Classification of Elements and Periodicity in Properties
89521
Which of the following is not true for oxidation?
1 Addition of oxygen
2 addition of electronegative element
3 removal of hydrogen
4 removal of electronegative element
Explanation:
In the oxidation reaction, addition of oxygen and removal of hydrogen takes place. As we know, oxygen is more electronegative atom if we remove any electronegative atom then this process is termed reduction.
Kerala-CEE-29.08.2021
Classification of Elements and Periodicity in Properties
89518
Which of the following indicates the correct variation in electronegativities?
1 F $>$ N $<$ O $>\mathrm{C}$
2 F $>$ O $>\mathrm{N}>\mathrm{C}$
3 F $<$ N $<$ O $<\mathrm{C}$
4 $\mathrm{F}>\mathrm{N}>\mathrm{O}<\mathrm{C}$
Explanation:
On moving from left to right in a $2^{\text {nd }}$ period electronegativity increases. $\begin{array}{lllll}\text { Element } & \mathrm{C} & \mathrm{N} & \mathrm{O} & \mathrm{F} \\ \text { EN, Value } & 2.5 & 3.0 & 3.5 & 4\end{array}$ Hence, the order of $\mathrm{EN}$ are- $$ \mathrm{F}>\mathrm{O}>\mathrm{N}>\mathrm{C} $$
CG PET - 2018
Classification of Elements and Periodicity in Properties
89519
Pauling's equation for determining the electronegativity of an element is
According to Pauiling equation. $$ \begin{aligned} & \mathrm{X}_{\mathrm{A}}-\mathrm{X}_{\mathrm{B}}=0.208\left[\mathrm{E}_{\mathrm{A}-\mathrm{B}}-\left(\mathrm{E}_{\mathrm{A}-\mathrm{A}} \times \mathrm{E}_{\mathrm{B}-\mathrm{B}}\right)^{1 / 2}\right]^{1 / 2} \\ & \mathrm{X}_{\mathrm{A}}-\mathrm{X}_{\mathrm{B}}=0.208 \sqrt{\Delta} \end{aligned} $$ Where, $\Delta$ is the difference of bond dissociation energies of $\mathrm{A}-\mathrm{A}$ and B-B bonds. $\mathrm{X}_{\mathrm{A}}$ and $\mathrm{X}_{\mathrm{B}}$ are the electronegativities of $\mathrm{A}$ and $\mathrm{B}$ respectively. The factor 0.208 arises from the conversion of Kcals to electron volt.
J and K CET-(2006)
Classification of Elements and Periodicity in Properties
89526
The electronegativity of the given elements increases in the order
Electro negativity of given element is- $\begin{aligned} & \mathrm{Si}=1.9 \\ & \mathrm{P}=2.19 \\ & \mathrm{C}=2.55 \\ & \mathrm{~N}=3.04 \end{aligned}$ Therefore the increasing order is $-\mathrm{Si}<\mathrm{P}<\mathrm{C}<\mathrm{N}$.
AP EAPCET 20.08.2021 Shift-II
Classification of Elements and Periodicity in Properties
89528
Which pair of elements has maximum electro negativity difference?
1 Li \& F
2 $\mathrm{Na} \& \mathrm{~F}$
3 $\mathrm{Na} \& \mathrm{Br}$
4 $\mathrm{Na} \& \mathrm{Cl}$
Explanation:
Among the following $\mathrm{Li}$ and $\mathrm{Na}$ are the most electropositive and $\mathrm{F}, \mathrm{Cl}$ and $\mathrm{Br}$ are most electronegative. So, the electronegative of the species are- | Atom | $F$ | $Cl$ | $Br$ | $I$ | $Li$ | $Na$ | |---|---|---|---|---|---|---| |EN Value | 4 | 3 | 2.8 | 2.5 | 1 | 0.9 | | The difference in electronegativity are. (i) $\mathrm{F}-\mathrm{Li}=4-2.5=1.5$ (ii) $\mathrm{F}-\mathrm{Na}=4-0.9=3.1$ (iii) $\mathrm{Br}-\mathrm{Na}=2.8-0.9=1.9$ (iv) $\mathrm{Cl}-\mathrm{Na}=3-0.9=2.1$ Hence, the $\mathrm{Na}$ and $\mathrm{F}$ pairs of elements has maximum electronegativity difference.
Classification of Elements and Periodicity in Properties
89521
Which of the following is not true for oxidation?
1 Addition of oxygen
2 addition of electronegative element
3 removal of hydrogen
4 removal of electronegative element
Explanation:
In the oxidation reaction, addition of oxygen and removal of hydrogen takes place. As we know, oxygen is more electronegative atom if we remove any electronegative atom then this process is termed reduction.
Kerala-CEE-29.08.2021
Classification of Elements and Periodicity in Properties
89518
Which of the following indicates the correct variation in electronegativities?
1 F $>$ N $<$ O $>\mathrm{C}$
2 F $>$ O $>\mathrm{N}>\mathrm{C}$
3 F $<$ N $<$ O $<\mathrm{C}$
4 $\mathrm{F}>\mathrm{N}>\mathrm{O}<\mathrm{C}$
Explanation:
On moving from left to right in a $2^{\text {nd }}$ period electronegativity increases. $\begin{array}{lllll}\text { Element } & \mathrm{C} & \mathrm{N} & \mathrm{O} & \mathrm{F} \\ \text { EN, Value } & 2.5 & 3.0 & 3.5 & 4\end{array}$ Hence, the order of $\mathrm{EN}$ are- $$ \mathrm{F}>\mathrm{O}>\mathrm{N}>\mathrm{C} $$
CG PET - 2018
Classification of Elements and Periodicity in Properties
89519
Pauling's equation for determining the electronegativity of an element is
According to Pauiling equation. $$ \begin{aligned} & \mathrm{X}_{\mathrm{A}}-\mathrm{X}_{\mathrm{B}}=0.208\left[\mathrm{E}_{\mathrm{A}-\mathrm{B}}-\left(\mathrm{E}_{\mathrm{A}-\mathrm{A}} \times \mathrm{E}_{\mathrm{B}-\mathrm{B}}\right)^{1 / 2}\right]^{1 / 2} \\ & \mathrm{X}_{\mathrm{A}}-\mathrm{X}_{\mathrm{B}}=0.208 \sqrt{\Delta} \end{aligned} $$ Where, $\Delta$ is the difference of bond dissociation energies of $\mathrm{A}-\mathrm{A}$ and B-B bonds. $\mathrm{X}_{\mathrm{A}}$ and $\mathrm{X}_{\mathrm{B}}$ are the electronegativities of $\mathrm{A}$ and $\mathrm{B}$ respectively. The factor 0.208 arises from the conversion of Kcals to electron volt.
J and K CET-(2006)
Classification of Elements and Periodicity in Properties
89526
The electronegativity of the given elements increases in the order
Electro negativity of given element is- $\begin{aligned} & \mathrm{Si}=1.9 \\ & \mathrm{P}=2.19 \\ & \mathrm{C}=2.55 \\ & \mathrm{~N}=3.04 \end{aligned}$ Therefore the increasing order is $-\mathrm{Si}<\mathrm{P}<\mathrm{C}<\mathrm{N}$.
AP EAPCET 20.08.2021 Shift-II
Classification of Elements and Periodicity in Properties
89528
Which pair of elements has maximum electro negativity difference?
1 Li \& F
2 $\mathrm{Na} \& \mathrm{~F}$
3 $\mathrm{Na} \& \mathrm{Br}$
4 $\mathrm{Na} \& \mathrm{Cl}$
Explanation:
Among the following $\mathrm{Li}$ and $\mathrm{Na}$ are the most electropositive and $\mathrm{F}, \mathrm{Cl}$ and $\mathrm{Br}$ are most electronegative. So, the electronegative of the species are- | Atom | $F$ | $Cl$ | $Br$ | $I$ | $Li$ | $Na$ | |---|---|---|---|---|---|---| |EN Value | 4 | 3 | 2.8 | 2.5 | 1 | 0.9 | | The difference in electronegativity are. (i) $\mathrm{F}-\mathrm{Li}=4-2.5=1.5$ (ii) $\mathrm{F}-\mathrm{Na}=4-0.9=3.1$ (iii) $\mathrm{Br}-\mathrm{Na}=2.8-0.9=1.9$ (iv) $\mathrm{Cl}-\mathrm{Na}=3-0.9=2.1$ Hence, the $\mathrm{Na}$ and $\mathrm{F}$ pairs of elements has maximum electronegativity difference.
Classification of Elements and Periodicity in Properties
89521
Which of the following is not true for oxidation?
1 Addition of oxygen
2 addition of electronegative element
3 removal of hydrogen
4 removal of electronegative element
Explanation:
In the oxidation reaction, addition of oxygen and removal of hydrogen takes place. As we know, oxygen is more electronegative atom if we remove any electronegative atom then this process is termed reduction.
Kerala-CEE-29.08.2021
Classification of Elements and Periodicity in Properties
89518
Which of the following indicates the correct variation in electronegativities?
1 F $>$ N $<$ O $>\mathrm{C}$
2 F $>$ O $>\mathrm{N}>\mathrm{C}$
3 F $<$ N $<$ O $<\mathrm{C}$
4 $\mathrm{F}>\mathrm{N}>\mathrm{O}<\mathrm{C}$
Explanation:
On moving from left to right in a $2^{\text {nd }}$ period electronegativity increases. $\begin{array}{lllll}\text { Element } & \mathrm{C} & \mathrm{N} & \mathrm{O} & \mathrm{F} \\ \text { EN, Value } & 2.5 & 3.0 & 3.5 & 4\end{array}$ Hence, the order of $\mathrm{EN}$ are- $$ \mathrm{F}>\mathrm{O}>\mathrm{N}>\mathrm{C} $$
CG PET - 2018
Classification of Elements and Periodicity in Properties
89519
Pauling's equation for determining the electronegativity of an element is
According to Pauiling equation. $$ \begin{aligned} & \mathrm{X}_{\mathrm{A}}-\mathrm{X}_{\mathrm{B}}=0.208\left[\mathrm{E}_{\mathrm{A}-\mathrm{B}}-\left(\mathrm{E}_{\mathrm{A}-\mathrm{A}} \times \mathrm{E}_{\mathrm{B}-\mathrm{B}}\right)^{1 / 2}\right]^{1 / 2} \\ & \mathrm{X}_{\mathrm{A}}-\mathrm{X}_{\mathrm{B}}=0.208 \sqrt{\Delta} \end{aligned} $$ Where, $\Delta$ is the difference of bond dissociation energies of $\mathrm{A}-\mathrm{A}$ and B-B bonds. $\mathrm{X}_{\mathrm{A}}$ and $\mathrm{X}_{\mathrm{B}}$ are the electronegativities of $\mathrm{A}$ and $\mathrm{B}$ respectively. The factor 0.208 arises from the conversion of Kcals to electron volt.
J and K CET-(2006)
Classification of Elements and Periodicity in Properties
89526
The electronegativity of the given elements increases in the order
Electro negativity of given element is- $\begin{aligned} & \mathrm{Si}=1.9 \\ & \mathrm{P}=2.19 \\ & \mathrm{C}=2.55 \\ & \mathrm{~N}=3.04 \end{aligned}$ Therefore the increasing order is $-\mathrm{Si}<\mathrm{P}<\mathrm{C}<\mathrm{N}$.
AP EAPCET 20.08.2021 Shift-II
Classification of Elements and Periodicity in Properties
89528
Which pair of elements has maximum electro negativity difference?
1 Li \& F
2 $\mathrm{Na} \& \mathrm{~F}$
3 $\mathrm{Na} \& \mathrm{Br}$
4 $\mathrm{Na} \& \mathrm{Cl}$
Explanation:
Among the following $\mathrm{Li}$ and $\mathrm{Na}$ are the most electropositive and $\mathrm{F}, \mathrm{Cl}$ and $\mathrm{Br}$ are most electronegative. So, the electronegative of the species are- | Atom | $F$ | $Cl$ | $Br$ | $I$ | $Li$ | $Na$ | |---|---|---|---|---|---|---| |EN Value | 4 | 3 | 2.8 | 2.5 | 1 | 0.9 | | The difference in electronegativity are. (i) $\mathrm{F}-\mathrm{Li}=4-2.5=1.5$ (ii) $\mathrm{F}-\mathrm{Na}=4-0.9=3.1$ (iii) $\mathrm{Br}-\mathrm{Na}=2.8-0.9=1.9$ (iv) $\mathrm{Cl}-\mathrm{Na}=3-0.9=2.1$ Hence, the $\mathrm{Na}$ and $\mathrm{F}$ pairs of elements has maximum electronegativity difference.
NEET Test Series from KOTA - 10 Papers In MS WORD
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Classification of Elements and Periodicity in Properties
89521
Which of the following is not true for oxidation?
1 Addition of oxygen
2 addition of electronegative element
3 removal of hydrogen
4 removal of electronegative element
Explanation:
In the oxidation reaction, addition of oxygen and removal of hydrogen takes place. As we know, oxygen is more electronegative atom if we remove any electronegative atom then this process is termed reduction.
Kerala-CEE-29.08.2021
Classification of Elements and Periodicity in Properties
89518
Which of the following indicates the correct variation in electronegativities?
1 F $>$ N $<$ O $>\mathrm{C}$
2 F $>$ O $>\mathrm{N}>\mathrm{C}$
3 F $<$ N $<$ O $<\mathrm{C}$
4 $\mathrm{F}>\mathrm{N}>\mathrm{O}<\mathrm{C}$
Explanation:
On moving from left to right in a $2^{\text {nd }}$ period electronegativity increases. $\begin{array}{lllll}\text { Element } & \mathrm{C} & \mathrm{N} & \mathrm{O} & \mathrm{F} \\ \text { EN, Value } & 2.5 & 3.0 & 3.5 & 4\end{array}$ Hence, the order of $\mathrm{EN}$ are- $$ \mathrm{F}>\mathrm{O}>\mathrm{N}>\mathrm{C} $$
CG PET - 2018
Classification of Elements and Periodicity in Properties
89519
Pauling's equation for determining the electronegativity of an element is
According to Pauiling equation. $$ \begin{aligned} & \mathrm{X}_{\mathrm{A}}-\mathrm{X}_{\mathrm{B}}=0.208\left[\mathrm{E}_{\mathrm{A}-\mathrm{B}}-\left(\mathrm{E}_{\mathrm{A}-\mathrm{A}} \times \mathrm{E}_{\mathrm{B}-\mathrm{B}}\right)^{1 / 2}\right]^{1 / 2} \\ & \mathrm{X}_{\mathrm{A}}-\mathrm{X}_{\mathrm{B}}=0.208 \sqrt{\Delta} \end{aligned} $$ Where, $\Delta$ is the difference of bond dissociation energies of $\mathrm{A}-\mathrm{A}$ and B-B bonds. $\mathrm{X}_{\mathrm{A}}$ and $\mathrm{X}_{\mathrm{B}}$ are the electronegativities of $\mathrm{A}$ and $\mathrm{B}$ respectively. The factor 0.208 arises from the conversion of Kcals to electron volt.
J and K CET-(2006)
Classification of Elements and Periodicity in Properties
89526
The electronegativity of the given elements increases in the order
Electro negativity of given element is- $\begin{aligned} & \mathrm{Si}=1.9 \\ & \mathrm{P}=2.19 \\ & \mathrm{C}=2.55 \\ & \mathrm{~N}=3.04 \end{aligned}$ Therefore the increasing order is $-\mathrm{Si}<\mathrm{P}<\mathrm{C}<\mathrm{N}$.
AP EAPCET 20.08.2021 Shift-II
Classification of Elements and Periodicity in Properties
89528
Which pair of elements has maximum electro negativity difference?
1 Li \& F
2 $\mathrm{Na} \& \mathrm{~F}$
3 $\mathrm{Na} \& \mathrm{Br}$
4 $\mathrm{Na} \& \mathrm{Cl}$
Explanation:
Among the following $\mathrm{Li}$ and $\mathrm{Na}$ are the most electropositive and $\mathrm{F}, \mathrm{Cl}$ and $\mathrm{Br}$ are most electronegative. So, the electronegative of the species are- | Atom | $F$ | $Cl$ | $Br$ | $I$ | $Li$ | $Na$ | |---|---|---|---|---|---|---| |EN Value | 4 | 3 | 2.8 | 2.5 | 1 | 0.9 | | The difference in electronegativity are. (i) $\mathrm{F}-\mathrm{Li}=4-2.5=1.5$ (ii) $\mathrm{F}-\mathrm{Na}=4-0.9=3.1$ (iii) $\mathrm{Br}-\mathrm{Na}=2.8-0.9=1.9$ (iv) $\mathrm{Cl}-\mathrm{Na}=3-0.9=2.1$ Hence, the $\mathrm{Na}$ and $\mathrm{F}$ pairs of elements has maximum electronegativity difference.
Classification of Elements and Periodicity in Properties
89521
Which of the following is not true for oxidation?
1 Addition of oxygen
2 addition of electronegative element
3 removal of hydrogen
4 removal of electronegative element
Explanation:
In the oxidation reaction, addition of oxygen and removal of hydrogen takes place. As we know, oxygen is more electronegative atom if we remove any electronegative atom then this process is termed reduction.
Kerala-CEE-29.08.2021
Classification of Elements and Periodicity in Properties
89518
Which of the following indicates the correct variation in electronegativities?
1 F $>$ N $<$ O $>\mathrm{C}$
2 F $>$ O $>\mathrm{N}>\mathrm{C}$
3 F $<$ N $<$ O $<\mathrm{C}$
4 $\mathrm{F}>\mathrm{N}>\mathrm{O}<\mathrm{C}$
Explanation:
On moving from left to right in a $2^{\text {nd }}$ period electronegativity increases. $\begin{array}{lllll}\text { Element } & \mathrm{C} & \mathrm{N} & \mathrm{O} & \mathrm{F} \\ \text { EN, Value } & 2.5 & 3.0 & 3.5 & 4\end{array}$ Hence, the order of $\mathrm{EN}$ are- $$ \mathrm{F}>\mathrm{O}>\mathrm{N}>\mathrm{C} $$
CG PET - 2018
Classification of Elements and Periodicity in Properties
89519
Pauling's equation for determining the electronegativity of an element is
According to Pauiling equation. $$ \begin{aligned} & \mathrm{X}_{\mathrm{A}}-\mathrm{X}_{\mathrm{B}}=0.208\left[\mathrm{E}_{\mathrm{A}-\mathrm{B}}-\left(\mathrm{E}_{\mathrm{A}-\mathrm{A}} \times \mathrm{E}_{\mathrm{B}-\mathrm{B}}\right)^{1 / 2}\right]^{1 / 2} \\ & \mathrm{X}_{\mathrm{A}}-\mathrm{X}_{\mathrm{B}}=0.208 \sqrt{\Delta} \end{aligned} $$ Where, $\Delta$ is the difference of bond dissociation energies of $\mathrm{A}-\mathrm{A}$ and B-B bonds. $\mathrm{X}_{\mathrm{A}}$ and $\mathrm{X}_{\mathrm{B}}$ are the electronegativities of $\mathrm{A}$ and $\mathrm{B}$ respectively. The factor 0.208 arises from the conversion of Kcals to electron volt.
J and K CET-(2006)
Classification of Elements and Periodicity in Properties
89526
The electronegativity of the given elements increases in the order
Electro negativity of given element is- $\begin{aligned} & \mathrm{Si}=1.9 \\ & \mathrm{P}=2.19 \\ & \mathrm{C}=2.55 \\ & \mathrm{~N}=3.04 \end{aligned}$ Therefore the increasing order is $-\mathrm{Si}<\mathrm{P}<\mathrm{C}<\mathrm{N}$.
AP EAPCET 20.08.2021 Shift-II
Classification of Elements and Periodicity in Properties
89528
Which pair of elements has maximum electro negativity difference?
1 Li \& F
2 $\mathrm{Na} \& \mathrm{~F}$
3 $\mathrm{Na} \& \mathrm{Br}$
4 $\mathrm{Na} \& \mathrm{Cl}$
Explanation:
Among the following $\mathrm{Li}$ and $\mathrm{Na}$ are the most electropositive and $\mathrm{F}, \mathrm{Cl}$ and $\mathrm{Br}$ are most electronegative. So, the electronegative of the species are- | Atom | $F$ | $Cl$ | $Br$ | $I$ | $Li$ | $Na$ | |---|---|---|---|---|---|---| |EN Value | 4 | 3 | 2.8 | 2.5 | 1 | 0.9 | | The difference in electronegativity are. (i) $\mathrm{F}-\mathrm{Li}=4-2.5=1.5$ (ii) $\mathrm{F}-\mathrm{Na}=4-0.9=3.1$ (iii) $\mathrm{Br}-\mathrm{Na}=2.8-0.9=1.9$ (iv) $\mathrm{Cl}-\mathrm{Na}=3-0.9=2.1$ Hence, the $\mathrm{Na}$ and $\mathrm{F}$ pairs of elements has maximum electronegativity difference.