$\mathrm{Na}^{+2}-1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^5$ $$ \begin{aligned} & \mathrm{Ne}^{+2}-1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}^4 \\ & \mathrm{Mg}^{+2}-1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}^6, 3 \mathrm{~s}^0 \\ & \mathrm{Al}^{+2}-1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}^6, 3 \mathrm{~s}^1 \end{aligned} $$ Thus, the ionization energy decreases as the size of the atom increases. Since, $\mathrm{Mg}^{+2}$ is stable electronic configuration. Hence the ionization energy (I.E $\left.\mathrm{I}_2\right)$ is the highest than other species. $$ \mathrm{IE}_2 \Rightarrow \mathrm{Mg}<\mathrm{Al}<\mathrm{Ne}<\mathrm{Na} \text {. } $$
AP-EAMCET- (Engg.)-2011
Classification of Elements and Periodicity in Properties
89642
Which element is expected to have lowest first ionization energy?
1 $\mathrm{Sr}$
2 As
3 $\mathrm{Xe} \quad$
4 $\mathrm{S}$
Explanation:
The electronic configuration of given element are- $\begin{aligned} & { }_{38} \mathrm{Sr}=[\mathrm{Kr}] 5 \mathrm{~s}^2 \\ & { }_{33} \mathrm{As}=[\mathrm{Ar}] 3 \mathrm{~d}^{10} 4 \mathrm{~s}^2 4 \mathrm{p}^3 \\ & { }_{54} \mathrm{Xe}=[\mathrm{Kr}] 3 \mathrm{~d}^{10} 5 \mathrm{~s}^2 5 \mathrm{p}^6 \\ & { }_{16}^{\mathrm{S}}=[\mathrm{Ne}] 3 \mathrm{~s}^2 3 \mathrm{p}^4 \end{aligned}$ When the distance from outer electron from nucleus increases, it becomes very easy to remove an electron. Therefore, the ionisaiton energy decrease. So, $\mathrm{Sr}$ is lowest first ionisation energy among the given.
J and K CET-(2010)
Classification of Elements and Periodicity in Properties
89641
Assertion: Element has a tendency to lose the electron(s) to attain the stable configuration. Reason: Ionization enthalpy is the energy released to remove and electron from an isolated gaseous atom in its ground state.
1 If both Assertion and Reason are correct and the Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
The element has tendency to lose the electrons to attain the stable configuration. The atom with one valence electron will lose one electron to form a stable configuration. Ionization enthalpy is the energy required to remove an electron from an isolated gaseous atom in its ground state. Energy is required to overcame the attraction between nucleus and valence electron.
AIIMS-2011
Classification of Elements and Periodicity in Properties
89646
Assertion (A) K, Rb and Cs form superoxide's. Reason (R) the stability of the superoxide's increases from ' $K$ ' to ' $C$ 's' due to decrease in lattice energy. The correct answer is
1 Both (A) and (R) are true and (R) is the correct explanation of $\mathbf{( A )}$
2 Both (A) and (R) are true but (R) is not the correct explanation of $(\mathbf{A})$
3 (A) is true but (R) is not true
4 (A) is not true but (R) is not true
Explanation:
$\mathrm{R}^{+}, \mathrm{Rb}^{+}$and $\mathrm{Cs}^{+}$are large cations and superoxide ion is bigger than oxide and peroxide ions because of the higher lattice energies, a large cation Stabilizes a large anion, hence these metals form super oxides. As, we move down the group, the size of an atom from $\mathrm{K}$ to $\mathrm{Cs}$ increases so, lattice energy decreases, therefore the stability of superoxide also decreases.
$\mathrm{Na}^{+2}-1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^5$ $$ \begin{aligned} & \mathrm{Ne}^{+2}-1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}^4 \\ & \mathrm{Mg}^{+2}-1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}^6, 3 \mathrm{~s}^0 \\ & \mathrm{Al}^{+2}-1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}^6, 3 \mathrm{~s}^1 \end{aligned} $$ Thus, the ionization energy decreases as the size of the atom increases. Since, $\mathrm{Mg}^{+2}$ is stable electronic configuration. Hence the ionization energy (I.E $\left.\mathrm{I}_2\right)$ is the highest than other species. $$ \mathrm{IE}_2 \Rightarrow \mathrm{Mg}<\mathrm{Al}<\mathrm{Ne}<\mathrm{Na} \text {. } $$
AP-EAMCET- (Engg.)-2011
Classification of Elements and Periodicity in Properties
89642
Which element is expected to have lowest first ionization energy?
1 $\mathrm{Sr}$
2 As
3 $\mathrm{Xe} \quad$
4 $\mathrm{S}$
Explanation:
The electronic configuration of given element are- $\begin{aligned} & { }_{38} \mathrm{Sr}=[\mathrm{Kr}] 5 \mathrm{~s}^2 \\ & { }_{33} \mathrm{As}=[\mathrm{Ar}] 3 \mathrm{~d}^{10} 4 \mathrm{~s}^2 4 \mathrm{p}^3 \\ & { }_{54} \mathrm{Xe}=[\mathrm{Kr}] 3 \mathrm{~d}^{10} 5 \mathrm{~s}^2 5 \mathrm{p}^6 \\ & { }_{16}^{\mathrm{S}}=[\mathrm{Ne}] 3 \mathrm{~s}^2 3 \mathrm{p}^4 \end{aligned}$ When the distance from outer electron from nucleus increases, it becomes very easy to remove an electron. Therefore, the ionisaiton energy decrease. So, $\mathrm{Sr}$ is lowest first ionisation energy among the given.
J and K CET-(2010)
Classification of Elements and Periodicity in Properties
89641
Assertion: Element has a tendency to lose the electron(s) to attain the stable configuration. Reason: Ionization enthalpy is the energy released to remove and electron from an isolated gaseous atom in its ground state.
1 If both Assertion and Reason are correct and the Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
The element has tendency to lose the electrons to attain the stable configuration. The atom with one valence electron will lose one electron to form a stable configuration. Ionization enthalpy is the energy required to remove an electron from an isolated gaseous atom in its ground state. Energy is required to overcame the attraction between nucleus and valence electron.
AIIMS-2011
Classification of Elements and Periodicity in Properties
89646
Assertion (A) K, Rb and Cs form superoxide's. Reason (R) the stability of the superoxide's increases from ' $K$ ' to ' $C$ 's' due to decrease in lattice energy. The correct answer is
1 Both (A) and (R) are true and (R) is the correct explanation of $\mathbf{( A )}$
2 Both (A) and (R) are true but (R) is not the correct explanation of $(\mathbf{A})$
3 (A) is true but (R) is not true
4 (A) is not true but (R) is not true
Explanation:
$\mathrm{R}^{+}, \mathrm{Rb}^{+}$and $\mathrm{Cs}^{+}$are large cations and superoxide ion is bigger than oxide and peroxide ions because of the higher lattice energies, a large cation Stabilizes a large anion, hence these metals form super oxides. As, we move down the group, the size of an atom from $\mathrm{K}$ to $\mathrm{Cs}$ increases so, lattice energy decreases, therefore the stability of superoxide also decreases.
$\mathrm{Na}^{+2}-1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^5$ $$ \begin{aligned} & \mathrm{Ne}^{+2}-1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}^4 \\ & \mathrm{Mg}^{+2}-1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}^6, 3 \mathrm{~s}^0 \\ & \mathrm{Al}^{+2}-1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}^6, 3 \mathrm{~s}^1 \end{aligned} $$ Thus, the ionization energy decreases as the size of the atom increases. Since, $\mathrm{Mg}^{+2}$ is stable electronic configuration. Hence the ionization energy (I.E $\left.\mathrm{I}_2\right)$ is the highest than other species. $$ \mathrm{IE}_2 \Rightarrow \mathrm{Mg}<\mathrm{Al}<\mathrm{Ne}<\mathrm{Na} \text {. } $$
AP-EAMCET- (Engg.)-2011
Classification of Elements and Periodicity in Properties
89642
Which element is expected to have lowest first ionization energy?
1 $\mathrm{Sr}$
2 As
3 $\mathrm{Xe} \quad$
4 $\mathrm{S}$
Explanation:
The electronic configuration of given element are- $\begin{aligned} & { }_{38} \mathrm{Sr}=[\mathrm{Kr}] 5 \mathrm{~s}^2 \\ & { }_{33} \mathrm{As}=[\mathrm{Ar}] 3 \mathrm{~d}^{10} 4 \mathrm{~s}^2 4 \mathrm{p}^3 \\ & { }_{54} \mathrm{Xe}=[\mathrm{Kr}] 3 \mathrm{~d}^{10} 5 \mathrm{~s}^2 5 \mathrm{p}^6 \\ & { }_{16}^{\mathrm{S}}=[\mathrm{Ne}] 3 \mathrm{~s}^2 3 \mathrm{p}^4 \end{aligned}$ When the distance from outer electron from nucleus increases, it becomes very easy to remove an electron. Therefore, the ionisaiton energy decrease. So, $\mathrm{Sr}$ is lowest first ionisation energy among the given.
J and K CET-(2010)
Classification of Elements and Periodicity in Properties
89641
Assertion: Element has a tendency to lose the electron(s) to attain the stable configuration. Reason: Ionization enthalpy is the energy released to remove and electron from an isolated gaseous atom in its ground state.
1 If both Assertion and Reason are correct and the Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
The element has tendency to lose the electrons to attain the stable configuration. The atom with one valence electron will lose one electron to form a stable configuration. Ionization enthalpy is the energy required to remove an electron from an isolated gaseous atom in its ground state. Energy is required to overcame the attraction between nucleus and valence electron.
AIIMS-2011
Classification of Elements and Periodicity in Properties
89646
Assertion (A) K, Rb and Cs form superoxide's. Reason (R) the stability of the superoxide's increases from ' $K$ ' to ' $C$ 's' due to decrease in lattice energy. The correct answer is
1 Both (A) and (R) are true and (R) is the correct explanation of $\mathbf{( A )}$
2 Both (A) and (R) are true but (R) is not the correct explanation of $(\mathbf{A})$
3 (A) is true but (R) is not true
4 (A) is not true but (R) is not true
Explanation:
$\mathrm{R}^{+}, \mathrm{Rb}^{+}$and $\mathrm{Cs}^{+}$are large cations and superoxide ion is bigger than oxide and peroxide ions because of the higher lattice energies, a large cation Stabilizes a large anion, hence these metals form super oxides. As, we move down the group, the size of an atom from $\mathrm{K}$ to $\mathrm{Cs}$ increases so, lattice energy decreases, therefore the stability of superoxide also decreases.
$\mathrm{Na}^{+2}-1 \mathrm{~s}^2 2 \mathrm{~s}^2 2 \mathrm{p}^5$ $$ \begin{aligned} & \mathrm{Ne}^{+2}-1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}^4 \\ & \mathrm{Mg}^{+2}-1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}^6, 3 \mathrm{~s}^0 \\ & \mathrm{Al}^{+2}-1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}^6, 3 \mathrm{~s}^1 \end{aligned} $$ Thus, the ionization energy decreases as the size of the atom increases. Since, $\mathrm{Mg}^{+2}$ is stable electronic configuration. Hence the ionization energy (I.E $\left.\mathrm{I}_2\right)$ is the highest than other species. $$ \mathrm{IE}_2 \Rightarrow \mathrm{Mg}<\mathrm{Al}<\mathrm{Ne}<\mathrm{Na} \text {. } $$
AP-EAMCET- (Engg.)-2011
Classification of Elements and Periodicity in Properties
89642
Which element is expected to have lowest first ionization energy?
1 $\mathrm{Sr}$
2 As
3 $\mathrm{Xe} \quad$
4 $\mathrm{S}$
Explanation:
The electronic configuration of given element are- $\begin{aligned} & { }_{38} \mathrm{Sr}=[\mathrm{Kr}] 5 \mathrm{~s}^2 \\ & { }_{33} \mathrm{As}=[\mathrm{Ar}] 3 \mathrm{~d}^{10} 4 \mathrm{~s}^2 4 \mathrm{p}^3 \\ & { }_{54} \mathrm{Xe}=[\mathrm{Kr}] 3 \mathrm{~d}^{10} 5 \mathrm{~s}^2 5 \mathrm{p}^6 \\ & { }_{16}^{\mathrm{S}}=[\mathrm{Ne}] 3 \mathrm{~s}^2 3 \mathrm{p}^4 \end{aligned}$ When the distance from outer electron from nucleus increases, it becomes very easy to remove an electron. Therefore, the ionisaiton energy decrease. So, $\mathrm{Sr}$ is lowest first ionisation energy among the given.
J and K CET-(2010)
Classification of Elements and Periodicity in Properties
89641
Assertion: Element has a tendency to lose the electron(s) to attain the stable configuration. Reason: Ionization enthalpy is the energy released to remove and electron from an isolated gaseous atom in its ground state.
1 If both Assertion and Reason are correct and the Reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.
3 If Assertion is correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
The element has tendency to lose the electrons to attain the stable configuration. The atom with one valence electron will lose one electron to form a stable configuration. Ionization enthalpy is the energy required to remove an electron from an isolated gaseous atom in its ground state. Energy is required to overcame the attraction between nucleus and valence electron.
AIIMS-2011
Classification of Elements and Periodicity in Properties
89646
Assertion (A) K, Rb and Cs form superoxide's. Reason (R) the stability of the superoxide's increases from ' $K$ ' to ' $C$ 's' due to decrease in lattice energy. The correct answer is
1 Both (A) and (R) are true and (R) is the correct explanation of $\mathbf{( A )}$
2 Both (A) and (R) are true but (R) is not the correct explanation of $(\mathbf{A})$
3 (A) is true but (R) is not true
4 (A) is not true but (R) is not true
Explanation:
$\mathrm{R}^{+}, \mathrm{Rb}^{+}$and $\mathrm{Cs}^{+}$are large cations and superoxide ion is bigger than oxide and peroxide ions because of the higher lattice energies, a large cation Stabilizes a large anion, hence these metals form super oxides. As, we move down the group, the size of an atom from $\mathrm{K}$ to $\mathrm{Cs}$ increases so, lattice energy decreases, therefore the stability of superoxide also decreases.
