$$ \frac{\mathrm{Z}}{\mathrm{e}}=\frac{\text { nuclear charge }}{\text { no. of electrons }} $$ When $\mathrm{Z} / \mathrm{e}$ ratio decrease the size increase for $$ \left\{\begin{array}{l} \mathrm{O}^{2-}, \frac{\mathrm{Z}}{\mathrm{e}}=\frac{8}{10}=0.8 \\ \mathrm{~F}^{-}, \frac{\mathrm{Z}}{\mathrm{e}}=\frac{9}{10}=0.9 \\ \mathrm{Na}^{+}, \frac{\mathrm{Z}}{\mathrm{e}}=\frac{11}{10}=1.1 \\ \mathrm{Mg}^{2+}, \frac{\mathrm{Z}}{\mathrm{e}}=\frac{12}{10}=1.2 \\ \mathrm{Al}^{3+}, \frac{\mathrm{Z}}{\mathrm{e}}=\frac{13}{10}=1.3 \end{array}\right. $$ Hence, the decreasing order of ionic radii of the element are- $$ \mathrm{O}^{2-}>\mathrm{F}^{-}>\mathrm{Na}^{+}>\mathrm{Mg}^{2+}>\mathrm{Al}^{3+} $$
JEE Main-04.09.2020
Classification of Elements and Periodicity in Properties
89418
The set representing the correct order of ionic radius is
Ionic radii increases when moving down the group and its decrease as the positive charge on the element increases due to effective nuclear charge. In the cation, number of electrons are less than number of protons, hence its effective nuclear charge increase and ionic radii decrease. The correct order of ionic radii- $$ \mathrm{Na}_{95}^{+}>\underset{76}{\mathrm{Li}^{+}}>\underset{72}{\mathrm{Mg}^{2+}}>\mathrm{Be}_{31}^{2+} $$ Ionic radii $(\mathrm{pm})$
UPTU/UPSEE-2015
Classification of Elements and Periodicity in Properties
89452
What is the correct increasing order of ionic or atomic radii in the following?
$\mathrm{Zn}$ atomic radii is $137 \mathrm{pm}$. $\mathrm{Fe}$ atomic radii is $126 \mathrm{pm}$. $\mathrm{Fe}^{3+}$ ionic radius is $63 \mathrm{pm}$. $\mathrm{Fe}^{2+}$ has an ionic radius of $77 \mathrm{pm}$. Thus, the correct order of their size is- $$ \mathrm{Zn}>\mathrm{Fe}>\mathrm{Fe}^{2+}>\mathrm{Fe}^{3+} $$ Hence, the correct option is (a).
Shift-II
Classification of Elements and Periodicity in Properties
89449
Which has least covalent radius?
1 $\mathrm{Mn}$
2 $\mathrm{Cu}$
3 $\mathrm{Zn}$
4 $\mathrm{Ni}$
Explanation:
Covalent radius of an atom denotes half of the internuclear distance between the atoms of the monoatomic species. It is different from ionic and atomic radius. It depends on the bond order of the compound formed by the covalent bonding. Hence, the covalent radius in $\mathrm{Pm}$ are- $\mathrm{Ni}<\mathrm{Cu}<\mathrm{Zn}=\mathrm{Mn}$ $\begin{array}{llll}125 & 128 & 137 & 137\end{array}$ So, the least covalent radius is $\mathrm{Ni}$.
$$ \frac{\mathrm{Z}}{\mathrm{e}}=\frac{\text { nuclear charge }}{\text { no. of electrons }} $$ When $\mathrm{Z} / \mathrm{e}$ ratio decrease the size increase for $$ \left\{\begin{array}{l} \mathrm{O}^{2-}, \frac{\mathrm{Z}}{\mathrm{e}}=\frac{8}{10}=0.8 \\ \mathrm{~F}^{-}, \frac{\mathrm{Z}}{\mathrm{e}}=\frac{9}{10}=0.9 \\ \mathrm{Na}^{+}, \frac{\mathrm{Z}}{\mathrm{e}}=\frac{11}{10}=1.1 \\ \mathrm{Mg}^{2+}, \frac{\mathrm{Z}}{\mathrm{e}}=\frac{12}{10}=1.2 \\ \mathrm{Al}^{3+}, \frac{\mathrm{Z}}{\mathrm{e}}=\frac{13}{10}=1.3 \end{array}\right. $$ Hence, the decreasing order of ionic radii of the element are- $$ \mathrm{O}^{2-}>\mathrm{F}^{-}>\mathrm{Na}^{+}>\mathrm{Mg}^{2+}>\mathrm{Al}^{3+} $$
JEE Main-04.09.2020
Classification of Elements and Periodicity in Properties
89418
The set representing the correct order of ionic radius is
Ionic radii increases when moving down the group and its decrease as the positive charge on the element increases due to effective nuclear charge. In the cation, number of electrons are less than number of protons, hence its effective nuclear charge increase and ionic radii decrease. The correct order of ionic radii- $$ \mathrm{Na}_{95}^{+}>\underset{76}{\mathrm{Li}^{+}}>\underset{72}{\mathrm{Mg}^{2+}}>\mathrm{Be}_{31}^{2+} $$ Ionic radii $(\mathrm{pm})$
UPTU/UPSEE-2015
Classification of Elements and Periodicity in Properties
89452
What is the correct increasing order of ionic or atomic radii in the following?
$\mathrm{Zn}$ atomic radii is $137 \mathrm{pm}$. $\mathrm{Fe}$ atomic radii is $126 \mathrm{pm}$. $\mathrm{Fe}^{3+}$ ionic radius is $63 \mathrm{pm}$. $\mathrm{Fe}^{2+}$ has an ionic radius of $77 \mathrm{pm}$. Thus, the correct order of their size is- $$ \mathrm{Zn}>\mathrm{Fe}>\mathrm{Fe}^{2+}>\mathrm{Fe}^{3+} $$ Hence, the correct option is (a).
Shift-II
Classification of Elements and Periodicity in Properties
89449
Which has least covalent radius?
1 $\mathrm{Mn}$
2 $\mathrm{Cu}$
3 $\mathrm{Zn}$
4 $\mathrm{Ni}$
Explanation:
Covalent radius of an atom denotes half of the internuclear distance between the atoms of the monoatomic species. It is different from ionic and atomic radius. It depends on the bond order of the compound formed by the covalent bonding. Hence, the covalent radius in $\mathrm{Pm}$ are- $\mathrm{Ni}<\mathrm{Cu}<\mathrm{Zn}=\mathrm{Mn}$ $\begin{array}{llll}125 & 128 & 137 & 137\end{array}$ So, the least covalent radius is $\mathrm{Ni}$.
$$ \frac{\mathrm{Z}}{\mathrm{e}}=\frac{\text { nuclear charge }}{\text { no. of electrons }} $$ When $\mathrm{Z} / \mathrm{e}$ ratio decrease the size increase for $$ \left\{\begin{array}{l} \mathrm{O}^{2-}, \frac{\mathrm{Z}}{\mathrm{e}}=\frac{8}{10}=0.8 \\ \mathrm{~F}^{-}, \frac{\mathrm{Z}}{\mathrm{e}}=\frac{9}{10}=0.9 \\ \mathrm{Na}^{+}, \frac{\mathrm{Z}}{\mathrm{e}}=\frac{11}{10}=1.1 \\ \mathrm{Mg}^{2+}, \frac{\mathrm{Z}}{\mathrm{e}}=\frac{12}{10}=1.2 \\ \mathrm{Al}^{3+}, \frac{\mathrm{Z}}{\mathrm{e}}=\frac{13}{10}=1.3 \end{array}\right. $$ Hence, the decreasing order of ionic radii of the element are- $$ \mathrm{O}^{2-}>\mathrm{F}^{-}>\mathrm{Na}^{+}>\mathrm{Mg}^{2+}>\mathrm{Al}^{3+} $$
JEE Main-04.09.2020
Classification of Elements and Periodicity in Properties
89418
The set representing the correct order of ionic radius is
Ionic radii increases when moving down the group and its decrease as the positive charge on the element increases due to effective nuclear charge. In the cation, number of electrons are less than number of protons, hence its effective nuclear charge increase and ionic radii decrease. The correct order of ionic radii- $$ \mathrm{Na}_{95}^{+}>\underset{76}{\mathrm{Li}^{+}}>\underset{72}{\mathrm{Mg}^{2+}}>\mathrm{Be}_{31}^{2+} $$ Ionic radii $(\mathrm{pm})$
UPTU/UPSEE-2015
Classification of Elements and Periodicity in Properties
89452
What is the correct increasing order of ionic or atomic radii in the following?
$\mathrm{Zn}$ atomic radii is $137 \mathrm{pm}$. $\mathrm{Fe}$ atomic radii is $126 \mathrm{pm}$. $\mathrm{Fe}^{3+}$ ionic radius is $63 \mathrm{pm}$. $\mathrm{Fe}^{2+}$ has an ionic radius of $77 \mathrm{pm}$. Thus, the correct order of their size is- $$ \mathrm{Zn}>\mathrm{Fe}>\mathrm{Fe}^{2+}>\mathrm{Fe}^{3+} $$ Hence, the correct option is (a).
Shift-II
Classification of Elements and Periodicity in Properties
89449
Which has least covalent radius?
1 $\mathrm{Mn}$
2 $\mathrm{Cu}$
3 $\mathrm{Zn}$
4 $\mathrm{Ni}$
Explanation:
Covalent radius of an atom denotes half of the internuclear distance between the atoms of the monoatomic species. It is different from ionic and atomic radius. It depends on the bond order of the compound formed by the covalent bonding. Hence, the covalent radius in $\mathrm{Pm}$ are- $\mathrm{Ni}<\mathrm{Cu}<\mathrm{Zn}=\mathrm{Mn}$ $\begin{array}{llll}125 & 128 & 137 & 137\end{array}$ So, the least covalent radius is $\mathrm{Ni}$.
$$ \frac{\mathrm{Z}}{\mathrm{e}}=\frac{\text { nuclear charge }}{\text { no. of electrons }} $$ When $\mathrm{Z} / \mathrm{e}$ ratio decrease the size increase for $$ \left\{\begin{array}{l} \mathrm{O}^{2-}, \frac{\mathrm{Z}}{\mathrm{e}}=\frac{8}{10}=0.8 \\ \mathrm{~F}^{-}, \frac{\mathrm{Z}}{\mathrm{e}}=\frac{9}{10}=0.9 \\ \mathrm{Na}^{+}, \frac{\mathrm{Z}}{\mathrm{e}}=\frac{11}{10}=1.1 \\ \mathrm{Mg}^{2+}, \frac{\mathrm{Z}}{\mathrm{e}}=\frac{12}{10}=1.2 \\ \mathrm{Al}^{3+}, \frac{\mathrm{Z}}{\mathrm{e}}=\frac{13}{10}=1.3 \end{array}\right. $$ Hence, the decreasing order of ionic radii of the element are- $$ \mathrm{O}^{2-}>\mathrm{F}^{-}>\mathrm{Na}^{+}>\mathrm{Mg}^{2+}>\mathrm{Al}^{3+} $$
JEE Main-04.09.2020
Classification of Elements and Periodicity in Properties
89418
The set representing the correct order of ionic radius is
Ionic radii increases when moving down the group and its decrease as the positive charge on the element increases due to effective nuclear charge. In the cation, number of electrons are less than number of protons, hence its effective nuclear charge increase and ionic radii decrease. The correct order of ionic radii- $$ \mathrm{Na}_{95}^{+}>\underset{76}{\mathrm{Li}^{+}}>\underset{72}{\mathrm{Mg}^{2+}}>\mathrm{Be}_{31}^{2+} $$ Ionic radii $(\mathrm{pm})$
UPTU/UPSEE-2015
Classification of Elements and Periodicity in Properties
89452
What is the correct increasing order of ionic or atomic radii in the following?
$\mathrm{Zn}$ atomic radii is $137 \mathrm{pm}$. $\mathrm{Fe}$ atomic radii is $126 \mathrm{pm}$. $\mathrm{Fe}^{3+}$ ionic radius is $63 \mathrm{pm}$. $\mathrm{Fe}^{2+}$ has an ionic radius of $77 \mathrm{pm}$. Thus, the correct order of their size is- $$ \mathrm{Zn}>\mathrm{Fe}>\mathrm{Fe}^{2+}>\mathrm{Fe}^{3+} $$ Hence, the correct option is (a).
Shift-II
Classification of Elements and Periodicity in Properties
89449
Which has least covalent radius?
1 $\mathrm{Mn}$
2 $\mathrm{Cu}$
3 $\mathrm{Zn}$
4 $\mathrm{Ni}$
Explanation:
Covalent radius of an atom denotes half of the internuclear distance between the atoms of the monoatomic species. It is different from ionic and atomic radius. It depends on the bond order of the compound formed by the covalent bonding. Hence, the covalent radius in $\mathrm{Pm}$ are- $\mathrm{Ni}<\mathrm{Cu}<\mathrm{Zn}=\mathrm{Mn}$ $\begin{array}{llll}125 & 128 & 137 & 137\end{array}$ So, the least covalent radius is $\mathrm{Ni}$.
$$ \frac{\mathrm{Z}}{\mathrm{e}}=\frac{\text { nuclear charge }}{\text { no. of electrons }} $$ When $\mathrm{Z} / \mathrm{e}$ ratio decrease the size increase for $$ \left\{\begin{array}{l} \mathrm{O}^{2-}, \frac{\mathrm{Z}}{\mathrm{e}}=\frac{8}{10}=0.8 \\ \mathrm{~F}^{-}, \frac{\mathrm{Z}}{\mathrm{e}}=\frac{9}{10}=0.9 \\ \mathrm{Na}^{+}, \frac{\mathrm{Z}}{\mathrm{e}}=\frac{11}{10}=1.1 \\ \mathrm{Mg}^{2+}, \frac{\mathrm{Z}}{\mathrm{e}}=\frac{12}{10}=1.2 \\ \mathrm{Al}^{3+}, \frac{\mathrm{Z}}{\mathrm{e}}=\frac{13}{10}=1.3 \end{array}\right. $$ Hence, the decreasing order of ionic radii of the element are- $$ \mathrm{O}^{2-}>\mathrm{F}^{-}>\mathrm{Na}^{+}>\mathrm{Mg}^{2+}>\mathrm{Al}^{3+} $$
JEE Main-04.09.2020
Classification of Elements and Periodicity in Properties
89418
The set representing the correct order of ionic radius is
Ionic radii increases when moving down the group and its decrease as the positive charge on the element increases due to effective nuclear charge. In the cation, number of electrons are less than number of protons, hence its effective nuclear charge increase and ionic radii decrease. The correct order of ionic radii- $$ \mathrm{Na}_{95}^{+}>\underset{76}{\mathrm{Li}^{+}}>\underset{72}{\mathrm{Mg}^{2+}}>\mathrm{Be}_{31}^{2+} $$ Ionic radii $(\mathrm{pm})$
UPTU/UPSEE-2015
Classification of Elements and Periodicity in Properties
89452
What is the correct increasing order of ionic or atomic radii in the following?
$\mathrm{Zn}$ atomic radii is $137 \mathrm{pm}$. $\mathrm{Fe}$ atomic radii is $126 \mathrm{pm}$. $\mathrm{Fe}^{3+}$ ionic radius is $63 \mathrm{pm}$. $\mathrm{Fe}^{2+}$ has an ionic radius of $77 \mathrm{pm}$. Thus, the correct order of their size is- $$ \mathrm{Zn}>\mathrm{Fe}>\mathrm{Fe}^{2+}>\mathrm{Fe}^{3+} $$ Hence, the correct option is (a).
Shift-II
Classification of Elements and Periodicity in Properties
89449
Which has least covalent radius?
1 $\mathrm{Mn}$
2 $\mathrm{Cu}$
3 $\mathrm{Zn}$
4 $\mathrm{Ni}$
Explanation:
Covalent radius of an atom denotes half of the internuclear distance between the atoms of the monoatomic species. It is different from ionic and atomic radius. It depends on the bond order of the compound formed by the covalent bonding. Hence, the covalent radius in $\mathrm{Pm}$ are- $\mathrm{Ni}<\mathrm{Cu}<\mathrm{Zn}=\mathrm{Mn}$ $\begin{array}{llll}125 & 128 & 137 & 137\end{array}$ So, the least covalent radius is $\mathrm{Ni}$.
