Periodic Table & Periodicity

Element $E$ is in period 4 and group 16.

The element in period 4 and group 16 is Selenium $(Se)$.

The element just above Selenium in the same group is Sulphur $(S)$.

Sulphur belongs to period 3, and its valence shell electronic configuration is:

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

So, the correct option is:

Option A: $3s^2,\ 3p^4$

The size of the ion depends on number of shells, more are the number of shells, larger is the size of anion. When number of shells are same (like in $\mathrm{O}^{2-}$ and $\mathrm{F}^{-}$), more negative ion will have larger size. Therefore, the correct decreasing order of ionic radius will be

$ \mathrm{I}^{-}>\mathrm{Se}^{2-}>\mathrm{Br}^{-}>\mathrm{O}^{2-}>\mathrm{F}^{-} $

There is a sudden rise observed in fourth ionisation energy after 3rd ionisation energy. The element might have achieved inert gas configuration after the removal of 3rd electron and hence there is large energy required for removal of electron from a stable configuration. Thus, element belongs to thirteenth group of periodic table. The option possible from given options is boron which in all has 5 electrons.

∴ Correct option is boron (B).

(A) is true because the ionic radii goes on decreasing on moving left to right across a period as the nuclear charge goes on increasing from left to right. Thus, nucleus will hold the newly added electron more tightly. Hence ionic radii of alkaline earth metals is smaller than alkali metals.

$(\mathrm{R})$ is false because as the number of electron increases from left to right in a period and the atomic number also increases by a value of one. Therefore, alkaline earth metals have higher nuclear charge than the alkali metals.

Isoelectronic species means species having same number of electrons.

(a) $\mathrm{O}^{2-}=10$ electrons

(b) $\mathrm{F}^{-}=10$ electrons

(c) $\mathrm{Na}^{+}=10$ electrons

(d) $\mathrm{Mg}^{2+}=10$ electrons

∴ All are isoelectronic species.

Here all these species are isoelectronic species, i.e. possess 10 electrons.

The species which has maximum negative charge will have maximum ionic radii and species which have maximum positive charge will have minimum ionic radii. Thus order of ionic radii will be

$ \mathrm{O}^{2-}>\mathrm{F}^{-}>\mathrm{Na}^{+}>\mathrm{Mg}^{2+} $

The correct order of ionic radii is $\mathrm{K}^{+}>\mathrm{Na}^{+}>\mathrm{Mg}^{2+}>\mathrm{Al}^{3+}$.

In this $\mathrm{Na}^{+}, \mathrm{Mg}^{2+}$ and $\mathrm{Al}^{3+}$ are isoelectronic species so in these size decreases with rising atomic number because on rising atomic number $Z_{\text {eff }}$ increases ( $\mathrm{Z}_{\mathrm{eff}}=$ effective nuclear charge).

Also, $\mathrm{K}^{+}$has the highest ionic radii as it belongs to fourth period and has highest number of shells.

Metallic character decreases along a period due to decrease in atomic radius. As effective nuclear charge between electrons and nucleus increases. So, atomic radius decreases and hence, ability to loose electron increase and metallic character decreases.

On moving down the group, from S to Se electron gain enthalpy decreases. Oxygen has an unexpectedly low value of electron gain enthalpy. This is because of smaller size of oxygen atom due to which incoming electron faces more electronic repulsions which decreases its electron affinity. So, the correct order is $\mathrm{O}>\mathrm{Se} >\mathrm{S}$.

$ \left.\begin{array}{l} \mathrm{Se} \rightarrow-195.5 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ \mathrm{~S} \rightarrow-208.8 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ \mathrm{O} \rightarrow-141.4 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{array}\right\} \begin{aligned} & \text { electron gain } \\ & \text { enthalpy } \end{aligned} $

The acidity of oxides increases on moving from left to right in the periodic table. Oxides of group 1 are highly basic while those of group 17 are highly acidic. Therefore, the correct order of decreasing acidic nature of given oxides is as follows

$ \mathrm{N}_2 \mathrm{O}_3>\mathrm{CO}_2>\mathrm{B}_2 \mathrm{O}_3>\mathrm{BeO}>\mathrm{Li}_2 \mathrm{O} $

Helium (He) is the second-most abundant element in the universe, after hydrogen, and accounts for about 25 per cent of the elements in the universe. Most of the helium in the universe was created in the Big Bang, but it is also

obtained as the product of hydrogen fusion in stars. The Darmstadtium is a chemical element with the symbol Ds and atomic number 110 . The very small atomic radius of osmium results in a small metal-metal separation. This small atomic separation alongwith osmium's relatively high atomic number gives rise to osmium's highest density.

In group 1A, francium is more electropositive than ceasium but, not stable as it is highly radioactive (maximum half-life of only 22 minutes).

The electronic configuration of elements are as follows

$ \begin{aligned} & \mathrm{Li}=1 s^2 2 s^1 ; \quad \mathrm{Be}=1 s^2 2 s^2 \\ & \mathrm{~B}=1 s^2 2 s^2 2 p^1 ; \mathrm{N}=1 s^2 2 s^2 2 p^3 \end{aligned} $

All these element belong to second period. The ionisation energy generally increases as you move from left to right across a period. The first ionisation energy of beryllium is greater than that of boron since the order of attraction of electrons towards nucleus is $2 s>2 p$. The

beryllium has a stable complete electronic configuration $1 s^2 2 s^2$ whereas boron has the electronic configuration $1 s^2 2 s^2 2 p^1$ which needs lesser energy than that of beryllium to remove the valence electron. Therefore, the order of first ionisation energy is $\mathrm{Li}<\mathrm{B}<\mathrm{Be}<\mathrm{N}$.

According to given electronic configuration, $X$ is $\mathrm{Al}, Y$ is Cl and $Z$ is Mg . Along a period, the basic nature of oxides decreases and acidic nature of oxides increases. $\mathrm{Mg}, \mathrm{Al}$ and Cl belong to same period with Mg on left side of the period, Cl on the right side and Al in between them.

$\therefore$ Oxide of Cl is i.e. $Y$ will be most acidic oxide of Mg , i.e. Z will be basic and oxide of Al , i.e. $X$ in between the other two.

$\therefore$ Correct increasing order acidic nature will be $Z < X < Y$

Assertion (A): 16th group elements have higher ionisation enthalpy values than 15th group elements in the corresponding periods.

Reason (R): 15th group elements have half-filled stable electronic configurations.

To analyze this, let's review the concepts of ionisation enthalpy and electronic configurations:

Ionisation Enthalpy: Ionisation enthalpy is the energy required to remove an electron from an isolated gaseous atom. Higher ionisation enthalpy indicates it is more difficult to remove an electron.

Electronic Configuration: The 15th group elements (nitrogen group) have half-filled p orbitals, which are considered to be more stable due to exchange energy and symmetry. This stability can lead to slightly higher ionisation enthalpies compared to their neighboring elements.

However, the statement in Assertion (A) indicates that the 16th group elements have higher ionisation enthalpy values than the 15th group elements. Let's provide a detailed explanation:

In the 15th group, the stability due to the half-filled p orbital configuration results in relatively higher ionisation enthalpies compared to the 16th group, where such stability is not present. Thus, the 15th group elements typically have higher ionisation enthalpies than the 16th group elements.

Therefore, Assertion (A) is incorrect.

The Reason (R) correctly states that 15th group elements have half-filled stable electronic configurations, leading to relatively higher ionisation enthalpy.

Thus, Reason (R) is correct.

Correct option: D. A is incorrect but R is correct.

To evaluate the Assertion (A) and Reason (R) and determine the correct option, we need to understand the concepts of electron gain enthalpy and the factors affecting it.

Assertion (A): Fluorine has a smaller negative electron gain enthalpy than chlorine.

Electron gain enthalpy refers to the enthalpy change when an electron is added to a neutral atom to form a negative ion. A more negative electron gain enthalpy indicates a greater tendency to gain an electron. Fluorine has a smaller atomic radius than chlorine, which means that the added electron in fluorine is closer to the nucleus and experiences a stronger attraction. Therefore, technically, one would expect fluorine to have a more negative electron gain enthalpy. However, due to the very small size of the fluorine atom, the inter-electronic repulsion in the compact 2p orbital is quite high. On the other hand, these repulsions are slightly lesser in the 3p orbitals of chlorine, making the overall electron gain enthalpy of chlorine more negative than that of fluorine. Hence, Assertion (A) is correct.

Reason (R): The electron-electron repulsion is higher in chlorine than in fluorine.

As discussed, the electron-electron repulsion is actually higher in fluorine due to its smaller size and the compact nature of its orbitals. Chlorine, with its larger size, experiences lesser electron-electron repulsion when an additional electron is added. Hence, Reason (R) is incorrect.

Based on this discussion, the correct evaluation of Assertion (A) and Reason (R) is given by:

Option C

A is correct but R is incorrect

Electron gain enthalpy increases across a period on moving from left to right. So, F has maximum electron gain enthalpy whereas N has minimum electron gain enthalpy. On moving down the group, electron gain enthalpy decreases but from 2 nd period to 3 rd period, it increases because of high electronic repulsions in 2nd period. Hence, the correct order is

$\mathrm{F}>\mathrm{S}>\mathrm{O}>\mathrm{N}$

The order of first ionisation enthalpy across period 2 is

$\mathrm{Li}<\mathrm{B}<\mathrm{Be}<\mathrm{C}<\mathrm{O}<\mathrm{N}<\mathrm{F}<\mathrm{Ne}$

So, $X$ element having second lowest first ionisation enthalpy is boron (B).

$Y$ element having second highest first ionisation enthalpy is fluorine ( F ).

Diagonal relationship exists between certain pairs of diagonally adjacent elements in the second and third period of the periodic table. It occurs because of the directions of trends in various properties as we move across the period or down the periodic table.

Lithium shows diagonal relationship with magnesium whereas aluminium with beryllium. Hence, the element $X$ and $Y$ are Mg and Be respectively.

Metallic character decreases on moving from left to right across a period. So, sodium ( Na ) is more metallic than aluminium ( Al ).

Also, metallic character increases on moving down the group. Thus, potassium $(\mathrm{K})$ is more metallic than sodium (Na).

As Be lies in 2nd period, so it is least metallic.

So, the correct order of the metallic character is

$\mathrm{K}>\mathrm{Na}>\mathrm{Al}>\mathrm{Be}$

Tetrahalide, $\mathrm{PbI}_4$ does not exist because iodine is a good reducing agent and it reduces $\mathrm{Pb}^{4+}$ to $\mathrm{Pb}^{2+}$. So, the compound formed is $\mathrm{PbI}_2$ not $\mathrm{PbI}_4$.

We are told:

$\mathrm{IE}(\mathrm{Mg})$ is smaller than $\mathrm{IE}(X)$ and $\mathrm{IE}(Y)$.

$\mathrm{IE}(\mathrm{Mg})$ is greater than $\mathrm{IE}(Z)$.

Hence we want:

$ \mathrm{IE}(\mathrm{Mg}) < \mathrm{IE}(X), \quad \mathrm{IE}(\mathrm{Mg}) < \mathrm{IE}(Y), \quad \mathrm{IE}(\mathrm{Mg}) > \mathrm{IE}(Z). $

Approximate first ionization energies (in $\mathrm{kJ\,mol^{-1}}$) of the relevant elements:

$\mathrm{IE}(\mathrm{Ne}) \approx 2080$

$\mathrm{IE}(\mathrm{Na}) \approx 496$

$\mathrm{IE}(\mathrm{Mg}) \approx 738$

$\mathrm{IE}(\mathrm{Al}) \approx 578$ (not listed but for reference)

$\mathrm{IE}(\mathrm{Si}) \approx 787$ (not listed but for reference)

$\mathrm{IE}(\mathrm{P}) \approx 1012$ (not listed but for reference)

$\mathrm{IE}(\mathrm{S}) \approx 1000$ (not listed but for reference)

$\mathrm{IE}(\mathrm{Cl}) \approx 1251$

$\mathrm{IE}(\mathrm{Ar}) \approx 1521$

$\mathrm{IE}(\mathrm{Li}) \approx 520$

We test each option:

(A) neon, sodium, chlorine

$X=\mathrm{Ne}$: $\mathrm{IE}(\mathrm{Ne})=2080 > 738$ ✓

$Y=\mathrm{Na}$: $\mathrm{IE}(\mathrm{Na})=496 < 738$ ✗ (contradicts “$\mathrm{Mg}$ < $Y$”)

$Z=\mathrm{Cl}$: $\mathrm{IE}(\mathrm{Cl})=1251 > 738$ ✗ (contradicts “$\mathrm{Mg}$ > $Z$”)

So (A) does not fit the requirement.

(B) argon, chlorine, sodium

$X=\mathrm{Ar}$: $\mathrm{IE}(\mathrm{Ar})=1521 > 738$ ✓

$Y=\mathrm{Cl}$: $\mathrm{IE}(\mathrm{Cl})=1251 > 738$ ✓

$Z=\mathrm{Na}$: $\mathrm{IE}(\mathrm{Na})=496 < 738$ ✓

Thus $\mathrm{IE}(\mathrm{Mg}) < \mathrm{IE}(\mathrm{Ar}), \,\mathrm{IE}(\mathrm{Cl})$ and $\mathrm{IE}(\mathrm{Mg}) > \mathrm{IE}(\mathrm{Na})$.

This matches perfectly.

(C) chlorine, lithium, sodium

$X=\mathrm{Cl}$: $\mathrm{IE}(\mathrm{Cl})=1251 > 738$ ✓

$Y=\mathrm{Li}$: $\mathrm{IE}(\mathrm{Li})=520 < 738$ ✗ (contradicts “$\mathrm{Mg}$ < $Y$”)

$Z=\mathrm{Na}$: $\mathrm{IE}(\mathrm{Na})=496 < 738$, but the mismatch in $Y$ is enough to discard it.

(D) argon, lithium, sodium

$X=\mathrm{Ar}$: $\mathrm{IE}(\mathrm{Ar})=1521 > 738$ ✓

$Y=\mathrm{Li}$: $\mathrm{IE}(\mathrm{Li})=520 < 738$ ✗ (contradicts “$\mathrm{Mg}$ < $Y$”)

$Z=\mathrm{Na}$: $\mathrm{IE}(\mathrm{Na})=496 < 738$ is fine for $Z$, but $Y$ fails.

Hence only (B) satisfies all conditions:

$ \boxed{\text{(B) argon, chlorine, and sodium.}} $

The ionic radii (in $\mathop A\limits^o $) of $\text{Mg}^{2+}$ and $\text{Al}^{3+}$, respectively, are:

Option B:

$0.72 \mathop A\limits^o$ for $\text{Mg}^{2+}$

$0.54 \mathop A\limits^o$ for $\text{Al}^{3+}$

Among halogens (F₂, Cl₂, Br₂ and I₂), bond dissociation enthalpy ($\Delta$_dissH$^\circ$) of I₂, is minimum because of larger size of I-atom there is a steric repulsion between bonded I-atoms, which makes I-I bond weakest.

Whereas, smaller size and highest electronegativity of F-atom cause highest electron density on F-atom of F₂ molecule. As a result, F-F bond becomes weaker due to electrostatic repulsion between bonded F-atoms.

Thus, the order of $\Delta$_dissH$^\circ$ (in kJ mol^$-$1) is

$\mathop {Cl - Cl}\limits_{(242.6)} > \mathop {Br - Br}\limits_{(192.3)} > \mathop {F - F}\limits_{(158.8)\,Electrostatic\,repulsion} > \mathop {I - I}\limits_{(151.1)\,Steric\,repulsion} $

On moving left to right in a period of the periodic table, Ionisation energy (I.E.) increases due to increase in effective nuclear charge i.e. Z_eff.

But due to extra stability of fully filled and half-filled electronic configuration of Mg and P required ionisation enthalpy is more from neighbouring elements.

i.e. First ionisation enthalpy order is Al < Mg < Si < S < P.

Electronegativity is the tendency to attract shared pair of electrons.

Electronegativity of elements generally increases across the period and decreases down the group. Si = 1.8, P = 2.1, C = 2.5 and N = 3.0

So, the correct order is

Si < P < C < N

Magnesium ( $\mathrm{Mg}$ ) has the first ionisation enthalpy higher than that of Al. Due to fulfilled $s$-orbital of $\mathrm{Mg}$ it requires higher energy to remove the electron from outermost shell. On the other hand, $\mathrm{Al}$ can easily lose one electron as its $p$-orbital has only one electron.

Therefore,

$\begin{aligned} \mathrm{IE}_1(\mathrm{Mg}) & =737.7 \mathrm{kJmol}^{-1} \\ \mathrm{IE}_1(\mathrm{Al}) & =577.5 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}$