Electronegativity Questions in English

(A) Electronegativity generally decreases down a group in the periodic table due to the increase in atomic size and shielding effect.
Group $15$ elements are $N, P, As, Sb, Bi$.
As we move from $N$ to $Bi$,the electronegativity decreases.
Therefore,$Bi$ (Bismuth) has the lowest electronegativity among the given elements.

(B) According to the Pauling scale,the electronegativity of oxygen is $3.5$.

(B) Electronegativity increases across a period from left to right and decreases down a group.
Comparing the elements:
$F$ $(Z=9)$ is in period $2$,group $17$.
$O$ $(Z=8)$ is in period $2$,group $16$.
$N$ $(Z=7)$ is in period $2$,group $15$.
$P$ $(Z=15)$ is in period $3$,group $15$.
Since $F$,$O$,and $N$ are in the same period $(2)$,their electronegativity follows the order $F > O > N$.
$P$ is below $N$ in group $15$,so its electronegativity is lower than $N$.
Thus,the correct order is $F > O > N > P$.

(C) Electronegativity generally increases across a period from left to right and decreases down a group from top to bottom.
In the periodic table,$C$ and $N$ belong to the $2^{nd}$ period,while $Si$ and $P$ belong to the $3^{rd}$ period.
Comparing the values: $Si$ $(1.90)$ < $P$ $(2.19)$ < $C$ $(2.55)$ < $N$ $(3.04)$.
Thus,the correct increasing order of electronegativity is $Si < P < C < N$.

(A) The element with the highest electronegativity in the periodic table is Fluorine $(F)$.
Fluorine has an atomic number of $9$,and its electronic configuration is $1s^2, 2s^2, 2p^5$.
Since the highest principal quantum number is $n = 2$,it belongs to the $2^{nd}$ period.
It has $7$ valence electrons $(2s^2, 2p^5)$,which places it in Group $17$ (the Halogen group).
Therefore,the period and group numbers are $2$ and $17$ respectively.

(C) Fluorine $(F)$ is the element with the highest electronegativity,as it is located at the extreme right of the second period in the periodic table.
Francium $(Fr)$ is the most electropositive or least electronegative element,as it is located at the bottom left of the periodic table.

(B) Electronegativity is the tendency of an atom to attract a shared pair of electrons towards itself.
According to the Pauling scale,fluorine $(F)$ is the most electronegative element with a value of $4.0$.
Oxygen $(O)$ is the second most electronegative element with a value of $3.5$.

(A) According to the Mulliken scale,the electronegativity $(\chi)$ of an element is given by the average of its ionization potential $(IP)$ and electron gain enthalpy $(EA)$:
$\chi = \frac{IP + EA}{2}$
Given $IP = 13 \ eV$ and $EA = 4 \ eV$.
Substituting the values:
$\chi = \frac{13 + 4}{2} = \frac{17}{2} = 8.5 \ eV$.
Note: The formula $\chi = \frac{IP + EA}{5.6}$ is used to convert Mulliken electronegativity to the Pauling scale. Since the question asks for the value on the Mulliken scale,the correct calculation is $\frac{IP + EA}{2} = 8.5 \ eV$.

(A) Electronegativity generally increases across a period from left to right and decreases down a group in the periodic table.
Comparing the elements $Si$,$P$,$C$,and $N$:
$1$. In the second period,the order is $C < N$.
$2$. In the third period,the order is $Si < P$.
$3$. Comparing groups,$C > Si$ and $N > P$.
Combining these trends,the overall order of increasing electronegativity is $Si < P < C < N$.

(C) Statement-$I$ is correct. Due to the poor shielding effect of $d$ and $f$ electrons in $Tl$,the effective nuclear charge increases,resulting in a higher electronegativity for $Tl$ compared to $Al$ and $In$. The order is $B (2.04) > Tl (1.8) > Al (1.61) > In (1.5)$.
Statement-$II$ is incorrect. Boric acid ($H_3BO_3$ or $B(OH)_3$) is not a protonic acid (it does not release $H^+$ ions). Instead,it acts as a Lewis acid by accepting an $OH^-$ ion from water to form $[B(OH)_4]^-$,releasing $H^+$ ions from the water molecule.

(A) The electronegativity values for group $13$ elements are: $B (2.04)$,$Al (1.61)$,$Ga (1.81)$,$In (1.78)$,and $Tl (1.62)$.
Comparing these values,the order is $B > Ga > In > Tl > Al$.
However,considering the general trend and the specific values provided in standard textbooks,the electronegativity decreases from $B$ to $Al$,then increases slightly due to poor shielding of $d$ and $f$ electrons in $Ga$,$In$,and $Tl$.
The correct order is $B > Ga > In > Tl > Al$.

(D) In group $14$ (carbon family),electronegativity generally decreases down the group due to the increase in atomic size.
However,due to the poor shielding effect of $d$ and $f$ orbitals,the effective nuclear charge increases,leading to an anomaly in the trend.
The electronegativity values are approximately: $C$ $(2.55)$,$Si$ $(1.90)$,$Ge$ $(2.01)$,$Sn$ $(1.96)$,and $Pb$ $(2.33)$.
Comparing the given elements $C$,$Ge$,and $Pb$,the order is $C > Pb > Ge$.

(A) Electronegativity is the tendency of an atom to attract a shared pair of electrons towards itself in a chemical bond.
Across a period in the periodic table,from left to right,the effective nuclear charge increases and the atomic size decreases,leading to an increase in electronegativity.
The elements $C$,$N$,$O$,and $F$ belong to the second period of the periodic table.
Their positions from left to right are $C$ (Group $14$),$N$ (Group $15$),$O$ (Group $16$),and $F$ (Group $17$).
Therefore,the increasing order of electronegativity is $C < N < O < F$.

(C) Electronegativity is inversely proportional to the atomic size. Smaller the size,greater is the electronegativity.
Nitrogen $(N)$ is in the second period,so it has the smallest size and the highest electronegativity.
Iodine $(I)$ is a halogen,which are generally highly electronegative.
For sulfur $(S)$ and tellurium $(Te)$,as we move down the group,the atomic size increases,so the electronegativity decreases $(S > Te)$.
Comparing these,the electronegativity values are approximately: $N (3.04) > I (2.66) > S (2.58) > Te (2.10)$.
Thus,the correct order is $N > I > S > Te$.

(B) The ability of an atom to attract the shared pair of electrons towards itself is called electronegativity.
Fluorine $(F)$ has the highest electronegativity among the given species.
Due to its small atomic size and seven electrons in its outermost shell,it exhibits the highest electronegativity.
On the Pauling scale,its value is considered to be $4.0$.
Hence,option $(B)$ is the correct answer.

(A) Electronegativity is the tendency of an atom to attract a shared pair of electrons.
Based on the Pauling scale values:
$I$. $H (2.20)$ and $P (2.19)$ have similar values.
$II$. $Be (1.57)$ and $Al (1.61)$ have similar values.
$III$. $N (3.04)$ and $Cl (3.16)$ have similar values.
$IV$. $C (2.55)$ and $P (2.19)$ have different values.
Thus,the pairs $I, II,$ and $III$ have approximately the same electronegativity values.

(D) For group $13$ elements,electronegativity $(EN)$ decreases from $B$ to $Al$ due to the increase in atomic size.
From $Al$ to $Tl$,the electronegativity increases gradually due to the poor shielding effect of $d$ and $f$ orbitals,which increases the effective nuclear charge.
The values of $EN$ (on the Pauling scale) are: $B (2.0)$,$Al (1.5)$,$Ga (1.6)$,$In (1.7)$,and $Tl (1.8)$.
Thus,the correct variation shows a decrease from $B$ to $Al$ followed by an increase from $Al$ to $Tl$.

(C) Electronegativity generally increases across a period from left to right and decreases down a group.
Comparing the elements $N$ (Nitrogen),$O$ (Oxygen),and $P$ (Phosphorus):
$1$. $N$ and $O$ are in the $2^{nd}$ period,while $P$ is in the $3^{rd}$ period.
$2$. In the $2^{nd}$ period,electronegativity increases as $N < O$.
$3$. $P$ is below $N$ in the same group ($15^{th}$ group),so $N > P$.
$4$. Combining these,the electronegativity values on the Pauling scale are approximately: $O (3.44) > N (3.04) > P (2.19)$.
Therefore,the correct order is $O > N > P$.

(C) According to the Mulliken scale,the electronegativity $(EN)$ of an element is defined as the arithmetic mean of its ionisation potential $(IP)$ and electron affinity $(EA)$.
Mathematically,this is expressed as: $EN = \frac{IP + EA}{2}$.

(A) The electronegativity of an atom depends on the percentage of $s$-character in its hybrid orbitals. Higher $s$-character leads to higher electronegativity.
In $Me_{3}N$,the nitrogen atom is $sp^{3}$-hybridized ($25\% \ s$-character).
In $C_{5}H_{5}N$ (pyridine),the nitrogen atom is $sp^{2}$-hybridized ($33.3\% \ s$-character).
In $MeCN$ (acetonitrile),the nitrogen atom is $sp$-hybridized ($50\% \ s$-character).
Therefore,the order of electronegativity of the nitrogen atom is $MeCN > C_{5}H_{5}N > Me_{3}N$.

(A) $1$. Electronegativity trend: According to the Pauling scale,the electronegativity values are $F$ $(4.0)$,$O$ $(3.5)$,and $N$ $(3.0)$. Thus,the order $F > O > N$ is correct. Statement $I$ is true.
$2$. Oxidation states: In $OF_2$,fluorine is more electronegative than oxygen,so oxygen is $+2$. In $Na_2O$,sodium is $+1$,so oxygen is $-2$. Statement $II$ is true.
Therefore,both statements are true.