Classification of Elements and Periodicity in Properties
89699
Which one of the following arrangements represent the correct order of electron gain enthalpy (with negative sign) of the given atomic species?
1 $\mathrm{F}<\mathrm{Cl}<\mathrm{O}<\mathrm{S}$
2 S $<$ O $<\mathrm{Cl}<\mathrm{F}$
3 $\mathrm{Cl}<\mathrm{F}<\mathrm{S}<\mathrm{O}$
4 $\mathrm{O}<\mathrm{S}<\mathrm{F}<\mathrm{Cl}$
Explanation:
In general, on moving down in a group, EA values decreases due to increase in size, because greater is the size of valence shell, lesser is the attraction and in turn lesser is the EA. But in period II and III, this order is reversed due to the small size of period II elements. Thus, $$ \mathrm{O}<\mathrm{S} \text { and } \mathrm{F}<\mathrm{Cl} $$ Because $\mathrm{O}$ and $\mathrm{F}$ atoms have high electron density $\left(\frac{\text { charge }}{\text { volume }}\right)$ and so repel the test electrons. Further, EA values increases as moving left to right along a period. It is because effective nuclear charge increases on moving left to right and more is the effective nuclear charge, more is the attraction of nucleus towards test electrons and thus more will be electron affinity. Thus, the overall increasing order of EA would be $$ \mathrm{O}<\mathrm{S}<\mathrm{F}<\mathrm{Cl} $$
BITSAT-2014
Classification of Elements and Periodicity in Properties
89704
The correct order of electron affinity of oxygen, sulphur, chlorine, fluorine is
1 S $<$ O $<\mathrm{Cl}<\mathrm{F}$
2 $\mathrm{S}<\mathrm{O}<\mathrm{F}<\mathrm{Cl}$
3 O $<$ S $<\mathrm{F}<\mathrm{Cl}$
4 $\mathrm{O}<\mathrm{S}<\mathrm{Cl}<\mathrm{F}$
Explanation:
On moving from top to bottom in a group the electron affinity became less negative and on moving from left to right across a period, the electron affinity became more negative. Chlorine has more negative electron affinity than fluorine because on adding on electron to fluorine $2 \mathrm{p}$ orbital causes greater repulsion than adding on electron to chlorine $3 \mathrm{p}$ orbital which is larger size. Therefore, the order is- $$ \mathrm{O}<\mathrm{S}<\mathrm{F}<\mathrm{Cl} $$
WB JEE-30.04.2022 UPTU/UPSEE-2010
Classification of Elements and Periodicity in Properties
89705
The correct order of electron gain enthalpy is:
1 $\mathrm{O}>\mathrm{S}>\mathrm{Se}>\mathrm{Te}$
2 $\mathrm{Te}>\mathrm{Se}>\mathrm{S}>\mathrm{O}$
3 S $>\mathrm{O}>\mathrm{Se}>\mathrm{Te}$
4 $\mathrm{S}>\mathrm{Se}>\mathrm{Te}>\mathrm{O}$
Explanation:
Electron gain enthalpy of oxygen is very low due to small size and compact nature of oxygen atom. Sulphur has highest electron gain enthalpy because sulphur has vacant d-orbital. So the order of electron gain enthalpy is - $$ \mathrm{S}>\mathrm{Se}>\mathrm{Te}>\mathrm{O} $$
JEE Main 26.02.2021
Classification of Elements and Periodicity in Properties
89709
The correct order of electron gain enthalpy with negative sign of $\mathrm{F}, \mathrm{Cl}, \mathrm{Br}$ and $\mathrm{I}$, having atomic number $9,17,35$ and 53 respectively, is
1 I $>\mathrm{Br}>\mathrm{Cl}>\mathrm{F}$
2 $\mathrm{F}>\mathrm{Cl}>\mathrm{Br}>\mathrm{I}$
3 $\mathrm{Cl}>\mathrm{F}>\mathrm{Br}>$ I
4 $\mathrm{Br}>\mathrm{Cl}>$ I $>\mathrm{F}$
Explanation:
Electron gain enthalpy has maximum in corresponding period of halogen and it becomes less negative down the group. However, the negative electron gain enthalpy of fluorine is less than that of chlorine due to the small size of fluorine atom. Therefore, there are strong inter electronic repulsion in the relatively small 2 p-orbital of fluorine thus, The incoming electron does not experience much attraction. The correct order of electron gain enthalpy are- $$ \mathrm{Cl}>\mathrm{F}>\mathrm{Br}>\mathrm{I} $$
Classification of Elements and Periodicity in Properties
89699
Which one of the following arrangements represent the correct order of electron gain enthalpy (with negative sign) of the given atomic species?
1 $\mathrm{F}<\mathrm{Cl}<\mathrm{O}<\mathrm{S}$
2 S $<$ O $<\mathrm{Cl}<\mathrm{F}$
3 $\mathrm{Cl}<\mathrm{F}<\mathrm{S}<\mathrm{O}$
4 $\mathrm{O}<\mathrm{S}<\mathrm{F}<\mathrm{Cl}$
Explanation:
In general, on moving down in a group, EA values decreases due to increase in size, because greater is the size of valence shell, lesser is the attraction and in turn lesser is the EA. But in period II and III, this order is reversed due to the small size of period II elements. Thus, $$ \mathrm{O}<\mathrm{S} \text { and } \mathrm{F}<\mathrm{Cl} $$ Because $\mathrm{O}$ and $\mathrm{F}$ atoms have high electron density $\left(\frac{\text { charge }}{\text { volume }}\right)$ and so repel the test electrons. Further, EA values increases as moving left to right along a period. It is because effective nuclear charge increases on moving left to right and more is the effective nuclear charge, more is the attraction of nucleus towards test electrons and thus more will be electron affinity. Thus, the overall increasing order of EA would be $$ \mathrm{O}<\mathrm{S}<\mathrm{F}<\mathrm{Cl} $$
BITSAT-2014
Classification of Elements and Periodicity in Properties
89704
The correct order of electron affinity of oxygen, sulphur, chlorine, fluorine is
1 S $<$ O $<\mathrm{Cl}<\mathrm{F}$
2 $\mathrm{S}<\mathrm{O}<\mathrm{F}<\mathrm{Cl}$
3 O $<$ S $<\mathrm{F}<\mathrm{Cl}$
4 $\mathrm{O}<\mathrm{S}<\mathrm{Cl}<\mathrm{F}$
Explanation:
On moving from top to bottom in a group the electron affinity became less negative and on moving from left to right across a period, the electron affinity became more negative. Chlorine has more negative electron affinity than fluorine because on adding on electron to fluorine $2 \mathrm{p}$ orbital causes greater repulsion than adding on electron to chlorine $3 \mathrm{p}$ orbital which is larger size. Therefore, the order is- $$ \mathrm{O}<\mathrm{S}<\mathrm{F}<\mathrm{Cl} $$
WB JEE-30.04.2022 UPTU/UPSEE-2010
Classification of Elements and Periodicity in Properties
89705
The correct order of electron gain enthalpy is:
1 $\mathrm{O}>\mathrm{S}>\mathrm{Se}>\mathrm{Te}$
2 $\mathrm{Te}>\mathrm{Se}>\mathrm{S}>\mathrm{O}$
3 S $>\mathrm{O}>\mathrm{Se}>\mathrm{Te}$
4 $\mathrm{S}>\mathrm{Se}>\mathrm{Te}>\mathrm{O}$
Explanation:
Electron gain enthalpy of oxygen is very low due to small size and compact nature of oxygen atom. Sulphur has highest electron gain enthalpy because sulphur has vacant d-orbital. So the order of electron gain enthalpy is - $$ \mathrm{S}>\mathrm{Se}>\mathrm{Te}>\mathrm{O} $$
JEE Main 26.02.2021
Classification of Elements and Periodicity in Properties
89709
The correct order of electron gain enthalpy with negative sign of $\mathrm{F}, \mathrm{Cl}, \mathrm{Br}$ and $\mathrm{I}$, having atomic number $9,17,35$ and 53 respectively, is
1 I $>\mathrm{Br}>\mathrm{Cl}>\mathrm{F}$
2 $\mathrm{F}>\mathrm{Cl}>\mathrm{Br}>\mathrm{I}$
3 $\mathrm{Cl}>\mathrm{F}>\mathrm{Br}>$ I
4 $\mathrm{Br}>\mathrm{Cl}>$ I $>\mathrm{F}$
Explanation:
Electron gain enthalpy has maximum in corresponding period of halogen and it becomes less negative down the group. However, the negative electron gain enthalpy of fluorine is less than that of chlorine due to the small size of fluorine atom. Therefore, there are strong inter electronic repulsion in the relatively small 2 p-orbital of fluorine thus, The incoming electron does not experience much attraction. The correct order of electron gain enthalpy are- $$ \mathrm{Cl}>\mathrm{F}>\mathrm{Br}>\mathrm{I} $$
Classification of Elements and Periodicity in Properties
89699
Which one of the following arrangements represent the correct order of electron gain enthalpy (with negative sign) of the given atomic species?
1 $\mathrm{F}<\mathrm{Cl}<\mathrm{O}<\mathrm{S}$
2 S $<$ O $<\mathrm{Cl}<\mathrm{F}$
3 $\mathrm{Cl}<\mathrm{F}<\mathrm{S}<\mathrm{O}$
4 $\mathrm{O}<\mathrm{S}<\mathrm{F}<\mathrm{Cl}$
Explanation:
In general, on moving down in a group, EA values decreases due to increase in size, because greater is the size of valence shell, lesser is the attraction and in turn lesser is the EA. But in period II and III, this order is reversed due to the small size of period II elements. Thus, $$ \mathrm{O}<\mathrm{S} \text { and } \mathrm{F}<\mathrm{Cl} $$ Because $\mathrm{O}$ and $\mathrm{F}$ atoms have high electron density $\left(\frac{\text { charge }}{\text { volume }}\right)$ and so repel the test electrons. Further, EA values increases as moving left to right along a period. It is because effective nuclear charge increases on moving left to right and more is the effective nuclear charge, more is the attraction of nucleus towards test electrons and thus more will be electron affinity. Thus, the overall increasing order of EA would be $$ \mathrm{O}<\mathrm{S}<\mathrm{F}<\mathrm{Cl} $$
BITSAT-2014
Classification of Elements and Periodicity in Properties
89704
The correct order of electron affinity of oxygen, sulphur, chlorine, fluorine is
1 S $<$ O $<\mathrm{Cl}<\mathrm{F}$
2 $\mathrm{S}<\mathrm{O}<\mathrm{F}<\mathrm{Cl}$
3 O $<$ S $<\mathrm{F}<\mathrm{Cl}$
4 $\mathrm{O}<\mathrm{S}<\mathrm{Cl}<\mathrm{F}$
Explanation:
On moving from top to bottom in a group the electron affinity became less negative and on moving from left to right across a period, the electron affinity became more negative. Chlorine has more negative electron affinity than fluorine because on adding on electron to fluorine $2 \mathrm{p}$ orbital causes greater repulsion than adding on electron to chlorine $3 \mathrm{p}$ orbital which is larger size. Therefore, the order is- $$ \mathrm{O}<\mathrm{S}<\mathrm{F}<\mathrm{Cl} $$
WB JEE-30.04.2022 UPTU/UPSEE-2010
Classification of Elements and Periodicity in Properties
89705
The correct order of electron gain enthalpy is:
1 $\mathrm{O}>\mathrm{S}>\mathrm{Se}>\mathrm{Te}$
2 $\mathrm{Te}>\mathrm{Se}>\mathrm{S}>\mathrm{O}$
3 S $>\mathrm{O}>\mathrm{Se}>\mathrm{Te}$
4 $\mathrm{S}>\mathrm{Se}>\mathrm{Te}>\mathrm{O}$
Explanation:
Electron gain enthalpy of oxygen is very low due to small size and compact nature of oxygen atom. Sulphur has highest electron gain enthalpy because sulphur has vacant d-orbital. So the order of electron gain enthalpy is - $$ \mathrm{S}>\mathrm{Se}>\mathrm{Te}>\mathrm{O} $$
JEE Main 26.02.2021
Classification of Elements and Periodicity in Properties
89709
The correct order of electron gain enthalpy with negative sign of $\mathrm{F}, \mathrm{Cl}, \mathrm{Br}$ and $\mathrm{I}$, having atomic number $9,17,35$ and 53 respectively, is
1 I $>\mathrm{Br}>\mathrm{Cl}>\mathrm{F}$
2 $\mathrm{F}>\mathrm{Cl}>\mathrm{Br}>\mathrm{I}$
3 $\mathrm{Cl}>\mathrm{F}>\mathrm{Br}>$ I
4 $\mathrm{Br}>\mathrm{Cl}>$ I $>\mathrm{F}$
Explanation:
Electron gain enthalpy has maximum in corresponding period of halogen and it becomes less negative down the group. However, the negative electron gain enthalpy of fluorine is less than that of chlorine due to the small size of fluorine atom. Therefore, there are strong inter electronic repulsion in the relatively small 2 p-orbital of fluorine thus, The incoming electron does not experience much attraction. The correct order of electron gain enthalpy are- $$ \mathrm{Cl}>\mathrm{F}>\mathrm{Br}>\mathrm{I} $$
Classification of Elements and Periodicity in Properties
89699
Which one of the following arrangements represent the correct order of electron gain enthalpy (with negative sign) of the given atomic species?
1 $\mathrm{F}<\mathrm{Cl}<\mathrm{O}<\mathrm{S}$
2 S $<$ O $<\mathrm{Cl}<\mathrm{F}$
3 $\mathrm{Cl}<\mathrm{F}<\mathrm{S}<\mathrm{O}$
4 $\mathrm{O}<\mathrm{S}<\mathrm{F}<\mathrm{Cl}$
Explanation:
In general, on moving down in a group, EA values decreases due to increase in size, because greater is the size of valence shell, lesser is the attraction and in turn lesser is the EA. But in period II and III, this order is reversed due to the small size of period II elements. Thus, $$ \mathrm{O}<\mathrm{S} \text { and } \mathrm{F}<\mathrm{Cl} $$ Because $\mathrm{O}$ and $\mathrm{F}$ atoms have high electron density $\left(\frac{\text { charge }}{\text { volume }}\right)$ and so repel the test electrons. Further, EA values increases as moving left to right along a period. It is because effective nuclear charge increases on moving left to right and more is the effective nuclear charge, more is the attraction of nucleus towards test electrons and thus more will be electron affinity. Thus, the overall increasing order of EA would be $$ \mathrm{O}<\mathrm{S}<\mathrm{F}<\mathrm{Cl} $$
BITSAT-2014
Classification of Elements and Periodicity in Properties
89704
The correct order of electron affinity of oxygen, sulphur, chlorine, fluorine is
1 S $<$ O $<\mathrm{Cl}<\mathrm{F}$
2 $\mathrm{S}<\mathrm{O}<\mathrm{F}<\mathrm{Cl}$
3 O $<$ S $<\mathrm{F}<\mathrm{Cl}$
4 $\mathrm{O}<\mathrm{S}<\mathrm{Cl}<\mathrm{F}$
Explanation:
On moving from top to bottom in a group the electron affinity became less negative and on moving from left to right across a period, the electron affinity became more negative. Chlorine has more negative electron affinity than fluorine because on adding on electron to fluorine $2 \mathrm{p}$ orbital causes greater repulsion than adding on electron to chlorine $3 \mathrm{p}$ orbital which is larger size. Therefore, the order is- $$ \mathrm{O}<\mathrm{S}<\mathrm{F}<\mathrm{Cl} $$
WB JEE-30.04.2022 UPTU/UPSEE-2010
Classification of Elements and Periodicity in Properties
89705
The correct order of electron gain enthalpy is:
1 $\mathrm{O}>\mathrm{S}>\mathrm{Se}>\mathrm{Te}$
2 $\mathrm{Te}>\mathrm{Se}>\mathrm{S}>\mathrm{O}$
3 S $>\mathrm{O}>\mathrm{Se}>\mathrm{Te}$
4 $\mathrm{S}>\mathrm{Se}>\mathrm{Te}>\mathrm{O}$
Explanation:
Electron gain enthalpy of oxygen is very low due to small size and compact nature of oxygen atom. Sulphur has highest electron gain enthalpy because sulphur has vacant d-orbital. So the order of electron gain enthalpy is - $$ \mathrm{S}>\mathrm{Se}>\mathrm{Te}>\mathrm{O} $$
JEE Main 26.02.2021
Classification of Elements and Periodicity in Properties
89709
The correct order of electron gain enthalpy with negative sign of $\mathrm{F}, \mathrm{Cl}, \mathrm{Br}$ and $\mathrm{I}$, having atomic number $9,17,35$ and 53 respectively, is
1 I $>\mathrm{Br}>\mathrm{Cl}>\mathrm{F}$
2 $\mathrm{F}>\mathrm{Cl}>\mathrm{Br}>\mathrm{I}$
3 $\mathrm{Cl}>\mathrm{F}>\mathrm{Br}>$ I
4 $\mathrm{Br}>\mathrm{Cl}>$ I $>\mathrm{F}$
Explanation:
Electron gain enthalpy has maximum in corresponding period of halogen and it becomes less negative down the group. However, the negative electron gain enthalpy of fluorine is less than that of chlorine due to the small size of fluorine atom. Therefore, there are strong inter electronic repulsion in the relatively small 2 p-orbital of fluorine thus, The incoming electron does not experience much attraction. The correct order of electron gain enthalpy are- $$ \mathrm{Cl}>\mathrm{F}>\mathrm{Br}>\mathrm{I} $$
