Electronegativity Questions in English

(A) Electronegativity $(EN)$ is a dimensionless property defined as the relative tendency of an atom to attract a shared pair of electrons towards itself in a covalent bond.
Electron Affinity $(EA)$ and Ionization Potential $(IP)$ are measured in units of energy (e.g., $kJ \ mol^{-1}$ or $eV \ atom^{-1}$).
Atomic Radius $(AR)$ is measured in units of length (e.g., $pm$ or $\mathring{A}$).
Therefore, $EN$ has no unit.

(C) Fluorine is the most electronegative element in the periodic table.
Due to its highest electronegativity,it always attracts electrons and never loses them to form positive oxidation states.
Therefore,it exhibits an oxidation state of $-1$ only.

(C) The electronegativity ($E$.$N$.) of $Al$ is $1.5$,which is similar to that of $Be$ $(1.5)$.
$Al$ and $Be$ exhibit a diagonal relationship in the periodic table,which explains the similarity in their properties.

(C) Electronegativity generally increases from left to right across a period and decreases down a group in the periodic table.
Comparing the given pairs:
$(A)$ $Te$ $(2.1)$ and $Xe$ $(2.6)$: $Xe > Te$,so $Te > Xe$ is incorrect.
$(B)$ $Ga$ $(1.8)$ and $Ge$ $(2.0)$: $Ge > Ga$,so $Ga > Ge$ is incorrect.
$(C)$ $Si$ $(1.9)$ and $Al$ $(1.6)$: $Si > Al$ is correct.
$(D)$ $P$ $(2.1)$ and $S$ $(2.5)$: $S > P$,so $P > S$ is incorrect.
Therefore,the correct option is $C$.

(D) The elements of the $3^{rd}$ period of the $p-$block are $Al$,$Si$,$P$,$S$,and $Cl$.
Given that one is a metal and the rest are non-metals,and the electronegativity order is $P < Q < R < S$,we identify the elements as $Al$ (metal),$Si$,$P$,and $S$ (non-metals).
Electronegativity values for these elements are: $Al (1.61) < Si (1.90) < P (2.19) < S (2.58)$.
Thus,$P = Al$,$Q = Si$,$R = P$,and $S = S$.
$A$ hypovalent compound is one where the central atom has fewer than $8$ electrons in its valence shell (electron-deficient).
$AlCl_3$ is a well-known hypovalent compound because the $Al$ atom has only $6$ electrons in its valence shell.
Therefore,the element $P$ $(Al)$ forms a hypovalent chloride.

(D) $1$. Ionisation potential $(IP)$ decreases down the group. Thus,$IP$ of $Cl^{-} < F^{-}$ is incorrect; actually,$IP$ of $F^{-} > Cl^{-}$. However,comparing the options,let us evaluate $D$.
$2$. Electronegativity depends on the percentage of $s$-character. $s$-character in $sp^3$ is $25\%$,while in $sp$ it is $50\%$. Therefore,electronegativity order is $sp > sp^2 > sp^3$. The statement $sp^3 > sp$ is definitely incorrect.
$3$. $F < Cl$ is correct for electron affinity due to inter-electronic repulsion in the small $2p$ orbital of $F$.
$4$. $O^{-} < S^{-}$ is correct for $IP$ as $S$ is larger than $O$.

(C) $1$. Pauling's electronegativity $(EN)$ scale is based on bond energy differences.
$2$. Allred-Rochow $EN$ is defined as $0.359 \times (Z_{eff} / r^2) + 0.744$,so it increases with increasing $Z_{eff}$.
$3$. Mulliken $EN$ is defined as the arithmetic mean of ionization potential $(I.P.)$ and electron affinity $(EA)$,i.e.,$EN_M = (I.P. + EA) / 2$. It is not the geometrical mean.
$4$. Electronegativity generally decreases on moving down a group.

(C) Electronegativity $(E.N.)$ depends on the oxidation state and hybridization of the atom.
$A$. $F$ $(4.0)$ is more electronegative than $N$ $(3.04)$ on the Pauling scale. This is correct.
$B$. As the oxidation state of an atom increases,its electronegativity increases. In $Cl_2O_7$,the oxidation state of $Cl$ is $+7$,while in $Cl_2O_5$,it is $+5$. Thus,$E.N.$ of $Cl$ in $Cl_2O_7$ $>$ $E.N.$ of $Cl$ in $Cl_2O_5$. This is correct.
$C$. In $CH_4$,$C$ is $sp^3$ hybridized ($25\% \ s$-character),while in $CO_2$,$C$ is $sp$ hybridized ($50\% \ s$-character). Higher $s$-character leads to higher electronegativity. Therefore,$E.N.$ of $C$ in $CO_2$ $>$ $E.N.$ of $C$ in $CH_4$. The given statement is incorrect.
$D$. Higher positive charge on an ion increases its electronegativity. $Cu^{2+}$ has a higher positive charge than $Cu^+$,so $E.N.$ of $Cu^{2+}$ $>$ $E.N.$ of $Cu^+$. This is correct.
Therefore,the incorrect statement is $C$.

(A) The nature of the solution depends on the bond cleavage in $M-O-H$. If the $M-O$ bond breaks,it releases $OH-$,making it basic. If the $O-H$ bond breaks,it releases $H $,making it acidic.
Bond cleavage occurs where the difference in electronegativity $(\Delta EN)$ is higher.
For $M_1OH$: $\Delta EN(M_1-O) = |3.4 - 3.5| = 0.1$ and $\Delta EN(O-H) = |3.5 - 2.1| = 1.4$. Since $\Delta EN(O-H) > \Delta EN(M_1-O)$,the $O-H$ bond breaks,releasing $H $,so the solution is acidic.
For $M_2OH$: $\Delta EN(M_2-O) = |1.2 - 3.5| = 2.3$ and $\Delta EN(O-H) = |3.5 - 2.1| = 1.4$. Since $\Delta EN(M_2-O) > \Delta EN(O-H)$,the $M_2-O$ bond breaks,releasing $OH^-$,so the solution is basic.
Therefore,the solutions are acidic and basic respectively.

(B) The electronegativity values on the Pauling scale are approximately: $Si (1.90)$,$P (2.19)$,$C (2.55)$,and $N (3.04)$.
Thus,the order of increasing electronegativity is $Si < P < C < N$.
In a period,electronegativity increases from left to right due to an increase in effective nuclear charge.
In a group,electronegativity decreases down the group due to an increase in atomic size and shielding effect.

(D) The Allred-Rochow scale defines electronegativity based on the electrostatic force exerted by the nucleus on the valence electrons.
The formula is given by: $EN_{AR} = 0.359 \frac{Z_{eff.}}{r^2} + 0.744$,where $Z_{eff.}$ is the effective nuclear charge and $r$ is the covalent radius in $\mathring{A}$.

(C) In the compound $M-O-H$,the cleavage of the bond depends on the electronegativity difference.
If the electronegativity difference between $M$ and $O$ is greater than the electronegativity difference between $O$ and $H$,the $M-O$ bond is more polar and breaks to form $M^+$ and $OH^-$ ions.
Therefore,the $M-O$ bond breaks if $\Delta (E.N.)$ of $M$ and $O > \Delta (E.N.)$ of $O$ and $H$.

(A) The electronegativity $(X)$ on the Pauling scale is related to the sum of ionization enthalpy $(I.E.)$ and electron gain enthalpy $(E.A.)$ by the formula: $X = \frac{I.E. + E.A.}{540}$ (where energies are in $kJ \ mol^{-1}$).
Given: $I.E. = 275 \ kcal \ mol^{-1}$ and $E.A. = 86 \ kcal \ mol^{-1}$.
Sum $= 275 + 86 = 361 \ kcal \ mol^{-1}$.
Convert to $kJ \ mol^{-1}$ using $1 \ kcal = 4.184 \ kJ$: $361 \times 4.184 = 1510.42 \ kJ \ mol^{-1}$.
Electronegativity $= \frac{1510.42}{540} \approx 2.797 \approx 2.8$.

(C) The partial charge on an atom in a covalent bond depends on the relative electronegativity $(EN)$ of the bonded atoms. The atom with higher electronegativity acquires a partial negative charge $(\delta-)$,while the atom with lower electronegativity acquires a partial positive charge $(\delta )$.
$(a)$ In $HCl$,$EN_{Cl} (3.16) > EN_H (2.20)$,so $Cl$ has a partial negative charge.
$(b)$ In $BrCl$,$EN_{Cl} (3.16) > EN_{Br} (2.96)$,so $Cl$ has a partial negative charge.
$(c)$ In $OCl_2$,$EN_O (3.44) > EN_{Cl} (3.16)$,so $Cl$ has a partial positive charge $(\delta )$.
$(d)$ In $SCl_2$,$EN_{Cl} (3.16) > EN_S (2.58)$,so $Cl$ has a partial negative charge.
Therefore,the correct option is $C$.

(C) The polarity of a bond depends on the electronegativity difference between the bonded atoms. The electronegativity values are: $N = 3.04$,$F = 3.98$,$O = 3.44$,$Cl = 3.16$.
Calculating the differences:
$|N - Cl| = |3.04 - 3.16| = 0.12$
$|O - F| = |3.44 - 3.98| = 0.54$
$|N - F| = |3.04 - 3.98| = 0.94$
$|N - N| = |3.04 - 3.04| = 0$
Since $N - F$ has the largest electronegativity difference,it is the most polar bond.

(C) Due to the high electronegativity of fluorine,it has a very high tendency to accept electrons and undergo reduction. Therefore,fluorine acts as a stronger oxidizing agent compared to $Br_2$.

(A) The electronegativity values on the Pauling scale are: $F (4.0)$,$O (3.44)$,$Cl (3.16)$,and $Br (2.96)$.
Therefore,the correct order of electronegativity is $F > O > Cl > Br$.

(C) $1$. Option $A$: $S^{2-}, Cl^{-}, K^{+}, Ca^{2+}$ are isoelectronic species with $18$ electrons. Ionisation energy increases as the nuclear charge $(Z)$ increases. The order should be $Ca^{2+} > K^{+} > Cl^{-} > S^{2-}$. Thus,$A$ is incorrect.
$2$. Option $B$: The $2^{nd}$ ionisation energy involves removing an electron from a cation. For $C^{+}$,$N^{+}$,$O^{+}$,$F^{+}$,the order is $C < N < O < F$. The given order $C < N < F < O$ is incorrect.
$3$. Option $C$: Electronegativity generally decreases down a group. The order $B > Al > Ga > In > Tl$ is correct as it follows the group trend.
$4$. Option $D$: Ionic radius for isoelectronic species decreases as nuclear charge increases. The correct order is $Na^{+} > Mg^{2+} > Al^{3+}$ and $Li^{+} > Be^{2+}$. The given order is incorrect.

(A) The ionic character of a covalent bond is directly proportional to the difference in electronegativity between the bonded atoms.
The electronegativity of hydrogen $(H)$ is $2.1$.
The electronegativity differences $(\Delta EN)$ for the hydrogen halides are:
$HF: |4.0 - 2.1| = 1.9$
$HCl: |3.0 - 2.1| = 0.9$
$HBr: |2.8 - 2.1| = 0.7$
$HI: |2.5 - 2.1| = 0.4$
Since $HF$ has the largest electronegativity difference,it possesses the maximum ionic character.

(C) According to the Mulliken scale,the electronegativity $(EN)$ of an element is defined as the arithmetic mean of its ionization energy $(IE)$ and electron affinity $(EA)$.
Mathematically,it is expressed as: $EN = \frac{IE + EA}{2}$.
Therefore,both ionization energy and electron affinity are required to determine electronegativity.

(C) Electronegativity generally increases across a period from left to right and decreases down a group.
For the given elements: $Si$ (Group $14$,Period $3$),$P$ (Group $15$,Period $3$),$C$ (Group $14$,Period $2$),and $N$ (Group $15$,Period $2$).
Comparing their values on the Pauling scale: $Si \approx 1.90$,$P \approx 2.19$,$C \approx 2.55$,and $N \approx 3.04$.
Therefore,the correct order of increasing electronegativity is $Si < P < C < N$.

(C) The electronegativity values on the Pauling scale are as follows:
$H = 2.1$
$Li = 1.0$
$Na = 0.9$
Comparing these values,the order is $H (2.1) > Li (1.0) > Na (0.9)$.
Therefore,the correct decreasing order is $H, Li, Na$.

(D) Electronegativity generally increases across a period from left to right in the periodic table.
The elements $C$,$N$,$O$,and $F$ belong to the second period.
Their atomic numbers are $C (6)$,$N (7)$,$O (8)$,and $F (9)$.
Following the periodic trend,the electronegativity increases as: $C < N < O < F$.
Therefore,the decreasing order is $F > O > N > C$.

(A) The most electronegative element in the periodic table is Fluorine $(F)$.
Its atomic number is $9$.
The electronic configuration of Fluorine is $1s^2 2s^2 2p^5$.
The valence shell configuration is $ns^2 np^5$ (where $n=2$).
Therefore,the correct option is $A$.

(B) In the hydroxide $E-O-H$,the nature depends on the electronegativity difference between $E$ and $O$ and between $O$ and $H$.
If the electronegativity difference between $E$ and $O$ is large (greater than $1.4$),the $E-O$ bond is highly polar and breaks to release $OH^-$ ions.
Therefore,the compound acts as a base.
Conversely,if the electronegativity difference between $O$ and $H$ is greater,the $O-H$ bond breaks to release $H^+$ ions,making the compound acidic.

(C) The acidic nature of oxides increases with the non-metallic character of the element,which in turn increases with electronegativity $(EN)$.
$X$ forms an acidic oxide,so it is a non-metal with high $EN$.
$Z$ forms an amphoteric oxide,which is intermediate in character.
$Y$ forms a basic oxide,so it is a metal with low $EN$.
Therefore,the order of electronegativity is $X > Z > Y$.

(B) Electronegativity $(EN)$ of an atom increases with an increase in the oxidation state of the atom because the effective nuclear charge increases.
In $N_2O_5$,the oxidation state of $N$ is $+5$,while in $NO_2$,it is $+4$. Thus,$EN$ of $N$ in $N_2O_5 > NO_2$ is correct.
In $CH_4$,the oxidation state of $C$ is $-4$,while in $CO_2$,it is $+4$. Thus,$EN$ of $C$ in $CO_2 > CH_4$. Therefore,the order $CH_4 > CO_2$ is incorrect.
For $Cu^{2+}$ and $Cu^+$,the higher positive charge leads to higher $EN$,so $Cu^{2+} > Cu^+$ is correct.
Bond polarity depends on the difference in electronegativity. The difference between $P$ and $F$ is greater than the difference between $O$ and $F$,making $P-F$ more polar than $O-F$.

(A) According to the Mulliken scale,the electronegativity $(EN)$ of an element is the average of its ionization potential $(IP)$ and electron affinity $(EA)$.
The formula is given by: $EN = \frac{IP + EA}{2}$.
Given that $EN = E_1$ and $IP = E_2$,we substitute these values into the formula:
$E_1 = \frac{E_2 + EA}{2}$.
Multiplying both sides by $2$ gives:
$2E_1 = E_2 + EA$.
Rearranging to solve for $EA$:
$EA = 2E_1 - E_2$.

(D) The polarity of a bond depends on the difference in electronegativity between the bonded atoms. The electronegativity values (Pauling scale) are: $O = 3.44$,$F = 3.98$,$S = 2.58$,$Cl = 3.16$,and $Si = 1.90$.
Calculating the differences:
$OF_2$: $|3.98 - 3.44| = 0.54$
$SF_4$: $|3.98 - 2.58| = 1.40$
$ClF$: $|3.98 - 3.16| = 0.82$
$SiF_4$: $|3.98 - 1.90| = 2.08$
The bond with the maximum difference in electronegativity is the $Si-F$ bond in $SiF_4$.

(C) The Assertion is correct because electron affinity is the energy released when an electron is added to an isolated gaseous atom,whereas electronegativity is the tendency of a bonded atom to attract shared electrons.
The Reason is incorrect because electron affinity is an experimentally measurable quantity (in $kJ \ mol^{-1}$),whereas electronegativity is a relative scale (like the $Pauling$ scale) and is not directly measurable.

(C) The assertion is true because $F$ is the most electronegative element in the periodic table.
The reason is false because $F$ has a lower electron affinity than $Cl$.
This is due to the small size of the $F$ atom,which leads to strong inter-electronic repulsions among the electrons in its $2p$ subshell,making it less favorable to add an extra electron compared to $Cl$.

(N/A) The electronegativity of an element is not a fixed value; it is a variable property that depends on the oxidation state and the chemical environment of the atom in a molecule.
Therefore,the statement that the electronegativity of $N$ on the Pauling scale is $3.0$ in all nitrogen compounds is incorrect.
For example,the electronegativity of $N$ varies depending on its hybridization and the nature of the atoms bonded to it,such as in $NH_3$ versus $NO_2$.

(N/A) Electronegativity is defined as the tendency of an atom in a chemical compound to attract a shared pair of electrons towards itself.
Differences between electronegativity and electron gain enthalpy:
$1$. Electronegativity is a relative tendency of an atom to attract electrons in a bonded state,whereas electron gain enthalpy is the energy change when an electron is added to an isolated gaseous atom.
$2$. Electronegativity is not a measurable quantity,while electron gain enthalpy can be measured experimentally.
$3$. Electronegativity is not a constant value for an element as it depends on the bonded atom,whereas electron gain enthalpy is a characteristic property of an element.

(N/A) The statement is incorrect.
The electronegativity of an element is not a constant value; it depends on the chemical environment.
Specifically,it depends on the hybridization state of the nitrogen atom (e.g.,$sp$,$sp^2$,or $sp^3$ hybridization).
It also depends on the oxidation state of the nitrogen atom and the nature of the atoms or groups attached to it.

Electronegativity:
- The tendency of an atom in a chemical compound to attract a shared pair of electrons towards itself is known as electronegativity.
$\rightarrow$ It is not a directly measurable quantity.
$\rightarrow$ Linus Pauling,an American scientist,in $1922$ assigned an arbitrary value of $4.0$ to fluorine.
Periodicity:
- As we move from left to right in a period,electronegativity increases due to an increase in effective nuclear charge and a decrease in atomic size.
Example: Second Period

Element	Electronegativity
$Li$	$1.0$
$Be$	$1.5$
$B$	$2.0$
$C$	$2.5$
$N$	$3.0$
$O$	$3.5$
$F$	$4.0$

- As we move from top to bottom in a group,electronegativity decreases due to an increase in atomic size and shielding effect.
Example: Group-$17$

Element	Electronegativity
$F$	$4.0$
$Cl$	$3.0$
$Br$	$2.8$
$I$	$2.5$
$At$	$2.2$

(N/A) The ionic character of a bond is directly proportional to the electronegativity difference between the two bonded atoms. The greater the difference,the more polar (ionic) the bond is.

Bond	Electronegativity difference
$C-H$	$2.5 - 2.1 = 0.4$
$N-H$	$3.0 - 2.1 = 0.9$
$O-H$	$3.5 - 2.1 = 1.4$
$F-H$	$4.0 - 2.1 = 1.9$

Based on the electronegativity values,the increasing order of ionic character is:
$C-H < N-H < O-H < F-H$

(A) The ionic character of a covalent bond depends on the difference in electronegativity between the bonded atoms.
Greater the electronegativity difference,higher is the ionic character.
The electronegativity values on the Pauling scale are: $C = 2.5, N = 3.0, O = 3.5$.
$1$. For $C-C$: Electronegativity difference = $|2.5 - 2.5| = 0$.
$2$. For $C-N$: Electronegativity difference = $|3.0 - 2.5| = 0.5$.
$3$. For $C-O$: Electronegativity difference = $|3.5 - 2.5| = 1.0$.
Since the electronegativity difference follows the order $0 < 0.5 < 1.0$,the order of increasing ionic character is $C-C < C-N < C-O$.

(B) The polarity of a covalent bond depends on the electronegativity difference between the bonded atoms.
Electronegativity values are: $N = 3.04$,$H = 2.20$,and $F = 3.98$.
The electronegativity difference for the $N-H$ bond is $|3.04 - 2.20| = 0.84$.
The electronegativity difference for the $N-F$ bond is $|3.04 - 3.98| = 0.94$.
Since $0.94 > 0.84$,the $N-F$ bond has a larger electronegativity difference and is therefore more polar than the $N-H$ bond.

(B) Electronegativity $(E.N.)$ increases across a period and decreases down a group. Fluorine $(F)$ is the most electronegative element $(E.N. = 4.0)$,while alkali metals have the lowest $E.N.$ values.
Comparing the given pairs:
$1$. $Li$ $(1.0)$ and $F$ $(4.0)$: Difference = $3.0$
$2$. $Na$ $(0.9)$ and $F$ $(4.0)$: Difference = $3.1$
$3$. $Na$ $(0.9)$ and $Br$ $(2.8)$: Difference = $1.9$
$4$. $Na$ $(0.9)$ and $Cl$ $(3.0)$: Difference = $2.1$
Thus,the pair with the maximum electronegativity difference is $Na \& F$.

(D) The electronegativity $(E.N.)$ values on the Pauling scale for the given elements are as follows:

Element	$E.N.$ Value
$Br$	$3.0$
$C$	$2.5$
$At$	$2.2$
$P$	$2.1$

Based on these values,the correct order of electronegativity is $Br > C > At > P$. Therefore,the correct option is $D$.

(A) The electronegativity values for group $14$ elements are as follows:
$C = 2.5$,$Si = 1.8$,$Ge = 1.8$,$Sn = 1.8$,$Pb = 1.9$.
From $Si$ to $Pb$,the electronegativity values do not decrease gradually; they remain almost constant ($1.8$ to $1.9$). Thus,Statement $I$ is false.
Group $14$ includes Carbon (non-metal),Silicon and Germanium (metalloids),and Tin and Lead (metals). Thus,Statement $II$ is true.

(D) Electronegativity generally increases across a period from left to right and decreases down a group.
$Si$ is in period $3$,group $14$.
$C$ is in period $2$,group $14$.
$N, O, F$ are in period $2$,groups $15, 16, 17$ respectively.
Comparing $Si$ and $C$: $Si$ is below $C$,so $Si < C$.
Comparing elements in period $2$: Electronegativity increases as $C < N < O < F$.
Combining these,the order of increasing electronegativity is $Si < C < N < O < F$.
The correct option is $D$.

(A) Electronegativity generally increases across a period from left to right and decreases down a group.
The electronegativity values (on the Pauling scale) for the elements in option $A$ are:
$Mg = 1.2$,$Al = 1.5$,$B = 2.0$,$N = 3.0$.
Therefore,the correct increasing order is $Mg < Al < B < N$.
The given order $Al < Mg < B < N$ is incorrect.

(D) Identify the elements based on their electronic configurations:
$A: 1s^2 2s^2 2p^3$ is Nitrogen $(N)$,Electronegativity $\approx 3.04$.
$B: 1s^2 2s^2 2p^4$ is Oxygen $(O)$,Electronegativity $\approx 3.44$.
$C: 1s^2 2s^2 2p^5$ is Fluorine $(F)$,Electronegativity $\approx 3.98$.
$D: 1s^2 2s^2 2p^2$ is Carbon $(C)$,Electronegativity $\approx 2.55$.
Electronegativity increases across a period from left to right in the periodic table.
The order of increasing electronegativity is $C < N < O < F$,which corresponds to $D < A < B < C$.

(C) The electronegativity of $Al$ is $1.61$ and that of $Tl$ is $1.62$. Thus,the electronegativity of $Tl$ is slightly greater than $Al$. Therefore,the Assertion is incorrect.
In the Boron family,due to the poor shielding effect of $d$ and $f$ orbitals (lanthanoid contraction),the atomic size does not increase regularly down the group. The size of $Ga$ is almost equal to $Al$. Therefore,the Reason is also incorrect.

(C) The electronic configuration $1s^2$ corresponds to Helium $(He)$.
Noble gases have a stable octet (or duplet in the case of $He$) configuration,which results in a very high ionization energy and effectively zero electron affinity and electronegativity.

(A) The electronegativity of Group $1$ elements decreases down the group as the atomic size increases and the effective nuclear charge decreases.
Therefore,the decreasing order of electronegativity is $Li > Na > K > Cs$.
Thus,$Li$ has the highest electronegativity among the given elements.

(D) Electronegativity generally increases as we move up a group in the periodic table due to a decrease in atomic size and an increase in effective nuclear charge.
All the given elements ($Be$,$Mg$,$Ca$,$Sr$) belong to Group $2$ (Alkaline Earth Metals).
The order of elements from top to bottom in Group $2$ is $Be$,$Mg$,$Ca$,$Sr$.
Since $Be$ is at the top of this group,it has the smallest atomic size and the highest electronegativity among the given options.

(A) Electronegativity is the tendency of an atom to attract a shared pair of electrons towards itself.
In the periodic table,electronegativity generally decreases down a group as the atomic size increases and the effective nuclear charge decreases.
All the given elements ($Li$,$Na$,$K$,$Rb$) belong to Group $1$ (alkali metals).
Since $Li$ is at the top of the group,it has the smallest atomic size and the highest effective nuclear charge among the given options.
Therefore,$Li$ has the highest electronegativity.

(B) Electronegativity generally increases across a period from left to right and decreases down a group.
Comparing the given elements: $O$ and $F$ are in the $2^{nd}$ period,while $S$ and $Cl$ are in the $3^{rd}$ period.
Since electronegativity decreases down a group,the elements in the $3^{rd}$ period ($S$ and $Cl$) have lower electronegativity than those in the $2^{nd}$ period ($O$ and $F$).
Between $S$ and $Cl$,$S$ is to the left of $Cl$ in the $3^{rd}$ period,therefore $S$ has the lowest electronegativity among the given options.