Periodic Table and Periodicity

Both Ga and Cs have low melting points.

$\bullet$ On Pauling scale, the electronegativity value of N and Cl have same (3.0).

$\bullet$ $\mathrm{Ar}, \mathrm{K}^{+}, \mathrm{Cl}^{-}, \mathrm{Ca}^{2+}$ and $\mathrm{S}^{2-}$ have 18 electrons. So these are isoelectronic species.

$\bullet$ The correct order of first ionization enthalpy is $\mathrm{Si}>\mathrm{Mg}>\mathrm{Al}>\mathrm{Na}$

First ionisation enthalpy of Mg is higher than Al because the penetration of a 3 s -electron to the nucleus is more than that of a 2 p -electron.

$\bullet$ Generally down the group atomic radii increases

(A) $[\mathrm{Ne}] 3 \mathrm{~s}^1 ; \quad \mathrm{Na}$ (s-block)

(B) $[$ Ar $] 3 \mathrm{~d}^3 4 \mathrm{~s}^2 ; \quad \mathrm{V}$ (d-block)

(C) $[\mathrm{Kr}] 4 \mathrm{~d}^{10} 5 \mathrm{~s}^2 5 \mathrm{p}^5 ; \mathrm{I}$ (p-block)

(D) $[\mathrm{Ar}] 3 \mathrm{~d}^{10} 4 \mathrm{~s}^1 ; \quad \mathrm{Cu}$ (d-block)

(E) $[R n] 5 f^0 6 d^2 7 s^2 ; \quad$ Th (f-block)

Main group elements (A and C only)

As the atomic number in a period increases, the effective nuclear charge also increases hence, atomic radii along the period decreases.

Correct order of atomic radii

$\mathrm{Li}>\mathrm{Be}>\mathrm{B}>\mathrm{C}$

The correct match is A-III, B-IV, C-I, D-II

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:

$\mathrm{Li} < \mathrm{B} < \mathrm{Be} < \mathrm{C} < \mathrm{N}$

This matches with Option B. Thus, the correct answer is:

Option B

$\mathrm{Li}<\mathrm{B}<\mathrm{Be}<\mathrm{C}<\mathrm{N}$

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:

$ \text{Si} < \text{C} < \text{N} < \text{O} < \text{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.

$\mathrm{Mn}_2 \mathrm{O}_7, \mathrm{SO}_2, \mathrm{TeO}_3$ are acidic oxides.

NO $\to$ neutral

CaO $\to$ Basic

CO$_2$ $\to$ Acidic

ZnO $\to$ Amphoteric

$\bullet$ Generally, on moving left to right in a period. First ionization enthalpy of elements increases due to increase in effective nuclear charge.

$\bullet$ 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

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

As it can be observed from given data of question, in case of element 'X' there is huge difference between IP₁ and IP₂ hence it will have one electron in outermost shell and will be alkali metal.

While for 'Y' difference is not that high hence it will be alkaline earth metal.

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.