$\mathrm{F}, \mathrm{F}^{-}, \mathrm{O}$ and $\mathrm{O}^{2-}$ are of the same period. As the number of anions increases of isolated atom, the size of anion increases because of extra addition of electron in outer orbital. The size of anions are in the order- $$ \mathrm{O}^{2-}>\mathrm{F}^{-}>\mathrm{O}>\mathrm{F} $$
BITSAT 2005
Classification of Elements and Periodicity in Properties
89414
Ionic radii are
1 inversely proportional to effective number charge
2 inversely proportional to square of effective nuclear charge
3 directly proportional to effective nuclear charge
4 directly proportional to square of effective nuclear charge
Explanation:
Ionic radii varies with effective nuclear charge $\left(Z_{\text {eff }}\right)$ and screening effects. Ionic radii is inversely proportional to the effective nuclear charge as if the radii increases nuclear attractions decreases gradually. $$ \mathrm{r}_{\text {ionic }}=\propto \frac{1}{Z_{\text {eff }}} $$
JCECE - 2008
Classification of Elements and Periodicity in Properties
89416
Which of the following orders of ionic radii is correctly represented?
It is known that the radius of cation is always smaller than that of a neutral atom due to decrease in the number of orbits. Where as the radius of anion is always greater than a cation due to decrease in effective nuclear charge hence, the order are- $$ \mathrm{H}^{-}>\mathrm{H}>\mathrm{H}^{+} $$
Karnataka-CET-2018
Classification of Elements and Periodicity in Properties
89417
The increasing order of the atomic radii of the following elements is
1 (A) $<$ (B) $<$ (C) $<$ (D) $<$ (E)
2 (C) $<$ (B) $<$ (A) $<$ (D) $<$ (E)
3 (D) $<$ (C) $<$ (B) $<$ (A) $<$ (E)
4 (B) $<$ (C) $<$ (D) $<$ (A) $<$ (E)
5 $\mathrm{Br}$
Explanation:
On moving from left to right in period atomic radii decrease and moving from top to bottom in a group atomic radii increases. $\therefore$ Thus, the increasing order of atomic radii are- $$ \mathrm{F}<\mathrm{O}<\mathrm{C}<\mathrm{Cl}<\mathrm{Br} $$
$\mathrm{F}, \mathrm{F}^{-}, \mathrm{O}$ and $\mathrm{O}^{2-}$ are of the same period. As the number of anions increases of isolated atom, the size of anion increases because of extra addition of electron in outer orbital. The size of anions are in the order- $$ \mathrm{O}^{2-}>\mathrm{F}^{-}>\mathrm{O}>\mathrm{F} $$
BITSAT 2005
Classification of Elements and Periodicity in Properties
89414
Ionic radii are
1 inversely proportional to effective number charge
2 inversely proportional to square of effective nuclear charge
3 directly proportional to effective nuclear charge
4 directly proportional to square of effective nuclear charge
Explanation:
Ionic radii varies with effective nuclear charge $\left(Z_{\text {eff }}\right)$ and screening effects. Ionic radii is inversely proportional to the effective nuclear charge as if the radii increases nuclear attractions decreases gradually. $$ \mathrm{r}_{\text {ionic }}=\propto \frac{1}{Z_{\text {eff }}} $$
JCECE - 2008
Classification of Elements and Periodicity in Properties
89416
Which of the following orders of ionic radii is correctly represented?
It is known that the radius of cation is always smaller than that of a neutral atom due to decrease in the number of orbits. Where as the radius of anion is always greater than a cation due to decrease in effective nuclear charge hence, the order are- $$ \mathrm{H}^{-}>\mathrm{H}>\mathrm{H}^{+} $$
Karnataka-CET-2018
Classification of Elements and Periodicity in Properties
89417
The increasing order of the atomic radii of the following elements is
1 (A) $<$ (B) $<$ (C) $<$ (D) $<$ (E)
2 (C) $<$ (B) $<$ (A) $<$ (D) $<$ (E)
3 (D) $<$ (C) $<$ (B) $<$ (A) $<$ (E)
4 (B) $<$ (C) $<$ (D) $<$ (A) $<$ (E)
5 $\mathrm{Br}$
Explanation:
On moving from left to right in period atomic radii decrease and moving from top to bottom in a group atomic radii increases. $\therefore$ Thus, the increasing order of atomic radii are- $$ \mathrm{F}<\mathrm{O}<\mathrm{C}<\mathrm{Cl}<\mathrm{Br} $$
$\mathrm{F}, \mathrm{F}^{-}, \mathrm{O}$ and $\mathrm{O}^{2-}$ are of the same period. As the number of anions increases of isolated atom, the size of anion increases because of extra addition of electron in outer orbital. The size of anions are in the order- $$ \mathrm{O}^{2-}>\mathrm{F}^{-}>\mathrm{O}>\mathrm{F} $$
BITSAT 2005
Classification of Elements and Periodicity in Properties
89414
Ionic radii are
1 inversely proportional to effective number charge
2 inversely proportional to square of effective nuclear charge
3 directly proportional to effective nuclear charge
4 directly proportional to square of effective nuclear charge
Explanation:
Ionic radii varies with effective nuclear charge $\left(Z_{\text {eff }}\right)$ and screening effects. Ionic radii is inversely proportional to the effective nuclear charge as if the radii increases nuclear attractions decreases gradually. $$ \mathrm{r}_{\text {ionic }}=\propto \frac{1}{Z_{\text {eff }}} $$
JCECE - 2008
Classification of Elements and Periodicity in Properties
89416
Which of the following orders of ionic radii is correctly represented?
It is known that the radius of cation is always smaller than that of a neutral atom due to decrease in the number of orbits. Where as the radius of anion is always greater than a cation due to decrease in effective nuclear charge hence, the order are- $$ \mathrm{H}^{-}>\mathrm{H}>\mathrm{H}^{+} $$
Karnataka-CET-2018
Classification of Elements and Periodicity in Properties
89417
The increasing order of the atomic radii of the following elements is
1 (A) $<$ (B) $<$ (C) $<$ (D) $<$ (E)
2 (C) $<$ (B) $<$ (A) $<$ (D) $<$ (E)
3 (D) $<$ (C) $<$ (B) $<$ (A) $<$ (E)
4 (B) $<$ (C) $<$ (D) $<$ (A) $<$ (E)
5 $\mathrm{Br}$
Explanation:
On moving from left to right in period atomic radii decrease and moving from top to bottom in a group atomic radii increases. $\therefore$ Thus, the increasing order of atomic radii are- $$ \mathrm{F}<\mathrm{O}<\mathrm{C}<\mathrm{Cl}<\mathrm{Br} $$
$\mathrm{F}, \mathrm{F}^{-}, \mathrm{O}$ and $\mathrm{O}^{2-}$ are of the same period. As the number of anions increases of isolated atom, the size of anion increases because of extra addition of electron in outer orbital. The size of anions are in the order- $$ \mathrm{O}^{2-}>\mathrm{F}^{-}>\mathrm{O}>\mathrm{F} $$
BITSAT 2005
Classification of Elements and Periodicity in Properties
89414
Ionic radii are
1 inversely proportional to effective number charge
2 inversely proportional to square of effective nuclear charge
3 directly proportional to effective nuclear charge
4 directly proportional to square of effective nuclear charge
Explanation:
Ionic radii varies with effective nuclear charge $\left(Z_{\text {eff }}\right)$ and screening effects. Ionic radii is inversely proportional to the effective nuclear charge as if the radii increases nuclear attractions decreases gradually. $$ \mathrm{r}_{\text {ionic }}=\propto \frac{1}{Z_{\text {eff }}} $$
JCECE - 2008
Classification of Elements and Periodicity in Properties
89416
Which of the following orders of ionic radii is correctly represented?
It is known that the radius of cation is always smaller than that of a neutral atom due to decrease in the number of orbits. Where as the radius of anion is always greater than a cation due to decrease in effective nuclear charge hence, the order are- $$ \mathrm{H}^{-}>\mathrm{H}>\mathrm{H}^{+} $$
Karnataka-CET-2018
Classification of Elements and Periodicity in Properties
89417
The increasing order of the atomic radii of the following elements is
1 (A) $<$ (B) $<$ (C) $<$ (D) $<$ (E)
2 (C) $<$ (B) $<$ (A) $<$ (D) $<$ (E)
3 (D) $<$ (C) $<$ (B) $<$ (A) $<$ (E)
4 (B) $<$ (C) $<$ (D) $<$ (A) $<$ (E)
5 $\mathrm{Br}$
Explanation:
On moving from left to right in period atomic radii decrease and moving from top to bottom in a group atomic radii increases. $\therefore$ Thus, the increasing order of atomic radii are- $$ \mathrm{F}<\mathrm{O}<\mathrm{C}<\mathrm{Cl}<\mathrm{Br} $$