Electronegativity Questions in English

(C) The $HCl$ molecule is a covalent compound that exhibits partial ionic character.
This occurs because the electronegativity of chlorine $(3.0)$ is significantly greater than that of hydrogen $(2.1)$.
As a result,the shared electron pair is pulled more strongly towards the chlorine atom,creating a dipole moment where chlorine acquires a partial negative charge $(\delta^-)$ and hydrogen acquires a partial positive charge $(\delta^+)$.

(B) The polarity of a covalent bond is determined by the difference in the electronegativities of the two atoms forming the bond.
Greater the difference in electronegativity between the bonded atoms,the greater is the polarity of the bond due to the unequal sharing of the shared electron pair.

(B) The polarity of a bond in a diatomic molecule arises due to the difference in electronegativity between the two bonded atoms. When two atoms with different electronegativities form a covalent bond,the shared electron pair is attracted more towards the more electronegative atom,creating a partial positive charge $(\delta^+)$ on one atom and a partial negative charge $(\delta^-)$ on the other.

(D) The percentage of ionic character can be calculated using the Hannay-Smith equation:
Percentage of ionic character $= 16|E_A - E_B| + 3.5|E_A - E_B|^2$
Given,electronegativity of $A$ $(E_A)$ $= 1.20$ and electronegativity of $B$ $(E_B)$ $= 4.0$.
Difference in electronegativity $= |1.20 - 4.0| = 2.8$.
Substituting the values in the formula:
Percentage of ionic character $= 16(2.8) + 3.5(2.8)^2$
$= 44.8 + 3.5(7.84)$
$= 44.8 + 27.44$
$= 72.24\%$

(D) Fluorine is the most electronegative element in the periodic table. Due to its highest electronegativity,it always exhibits a $-1$ oxidation state in all its compounds.

(D) Fluorine $(F)$ is the most electronegative element in the periodic table.
Due to its highest electronegativity,it always attracts electrons towards itself in chemical bonds and never exhibits a positive oxidation state.

(B) In a period,as we move from left to right,the atomic size decreases and the effective nuclear charge increases,leading to an increase in electronegativity. \\ Conversely,on going from right to left in a period,the atomic size increases and the effective nuclear charge decreases,which causes the electronegativity to decrease. \\ Therefore,the correct option is $(B)$.

(D) The electronegativity values on the Pauling scale are as follows:
$O = 3.44$
$N = 3.04$
$Cl = 3.16$
$Br = 2.96$
Since the electronegativity of $Br$ is $2.96$,which is less than $3.0$,the correct option is $D$.

(A) The correct answer is $A$.
Fluorine $(F)$ has the highest electronegativity among all elements in the periodic table.
This is due to its small atomic size and high effective nuclear charge,which allows it to attract shared electrons in a covalent bond most strongly.

(C) The correct answer is $C$ (Oxygen).
Electronegativity increases across a period from left to right and decreases down a group.
Among the given elements ($C$,$Mg$,$O$,$S$),$O$ (Oxygen) is located in the second period and group $16$,making it the most electronegative due to its small atomic size and high effective nuclear charge.

(A) Electronegativity is defined as the tendency of an atom to attract a shared pair of electrons towards itself.
In the periodic table,electronegativity generally decreases as we move down a group due to an increase in atomic size and shielding effect.
All the given elements $(P, As, Sb, Bi)$ belong to Group $15$ of the periodic table.
The order of these elements in the group is $P > As > Sb > Bi$.
Therefore,$P$ (Phosphorus) is at the top of this group among the given options and possesses the maximum electronegative character.

(C) The most electronegative element in the periodic table is Fluorine $(F)$.
Fluorine belongs to the halogen group (Group $17$),which has the general valence shell electronic configuration of $ns^2 np^5$.
Therefore,the correct option is $(C)$.

(B) Electronegativity decreases down the group as the atomic size increases and the valence electrons are further away from the nucleus.

(D) The correct answer is $(D)$.
Electronegativity generally increases across a period from left to right and decreases down a group.
Among the halogens $(F, Cl, Br, I)$,fluorine $(F)$ is at the top of the group.
Due to its smallest atomic size and highest effective nuclear charge,$F$ has the highest electronegativity among all elements in the periodic table.

(A) The elements given belong to Group $14$ of the periodic table.
Electronegativity generally decreases as we move down a group due to an increase in atomic size and shielding effect.
The order of electronegativity for Group $14$ elements is $C > Si > Ge > Sn > Pb$.
Therefore,$Carbon$ $(C)$ is the most electronegative element among the given options.

(C) Electronegativity is defined as the tendency of an atom in a chemical compound to attract a shared pair of electrons towards itself.
Since the question specifies the attraction of electrons by an atom within a molecule,it refers to electronegativity.
Electron affinity,on the other hand,is the energy change when an electron is added to a neutral gaseous atom.

(A) . Electronegativity increases as we move from left to right across a period because the atomic size decreases and the effective nuclear charge increases.

(A) The correct order of increasing electronegativity is $Si < P < S$.
Across a period in the periodic table,the effective nuclear charge increases while the atomic radius decreases,which leads to an increase in electronegativity.
Since $Si$,$P$,and $S$ belong to the same period ($3^{rd}$ period) and are arranged in the order of increasing atomic number,their electronegativity increases in the order $Si < P < S$.

(A) Across a period,the atomic size decreases due to an increase in effective nuclear charge.
As a result,the attraction for electrons increases,leading to an increase in both $Electronegativity$ and $Ionization \text{ } Potential$.
$Electronegativity$ increases continuously across a period from left to right.
While $Electron \text{ } affinity$ generally increases,it does not show a perfectly progressive increase due to stable electronic configurations (e.g.,noble gases).
Therefore,$Electronegativity$ is the most accurate answer for a progressive increase.

(A) The electronegativity of chlorine $(Cl)$ is $3.16$ on the Pauling scale,while the electronegativity of hydrogen $(H)$ is $2.20$.
Since the electronegativity of $Cl$ is greater than that of $H$,the shared pair of electrons in the $H-Cl$ bond is shifted towards $Cl$.
Therefore,$H$ carries a partial positive charge $(\delta+)$ and $Cl$ carries a partial negative charge $(\delta-)$.
Thus,with respect to chlorine,hydrogen acts as the electropositive element.

(A) Electronegativity is defined as the tendency of an atom in a chemical compound to attract a shared pair of electrons towards itself.

(D) As we move from left to right in a period,the atomic size decreases due to an increase in effective nuclear charge.
Consequently,the tendency to attract shared electrons increases,which means the electronegative character increases.

(B) The electronegativity of elements increases as we move from left to right across a period in the periodic table.
All the given elements ($Al$,$Si$,$P$,and $S$) belong to the $3^{rd}$ period.
The order of electronegativity for these elements is $Al < Si < P < S$.
Therefore,$S$ (Sulfur) has the maximum electronegativity among the given options.

(A) Electronegativity is the tendency of an atom to attract a shared pair of electrons towards itself.
Among the given elements,$Li$ (Lithium) is an alkali metal located in Group $1$.
Alkali metals have the lowest electronegativity values in their respective periods because they have a strong tendency to lose their valence electron to achieve a stable noble gas configuration.
$F$ (Fluorine) is the most electronegative element in the periodic table.
Therefore,$Li$ has the lowest electronegativity among the options provided.

(B) The tendency of an atom in a chemical compound to attract a shared pair of electrons towards itself is defined as $Electronegativity$.
$Electron \ affinity$ refers to the energy released when an electron is added to a neutral gaseous atom.
$Ionisation \ energy$ is the energy required to remove an electron from a neutral gaseous atom.
$Valence$ refers to the combining capacity of an element.
Therefore,the correct term is $Electronegativity$.

(C) The electronegativity values (Pauling scale) for the given elements are approximately: $Si (1.90) < P (2.19) < C (2.55) < N (3.04)$.
In the periodic table,electronegativity increases from left to right across a period and decreases down a group.
Comparing the elements: $Si$ and $P$ are in the $3^{rd}$ period,while $C$ and $N$ are in the $2^{nd}$ period.
Since $Si$ and $P$ are below $C$ and $N$,they have lower electronegativity values.
Within the same period,electronegativity increases from left to right ($Si < P$ and $C < N$).
Therefore,the correct order is $Si < P < C < N$.

(B) The correct statement is $(B)$.
Electronegativity decreases down a group because the atomic radius increases,which leads to a decrease in the attraction between the nucleus and the shared pair of electrons.

(B) Electronegativity increases across a period from left to right due to an increase in effective nuclear charge and a decrease in atomic size.
For the elements $C$,$N$,$O$,and $F$,which belong to the same period ($2^{nd}$ period),the electronegativity increases as we move from $C$ to $F$.
Therefore,the order is $C < N < O < F$.

(A) The electronegativity values on the Pauling scale for the given elements are: $F (4.0) > O (3.5) > N (3.0) > C (2.5)$.
Therefore,the correct order of electronegativity is $F > O > N > C$.

(B) The electronegativity of elements increases across a period from left to right and decreases down a group.
Comparing the given elements: $Si$ (Group $14$,Period $3$),$P$ (Group $15$,Period $3$),$Cl$ (Group $17$,Period $3$),and $Br$ (Group $17$,Period $4$).
Among these,$Cl$ is in the same period as $Si$ and $P$ but further to the right,and it is in the same group as $Br$ but in a higher period (smaller atomic size).
Therefore,$Cl$ has the highest electronegativity.

(A) Electronegativity is a relative measure of the tendency of an atom to attract a shared pair of electrons towards itself in a chemical bond. Since it is a relative value,it is a dimensionless quantity and has no unit.
$B$,$C$,and $D$ represent energy or potential values,which are measured in units like $kJ \ mol^{-1}$ or $eV$.

(A) The electronegativity of elements generally increases across a period from left to right and decreases down a group.
Comparing the given elements:
$Bi$ (Bismuth) is in Group $15$,Period $6$.
$P$ (Phosphorus) is in Group $15$,Period $3$.
$S$ (Sulfur) is in Group $16$,Period $3$.
$Cl$ (Chlorine) is in Group $17$,Period $3$.
Since $Bi$ is at the bottom of Group $15$,it has the lowest electronegativity.
In Period $3$,the order of electronegativity is $P < S < Cl$.
Therefore,the overall increasing order is $Bi < P < S < Cl$.

(D) The electronegativity values on the Pauling scale are: $F = 4.0$,$O = 3.5$,$N = 3.0$,and $Cl = 3.0$.
However,considering the periodic trends and specific values,$F$ is the most electronegative element.
Comparing $O$ $(3.5)$ and $N$ $(3.0)$,$O > N$.
Comparing $N$ and $Cl$,both are approximately $3.0$,but $N$ is generally considered slightly more electronegative or equal in some scales,while $Cl$ is less electronegative than $O$ and $N$ in the context of these specific elements.
The correct decreasing order is $F > O > N > Cl$.

(A) The electronegativity of beryllium $(Be)$ is $1.5$ on the Pauling scale. Among the given options,aluminium $(Al)$ also has an electronegativity value of approximately $1.5$. This similarity is due to the diagonal relationship between $Be$ and $Al$ in the periodic table.

Element	Electronegativity
$Be$	$1.5$
$Al$	$1.5$

(A) The correct answer is $A$. Electronegativity decreases down a group in the periodic table as the atomic size increases and the valence electrons are further from the nucleus. For Group $16$ elements,the electronegativity values are as follows:

Element	Electronegativity
$O$	$3.5$
$S$	$2.5$
$Se$	$2.4$
$Te$	$2.1$
$Po$	$2.0$

Thus,$O$ (Oxygen) has the highest electronegativity.

(C) The oxidising power of halogens depends on their ability to gain electrons,which is determined by the standard electrode potential $(E^{\circ})$.
Fluorine $(F_{2})$ has a very high positive standard reduction potential due to its high electronegativity,low dissociation energy of the $F-F$ bond,and high hydration energy of the $F^{-}$ ion.
Therefore,$F_{2}$ is a stronger oxidising agent than $Br_{2}$.

(A) The Pauling electronegativity values of elements are primarily used to predict the polarity of covalent bonds in a molecule. $A$ difference in electronegativity between two bonded atoms leads to an unequal sharing of electrons,resulting in bond polarity.

(B) Electronegativity is the tendency of an atom to attract a shared pair of electrons towards itself. Atoms with high electronegativity generally have a small atomic size and a high effective nuclear charge,which results in a high ionization energy.

(A) Electronegativity increases as we move from left to right across a period in the periodic table.
$Si$ (Group $14$) and $P$ (Group $15$) are in the same period.
Since $P$ is to the right of $Si$,its electronegativity is higher than that of $Si$.
Therefore,the correct order is $P > Si$.

(B) In the periodic table,electronegativity increases from left to right across a period and decreases from top to bottom down a group.
Comparing the elements: $Si$ (Group $14$,Period $3$),$P$ (Group $15$,Period $3$),$C$ (Group $14$,Period $2$),and $N$ (Group $15$,Period $2$).
Electronegativity values are approximately: $Si (1.9) < P (2.1) < C (2.5) < N (3.0)$.
Therefore,the correct increasing order is $Si < P < C < N$.

(A) The ionic character of a bond is directly proportional to the difference in electronegativity between the bonded atoms.
Electronegativity of $H$ is $2.1$.
The electronegativity differences 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 highest percentage of ionic character.

(B) If the two elements have similar electronegativities,the bond between them will be covalent,while a large difference in electronegativities leads to an ionic bond.

(A) The polarity of a bond depends on the difference in electronegativity between the two bonded atoms. The greater the difference,the more polar the bond.
$1$. For $O-H$: $\Delta EN = |3.5 - 2.1| = 1.4$
$2$. For $S-H$: $\Delta EN = |2.5 - 2.1| = 0.4$
$3$. For $N-H$: $\Delta EN = |3.0 - 2.1| = 0.9$
$4$. For $C-H$: $\Delta EN = |2.5 - 2.1| = 0.4$
Since the electronegativity difference is highest for the $O-H$ bond,it is the most polar bond.

(D) The polarity of a covalent bond depends on the difference in electronegativity between the two bonded atoms.
Greater the difference in electronegativity,higher is the polarity of the bond.
The electronegativity values of the atoms involved are: $C = 2.55$,$O = 3.44$,$Br = 2.96$,$S = 2.58$,and $F = 3.98$.
The differences in electronegativity are:
$C-O: |3.44 - 2.55| = 0.89$
$C-Br: |2.96 - 2.55| = 0.41$
$C-S: |2.58 - 2.55| = 0.03$
$C-F: |3.98 - 2.55| = 1.43$
Since the difference in electronegativity is highest for the $C-F$ bond,it has the most polar character.
Therefore,the correct option is $(D)$.

(D) The nature of a chemical bond depends on the electronegativity difference between the bonded atoms.
If the difference in electronegativities of two elements is very large,the electron pair is shifted significantly towards the more electronegative atom,resulting in a bond that is more ionic than covalent.

(D) The graph shows that the value of $Y$ increases as the atomic number $(X)$ increases for the alkali metals $Li$,$Na$,and $K$.
$1$. $IE_1$ (first ionization energy) decreases down the group as atomic size increases.
$2$. Electron affinity generally decreases down the group.
$3$. $Z_{eff}$ (effective nuclear charge) calculated using Slater's rule for valence electrons in alkali metals remains approximately constant ($Z_{eff} \approx 1.30$ for $Li$,$Na$,$K$).
$4$. Electronegativity decreases down the group,so $\frac{1}{\text{electronegativity}}$ increases down the group.
Therefore,the value of $\frac{1}{\text{electronegativity}}$ increases as we move from $Li$ to $Na$ to $K$.

(A) The nature of a hydroxide $M-OH$ depends on the electronegativity difference between the metal/non-metal $M$ and oxygen $O$,or the electronegativity of $M$ itself.
According to the rule,if the electronegativity of $M$ is greater than $1.7$,the $O-H$ bond breaks,making the solution acidic.
If the electronegativity of $M$ is less than $1.7$,the $M-O$ bond breaks,releasing $OH^-$ ions,making the solution basic.
For $M_1-OH$: Electronegativity of $M_1 = 3.4 > 1.7$,so it is acidic.
For $M_2-OH$: Electronegativity of $M_2 = 1.2 < 1.7$,so it is basic.
Therefore,the nature of the solutions is acidic and basic respectively.

(C) $50 \,\%$ ionic character occurs when the electronegativity difference between the atoms is equal to $1.7$.
This is evident from curve $C$,which shows that as the electronegativity difference increases,the percent ionic character increases,reaching $50 \,\%$ at $\Delta E.N. = 1.7$.
$\Delta E.N. = 1.7 \Rightarrow 50 \,\% \text{ ionic and } 50 \,\% \text{ covalent}$
$\Delta E.N. > 1.7 \Rightarrow \text{more than } 50 \,\% \text{ ionic and less than } 50 \,\% \text{ covalent}$
$\Delta E.N. < 1.7 \Rightarrow \text{less than } 50 \,\% \text{ ionic and more than } 50 \,\% \text{ covalent}$

(A) Pauling introduced the concept of electronegativity to quantify the tendency of an atom to attract shared electrons in a chemical bond.
He established that the ionic character of a bond is directly related to the difference in electronegativity between the bonded atoms.
$A$ large difference in electronegativity leads to a bond with a high degree of polar character,meaning the bond is predominantly ionic,whereas a small difference results in a covalent bond.
Therefore,Pauling's values are primarily used to predict the polarity of bonds in molecules.

(A) Electronegativity $(EN)$ generally decreases down a group and increases across a period.
For $K$ and $Rb$,$K$ is above $Rb$ in Group $1$,so $EN(K) > EN(Rb)$.
For $Cl$ and $S$,$Cl$ is to the right of $S$ in Period $3$,so $EN(Cl) > EN(S)$.
For $Zr$ and $Hf$,due to lanthanoid contraction,the $EN$ of $Hf$ $(1.3)$ is slightly higher than that of $Zr$ $(1.22)$.
Therefore,the pair where the first element has lower $EN$ than the second is $Zr, Hf$.