

• This statement is incorrect. The correct order of ionization enthalpies is Si > Al > Mg > Na. Magnesium (Mg) has a higher ionization enthalpy than Aluminum (Al), and Sodium (Na) has the lowest ionization enthalpy due to its larger atomic size and lower nuclear charge compared to others.

Statement E: "The atomic radius of Cs is greater than that of Li and Rb."

• This statement is correct. While Cesium (Cs) has a larger atomic radius than Lithium (Li), it is also larger than Rubidium (Rb). The atomic radius increases as you move down a group in the periodic table, so Li < Na < K < Rb < Cs..

​Therefore, the correct answer is: Option d) C and E only.

​Analyze the given electronic configurations to determine whether they belong to main group elements:

• A. [Ne]3s¹: This corresponds to sodium (Na), which is in Group 1. It is a main group element.
• B. [Ar]3d³4s²: This corresponds to vanadium (V), which is a transition metal (Group 5). It is not a main group element.
• C. [Kr]4d¹⁰5s²5p⁵: This corresponds to iodine (I), which is in Group 17. It is a main group element.
• D. [Ar]3d¹⁰4s¹: This corresponds to copper (Cu), which is a transition metal (Group 11). It is not a main group element.
• E. [Rn]5f⁰6d²7s²: This corresponds to thorium (Th), which is an actinide and not a main group element.

So, only A and C belong to main group elements.

Electronegativity is a chemical property that describes the tendency of an atom to attract a shared pair of electrons (or electron density) towards itself in a chemical bond. The Pauling scale is the most commonly used scale to measure electronegativity. According to this scale:
The electronegativity of Fluorine (F) is the highest among the elements at around 3.98.
Oxygen (O) follows next with an electronegativity of about 3.44.
Nitrogen (N) has an electronegativity of approximately 3.04.
Carbon (C) has an electronegativity value close to 2.55.
Silicon (Si), being further down the group in the periodic table than Carbon, has a lower electronegativity of about 1.90.
Based on these values, we can arrange the elements in order of increasing electronegativity as follows:
Si < C < N < O < F
Thus, considering the options provided:
Option A: Si < C < N < O < F is the correct answer since it correctly ranks the elements from the lowest to the highest electronegativity.

The first ionization enthalpy, also known as ionization energy, is the energy required to remove the most loosely bound electron from a neutral atom in the gaseous phase to form a cation. The trend of first ionization energies generally increases across a period from left to right in the periodic table. This is due to the increasing nuclear charge and the decreasing atomic radius, which cause the valence electrons to be attracted more strongly to the nucleus.
However, there are notable exceptions based on the electron configuration stability and electron pairing in orbitals. Let's analyze the given elements:
Lithium (Li): Being the first element in the period, it has the smallest nuclear charge and only one electron in its outer shell, which makes it easy to remove an electron.
Beryllium (Be): This element has two electrons in the 2s orbital. The removal of one electron slightly disturbs the fully filled 2s sub-shell, creating more stability than having an unpaired electron. Therefore, Be has a higher ionization energy than Li.
Boron (B): This element has a half-filled 2p orbital configuration (one electron in the 2p orbital), which is relatively less stable compared to a full or empty p orbital, leading to a slightly lower ionization energy than Be.
Carbon (C): With two electrons in separate 2p orbitals (following Hund's rule), C experiences more effective nuclear shielding and electron-electron repulsion compared to a single electron in Boron's 2p orbital. This makes it relatively easier to remove an electron from B than from C, but harder than removing one from Be.
Nitrogen (N): It has exactly half-filled 2p orbitals, which provides extra stability and hence has a higher ionization energy than Carbon. Contrarily, the configuration of three p electrons is stable owing to the exchange energy and symmetric distribution in space.
Given these points, we can order the elements by increasing first ionization enthalpy as follows:
Li < B < Be < C < N
This matches with Option B. Thus, the correct answer is:
Option B
Li < B < Be < C < N

Statement 1: The oxidation state of Al in [AlCl(H₂O)₅]²⁺ is +3, and the coordination number (covalency) is 6, which is correct for this complex. Thus, Statement 1 is correct.
Statement 2: Na₂O is indeed a basic oxide (since Na is an alkali metal), and Cl₂O₇ is an acidic oxide (since it is formed by a non-metal). Therefore, Statement 2 is correct.
Statement 3: O^2-, F^-, Na⁺, and Mg²⁺ all have the same number of electrons (10 electrons), making them isoelectronic species. Therefore, Statement 3 is correct.
Statement 4: Mg has an atomic radius larger than Mg²⁺, Al is larger than Al³⁺. Therefore, the smallest ion among Mg, Al, Mg²⁺, and Al³⁺ is Al³⁺, not Al. Hence, Statement 4 is incorrect.

A. Mendelevium
The symbol for Mendelevium is Md.
B. Meitnerium
The symbol for Meitnerium is Mt.
C. Moscovium
The symbol for Moscovium is Mc.
D. Nobelium
The symbol for Nobelium is No.
Thus, the correct match is: A-IV, B-I, C-II, D-III
So, the correct answer is (c).

The atomic radius generally decreases as we move across a period from left to right, due to the increase in nuclear charge, which pulls the electrons closer to the nucleus. On the other hand, the atomic radius increases as we move down a group because the number of electron shells increases, making the atom larger.
Now, let's look at the elements in question:

• Li (Lithium): Group 1, Period 2
• Be (Beryllium): Group 2, Period 2
• B (Boron): Group 13, Period 2
• C (Carbon): Group 14, Period 2
• Li has the largest atomic radius because it's in Group 1 and Period 2, so it has the least nuclear charge compared to the others.
• Be has a smaller atomic radius than Li because it is in Group 2, and its increased nuclear charge pulls the electrons closer.
• B has a smaller atomic radius than Be, as it is further across the period (Group 13).
• C has the smallest atomic radius, as it is in Group 14 and further along the period, with the highest nuclear charge.

Thus, the correct order is: Li > Be > B > C.
Therefore, the correct answer is (b).

A. Lanthanoid
Lanthanoids are elements in the f-block of the periodic table. The element Ce (Cerium) is a lanthanoid.
So, A-I: Ce.

B. d-Block element
d-block elements are transition metals. The element Mn (Manganese) belongs to the d-block.
So, B-IV: Mn.

C. p-Block element
p-block elements are located in groups 13 to 18. The element As (Arsenic) is a p-block element.
So, C-II: As.

D. s-Block element
s-block elements are found in groups 1 and 2. The element Cs (Cesium) is an s-block element.
So, D-III: Cs.
Thus, the correct answer is (C) A-I, B-IV, C-II, D-III.

To determine which element forms the largest ion to achieve the nearest noble gas configuration, we need to consider how each element achieves a stable electron configuration and the size of the resulting ion.

1. Sodium (Na): Sodium has an atomic number of 11 and typically forms a Na⁺ ion by losing 1 electron. The electron configuration of Na is [Ne]3s¹, and when it loses one electron, it achieves the electron configuration of Ne (a noble gas). The ion Na⁺ is smaller compared to the neutral atom.
2. Oxygen (O): Oxygen has an atomic number of 8 and typically forms an O^2- ion by gaining 2 electrons. The electron configuration of O is 1s² 2s² 2p⁴, and when it gains 2 electrons, it achieves the electron configuration of Ne (a noble gas). The ion O^2- is larger than the neutral oxygen atom due to the increased electron-electron repulsion after gaining electrons.
3. Fluorine (F): Fluorine has an atomic number of 9 and typically forms an F^- ion by gaining 1 electron. The electron configuration of F is 1s² 2s² 2p⁵, and when it gains 1 electron, it achieves the electron configuration of Ne (a noble gas). The ion F^- is also larger than the neutral fluorine atom due to the increased electron-electron repulsion.
4. Nitrogen (N): Nitrogen has an atomic number of 7 and typically forms a N^3- ion by gaining 3 electrons. The electron configuration of N is 1s² 2s² 2p³, and when it gains 3 electrons, it achieves the electron configuration of Ne (a noble gas). The N^3- ion is the largest among the ions because it gains 3 electrons, which causes the largest increase in size due to electron-electron repulsion.

Thus, the N^3- ion formed by Nitrogen (N) is the largest, as it gains 3 electrons and achieves the noble gas configuration of Ne.
Therefore, the correct answer is (D) Nitrogen (N).

To determine which set of species are isoelectronic, we need to check if they all have the same number of electrons.
Isoelectronic species:These are species that have the same number of electrons but may differ in their nuclear charge.Let's check the electron count for each option:Option (a): Na⁺, Cl^-, O^-, NO⁺

• Na⁺: Atomic number of Na = 11, so Na⁺ has 10 electrons.
• Cl^-: Atomic number of Cl = 17, so Cl^- has 18 electrons.
• O^-: Atomic number of O = 8, so O^- has 9 electrons.
• NO⁺: Nitrogen (N) has an atomic number of 7 and oxygen (O) has an atomic number of 8. NO⁺ has 9 electrons (8 from oxygen and 1 from nitrogen, after losing an electron).
  So, these species are not isoelectronic.

Option (b): N₂O, N₂O₄, NO⁺, NO

• The electron count for each of these species would need to be checked individually, but from the atomic structure, it's clear that they do not all have the same number of electrons. So, they are not isoelectronic.

Option (c): Na⁺, Mg²⁺, O^-, F^-

• Na⁺: 10 electrons (from earlier)
• Mg²⁺: Atomic number of Mg = 12, so Mg²⁺ has 10 electrons.
• O^-: 9 electrons.
• F^-: Atomic number of F = 9, so F^- has 10 electrons.
  These species are not all isoelectronic, as O^- has 9 electrons, while the others have 10.

Option (d): Ca²⁺, Ar, K⁺, Cl^-

• Ca²⁺: Atomic number of Ca = 20, so Ca²⁺ has 18 electrons.
• Ar: Atomic number of Ar = 18, so Ar has 18 electrons.
• K⁺: Atomic number of K = 19, so K⁺ has 18 electrons.
• Cl^-: Atomic number of Cl = 17, so Cl^- has 18 electrons.
  All these species have 18 electrons and are isoelectronic with each other.

Conclusion:
The correct answer is (d) Ca²⁺, Ar, K⁺, Cl^- because they all have the same number of electrons (18) and are isoelectronic.

Option C is correct.

Definitions: Basic oxides are oxides that react with acids to give salts and water; acidic oxides react with water or bases to give acids or salts; neutral oxides do not show marked acidic or basic behaviour; amphoteric oxides react with both acids and bases.

• Option A: K₂O is a basic oxide. However, In₂O₃ and SnO₂ show amphoteric behaviour (they react with both acids and bases). Hence this set is not purely basic and option A is incorrect.

• Option B: CO and N₂O are neutral oxides, but NO₂ is acidic; it reacts with water to give nitrous and nitric acids: 2 NO₂ + H₂O → HNO₂ + HNO₃. Thus option B is incorrect.

• Option C: All three listed oxides are acidic. For example, Mn₂O₇ + H₂O → 2 HMnO₄ and SO₂ + H₂O → H₂SO₃. TeO_{3 also behaves as an acidic oxide (forming oxyacids of tellurium). Therefore option C is correct.}

• Option D: BeO and Ga₂O₃ are amphoteric. The entry GeO as written is not the standard amphoteric oxide normally cited; GeO₂ is the commonly listed amphoteric oxide of germanium. Because of this incorrect species, option D is not fully correct.

Final answer: C.

• NO (Nitric oxide) is a neutral oxide, as it does not react significantly with acids or bases.
• CO₂ (Carbon dioxide) is an acidic oxide, as it reacts with water to form an acid (H₂CO₃) and with bases to form salts.
• CaO (Calcium oxide) is a basic oxide, as it reacts with acids to form salts and water.
• ZnO (Zinc oxide) is amphoteric, as it can react with both acids and bases to form salts.

So, the correct sequence is: NO (neutral), CO₂ (acidic), CaO (basic), ZnO (amphoteric).
Thus, the correct answer is (d) NO, CO₂, CaO, ZnO.

IUPAC name of element : 119 : ununennium

Electronic configuration of Gadolinium
Gd :- [Xe] 4f⁷ 5d¹ 6s²
In case of 3^rd ionisation enthalpy electron will be removed from 5d and resultant configuration will be [Xe]4f⁷ that is stable electronic configuration as it will have high exchange energy, hence less energy will be required to remove 3^rd electron.

• Generally, on moving left to right in a period. First ionization enthalpy of elements increases due to increase in effective nuclear charge.
• Due to more stable half-filled outer electronic configuration (2s²2p³) of N, its first ionization enthalpy is more than O. 

So, correct order of IP is : C < O < N < F

Due to more diffused nature of 5f orbitals as compared to 4f orbitals the shielding effect of 5f is poor, resulting in the decrease in size from left to right in actinoid series which is greater and gradual than that in lanthanoid series

Fluorine is a stronger oxidising agent than chlorine due to
(i) Low dissociation enthalpy of F-F bond
(ii) High hydration enthalpy of F− ion

101 - Unnilunium - Mendelevium
103 - Unniltrium - Lawrencium
106 - Unnilhexium - Seaborgium
111 - Unununium - Roentgenium
110 - Ununnilium - Darmstadtium

CO : Neutral oxide
BaO : Basic oxide
Al₂O₃ : Amphoteric oxide
Cl₂O₇ : Acidic oxide

Ionisation enthalpy increases in a period from left to right, because of increased nuclear charge and decrease in atomic radii. But because of half filled orbitals of Be, N and fully filled orbitals of Ne, stability of those atoms inceases.
So, the correct increasing order of first ionisation enthalpy will be :
Li < B < Be < C < O < N < F < Ne

Ga is slightly smaller than Al due poor shielding of d e– so Zeff increasing.
So, Atomic size : B < Ga < Al < In < Tl

In metals, on moving down the group, metallic character increases, so basic nature increases hence most acidic will be BeO. In fact, BeO is amphoteric oxide while other given oxides are basic oxides.

⇒ Ca²⁺ < K⁺ < Ar
Ar, K⁺ and Ca²⁺ are isoelectronic i.e with the same number of electrons, 18. For isoelectronic species, ionic radii decrease with increases in effective (relative) positive charge. Also Ar, K and Ca belong to the same period.

Cation loose electrons are smaller in size than the parent atom, where anions gain electrons are larger in size than the parent atom.
Hence the order is H^- > H^. > H⁺

As the nuclear charge increases, the force of attraction between the nucleus and the incoming electron increses and hence the elecron gain enthalpy becomes more negative, hence the correct order is Ca   <   Al    <   C    <   O   <    F

As the positive charge increases on metal cation, radius decreases. This is due to the fact that nuclear charge in the case of a cation is acting on lesser number of electrons and pulls them closer.

IE₁ of Na = – Electron gain enthalpy of Na⁺                    = – 5.1 Volt.

O < S < F < CI

Electron gain enthalpy – 141 – 200 – 333 – 349 kJ mol^–1