Classification of Elements and Periodicity in Properties
89587
The successive ionization energies of an element in $\mathrm{kJ} \mathrm{mol}^{-1}$ are $1060,1900,2920,6280$ and 21200 . This element would belong to group-
1 $\mathrm{V}$
2 VI
3 VII
4 $\mathrm{IV}$
Explanation:
The successive ionization energies of an element in $\mathrm{kJmole}^{-1}$ are $1060,1900,2920,6280$ and 21200. The element belong to group V.
Shift-I
Classification of Elements and Periodicity in Properties
89589
The first ionization energies (in $\mathrm{kJ} \mathrm{mol}^{-1}$ ) of four consecutive elements of the second period are given in the options. The first ionization energy of nitrogen is
1 1086
2 1402
3 1681
4 1314
Explanation:
The amount of energy required to remove an electron from an isolated atom or molecules is called ionization energy. The first ionization energy of nitrogen is $1402 \mathrm{~kJ} / \mathrm{mole}$ energy.
Shift-I
Classification of Elements and Periodicity in Properties
89672
Which one of the following is correct about stability of the given ions?
1 $\mathrm{Pb}^{2+}>\mathrm{Pb}^{4+}$
2 $\mathrm{Pb}^{4+}>\mathrm{Pb}^{2+}$
3 $\mathrm{Si}^{2+}>\mathrm{Si}^{4+}$
4 $\mathrm{Sn}^{4+}>\mathrm{Sn}^{2+}$
Explanation:
On going from top to bottom in a group, the stability of lower oxidation state increases due to inert pair effect. So, the correct order is $\mathrm{Pb}^{2+}>\mathrm{Pb}^{4+}$.
AP EAMCET- 2003
Classification of Elements and Periodicity in Properties
89599
Which element of the $3 d$-series has highest third ionisation enthalpy ?
1 $\mathrm{Mn}$
2 $\mathrm{Zn}$
3 $\mathrm{Fe}$
4 $\mathrm{Cu}$
Explanation:
The electronic configuration of zinc metal is $\mathrm{Zn}=[\mathrm{Ar}] 3 \mathrm{~d}^{10}, 4 \mathrm{~s}^2$ The third ionisation enthalpy of $\mathrm{Zn}$ means removal of electron from the stable configuration of $3 \mathrm{~d}^{10}$. The metal having the highest third ionisation enthalpy is $\mathrm{Zn}$. Hence, the correct option is (b).
Shift-II
Classification of Elements and Periodicity in Properties
89609
The successive ionization energies in $\mathrm{kJ} / \mathrm{mol}$ of an element $P$ are $740,1500,7000,10500,13600$, 18000 and 21700 . Which ion is the most likely to be formed when $P$ reacts with chloride?
1 $\mathrm{p}_{2+}^{2-}$
2 $\mathrm{p}_{3+}^{+}$
3 $\mathrm{p}^{2+}$
4 $\mathrm{p}^{3+}$
Explanation:
After losing two electrons, there is sudden huge increase in IE from 1500 to $7000 \mathrm{~kJ} / \mathrm{mol}$. It means that after losing two electrons, some stable configuration is achieved. So, element $\mathrm{P}$ forms $\mathrm{P}^{2+}$ ion.
Classification of Elements and Periodicity in Properties
89587
The successive ionization energies of an element in $\mathrm{kJ} \mathrm{mol}^{-1}$ are $1060,1900,2920,6280$ and 21200 . This element would belong to group-
1 $\mathrm{V}$
2 VI
3 VII
4 $\mathrm{IV}$
Explanation:
The successive ionization energies of an element in $\mathrm{kJmole}^{-1}$ are $1060,1900,2920,6280$ and 21200. The element belong to group V.
Shift-I
Classification of Elements and Periodicity in Properties
89589
The first ionization energies (in $\mathrm{kJ} \mathrm{mol}^{-1}$ ) of four consecutive elements of the second period are given in the options. The first ionization energy of nitrogen is
1 1086
2 1402
3 1681
4 1314
Explanation:
The amount of energy required to remove an electron from an isolated atom or molecules is called ionization energy. The first ionization energy of nitrogen is $1402 \mathrm{~kJ} / \mathrm{mole}$ energy.
Shift-I
Classification of Elements and Periodicity in Properties
89672
Which one of the following is correct about stability of the given ions?
1 $\mathrm{Pb}^{2+}>\mathrm{Pb}^{4+}$
2 $\mathrm{Pb}^{4+}>\mathrm{Pb}^{2+}$
3 $\mathrm{Si}^{2+}>\mathrm{Si}^{4+}$
4 $\mathrm{Sn}^{4+}>\mathrm{Sn}^{2+}$
Explanation:
On going from top to bottom in a group, the stability of lower oxidation state increases due to inert pair effect. So, the correct order is $\mathrm{Pb}^{2+}>\mathrm{Pb}^{4+}$.
AP EAMCET- 2003
Classification of Elements and Periodicity in Properties
89599
Which element of the $3 d$-series has highest third ionisation enthalpy ?
1 $\mathrm{Mn}$
2 $\mathrm{Zn}$
3 $\mathrm{Fe}$
4 $\mathrm{Cu}$
Explanation:
The electronic configuration of zinc metal is $\mathrm{Zn}=[\mathrm{Ar}] 3 \mathrm{~d}^{10}, 4 \mathrm{~s}^2$ The third ionisation enthalpy of $\mathrm{Zn}$ means removal of electron from the stable configuration of $3 \mathrm{~d}^{10}$. The metal having the highest third ionisation enthalpy is $\mathrm{Zn}$. Hence, the correct option is (b).
Shift-II
Classification of Elements and Periodicity in Properties
89609
The successive ionization energies in $\mathrm{kJ} / \mathrm{mol}$ of an element $P$ are $740,1500,7000,10500,13600$, 18000 and 21700 . Which ion is the most likely to be formed when $P$ reacts with chloride?
1 $\mathrm{p}_{2+}^{2-}$
2 $\mathrm{p}_{3+}^{+}$
3 $\mathrm{p}^{2+}$
4 $\mathrm{p}^{3+}$
Explanation:
After losing two electrons, there is sudden huge increase in IE from 1500 to $7000 \mathrm{~kJ} / \mathrm{mol}$. It means that after losing two electrons, some stable configuration is achieved. So, element $\mathrm{P}$ forms $\mathrm{P}^{2+}$ ion.
Classification of Elements and Periodicity in Properties
89587
The successive ionization energies of an element in $\mathrm{kJ} \mathrm{mol}^{-1}$ are $1060,1900,2920,6280$ and 21200 . This element would belong to group-
1 $\mathrm{V}$
2 VI
3 VII
4 $\mathrm{IV}$
Explanation:
The successive ionization energies of an element in $\mathrm{kJmole}^{-1}$ are $1060,1900,2920,6280$ and 21200. The element belong to group V.
Shift-I
Classification of Elements and Periodicity in Properties
89589
The first ionization energies (in $\mathrm{kJ} \mathrm{mol}^{-1}$ ) of four consecutive elements of the second period are given in the options. The first ionization energy of nitrogen is
1 1086
2 1402
3 1681
4 1314
Explanation:
The amount of energy required to remove an electron from an isolated atom or molecules is called ionization energy. The first ionization energy of nitrogen is $1402 \mathrm{~kJ} / \mathrm{mole}$ energy.
Shift-I
Classification of Elements and Periodicity in Properties
89672
Which one of the following is correct about stability of the given ions?
1 $\mathrm{Pb}^{2+}>\mathrm{Pb}^{4+}$
2 $\mathrm{Pb}^{4+}>\mathrm{Pb}^{2+}$
3 $\mathrm{Si}^{2+}>\mathrm{Si}^{4+}$
4 $\mathrm{Sn}^{4+}>\mathrm{Sn}^{2+}$
Explanation:
On going from top to bottom in a group, the stability of lower oxidation state increases due to inert pair effect. So, the correct order is $\mathrm{Pb}^{2+}>\mathrm{Pb}^{4+}$.
AP EAMCET- 2003
Classification of Elements and Periodicity in Properties
89599
Which element of the $3 d$-series has highest third ionisation enthalpy ?
1 $\mathrm{Mn}$
2 $\mathrm{Zn}$
3 $\mathrm{Fe}$
4 $\mathrm{Cu}$
Explanation:
The electronic configuration of zinc metal is $\mathrm{Zn}=[\mathrm{Ar}] 3 \mathrm{~d}^{10}, 4 \mathrm{~s}^2$ The third ionisation enthalpy of $\mathrm{Zn}$ means removal of electron from the stable configuration of $3 \mathrm{~d}^{10}$. The metal having the highest third ionisation enthalpy is $\mathrm{Zn}$. Hence, the correct option is (b).
Shift-II
Classification of Elements and Periodicity in Properties
89609
The successive ionization energies in $\mathrm{kJ} / \mathrm{mol}$ of an element $P$ are $740,1500,7000,10500,13600$, 18000 and 21700 . Which ion is the most likely to be formed when $P$ reacts with chloride?
1 $\mathrm{p}_{2+}^{2-}$
2 $\mathrm{p}_{3+}^{+}$
3 $\mathrm{p}^{2+}$
4 $\mathrm{p}^{3+}$
Explanation:
After losing two electrons, there is sudden huge increase in IE from 1500 to $7000 \mathrm{~kJ} / \mathrm{mol}$. It means that after losing two electrons, some stable configuration is achieved. So, element $\mathrm{P}$ forms $\mathrm{P}^{2+}$ ion.
Classification of Elements and Periodicity in Properties
89587
The successive ionization energies of an element in $\mathrm{kJ} \mathrm{mol}^{-1}$ are $1060,1900,2920,6280$ and 21200 . This element would belong to group-
1 $\mathrm{V}$
2 VI
3 VII
4 $\mathrm{IV}$
Explanation:
The successive ionization energies of an element in $\mathrm{kJmole}^{-1}$ are $1060,1900,2920,6280$ and 21200. The element belong to group V.
Shift-I
Classification of Elements and Periodicity in Properties
89589
The first ionization energies (in $\mathrm{kJ} \mathrm{mol}^{-1}$ ) of four consecutive elements of the second period are given in the options. The first ionization energy of nitrogen is
1 1086
2 1402
3 1681
4 1314
Explanation:
The amount of energy required to remove an electron from an isolated atom or molecules is called ionization energy. The first ionization energy of nitrogen is $1402 \mathrm{~kJ} / \mathrm{mole}$ energy.
Shift-I
Classification of Elements and Periodicity in Properties
89672
Which one of the following is correct about stability of the given ions?
1 $\mathrm{Pb}^{2+}>\mathrm{Pb}^{4+}$
2 $\mathrm{Pb}^{4+}>\mathrm{Pb}^{2+}$
3 $\mathrm{Si}^{2+}>\mathrm{Si}^{4+}$
4 $\mathrm{Sn}^{4+}>\mathrm{Sn}^{2+}$
Explanation:
On going from top to bottom in a group, the stability of lower oxidation state increases due to inert pair effect. So, the correct order is $\mathrm{Pb}^{2+}>\mathrm{Pb}^{4+}$.
AP EAMCET- 2003
Classification of Elements and Periodicity in Properties
89599
Which element of the $3 d$-series has highest third ionisation enthalpy ?
1 $\mathrm{Mn}$
2 $\mathrm{Zn}$
3 $\mathrm{Fe}$
4 $\mathrm{Cu}$
Explanation:
The electronic configuration of zinc metal is $\mathrm{Zn}=[\mathrm{Ar}] 3 \mathrm{~d}^{10}, 4 \mathrm{~s}^2$ The third ionisation enthalpy of $\mathrm{Zn}$ means removal of electron from the stable configuration of $3 \mathrm{~d}^{10}$. The metal having the highest third ionisation enthalpy is $\mathrm{Zn}$. Hence, the correct option is (b).
Shift-II
Classification of Elements and Periodicity in Properties
89609
The successive ionization energies in $\mathrm{kJ} / \mathrm{mol}$ of an element $P$ are $740,1500,7000,10500,13600$, 18000 and 21700 . Which ion is the most likely to be formed when $P$ reacts with chloride?
1 $\mathrm{p}_{2+}^{2-}$
2 $\mathrm{p}_{3+}^{+}$
3 $\mathrm{p}^{2+}$
4 $\mathrm{p}^{3+}$
Explanation:
After losing two electrons, there is sudden huge increase in IE from 1500 to $7000 \mathrm{~kJ} / \mathrm{mol}$. It means that after losing two electrons, some stable configuration is achieved. So, element $\mathrm{P}$ forms $\mathrm{P}^{2+}$ ion.
Classification of Elements and Periodicity in Properties
89587
The successive ionization energies of an element in $\mathrm{kJ} \mathrm{mol}^{-1}$ are $1060,1900,2920,6280$ and 21200 . This element would belong to group-
1 $\mathrm{V}$
2 VI
3 VII
4 $\mathrm{IV}$
Explanation:
The successive ionization energies of an element in $\mathrm{kJmole}^{-1}$ are $1060,1900,2920,6280$ and 21200. The element belong to group V.
Shift-I
Classification of Elements and Periodicity in Properties
89589
The first ionization energies (in $\mathrm{kJ} \mathrm{mol}^{-1}$ ) of four consecutive elements of the second period are given in the options. The first ionization energy of nitrogen is
1 1086
2 1402
3 1681
4 1314
Explanation:
The amount of energy required to remove an electron from an isolated atom or molecules is called ionization energy. The first ionization energy of nitrogen is $1402 \mathrm{~kJ} / \mathrm{mole}$ energy.
Shift-I
Classification of Elements and Periodicity in Properties
89672
Which one of the following is correct about stability of the given ions?
1 $\mathrm{Pb}^{2+}>\mathrm{Pb}^{4+}$
2 $\mathrm{Pb}^{4+}>\mathrm{Pb}^{2+}$
3 $\mathrm{Si}^{2+}>\mathrm{Si}^{4+}$
4 $\mathrm{Sn}^{4+}>\mathrm{Sn}^{2+}$
Explanation:
On going from top to bottom in a group, the stability of lower oxidation state increases due to inert pair effect. So, the correct order is $\mathrm{Pb}^{2+}>\mathrm{Pb}^{4+}$.
AP EAMCET- 2003
Classification of Elements and Periodicity in Properties
89599
Which element of the $3 d$-series has highest third ionisation enthalpy ?
1 $\mathrm{Mn}$
2 $\mathrm{Zn}$
3 $\mathrm{Fe}$
4 $\mathrm{Cu}$
Explanation:
The electronic configuration of zinc metal is $\mathrm{Zn}=[\mathrm{Ar}] 3 \mathrm{~d}^{10}, 4 \mathrm{~s}^2$ The third ionisation enthalpy of $\mathrm{Zn}$ means removal of electron from the stable configuration of $3 \mathrm{~d}^{10}$. The metal having the highest third ionisation enthalpy is $\mathrm{Zn}$. Hence, the correct option is (b).
Shift-II
Classification of Elements and Periodicity in Properties
89609
The successive ionization energies in $\mathrm{kJ} / \mathrm{mol}$ of an element $P$ are $740,1500,7000,10500,13600$, 18000 and 21700 . Which ion is the most likely to be formed when $P$ reacts with chloride?
1 $\mathrm{p}_{2+}^{2-}$
2 $\mathrm{p}_{3+}^{+}$
3 $\mathrm{p}^{2+}$
4 $\mathrm{p}^{3+}$
Explanation:
After losing two electrons, there is sudden huge increase in IE from 1500 to $7000 \mathrm{~kJ} / \mathrm{mol}$. It means that after losing two electrons, some stable configuration is achieved. So, element $\mathrm{P}$ forms $\mathrm{P}^{2+}$ ion.
