NEET Test Series from KOTA - 10 Papers In MS WORD
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Classification of Elements and Periodicity in Properties
89488
Atomic radii of $F$ and $N e$, in $\AA$, are given by :
1 $0.72,0.71$
2 $0.72,1.6$
3 $1.6,1.58$
4 $0.71,0.72$
Explanation:
Atomic radii is the total distance from the nucleus of an atom of the outermost orbital of its electron. On moving from left to right in a period generally atomic radii decrease due to increase in the effective nuclear charge with increase in atomic number. And on moving down in the group shells are increases resulting in a large size. Thus, the atomic radii of fluorine and neon in are $0.75 \AA$ and $1.60 \AA$ respectively.
BCECE-2006
Classification of Elements and Periodicity in Properties
89494
Which of the following has largest size?
1 Al
2 $\mathrm{Al}^{+}$
3 $\mathrm{Al}^{2+}$
4 $\mathrm{Al}^{3+}$
Explanation:
The atom contain more the +ve charged less the size of metal. The more electrostatic force act in small size of metal. Hence, the largest size is $\mathrm{Al}$. So, the decreasing order of size are - $\mathrm{Al}>\mathrm{Al}^{+}>\mathrm{Al}^{2+}>\mathrm{Al}^{+3}$
CG PET -2004
Classification of Elements and Periodicity in Properties
89497
Consider the ions $\mathrm{K}^{+}, \mathrm{S}^{2-}, \mathrm{Cl}^{-}$and $\mathrm{Ca}^{2+}$. The radii of these ionic species follow the order:
Ionic radii $\propto \frac{1}{\text { charge }}$ The charge of species is increases then the ionic radii will be decreases hence the decreasing order of the ionic radii will be- $\mathrm{S}^{2-}>\mathrm{Cl}^{-}>\mathrm{K}^{+}>\mathrm{Ca}^{2+}$
Kerala-CEE-2004
Classification of Elements and Periodicity in Properties
89498
Which one of the following is expected to have largest size?
1 $\mathrm{F}^{-}$
2 $\mathrm{O}^{2-}$
3 $\mathrm{A} l^{3+}$
4 $\mathrm{N}^{3-}$
Explanation:
Radius $\propto \frac{1}{Z}$ All the ions are isoelectronic but they have different size due to different values of $\mathrm{Z} . \mathrm{In}, \mathrm{Na}^{+}, \mathrm{O}^{2-}, \mathrm{N}^{3-}, \mathrm{F}^{-}$, the number of electron 10 . Thus greater the value of $\frac{e}{Z}$ greater the size of isoelectronic species. Hence $\mathrm{N}^{3-}$ is largest in size.
Classification of Elements and Periodicity in Properties
89488
Atomic radii of $F$ and $N e$, in $\AA$, are given by :
1 $0.72,0.71$
2 $0.72,1.6$
3 $1.6,1.58$
4 $0.71,0.72$
Explanation:
Atomic radii is the total distance from the nucleus of an atom of the outermost orbital of its electron. On moving from left to right in a period generally atomic radii decrease due to increase in the effective nuclear charge with increase in atomic number. And on moving down in the group shells are increases resulting in a large size. Thus, the atomic radii of fluorine and neon in are $0.75 \AA$ and $1.60 \AA$ respectively.
BCECE-2006
Classification of Elements and Periodicity in Properties
89494
Which of the following has largest size?
1 Al
2 $\mathrm{Al}^{+}$
3 $\mathrm{Al}^{2+}$
4 $\mathrm{Al}^{3+}$
Explanation:
The atom contain more the +ve charged less the size of metal. The more electrostatic force act in small size of metal. Hence, the largest size is $\mathrm{Al}$. So, the decreasing order of size are - $\mathrm{Al}>\mathrm{Al}^{+}>\mathrm{Al}^{2+}>\mathrm{Al}^{+3}$
CG PET -2004
Classification of Elements and Periodicity in Properties
89497
Consider the ions $\mathrm{K}^{+}, \mathrm{S}^{2-}, \mathrm{Cl}^{-}$and $\mathrm{Ca}^{2+}$. The radii of these ionic species follow the order:
Ionic radii $\propto \frac{1}{\text { charge }}$ The charge of species is increases then the ionic radii will be decreases hence the decreasing order of the ionic radii will be- $\mathrm{S}^{2-}>\mathrm{Cl}^{-}>\mathrm{K}^{+}>\mathrm{Ca}^{2+}$
Kerala-CEE-2004
Classification of Elements and Periodicity in Properties
89498
Which one of the following is expected to have largest size?
1 $\mathrm{F}^{-}$
2 $\mathrm{O}^{2-}$
3 $\mathrm{A} l^{3+}$
4 $\mathrm{N}^{3-}$
Explanation:
Radius $\propto \frac{1}{Z}$ All the ions are isoelectronic but they have different size due to different values of $\mathrm{Z} . \mathrm{In}, \mathrm{Na}^{+}, \mathrm{O}^{2-}, \mathrm{N}^{3-}, \mathrm{F}^{-}$, the number of electron 10 . Thus greater the value of $\frac{e}{Z}$ greater the size of isoelectronic species. Hence $\mathrm{N}^{3-}$ is largest in size.
Classification of Elements and Periodicity in Properties
89488
Atomic radii of $F$ and $N e$, in $\AA$, are given by :
1 $0.72,0.71$
2 $0.72,1.6$
3 $1.6,1.58$
4 $0.71,0.72$
Explanation:
Atomic radii is the total distance from the nucleus of an atom of the outermost orbital of its electron. On moving from left to right in a period generally atomic radii decrease due to increase in the effective nuclear charge with increase in atomic number. And on moving down in the group shells are increases resulting in a large size. Thus, the atomic radii of fluorine and neon in are $0.75 \AA$ and $1.60 \AA$ respectively.
BCECE-2006
Classification of Elements and Periodicity in Properties
89494
Which of the following has largest size?
1 Al
2 $\mathrm{Al}^{+}$
3 $\mathrm{Al}^{2+}$
4 $\mathrm{Al}^{3+}$
Explanation:
The atom contain more the +ve charged less the size of metal. The more electrostatic force act in small size of metal. Hence, the largest size is $\mathrm{Al}$. So, the decreasing order of size are - $\mathrm{Al}>\mathrm{Al}^{+}>\mathrm{Al}^{2+}>\mathrm{Al}^{+3}$
CG PET -2004
Classification of Elements and Periodicity in Properties
89497
Consider the ions $\mathrm{K}^{+}, \mathrm{S}^{2-}, \mathrm{Cl}^{-}$and $\mathrm{Ca}^{2+}$. The radii of these ionic species follow the order:
Ionic radii $\propto \frac{1}{\text { charge }}$ The charge of species is increases then the ionic radii will be decreases hence the decreasing order of the ionic radii will be- $\mathrm{S}^{2-}>\mathrm{Cl}^{-}>\mathrm{K}^{+}>\mathrm{Ca}^{2+}$
Kerala-CEE-2004
Classification of Elements and Periodicity in Properties
89498
Which one of the following is expected to have largest size?
1 $\mathrm{F}^{-}$
2 $\mathrm{O}^{2-}$
3 $\mathrm{A} l^{3+}$
4 $\mathrm{N}^{3-}$
Explanation:
Radius $\propto \frac{1}{Z}$ All the ions are isoelectronic but they have different size due to different values of $\mathrm{Z} . \mathrm{In}, \mathrm{Na}^{+}, \mathrm{O}^{2-}, \mathrm{N}^{3-}, \mathrm{F}^{-}$, the number of electron 10 . Thus greater the value of $\frac{e}{Z}$ greater the size of isoelectronic species. Hence $\mathrm{N}^{3-}$ is largest in size.
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Classification of Elements and Periodicity in Properties
89488
Atomic radii of $F$ and $N e$, in $\AA$, are given by :
1 $0.72,0.71$
2 $0.72,1.6$
3 $1.6,1.58$
4 $0.71,0.72$
Explanation:
Atomic radii is the total distance from the nucleus of an atom of the outermost orbital of its electron. On moving from left to right in a period generally atomic radii decrease due to increase in the effective nuclear charge with increase in atomic number. And on moving down in the group shells are increases resulting in a large size. Thus, the atomic radii of fluorine and neon in are $0.75 \AA$ and $1.60 \AA$ respectively.
BCECE-2006
Classification of Elements and Periodicity in Properties
89494
Which of the following has largest size?
1 Al
2 $\mathrm{Al}^{+}$
3 $\mathrm{Al}^{2+}$
4 $\mathrm{Al}^{3+}$
Explanation:
The atom contain more the +ve charged less the size of metal. The more electrostatic force act in small size of metal. Hence, the largest size is $\mathrm{Al}$. So, the decreasing order of size are - $\mathrm{Al}>\mathrm{Al}^{+}>\mathrm{Al}^{2+}>\mathrm{Al}^{+3}$
CG PET -2004
Classification of Elements and Periodicity in Properties
89497
Consider the ions $\mathrm{K}^{+}, \mathrm{S}^{2-}, \mathrm{Cl}^{-}$and $\mathrm{Ca}^{2+}$. The radii of these ionic species follow the order:
Ionic radii $\propto \frac{1}{\text { charge }}$ The charge of species is increases then the ionic radii will be decreases hence the decreasing order of the ionic radii will be- $\mathrm{S}^{2-}>\mathrm{Cl}^{-}>\mathrm{K}^{+}>\mathrm{Ca}^{2+}$
Kerala-CEE-2004
Classification of Elements and Periodicity in Properties
89498
Which one of the following is expected to have largest size?
1 $\mathrm{F}^{-}$
2 $\mathrm{O}^{2-}$
3 $\mathrm{A} l^{3+}$
4 $\mathrm{N}^{3-}$
Explanation:
Radius $\propto \frac{1}{Z}$ All the ions are isoelectronic but they have different size due to different values of $\mathrm{Z} . \mathrm{In}, \mathrm{Na}^{+}, \mathrm{O}^{2-}, \mathrm{N}^{3-}, \mathrm{F}^{-}$, the number of electron 10 . Thus greater the value of $\frac{e}{Z}$ greater the size of isoelectronic species. Hence $\mathrm{N}^{3-}$ is largest in size.