Classification of Elements and Periodicity in Properties
89501
Which of the following has least ionic radius?
1 $\mathrm{I}^{+}$
2 $\mathrm{I}^{3+}$
3 $\mathrm{I}^{5+}$
4 $\mathrm{I}^{7+}$
Explanation:
Ionic radii $\propto \frac{1}{\text { charge }}$ More is the positive charge smaller the size of ion. So, the least ionic radius is $\mathrm{I}^{7+}$.
J and K CET-(2001)
Classification of Elements and Periodicity in Properties
89503
$\mathrm{O}^{2-}$ and $\mathrm{Si}^{4+}$ are isoelectronic ions. If the ionic radius of $\mathrm{O}^{2-}$ is $1.4 \AA$, the ionic radius of $\mathrm{Si}^{4+}$ will be
1 $1.4 \AA$
2 $0.41 \AA$
3 $2.8 \AA$
4 $1.5 \AA$
Explanation:
Given that, $\mathrm{O}^{2-}$ and $\mathrm{Si}^{4+}$ are isoelectronic ions i.e. $10 \mathrm{e}^{-}$. Nuclear charge $\propto \frac{1}{\text { ionic radius }}$ $\therefore$ Nuclear charge for $\mathrm{Si}$ is larger, so it should be less than $1.4 \AA$
A.P.EAMCET 1999
Classification of Elements and Periodicity in Properties
89410
The correct order of increasing ionic radii is
Ionic radii is the distance from the nucleus up to which it has influence on its electron bond. The ionic radii of isoelectronic species increases with a decreases in the magnitudes of nuclear charge. So, The correct order of increasing ionic radii is $\mathrm{Mg}^{2+}<$ $\mathrm{Na}^{+}<\mathrm{F}^{-}<\mathrm{O}^{2-}<\mathrm{N}^{3-}$.
Shift-II
Classification of Elements and Periodicity in Properties
89427
Among the following ions, the one with the highest value of ionic radius is
1 $\mathrm{O}^{2-}$
2 $\mathrm{Mg}^{2+}$
3 $\mathrm{L}^{+}$
4 $\mathrm{L}^{-}$
Explanation:
decreasing order of ionic radius of given species is- $$ \mathrm{O}^{2-}>\mathrm{F}^{-}>\mathrm{Li}^{+}>\mathrm{Mg}^{+2} $$ On increasing -ve charge among the electronic species, ionic radius increases.
Shift-II
Classification of Elements and Periodicity in Properties
89428
Identify the correct order of ionic radii of the given species:
Ionic radii is the distance between the nucleus and the electron in the outermost shell of an ions. Increasing the electron on an atom, increases the ionic radii. Decrease the $\mathrm{e}^{-}$on an atom will be decrease ionic radii. So the order is- $$ \mathrm{S}^{2-}>\mathrm{K}^{+}>\mathrm{Sc}^{3+}>\mathrm{V}^{5+}>\mathrm{Mn}^{7+} $$
Classification of Elements and Periodicity in Properties
89501
Which of the following has least ionic radius?
1 $\mathrm{I}^{+}$
2 $\mathrm{I}^{3+}$
3 $\mathrm{I}^{5+}$
4 $\mathrm{I}^{7+}$
Explanation:
Ionic radii $\propto \frac{1}{\text { charge }}$ More is the positive charge smaller the size of ion. So, the least ionic radius is $\mathrm{I}^{7+}$.
J and K CET-(2001)
Classification of Elements and Periodicity in Properties
89503
$\mathrm{O}^{2-}$ and $\mathrm{Si}^{4+}$ are isoelectronic ions. If the ionic radius of $\mathrm{O}^{2-}$ is $1.4 \AA$, the ionic radius of $\mathrm{Si}^{4+}$ will be
1 $1.4 \AA$
2 $0.41 \AA$
3 $2.8 \AA$
4 $1.5 \AA$
Explanation:
Given that, $\mathrm{O}^{2-}$ and $\mathrm{Si}^{4+}$ are isoelectronic ions i.e. $10 \mathrm{e}^{-}$. Nuclear charge $\propto \frac{1}{\text { ionic radius }}$ $\therefore$ Nuclear charge for $\mathrm{Si}$ is larger, so it should be less than $1.4 \AA$
A.P.EAMCET 1999
Classification of Elements and Periodicity in Properties
89410
The correct order of increasing ionic radii is
Ionic radii is the distance from the nucleus up to which it has influence on its electron bond. The ionic radii of isoelectronic species increases with a decreases in the magnitudes of nuclear charge. So, The correct order of increasing ionic radii is $\mathrm{Mg}^{2+}<$ $\mathrm{Na}^{+}<\mathrm{F}^{-}<\mathrm{O}^{2-}<\mathrm{N}^{3-}$.
Shift-II
Classification of Elements and Periodicity in Properties
89427
Among the following ions, the one with the highest value of ionic radius is
1 $\mathrm{O}^{2-}$
2 $\mathrm{Mg}^{2+}$
3 $\mathrm{L}^{+}$
4 $\mathrm{L}^{-}$
Explanation:
decreasing order of ionic radius of given species is- $$ \mathrm{O}^{2-}>\mathrm{F}^{-}>\mathrm{Li}^{+}>\mathrm{Mg}^{+2} $$ On increasing -ve charge among the electronic species, ionic radius increases.
Shift-II
Classification of Elements and Periodicity in Properties
89428
Identify the correct order of ionic radii of the given species:
Ionic radii is the distance between the nucleus and the electron in the outermost shell of an ions. Increasing the electron on an atom, increases the ionic radii. Decrease the $\mathrm{e}^{-}$on an atom will be decrease ionic radii. So the order is- $$ \mathrm{S}^{2-}>\mathrm{K}^{+}>\mathrm{Sc}^{3+}>\mathrm{V}^{5+}>\mathrm{Mn}^{7+} $$
Classification of Elements and Periodicity in Properties
89501
Which of the following has least ionic radius?
1 $\mathrm{I}^{+}$
2 $\mathrm{I}^{3+}$
3 $\mathrm{I}^{5+}$
4 $\mathrm{I}^{7+}$
Explanation:
Ionic radii $\propto \frac{1}{\text { charge }}$ More is the positive charge smaller the size of ion. So, the least ionic radius is $\mathrm{I}^{7+}$.
J and K CET-(2001)
Classification of Elements and Periodicity in Properties
89503
$\mathrm{O}^{2-}$ and $\mathrm{Si}^{4+}$ are isoelectronic ions. If the ionic radius of $\mathrm{O}^{2-}$ is $1.4 \AA$, the ionic radius of $\mathrm{Si}^{4+}$ will be
1 $1.4 \AA$
2 $0.41 \AA$
3 $2.8 \AA$
4 $1.5 \AA$
Explanation:
Given that, $\mathrm{O}^{2-}$ and $\mathrm{Si}^{4+}$ are isoelectronic ions i.e. $10 \mathrm{e}^{-}$. Nuclear charge $\propto \frac{1}{\text { ionic radius }}$ $\therefore$ Nuclear charge for $\mathrm{Si}$ is larger, so it should be less than $1.4 \AA$
A.P.EAMCET 1999
Classification of Elements and Periodicity in Properties
89410
The correct order of increasing ionic radii is
Ionic radii is the distance from the nucleus up to which it has influence on its electron bond. The ionic radii of isoelectronic species increases with a decreases in the magnitudes of nuclear charge. So, The correct order of increasing ionic radii is $\mathrm{Mg}^{2+}<$ $\mathrm{Na}^{+}<\mathrm{F}^{-}<\mathrm{O}^{2-}<\mathrm{N}^{3-}$.
Shift-II
Classification of Elements and Periodicity in Properties
89427
Among the following ions, the one with the highest value of ionic radius is
1 $\mathrm{O}^{2-}$
2 $\mathrm{Mg}^{2+}$
3 $\mathrm{L}^{+}$
4 $\mathrm{L}^{-}$
Explanation:
decreasing order of ionic radius of given species is- $$ \mathrm{O}^{2-}>\mathrm{F}^{-}>\mathrm{Li}^{+}>\mathrm{Mg}^{+2} $$ On increasing -ve charge among the electronic species, ionic radius increases.
Shift-II
Classification of Elements and Periodicity in Properties
89428
Identify the correct order of ionic radii of the given species:
Ionic radii is the distance between the nucleus and the electron in the outermost shell of an ions. Increasing the electron on an atom, increases the ionic radii. Decrease the $\mathrm{e}^{-}$on an atom will be decrease ionic radii. So the order is- $$ \mathrm{S}^{2-}>\mathrm{K}^{+}>\mathrm{Sc}^{3+}>\mathrm{V}^{5+}>\mathrm{Mn}^{7+} $$
Classification of Elements and Periodicity in Properties
89501
Which of the following has least ionic radius?
1 $\mathrm{I}^{+}$
2 $\mathrm{I}^{3+}$
3 $\mathrm{I}^{5+}$
4 $\mathrm{I}^{7+}$
Explanation:
Ionic radii $\propto \frac{1}{\text { charge }}$ More is the positive charge smaller the size of ion. So, the least ionic radius is $\mathrm{I}^{7+}$.
J and K CET-(2001)
Classification of Elements and Periodicity in Properties
89503
$\mathrm{O}^{2-}$ and $\mathrm{Si}^{4+}$ are isoelectronic ions. If the ionic radius of $\mathrm{O}^{2-}$ is $1.4 \AA$, the ionic radius of $\mathrm{Si}^{4+}$ will be
1 $1.4 \AA$
2 $0.41 \AA$
3 $2.8 \AA$
4 $1.5 \AA$
Explanation:
Given that, $\mathrm{O}^{2-}$ and $\mathrm{Si}^{4+}$ are isoelectronic ions i.e. $10 \mathrm{e}^{-}$. Nuclear charge $\propto \frac{1}{\text { ionic radius }}$ $\therefore$ Nuclear charge for $\mathrm{Si}$ is larger, so it should be less than $1.4 \AA$
A.P.EAMCET 1999
Classification of Elements and Periodicity in Properties
89410
The correct order of increasing ionic radii is
Ionic radii is the distance from the nucleus up to which it has influence on its electron bond. The ionic radii of isoelectronic species increases with a decreases in the magnitudes of nuclear charge. So, The correct order of increasing ionic radii is $\mathrm{Mg}^{2+}<$ $\mathrm{Na}^{+}<\mathrm{F}^{-}<\mathrm{O}^{2-}<\mathrm{N}^{3-}$.
Shift-II
Classification of Elements and Periodicity in Properties
89427
Among the following ions, the one with the highest value of ionic radius is
1 $\mathrm{O}^{2-}$
2 $\mathrm{Mg}^{2+}$
3 $\mathrm{L}^{+}$
4 $\mathrm{L}^{-}$
Explanation:
decreasing order of ionic radius of given species is- $$ \mathrm{O}^{2-}>\mathrm{F}^{-}>\mathrm{Li}^{+}>\mathrm{Mg}^{+2} $$ On increasing -ve charge among the electronic species, ionic radius increases.
Shift-II
Classification of Elements and Periodicity in Properties
89428
Identify the correct order of ionic radii of the given species:
Ionic radii is the distance between the nucleus and the electron in the outermost shell of an ions. Increasing the electron on an atom, increases the ionic radii. Decrease the $\mathrm{e}^{-}$on an atom will be decrease ionic radii. So the order is- $$ \mathrm{S}^{2-}>\mathrm{K}^{+}>\mathrm{Sc}^{3+}>\mathrm{V}^{5+}>\mathrm{Mn}^{7+} $$
Classification of Elements and Periodicity in Properties
89501
Which of the following has least ionic radius?
1 $\mathrm{I}^{+}$
2 $\mathrm{I}^{3+}$
3 $\mathrm{I}^{5+}$
4 $\mathrm{I}^{7+}$
Explanation:
Ionic radii $\propto \frac{1}{\text { charge }}$ More is the positive charge smaller the size of ion. So, the least ionic radius is $\mathrm{I}^{7+}$.
J and K CET-(2001)
Classification of Elements and Periodicity in Properties
89503
$\mathrm{O}^{2-}$ and $\mathrm{Si}^{4+}$ are isoelectronic ions. If the ionic radius of $\mathrm{O}^{2-}$ is $1.4 \AA$, the ionic radius of $\mathrm{Si}^{4+}$ will be
1 $1.4 \AA$
2 $0.41 \AA$
3 $2.8 \AA$
4 $1.5 \AA$
Explanation:
Given that, $\mathrm{O}^{2-}$ and $\mathrm{Si}^{4+}$ are isoelectronic ions i.e. $10 \mathrm{e}^{-}$. Nuclear charge $\propto \frac{1}{\text { ionic radius }}$ $\therefore$ Nuclear charge for $\mathrm{Si}$ is larger, so it should be less than $1.4 \AA$
A.P.EAMCET 1999
Classification of Elements and Periodicity in Properties
89410
The correct order of increasing ionic radii is
Ionic radii is the distance from the nucleus up to which it has influence on its electron bond. The ionic radii of isoelectronic species increases with a decreases in the magnitudes of nuclear charge. So, The correct order of increasing ionic radii is $\mathrm{Mg}^{2+}<$ $\mathrm{Na}^{+}<\mathrm{F}^{-}<\mathrm{O}^{2-}<\mathrm{N}^{3-}$.
Shift-II
Classification of Elements and Periodicity in Properties
89427
Among the following ions, the one with the highest value of ionic radius is
1 $\mathrm{O}^{2-}$
2 $\mathrm{Mg}^{2+}$
3 $\mathrm{L}^{+}$
4 $\mathrm{L}^{-}$
Explanation:
decreasing order of ionic radius of given species is- $$ \mathrm{O}^{2-}>\mathrm{F}^{-}>\mathrm{Li}^{+}>\mathrm{Mg}^{+2} $$ On increasing -ve charge among the electronic species, ionic radius increases.
Shift-II
Classification of Elements and Periodicity in Properties
89428
Identify the correct order of ionic radii of the given species:
Ionic radii is the distance between the nucleus and the electron in the outermost shell of an ions. Increasing the electron on an atom, increases the ionic radii. Decrease the $\mathrm{e}^{-}$on an atom will be decrease ionic radii. So the order is- $$ \mathrm{S}^{2-}>\mathrm{K}^{+}>\mathrm{Sc}^{3+}>\mathrm{V}^{5+}>\mathrm{Mn}^{7+} $$
