Periodic Table and Periodicity

On moving left to right in a period, metallic character decreases and on moving top to bottom in a group, metallic character increases. So, the correct order of increasing metallic character is

$ \mathrm{P}<\mathrm{Si}<\mathrm{Be}<\mathrm{Mg}<\mathrm{Na} $

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

$ \begin{array}{llllll} & \mathrm{Na} & \mathrm{Mg} & \mathrm{Be} & \mathrm{Si} & \mathrm{P} \\ \text { Electronegativity (On Pauling scale) } & 0.9 & 1.2 & 1.5 & 1.8 & 2.1 \end{array} $

The largest species is Mg and the smallest one is $\mathrm{Al}^{3+}$ among $\mathrm{Mg}, \mathrm{Mg}^{2+}, \mathrm{Al}$ and $\mathrm{Al}^{3+}$.

Unnilseptium element has atomic number 107

$ \mathrm{Li}$ and Mg are diagonally related and have similar properties.

The oxidation state and covalency of Al in $\left[\mathrm{AlCl}\left(\mathrm{H}_2 \mathrm{O}\right)_5\right]^{2+}$ are +3 and 6 respectively

∴ Statement A is incorrect

$ \begin{array}{|c|c|} \hline \text { Group 17 Elements } & \text { Electron gain enthalpy (kJ/mol) } \\ \hline \mathrm{F} & -328 \\ \hline \mathrm{Cl} & -349 \\ \hline \mathrm{Br} & -325 \\ \hline \mathrm{I} & -295 \\ \hline \end{array} $

$ \text { So, the correct order of electron gain enthalpy is } \mathrm{Cl}>\mathrm{F}>\mathrm{Br}>\mathrm{I} \text {. } $

$ \begin{aligned} & N=1 s^2 2 s^2 2 p^3 \\ & F: 1 s^2 2 s^2 2 p^5 \end{aligned} $

First ionization enthalpy ( $\Delta_{\mathrm{i}} \mathrm{H}$ )

$ \left(\Delta_{\mathrm{i}} \mathrm{H}\right) \text { for } \mathrm{N}=1402 \mathrm{~kJ} \mathrm{~mol}^{-1} $

$\Delta_{\mathrm{i}} \mathrm{H}$ for $\mathrm{O}=1314 \mathrm{~kJ} \mathrm{~mol}^{-1}$

$\Delta_{\mathrm{i}} \mathrm{H}$ for $\mathrm{F}=1681 \mathrm{~kJ} / \mathrm{mol}$

Due to half filled stable electronic configuration of nitrogen, It's first ionization enthalpy is higher than that of oxygen

$ \begin{aligned} & \mathrm{O}(\mathrm{~g}) \longrightarrow \mathrm{O}^{+}(\mathrm{g})+\mathrm{e}^{-} \\ & 2 \mathrm{~s}^2 2 \mathrm{p}^4 \,\,\,\,\,\,\,\,\,\ 2 \mathrm{~s}^2 2 \mathrm{p}^3 \end{aligned} $

                        (half filled p subshell)

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.

In general, bond enthalpy or bond dissociation energy of halogen molecules $\left(X_2\right)$ decreases down the group from $\mathrm{Cl}_2$ to $\mathrm{I}_2$.

$\mathrm{Cl}_2>\mathrm{Br}_2>\mathrm{F}_2>\mathrm{I}_2$

The decrease in bond enthalpy from $\mathrm{Cl}_2$ to $\mathrm{I}_2$ is due to increase in atomic size that increases the distance between two atoms. Hence, it becomes easier to break them apart. The exceptionally low bond dissociation energy of fluorine molecule is due to very small size of fluorine atom. This brings the non - bonding electrons on fluorine nearer to each other, resulting in a much greater lone pair-lone pair repulsion, which weakens the covalent bond and lowers its dissociation energy. Hence, option (d) is correct.

Since, $\mathrm{Zn}$ and $\mathrm{Cd}$ both have completely filled outermost shells, i.e. have no unpaired electrons. Thus, they can not form covalent and metallic bonds strongly and show boiling point less than $500^{\circ} \mathrm{C}\left[\mathrm{Zn}\left(146^{\circ} \mathrm{C}\right)\right.$ and $\left.\mathrm{Cd}\left(321^{\circ} \mathrm{C}\right)\right]$

Hence, option (b) is the correct.

Due to lanthanoid contraction poor screening offered by $[n-1] d$-electrons, size of first series of $d$-block elements initially decreases and finally increases in last.

Thus, (Ni) has least while $\mathrm{Zn}$ has maximum radius among the given options.

Hence, option (d) is the correct.

Electronegativity decreases down the group and increases on moving left to right in a period. Thus, Li is more electronegative than Na and F is more electronegative than Cl and Br. Br is less electronegative than Cl because of its larger size. Also, lesser be the electronegative of metal element and more is the electronegativity of halogen atom, more be the electronegativity difference of a given species.

Thus, Na and F show the maximum electronegative difference and option(b) is the correct.

The correct increasing order of basic strength is

$\mathrm{NH}_3>\mathrm{PH}_3>\mathrm{AsH}_3>\mathrm{SbH}_3$

$\mathrm{NH}_3$ is the most basic because of its small size, the electron density of electron pair is concentrated over small region. As the size increases, the electron density gets diffused over a large surface area. Hence, the ability to donate the electron pair (basicity) decreases. The other three statements given are correct.

$\mathrm{C}^{4-}, \mathrm{N}^{3-}$ and $\mathrm{O}^{2-}$ are isoelectronic species.

The ionic radius of isoelectronic species decreases with increase in nuclear charge.

Hence, the order of ionic radius is

species : $\mathrm{C}^{4-}>\mathrm{N}^{3-}>\mathrm{O}^{2-}$

Ionic radii $(\mathop A\limits^o): 2.60>1.71>1.40$

Assertion and reason both are correct and reason is the correct explanation of the assertion. Electron gain enthalpy becomes less negative as the size of an atom increases down the group. This is because within a group screening effect increases on going downward and the added electron would be farther away from the nucleus.