Classification of Elements and Periodicity in Properties
89479
What type of structure does $\left(\mathrm{NPCl}_2\right)_4$ Have?
1 Linear
2 Hexagonal
3 Cyclic
4 Polymeric
Explanation:
Phosphonitrilic compounds are known as phosphazenes. Their general formula is $\left(\mathrm{NPCl}_2\right)_{\mathrm{n}}$.
$\mathrm{AMU}-2010$
Classification of Elements and Periodicity in Properties
89480
Which one of the following ions has the highest value of ionic radius?
1 $\mathrm{O}^{2-}$
2 $\mathrm{B}^{3+}$
3 $\mathrm{Li}^{+}$
4 $\mathrm{F}^{-}$
Explanation:
The ionic radii follows the order $\mathrm{O}^{2-}>\mathrm{F}^{-}>\mathrm{Li}^{+}>\mathrm{B}^{3+}$
BITSAT 2009
Classification of Elements and Periodicity in Properties
89481
Atomic radii of $T i, Z r$ and $H f$ vary
1 $\mathrm{Ti}>\mathrm{Zr}>\mathrm{Hf}$
2 $\mathrm{Ti}<\mathrm{Zr}<\mathrm{Hf}$
3 $\mathrm{Ti}<\mathrm{Hf}<\mathrm{Zr}$
4 $\mathrm{Ti}<\mathrm{Zr}=\mathrm{Hf}$
Explanation:
As we move down a group from top to bottom, radii increase but $\mathrm{Zr}$ and $\mathrm{Hf}$ have almost same radius due to poor shielding of f-orbitals. The atomic radii of $4 \mathrm{~d}$ and $5 \mathrm{~d}$ transition element are nearly same. This similarity in size is consequence of lanthanide contraction. Because of this lanthanide contraction the radii of $\mathrm{HF}$ becomes nearly equal to that of $\mathrm{Zr}$. Hence, the order are$\mathrm{Ti}<\mathrm{Zr}=\mathrm{Hf}$
CG PET -2009
Classification of Elements and Periodicity in Properties
89483
Difference between $\mathrm{S}$ and $\mathrm{S}^{2-}$ as $\mathrm{S}^{2-}$ has
1 larger radii and larger size
2 smaller radii and larger size
3 larger radii and smaller size
4 smaller radii and smaller size
Explanation:
The radii of anion is always larger than the atomic radii of its original atom. In an anion as electron are added to the neutral atom the nuclear charge acts more electron so that each electron is held less tightly and electron clouds are expand. Thus, $\mathrm{S}^{2-}$ has larger radii and larger size.
BCECE-2008
Classification of Elements and Periodicity in Properties
89484
$\mathbf{A l}^{3+}$ has a lower ionic radius than $\mathrm{Mg}^{2+}$ because
1 $\mathrm{Mg}$ atom has less number of neutrons than $\mathrm{Al}$
2 $\mathrm{Al}^{3+}$ has a higher nuclear charge than $\mathrm{Mg}^{2+}$
3 their electronegativities are different
4 $\mathrm{Al}$ has a lower ionisation potential than $\mathrm{Mg}$ atom.
Explanation:
$\mathrm{Al}^{3+}$ has a lower ionic radius than $\mathrm{Mg}^{2+}$ because $\mathrm{Al}^{3+}$ has higher nuclear charge than $\mathrm{Mg}^{2+}$.
Classification of Elements and Periodicity in Properties
89479
What type of structure does $\left(\mathrm{NPCl}_2\right)_4$ Have?
1 Linear
2 Hexagonal
3 Cyclic
4 Polymeric
Explanation:
Phosphonitrilic compounds are known as phosphazenes. Their general formula is $\left(\mathrm{NPCl}_2\right)_{\mathrm{n}}$.
$\mathrm{AMU}-2010$
Classification of Elements and Periodicity in Properties
89480
Which one of the following ions has the highest value of ionic radius?
1 $\mathrm{O}^{2-}$
2 $\mathrm{B}^{3+}$
3 $\mathrm{Li}^{+}$
4 $\mathrm{F}^{-}$
Explanation:
The ionic radii follows the order $\mathrm{O}^{2-}>\mathrm{F}^{-}>\mathrm{Li}^{+}>\mathrm{B}^{3+}$
BITSAT 2009
Classification of Elements and Periodicity in Properties
89481
Atomic radii of $T i, Z r$ and $H f$ vary
1 $\mathrm{Ti}>\mathrm{Zr}>\mathrm{Hf}$
2 $\mathrm{Ti}<\mathrm{Zr}<\mathrm{Hf}$
3 $\mathrm{Ti}<\mathrm{Hf}<\mathrm{Zr}$
4 $\mathrm{Ti}<\mathrm{Zr}=\mathrm{Hf}$
Explanation:
As we move down a group from top to bottom, radii increase but $\mathrm{Zr}$ and $\mathrm{Hf}$ have almost same radius due to poor shielding of f-orbitals. The atomic radii of $4 \mathrm{~d}$ and $5 \mathrm{~d}$ transition element are nearly same. This similarity in size is consequence of lanthanide contraction. Because of this lanthanide contraction the radii of $\mathrm{HF}$ becomes nearly equal to that of $\mathrm{Zr}$. Hence, the order are$\mathrm{Ti}<\mathrm{Zr}=\mathrm{Hf}$
CG PET -2009
Classification of Elements and Periodicity in Properties
89483
Difference between $\mathrm{S}$ and $\mathrm{S}^{2-}$ as $\mathrm{S}^{2-}$ has
1 larger radii and larger size
2 smaller radii and larger size
3 larger radii and smaller size
4 smaller radii and smaller size
Explanation:
The radii of anion is always larger than the atomic radii of its original atom. In an anion as electron are added to the neutral atom the nuclear charge acts more electron so that each electron is held less tightly and electron clouds are expand. Thus, $\mathrm{S}^{2-}$ has larger radii and larger size.
BCECE-2008
Classification of Elements and Periodicity in Properties
89484
$\mathbf{A l}^{3+}$ has a lower ionic radius than $\mathrm{Mg}^{2+}$ because
1 $\mathrm{Mg}$ atom has less number of neutrons than $\mathrm{Al}$
2 $\mathrm{Al}^{3+}$ has a higher nuclear charge than $\mathrm{Mg}^{2+}$
3 their electronegativities are different
4 $\mathrm{Al}$ has a lower ionisation potential than $\mathrm{Mg}$ atom.
Explanation:
$\mathrm{Al}^{3+}$ has a lower ionic radius than $\mathrm{Mg}^{2+}$ because $\mathrm{Al}^{3+}$ has higher nuclear charge than $\mathrm{Mg}^{2+}$.
Classification of Elements and Periodicity in Properties
89479
What type of structure does $\left(\mathrm{NPCl}_2\right)_4$ Have?
1 Linear
2 Hexagonal
3 Cyclic
4 Polymeric
Explanation:
Phosphonitrilic compounds are known as phosphazenes. Their general formula is $\left(\mathrm{NPCl}_2\right)_{\mathrm{n}}$.
$\mathrm{AMU}-2010$
Classification of Elements and Periodicity in Properties
89480
Which one of the following ions has the highest value of ionic radius?
1 $\mathrm{O}^{2-}$
2 $\mathrm{B}^{3+}$
3 $\mathrm{Li}^{+}$
4 $\mathrm{F}^{-}$
Explanation:
The ionic radii follows the order $\mathrm{O}^{2-}>\mathrm{F}^{-}>\mathrm{Li}^{+}>\mathrm{B}^{3+}$
BITSAT 2009
Classification of Elements and Periodicity in Properties
89481
Atomic radii of $T i, Z r$ and $H f$ vary
1 $\mathrm{Ti}>\mathrm{Zr}>\mathrm{Hf}$
2 $\mathrm{Ti}<\mathrm{Zr}<\mathrm{Hf}$
3 $\mathrm{Ti}<\mathrm{Hf}<\mathrm{Zr}$
4 $\mathrm{Ti}<\mathrm{Zr}=\mathrm{Hf}$
Explanation:
As we move down a group from top to bottom, radii increase but $\mathrm{Zr}$ and $\mathrm{Hf}$ have almost same radius due to poor shielding of f-orbitals. The atomic radii of $4 \mathrm{~d}$ and $5 \mathrm{~d}$ transition element are nearly same. This similarity in size is consequence of lanthanide contraction. Because of this lanthanide contraction the radii of $\mathrm{HF}$ becomes nearly equal to that of $\mathrm{Zr}$. Hence, the order are$\mathrm{Ti}<\mathrm{Zr}=\mathrm{Hf}$
CG PET -2009
Classification of Elements and Periodicity in Properties
89483
Difference between $\mathrm{S}$ and $\mathrm{S}^{2-}$ as $\mathrm{S}^{2-}$ has
1 larger radii and larger size
2 smaller radii and larger size
3 larger radii and smaller size
4 smaller radii and smaller size
Explanation:
The radii of anion is always larger than the atomic radii of its original atom. In an anion as electron are added to the neutral atom the nuclear charge acts more electron so that each electron is held less tightly and electron clouds are expand. Thus, $\mathrm{S}^{2-}$ has larger radii and larger size.
BCECE-2008
Classification of Elements and Periodicity in Properties
89484
$\mathbf{A l}^{3+}$ has a lower ionic radius than $\mathrm{Mg}^{2+}$ because
1 $\mathrm{Mg}$ atom has less number of neutrons than $\mathrm{Al}$
2 $\mathrm{Al}^{3+}$ has a higher nuclear charge than $\mathrm{Mg}^{2+}$
3 their electronegativities are different
4 $\mathrm{Al}$ has a lower ionisation potential than $\mathrm{Mg}$ atom.
Explanation:
$\mathrm{Al}^{3+}$ has a lower ionic radius than $\mathrm{Mg}^{2+}$ because $\mathrm{Al}^{3+}$ has higher nuclear charge than $\mathrm{Mg}^{2+}$.
Classification of Elements and Periodicity in Properties
89479
What type of structure does $\left(\mathrm{NPCl}_2\right)_4$ Have?
1 Linear
2 Hexagonal
3 Cyclic
4 Polymeric
Explanation:
Phosphonitrilic compounds are known as phosphazenes. Their general formula is $\left(\mathrm{NPCl}_2\right)_{\mathrm{n}}$.
$\mathrm{AMU}-2010$
Classification of Elements and Periodicity in Properties
89480
Which one of the following ions has the highest value of ionic radius?
1 $\mathrm{O}^{2-}$
2 $\mathrm{B}^{3+}$
3 $\mathrm{Li}^{+}$
4 $\mathrm{F}^{-}$
Explanation:
The ionic radii follows the order $\mathrm{O}^{2-}>\mathrm{F}^{-}>\mathrm{Li}^{+}>\mathrm{B}^{3+}$
BITSAT 2009
Classification of Elements and Periodicity in Properties
89481
Atomic radii of $T i, Z r$ and $H f$ vary
1 $\mathrm{Ti}>\mathrm{Zr}>\mathrm{Hf}$
2 $\mathrm{Ti}<\mathrm{Zr}<\mathrm{Hf}$
3 $\mathrm{Ti}<\mathrm{Hf}<\mathrm{Zr}$
4 $\mathrm{Ti}<\mathrm{Zr}=\mathrm{Hf}$
Explanation:
As we move down a group from top to bottom, radii increase but $\mathrm{Zr}$ and $\mathrm{Hf}$ have almost same radius due to poor shielding of f-orbitals. The atomic radii of $4 \mathrm{~d}$ and $5 \mathrm{~d}$ transition element are nearly same. This similarity in size is consequence of lanthanide contraction. Because of this lanthanide contraction the radii of $\mathrm{HF}$ becomes nearly equal to that of $\mathrm{Zr}$. Hence, the order are$\mathrm{Ti}<\mathrm{Zr}=\mathrm{Hf}$
CG PET -2009
Classification of Elements and Periodicity in Properties
89483
Difference between $\mathrm{S}$ and $\mathrm{S}^{2-}$ as $\mathrm{S}^{2-}$ has
1 larger radii and larger size
2 smaller radii and larger size
3 larger radii and smaller size
4 smaller radii and smaller size
Explanation:
The radii of anion is always larger than the atomic radii of its original atom. In an anion as electron are added to the neutral atom the nuclear charge acts more electron so that each electron is held less tightly and electron clouds are expand. Thus, $\mathrm{S}^{2-}$ has larger radii and larger size.
BCECE-2008
Classification of Elements and Periodicity in Properties
89484
$\mathbf{A l}^{3+}$ has a lower ionic radius than $\mathrm{Mg}^{2+}$ because
1 $\mathrm{Mg}$ atom has less number of neutrons than $\mathrm{Al}$
2 $\mathrm{Al}^{3+}$ has a higher nuclear charge than $\mathrm{Mg}^{2+}$
3 their electronegativities are different
4 $\mathrm{Al}$ has a lower ionisation potential than $\mathrm{Mg}$ atom.
Explanation:
$\mathrm{Al}^{3+}$ has a lower ionic radius than $\mathrm{Mg}^{2+}$ because $\mathrm{Al}^{3+}$ has higher nuclear charge than $\mathrm{Mg}^{2+}$.
Classification of Elements and Periodicity in Properties
89479
What type of structure does $\left(\mathrm{NPCl}_2\right)_4$ Have?
1 Linear
2 Hexagonal
3 Cyclic
4 Polymeric
Explanation:
Phosphonitrilic compounds are known as phosphazenes. Their general formula is $\left(\mathrm{NPCl}_2\right)_{\mathrm{n}}$.
$\mathrm{AMU}-2010$
Classification of Elements and Periodicity in Properties
89480
Which one of the following ions has the highest value of ionic radius?
1 $\mathrm{O}^{2-}$
2 $\mathrm{B}^{3+}$
3 $\mathrm{Li}^{+}$
4 $\mathrm{F}^{-}$
Explanation:
The ionic radii follows the order $\mathrm{O}^{2-}>\mathrm{F}^{-}>\mathrm{Li}^{+}>\mathrm{B}^{3+}$
BITSAT 2009
Classification of Elements and Periodicity in Properties
89481
Atomic radii of $T i, Z r$ and $H f$ vary
1 $\mathrm{Ti}>\mathrm{Zr}>\mathrm{Hf}$
2 $\mathrm{Ti}<\mathrm{Zr}<\mathrm{Hf}$
3 $\mathrm{Ti}<\mathrm{Hf}<\mathrm{Zr}$
4 $\mathrm{Ti}<\mathrm{Zr}=\mathrm{Hf}$
Explanation:
As we move down a group from top to bottom, radii increase but $\mathrm{Zr}$ and $\mathrm{Hf}$ have almost same radius due to poor shielding of f-orbitals. The atomic radii of $4 \mathrm{~d}$ and $5 \mathrm{~d}$ transition element are nearly same. This similarity in size is consequence of lanthanide contraction. Because of this lanthanide contraction the radii of $\mathrm{HF}$ becomes nearly equal to that of $\mathrm{Zr}$. Hence, the order are$\mathrm{Ti}<\mathrm{Zr}=\mathrm{Hf}$
CG PET -2009
Classification of Elements and Periodicity in Properties
89483
Difference between $\mathrm{S}$ and $\mathrm{S}^{2-}$ as $\mathrm{S}^{2-}$ has
1 larger radii and larger size
2 smaller radii and larger size
3 larger radii and smaller size
4 smaller radii and smaller size
Explanation:
The radii of anion is always larger than the atomic radii of its original atom. In an anion as electron are added to the neutral atom the nuclear charge acts more electron so that each electron is held less tightly and electron clouds are expand. Thus, $\mathrm{S}^{2-}$ has larger radii and larger size.
BCECE-2008
Classification of Elements and Periodicity in Properties
89484
$\mathbf{A l}^{3+}$ has a lower ionic radius than $\mathrm{Mg}^{2+}$ because
1 $\mathrm{Mg}$ atom has less number of neutrons than $\mathrm{Al}$
2 $\mathrm{Al}^{3+}$ has a higher nuclear charge than $\mathrm{Mg}^{2+}$
3 their electronegativities are different
4 $\mathrm{Al}$ has a lower ionisation potential than $\mathrm{Mg}$ atom.
Explanation:
$\mathrm{Al}^{3+}$ has a lower ionic radius than $\mathrm{Mg}^{2+}$ because $\mathrm{Al}^{3+}$ has higher nuclear charge than $\mathrm{Mg}^{2+}$.
