NEET Test Series from KOTA - 10 Papers In MS WORD
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Classification of Elements and Periodicity in Properties
89490
The ionic radii (A) of $\mathrm{C}^{4-}$ and $\mathrm{O}^{2-}$ respectively are 2.60 and 1.40 . The ionic radius of the isoelectronic ion $\mathrm{N}^{3-}$ would be
1 2.6
2 1.71
3 1.4
4 0.95
Explanation:
For the ionic radii of isoelectronic ions decrease with the increase in the magnitude of the nuclear charge. So, decreasing order of ionic radii is $\begin{aligned} & \mathrm{C}^{4-}>\mathrm{N}^{3-}>\mathrm{O}^{2-} \\ & 2.60 \quad 1.71 \quad 1.40(\mathrm{pm}) \end{aligned}$
J and K CET-(2006)
Classification of Elements and Periodicity in Properties
89491
The most probable radius (in pm) for finding the electron in $\mathrm{He}^{+}$is
1 0.0
2 52.9
3 26.5
4 105.8
Explanation:
Given that : For $\mathrm{He}^{+}(\mathrm{Z})=2$ Number of shell $(\mathrm{n})=1$ we know that Bohr's radius $\left(\mathrm{r}_{\mathrm{n}}\right)=\frac{59.2 \mathrm{n}^2}{\mathrm{Z}} \mathrm{pm}$ Where $\mathrm{n}=$ number of shell $\mathrm{Z}=$ atomic number $\mathrm{r}_{\mathrm{n}}=\frac{59.2 \times 1^2}{2}=26.45 \mathrm{pm}$
AIIMS-2005
Classification of Elements and Periodicity in Properties
89495
Which one of the following ions has the highest value of ionic radius?
1 $\mathrm{Li}^{+}$
2 $\mathrm{B}^{3+}$
3 $\mathrm{O}^{2-}$
4 $\mathrm{F}^{-}$
Explanation:
Z/e ratio increases the size decreases and when $Z / e$ ratio decrease the size increases. $\mathrm{Z} / \mathrm{e}=\frac{\text { nuclear charge }}{\text { no. of electrons }}$ For $\mathrm{Li}^{+}, \mathrm{Z} / \mathrm{e}=\frac{3}{2}=1.5$ For $\mathrm{B}^{3+}, \mathrm{Z} / \mathrm{e}=\frac{5}{2}=2.5$ For $\mathrm{O}^{2-}, \mathrm{Z} / \mathrm{e}=\frac{8}{10}=0.8$ For $\mathrm{F}^{-}, \mathrm{Z} / \mathrm{e}=\frac{9}{10}=0.9$ Therefore, $\mathrm{O}^{2-}$ has highest value of ionic radius
(AIEEE 2004)
Classification of Elements and Periodicity in Properties
89499
The radius of hydrogen atom is $0.53 \AA$. The radius of ${ }_3 \mathrm{Li}^{2+}$ is of
1 $1.27 \AA$
2 $0.17 \AA$
3 $0.57 \AA$
4 $0.99 \AA$
Explanation:
Radius of $\mathrm{n}^{\text {th }}$ orbit $=\frac{\mathrm{n}^2}{\mathrm{Z}} \times 0.53$ where $\mathrm{n}=$ number of orbit and $\mathrm{Z}=$ atomic number $=3$ $\therefore \mathrm{In}_3 \mathrm{Li}^{2+}$, no. of electrons $=1$ $\begin{aligned} \therefore \text { Radius } & =\frac{(1)^2 \times 0.53}{3} \\ & =\frac{0.53}{3}=0.17 \AA \end{aligned}$
Classification of Elements and Periodicity in Properties
89490
The ionic radii (A) of $\mathrm{C}^{4-}$ and $\mathrm{O}^{2-}$ respectively are 2.60 and 1.40 . The ionic radius of the isoelectronic ion $\mathrm{N}^{3-}$ would be
1 2.6
2 1.71
3 1.4
4 0.95
Explanation:
For the ionic radii of isoelectronic ions decrease with the increase in the magnitude of the nuclear charge. So, decreasing order of ionic radii is $\begin{aligned} & \mathrm{C}^{4-}>\mathrm{N}^{3-}>\mathrm{O}^{2-} \\ & 2.60 \quad 1.71 \quad 1.40(\mathrm{pm}) \end{aligned}$
J and K CET-(2006)
Classification of Elements and Periodicity in Properties
89491
The most probable radius (in pm) for finding the electron in $\mathrm{He}^{+}$is
1 0.0
2 52.9
3 26.5
4 105.8
Explanation:
Given that : For $\mathrm{He}^{+}(\mathrm{Z})=2$ Number of shell $(\mathrm{n})=1$ we know that Bohr's radius $\left(\mathrm{r}_{\mathrm{n}}\right)=\frac{59.2 \mathrm{n}^2}{\mathrm{Z}} \mathrm{pm}$ Where $\mathrm{n}=$ number of shell $\mathrm{Z}=$ atomic number $\mathrm{r}_{\mathrm{n}}=\frac{59.2 \times 1^2}{2}=26.45 \mathrm{pm}$
AIIMS-2005
Classification of Elements and Periodicity in Properties
89495
Which one of the following ions has the highest value of ionic radius?
1 $\mathrm{Li}^{+}$
2 $\mathrm{B}^{3+}$
3 $\mathrm{O}^{2-}$
4 $\mathrm{F}^{-}$
Explanation:
Z/e ratio increases the size decreases and when $Z / e$ ratio decrease the size increases. $\mathrm{Z} / \mathrm{e}=\frac{\text { nuclear charge }}{\text { no. of electrons }}$ For $\mathrm{Li}^{+}, \mathrm{Z} / \mathrm{e}=\frac{3}{2}=1.5$ For $\mathrm{B}^{3+}, \mathrm{Z} / \mathrm{e}=\frac{5}{2}=2.5$ For $\mathrm{O}^{2-}, \mathrm{Z} / \mathrm{e}=\frac{8}{10}=0.8$ For $\mathrm{F}^{-}, \mathrm{Z} / \mathrm{e}=\frac{9}{10}=0.9$ Therefore, $\mathrm{O}^{2-}$ has highest value of ionic radius
(AIEEE 2004)
Classification of Elements and Periodicity in Properties
89499
The radius of hydrogen atom is $0.53 \AA$. The radius of ${ }_3 \mathrm{Li}^{2+}$ is of
1 $1.27 \AA$
2 $0.17 \AA$
3 $0.57 \AA$
4 $0.99 \AA$
Explanation:
Radius of $\mathrm{n}^{\text {th }}$ orbit $=\frac{\mathrm{n}^2}{\mathrm{Z}} \times 0.53$ where $\mathrm{n}=$ number of orbit and $\mathrm{Z}=$ atomic number $=3$ $\therefore \mathrm{In}_3 \mathrm{Li}^{2+}$, no. of electrons $=1$ $\begin{aligned} \therefore \text { Radius } & =\frac{(1)^2 \times 0.53}{3} \\ & =\frac{0.53}{3}=0.17 \AA \end{aligned}$
Classification of Elements and Periodicity in Properties
89490
The ionic radii (A) of $\mathrm{C}^{4-}$ and $\mathrm{O}^{2-}$ respectively are 2.60 and 1.40 . The ionic radius of the isoelectronic ion $\mathrm{N}^{3-}$ would be
1 2.6
2 1.71
3 1.4
4 0.95
Explanation:
For the ionic radii of isoelectronic ions decrease with the increase in the magnitude of the nuclear charge. So, decreasing order of ionic radii is $\begin{aligned} & \mathrm{C}^{4-}>\mathrm{N}^{3-}>\mathrm{O}^{2-} \\ & 2.60 \quad 1.71 \quad 1.40(\mathrm{pm}) \end{aligned}$
J and K CET-(2006)
Classification of Elements and Periodicity in Properties
89491
The most probable radius (in pm) for finding the electron in $\mathrm{He}^{+}$is
1 0.0
2 52.9
3 26.5
4 105.8
Explanation:
Given that : For $\mathrm{He}^{+}(\mathrm{Z})=2$ Number of shell $(\mathrm{n})=1$ we know that Bohr's radius $\left(\mathrm{r}_{\mathrm{n}}\right)=\frac{59.2 \mathrm{n}^2}{\mathrm{Z}} \mathrm{pm}$ Where $\mathrm{n}=$ number of shell $\mathrm{Z}=$ atomic number $\mathrm{r}_{\mathrm{n}}=\frac{59.2 \times 1^2}{2}=26.45 \mathrm{pm}$
AIIMS-2005
Classification of Elements and Periodicity in Properties
89495
Which one of the following ions has the highest value of ionic radius?
1 $\mathrm{Li}^{+}$
2 $\mathrm{B}^{3+}$
3 $\mathrm{O}^{2-}$
4 $\mathrm{F}^{-}$
Explanation:
Z/e ratio increases the size decreases and when $Z / e$ ratio decrease the size increases. $\mathrm{Z} / \mathrm{e}=\frac{\text { nuclear charge }}{\text { no. of electrons }}$ For $\mathrm{Li}^{+}, \mathrm{Z} / \mathrm{e}=\frac{3}{2}=1.5$ For $\mathrm{B}^{3+}, \mathrm{Z} / \mathrm{e}=\frac{5}{2}=2.5$ For $\mathrm{O}^{2-}, \mathrm{Z} / \mathrm{e}=\frac{8}{10}=0.8$ For $\mathrm{F}^{-}, \mathrm{Z} / \mathrm{e}=\frac{9}{10}=0.9$ Therefore, $\mathrm{O}^{2-}$ has highest value of ionic radius
(AIEEE 2004)
Classification of Elements and Periodicity in Properties
89499
The radius of hydrogen atom is $0.53 \AA$. The radius of ${ }_3 \mathrm{Li}^{2+}$ is of
1 $1.27 \AA$
2 $0.17 \AA$
3 $0.57 \AA$
4 $0.99 \AA$
Explanation:
Radius of $\mathrm{n}^{\text {th }}$ orbit $=\frac{\mathrm{n}^2}{\mathrm{Z}} \times 0.53$ where $\mathrm{n}=$ number of orbit and $\mathrm{Z}=$ atomic number $=3$ $\therefore \mathrm{In}_3 \mathrm{Li}^{2+}$, no. of electrons $=1$ $\begin{aligned} \therefore \text { Radius } & =\frac{(1)^2 \times 0.53}{3} \\ & =\frac{0.53}{3}=0.17 \AA \end{aligned}$
Classification of Elements and Periodicity in Properties
89490
The ionic radii (A) of $\mathrm{C}^{4-}$ and $\mathrm{O}^{2-}$ respectively are 2.60 and 1.40 . The ionic radius of the isoelectronic ion $\mathrm{N}^{3-}$ would be
1 2.6
2 1.71
3 1.4
4 0.95
Explanation:
For the ionic radii of isoelectronic ions decrease with the increase in the magnitude of the nuclear charge. So, decreasing order of ionic radii is $\begin{aligned} & \mathrm{C}^{4-}>\mathrm{N}^{3-}>\mathrm{O}^{2-} \\ & 2.60 \quad 1.71 \quad 1.40(\mathrm{pm}) \end{aligned}$
J and K CET-(2006)
Classification of Elements and Periodicity in Properties
89491
The most probable radius (in pm) for finding the electron in $\mathrm{He}^{+}$is
1 0.0
2 52.9
3 26.5
4 105.8
Explanation:
Given that : For $\mathrm{He}^{+}(\mathrm{Z})=2$ Number of shell $(\mathrm{n})=1$ we know that Bohr's radius $\left(\mathrm{r}_{\mathrm{n}}\right)=\frac{59.2 \mathrm{n}^2}{\mathrm{Z}} \mathrm{pm}$ Where $\mathrm{n}=$ number of shell $\mathrm{Z}=$ atomic number $\mathrm{r}_{\mathrm{n}}=\frac{59.2 \times 1^2}{2}=26.45 \mathrm{pm}$
AIIMS-2005
Classification of Elements and Periodicity in Properties
89495
Which one of the following ions has the highest value of ionic radius?
1 $\mathrm{Li}^{+}$
2 $\mathrm{B}^{3+}$
3 $\mathrm{O}^{2-}$
4 $\mathrm{F}^{-}$
Explanation:
Z/e ratio increases the size decreases and when $Z / e$ ratio decrease the size increases. $\mathrm{Z} / \mathrm{e}=\frac{\text { nuclear charge }}{\text { no. of electrons }}$ For $\mathrm{Li}^{+}, \mathrm{Z} / \mathrm{e}=\frac{3}{2}=1.5$ For $\mathrm{B}^{3+}, \mathrm{Z} / \mathrm{e}=\frac{5}{2}=2.5$ For $\mathrm{O}^{2-}, \mathrm{Z} / \mathrm{e}=\frac{8}{10}=0.8$ For $\mathrm{F}^{-}, \mathrm{Z} / \mathrm{e}=\frac{9}{10}=0.9$ Therefore, $\mathrm{O}^{2-}$ has highest value of ionic radius
(AIEEE 2004)
Classification of Elements and Periodicity in Properties
89499
The radius of hydrogen atom is $0.53 \AA$. The radius of ${ }_3 \mathrm{Li}^{2+}$ is of
1 $1.27 \AA$
2 $0.17 \AA$
3 $0.57 \AA$
4 $0.99 \AA$
Explanation:
Radius of $\mathrm{n}^{\text {th }}$ orbit $=\frac{\mathrm{n}^2}{\mathrm{Z}} \times 0.53$ where $\mathrm{n}=$ number of orbit and $\mathrm{Z}=$ atomic number $=3$ $\therefore \mathrm{In}_3 \mathrm{Li}^{2+}$, no. of electrons $=1$ $\begin{aligned} \therefore \text { Radius } & =\frac{(1)^2 \times 0.53}{3} \\ & =\frac{0.53}{3}=0.17 \AA \end{aligned}$
