Classification of Elements and Periodicity in Properties
89720
How does electron affinity change when we move from left to right in a period in the periodic table?
1 It increases.
2 It decreases.
3 It remains unchanged.
4 It first increases and then decreases.
Explanation:
The electron affinity generally increases in a period from left to right due to the size of the atom decrease along the period.
J and K CET-(2013)
Classification of Elements and Periodicity in Properties
89723
Which one of the following arrangements represents the correct order of least negative to most negative electron gain enthalpy for $\mathrm{C}, \mathrm{Ca}$, Al, F and $O$ ?
Electron gain enthalpy becomes less negative from top to bottom in a group while is becomes more negative from left to right within a period.
KARNATAKA NEET 2013
Classification of Elements and Periodicity in Properties
89725
The electron affinity values of elements $A, B, C$ and $D$ are respectively $-135,-60,-200$ and $348 \mathrm{~kJ} \mathrm{~mol}^{-1}$. The outer electronic configuration of element $B$ is
1 $3 \mathrm{~s}^2 3 \mathrm{p}^5$
2 $3 \mathrm{~s}^2 3 \mathrm{p}^4$
3 $3 \mathrm{~s}^2 3 \mathrm{p}^3$
4 $3 \mathrm{~s}^2 3 \mathrm{p}^2$
Explanation:
Given that the electron affinity of elements A $=-135 \mathrm{~kJ} / \mathrm{mole}, \mathrm{B}=-60 \mathrm{~kJ} / \mathrm{mole}, \mathrm{C}=-200 \mathrm{~kJ} / \mathrm{mole}$ and $\mathrm{D}=-348 \mathrm{~kJ} / \mathrm{mole}$ then the element having half-filled or completely filled orbitals are stable electronic configuration, and have low negative value of electron affinity. Since, the electron affinity of $B$ is lowest and electronic configuration is $-1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^3$
AP-EAMCET- (Engg.) - 2010
Classification of Elements and Periodicity in Properties
89726
Carbon can reduce ferric oxide to iron at a temperature above $983 \mathrm{~K}$ because
1 carbon monoxide formed is thermodynamically less stable than ferric oxide
2 carbon has a higher affinity towards oxidation than iron
3 free energy change for the formation of carbon dioxide is less negative than that for ferric oxide
4 iron has a higher affinity towards oxygen than carbon
Explanation:
Above $983 \mathrm{~K}$, free energy change for the formation of $\mathrm{CO}_2$ is more negative than that for ferric oxide. Thus, above this temperature, carbon has a higher affinity towards oxidation than iron.
2010
Classification of Elements and Periodicity in Properties
89729
The electronic configuration of the element with maximum electron affinity is
The electronic configuration of $1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}^6$, $3 \mathrm{~s}^2, 3 \mathrm{p}^5$ is chlorine. The electron affinity is a measure the attraction between the incoming electrons to the nucleus. Electron affinity of $\mathrm{Cl}$ is greater than electron affinity of $F$ due to small size of $F$ atom. Hence, electron affinity is highest for chlorine.
Classification of Elements and Periodicity in Properties
89720
How does electron affinity change when we move from left to right in a period in the periodic table?
1 It increases.
2 It decreases.
3 It remains unchanged.
4 It first increases and then decreases.
Explanation:
The electron affinity generally increases in a period from left to right due to the size of the atom decrease along the period.
J and K CET-(2013)
Classification of Elements and Periodicity in Properties
89723
Which one of the following arrangements represents the correct order of least negative to most negative electron gain enthalpy for $\mathrm{C}, \mathrm{Ca}$, Al, F and $O$ ?
Electron gain enthalpy becomes less negative from top to bottom in a group while is becomes more negative from left to right within a period.
KARNATAKA NEET 2013
Classification of Elements and Periodicity in Properties
89725
The electron affinity values of elements $A, B, C$ and $D$ are respectively $-135,-60,-200$ and $348 \mathrm{~kJ} \mathrm{~mol}^{-1}$. The outer electronic configuration of element $B$ is
1 $3 \mathrm{~s}^2 3 \mathrm{p}^5$
2 $3 \mathrm{~s}^2 3 \mathrm{p}^4$
3 $3 \mathrm{~s}^2 3 \mathrm{p}^3$
4 $3 \mathrm{~s}^2 3 \mathrm{p}^2$
Explanation:
Given that the electron affinity of elements A $=-135 \mathrm{~kJ} / \mathrm{mole}, \mathrm{B}=-60 \mathrm{~kJ} / \mathrm{mole}, \mathrm{C}=-200 \mathrm{~kJ} / \mathrm{mole}$ and $\mathrm{D}=-348 \mathrm{~kJ} / \mathrm{mole}$ then the element having half-filled or completely filled orbitals are stable electronic configuration, and have low negative value of electron affinity. Since, the electron affinity of $B$ is lowest and electronic configuration is $-1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^3$
AP-EAMCET- (Engg.) - 2010
Classification of Elements and Periodicity in Properties
89726
Carbon can reduce ferric oxide to iron at a temperature above $983 \mathrm{~K}$ because
1 carbon monoxide formed is thermodynamically less stable than ferric oxide
2 carbon has a higher affinity towards oxidation than iron
3 free energy change for the formation of carbon dioxide is less negative than that for ferric oxide
4 iron has a higher affinity towards oxygen than carbon
Explanation:
Above $983 \mathrm{~K}$, free energy change for the formation of $\mathrm{CO}_2$ is more negative than that for ferric oxide. Thus, above this temperature, carbon has a higher affinity towards oxidation than iron.
2010
Classification of Elements and Periodicity in Properties
89729
The electronic configuration of the element with maximum electron affinity is
The electronic configuration of $1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}^6$, $3 \mathrm{~s}^2, 3 \mathrm{p}^5$ is chlorine. The electron affinity is a measure the attraction between the incoming electrons to the nucleus. Electron affinity of $\mathrm{Cl}$ is greater than electron affinity of $F$ due to small size of $F$ atom. Hence, electron affinity is highest for chlorine.
Classification of Elements and Periodicity in Properties
89720
How does electron affinity change when we move from left to right in a period in the periodic table?
1 It increases.
2 It decreases.
3 It remains unchanged.
4 It first increases and then decreases.
Explanation:
The electron affinity generally increases in a period from left to right due to the size of the atom decrease along the period.
J and K CET-(2013)
Classification of Elements and Periodicity in Properties
89723
Which one of the following arrangements represents the correct order of least negative to most negative electron gain enthalpy for $\mathrm{C}, \mathrm{Ca}$, Al, F and $O$ ?
Electron gain enthalpy becomes less negative from top to bottom in a group while is becomes more negative from left to right within a period.
KARNATAKA NEET 2013
Classification of Elements and Periodicity in Properties
89725
The electron affinity values of elements $A, B, C$ and $D$ are respectively $-135,-60,-200$ and $348 \mathrm{~kJ} \mathrm{~mol}^{-1}$. The outer electronic configuration of element $B$ is
1 $3 \mathrm{~s}^2 3 \mathrm{p}^5$
2 $3 \mathrm{~s}^2 3 \mathrm{p}^4$
3 $3 \mathrm{~s}^2 3 \mathrm{p}^3$
4 $3 \mathrm{~s}^2 3 \mathrm{p}^2$
Explanation:
Given that the electron affinity of elements A $=-135 \mathrm{~kJ} / \mathrm{mole}, \mathrm{B}=-60 \mathrm{~kJ} / \mathrm{mole}, \mathrm{C}=-200 \mathrm{~kJ} / \mathrm{mole}$ and $\mathrm{D}=-348 \mathrm{~kJ} / \mathrm{mole}$ then the element having half-filled or completely filled orbitals are stable electronic configuration, and have low negative value of electron affinity. Since, the electron affinity of $B$ is lowest and electronic configuration is $-1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^3$
AP-EAMCET- (Engg.) - 2010
Classification of Elements and Periodicity in Properties
89726
Carbon can reduce ferric oxide to iron at a temperature above $983 \mathrm{~K}$ because
1 carbon monoxide formed is thermodynamically less stable than ferric oxide
2 carbon has a higher affinity towards oxidation than iron
3 free energy change for the formation of carbon dioxide is less negative than that for ferric oxide
4 iron has a higher affinity towards oxygen than carbon
Explanation:
Above $983 \mathrm{~K}$, free energy change for the formation of $\mathrm{CO}_2$ is more negative than that for ferric oxide. Thus, above this temperature, carbon has a higher affinity towards oxidation than iron.
2010
Classification of Elements and Periodicity in Properties
89729
The electronic configuration of the element with maximum electron affinity is
The electronic configuration of $1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}^6$, $3 \mathrm{~s}^2, 3 \mathrm{p}^5$ is chlorine. The electron affinity is a measure the attraction between the incoming electrons to the nucleus. Electron affinity of $\mathrm{Cl}$ is greater than electron affinity of $F$ due to small size of $F$ atom. Hence, electron affinity is highest for chlorine.
Classification of Elements and Periodicity in Properties
89720
How does electron affinity change when we move from left to right in a period in the periodic table?
1 It increases.
2 It decreases.
3 It remains unchanged.
4 It first increases and then decreases.
Explanation:
The electron affinity generally increases in a period from left to right due to the size of the atom decrease along the period.
J and K CET-(2013)
Classification of Elements and Periodicity in Properties
89723
Which one of the following arrangements represents the correct order of least negative to most negative electron gain enthalpy for $\mathrm{C}, \mathrm{Ca}$, Al, F and $O$ ?
Electron gain enthalpy becomes less negative from top to bottom in a group while is becomes more negative from left to right within a period.
KARNATAKA NEET 2013
Classification of Elements and Periodicity in Properties
89725
The electron affinity values of elements $A, B, C$ and $D$ are respectively $-135,-60,-200$ and $348 \mathrm{~kJ} \mathrm{~mol}^{-1}$. The outer electronic configuration of element $B$ is
1 $3 \mathrm{~s}^2 3 \mathrm{p}^5$
2 $3 \mathrm{~s}^2 3 \mathrm{p}^4$
3 $3 \mathrm{~s}^2 3 \mathrm{p}^3$
4 $3 \mathrm{~s}^2 3 \mathrm{p}^2$
Explanation:
Given that the electron affinity of elements A $=-135 \mathrm{~kJ} / \mathrm{mole}, \mathrm{B}=-60 \mathrm{~kJ} / \mathrm{mole}, \mathrm{C}=-200 \mathrm{~kJ} / \mathrm{mole}$ and $\mathrm{D}=-348 \mathrm{~kJ} / \mathrm{mole}$ then the element having half-filled or completely filled orbitals are stable electronic configuration, and have low negative value of electron affinity. Since, the electron affinity of $B$ is lowest and electronic configuration is $-1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^3$
AP-EAMCET- (Engg.) - 2010
Classification of Elements and Periodicity in Properties
89726
Carbon can reduce ferric oxide to iron at a temperature above $983 \mathrm{~K}$ because
1 carbon monoxide formed is thermodynamically less stable than ferric oxide
2 carbon has a higher affinity towards oxidation than iron
3 free energy change for the formation of carbon dioxide is less negative than that for ferric oxide
4 iron has a higher affinity towards oxygen than carbon
Explanation:
Above $983 \mathrm{~K}$, free energy change for the formation of $\mathrm{CO}_2$ is more negative than that for ferric oxide. Thus, above this temperature, carbon has a higher affinity towards oxidation than iron.
2010
Classification of Elements and Periodicity in Properties
89729
The electronic configuration of the element with maximum electron affinity is
The electronic configuration of $1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}^6$, $3 \mathrm{~s}^2, 3 \mathrm{p}^5$ is chlorine. The electron affinity is a measure the attraction between the incoming electrons to the nucleus. Electron affinity of $\mathrm{Cl}$ is greater than electron affinity of $F$ due to small size of $F$ atom. Hence, electron affinity is highest for chlorine.
Classification of Elements and Periodicity in Properties
89720
How does electron affinity change when we move from left to right in a period in the periodic table?
1 It increases.
2 It decreases.
3 It remains unchanged.
4 It first increases and then decreases.
Explanation:
The electron affinity generally increases in a period from left to right due to the size of the atom decrease along the period.
J and K CET-(2013)
Classification of Elements and Periodicity in Properties
89723
Which one of the following arrangements represents the correct order of least negative to most negative electron gain enthalpy for $\mathrm{C}, \mathrm{Ca}$, Al, F and $O$ ?
Electron gain enthalpy becomes less negative from top to bottom in a group while is becomes more negative from left to right within a period.
KARNATAKA NEET 2013
Classification of Elements and Periodicity in Properties
89725
The electron affinity values of elements $A, B, C$ and $D$ are respectively $-135,-60,-200$ and $348 \mathrm{~kJ} \mathrm{~mol}^{-1}$. The outer electronic configuration of element $B$ is
1 $3 \mathrm{~s}^2 3 \mathrm{p}^5$
2 $3 \mathrm{~s}^2 3 \mathrm{p}^4$
3 $3 \mathrm{~s}^2 3 \mathrm{p}^3$
4 $3 \mathrm{~s}^2 3 \mathrm{p}^2$
Explanation:
Given that the electron affinity of elements A $=-135 \mathrm{~kJ} / \mathrm{mole}, \mathrm{B}=-60 \mathrm{~kJ} / \mathrm{mole}, \mathrm{C}=-200 \mathrm{~kJ} / \mathrm{mole}$ and $\mathrm{D}=-348 \mathrm{~kJ} / \mathrm{mole}$ then the element having half-filled or completely filled orbitals are stable electronic configuration, and have low negative value of electron affinity. Since, the electron affinity of $B$ is lowest and electronic configuration is $-1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^6 3 \mathrm{~s}^2 3 \mathrm{p}^3$
AP-EAMCET- (Engg.) - 2010
Classification of Elements and Periodicity in Properties
89726
Carbon can reduce ferric oxide to iron at a temperature above $983 \mathrm{~K}$ because
1 carbon monoxide formed is thermodynamically less stable than ferric oxide
2 carbon has a higher affinity towards oxidation than iron
3 free energy change for the formation of carbon dioxide is less negative than that for ferric oxide
4 iron has a higher affinity towards oxygen than carbon
Explanation:
Above $983 \mathrm{~K}$, free energy change for the formation of $\mathrm{CO}_2$ is more negative than that for ferric oxide. Thus, above this temperature, carbon has a higher affinity towards oxidation than iron.
2010
Classification of Elements and Periodicity in Properties
89729
The electronic configuration of the element with maximum electron affinity is
The electronic configuration of $1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}^6$, $3 \mathrm{~s}^2, 3 \mathrm{p}^5$ is chlorine. The electron affinity is a measure the attraction between the incoming electrons to the nucleus. Electron affinity of $\mathrm{Cl}$ is greater than electron affinity of $F$ due to small size of $F$ atom. Hence, electron affinity is highest for chlorine.
