Classification of Elements and Periodicity in Properties
89492
Identify the correct order in which the ionic radius of the following ions increases: (I) $\mathbf{F}^{-}$(II) $\mathrm{Na}^{+}$ (III) $\mathrm{N}^{3-}$
1 III, I, II
2 I, II, III
3 II, III, I
4 II, I, III
Explanation:
The given species are $\mathrm{F}^{-}$(I), $\mathrm{Na}^{+}$(II) and $\mathrm{N}^{3-}$ (III). These are iso-electronic species. As the-ve charge are increases in the element then the size of anion increases. More the -ve charge more will be size of element. Hence the correct increasing order of ionic sizes are- II $<$ I $<$ III
AP-EAMCET (Engg.)-2005
Classification of Elements and Periodicity in Properties
89493
Which of the following order is correct for the size of $\mathrm{Fe}^{3+}, \mathrm{Fe}$ and $\mathrm{Fe}^{2+}$ ?
1 $\mathrm{Fe}<\mathrm{Fe}^{2+}<\mathrm{Fe}^{3+}$
2 $\mathrm{Fe}^{2+}<\mathrm{Fe}^{3+}<\mathrm{Fe}$
3 $\mathrm{Fe}<\mathrm{Fe}^{3+}<\mathrm{Fe}^{2+}$
4 $\mathrm{Fe}^{3+}<\mathrm{Fe}^{2+}<\mathrm{Fe}$
Explanation:
contains more the +ve charge, less the size of metal. The more electrostatic force act in small size of metal. Hence, the correct order of ionic sizes are $-\mathrm{Fe}^{3+}<\mathrm{Fe}^{2+}$ $<\mathrm{Fe}$.
AP-EAMCET (Engg.)-2004
Classification of Elements and Periodicity in Properties
89496
The atomic size of cerium and promethium is quite close, due to:
1 they are in same period in periodic table
2 their electronic configuration is same
3 f-electrons have poor shielding effect
4 nuclear charge is higher on cerium than promethium
Explanation:
For the lanthanide is the contraction of the atomic size with increase in atomic number called as lanthanide contraction while moving along the lanthanide series from $\mathrm{Ce}^{58}$ to $\mathrm{Lu}^{71}$, a regular decrease in the size of the atom/ion and an increase in atomic number. This decrease in size is called as lanthanide contraction. Lanthanide contraction take place due to imperfect shielding of electron which increase by one by one more sub shell so, due to f-electrons have shielding effect.
JCECE - 2004
Classification of Elements and Periodicity in Properties
89500
Which of thr following is thr smallest anion?
1 $\mathrm{O}^{2-}$
2 $\mathrm{S}^{2-}$
3 $\mathrm{Cl}^{-}$
4 $\mathrm{Br}^{-}$
Explanation:
$\mathrm{O}^{2-}$ is belong the second period and $\mathrm{S}^{2-}, \mathrm{Cl}^{-}$ belong the third period and $\mathrm{Br}^{-}$belong to the fourth period on moving from top to bottom in a group atomic radii increase. Hence, the $\mathrm{O}^{2-}$ is the smallest anion.
Classification of Elements and Periodicity in Properties
89492
Identify the correct order in which the ionic radius of the following ions increases: (I) $\mathbf{F}^{-}$(II) $\mathrm{Na}^{+}$ (III) $\mathrm{N}^{3-}$
1 III, I, II
2 I, II, III
3 II, III, I
4 II, I, III
Explanation:
The given species are $\mathrm{F}^{-}$(I), $\mathrm{Na}^{+}$(II) and $\mathrm{N}^{3-}$ (III). These are iso-electronic species. As the-ve charge are increases in the element then the size of anion increases. More the -ve charge more will be size of element. Hence the correct increasing order of ionic sizes are- II $<$ I $<$ III
AP-EAMCET (Engg.)-2005
Classification of Elements and Periodicity in Properties
89493
Which of the following order is correct for the size of $\mathrm{Fe}^{3+}, \mathrm{Fe}$ and $\mathrm{Fe}^{2+}$ ?
1 $\mathrm{Fe}<\mathrm{Fe}^{2+}<\mathrm{Fe}^{3+}$
2 $\mathrm{Fe}^{2+}<\mathrm{Fe}^{3+}<\mathrm{Fe}$
3 $\mathrm{Fe}<\mathrm{Fe}^{3+}<\mathrm{Fe}^{2+}$
4 $\mathrm{Fe}^{3+}<\mathrm{Fe}^{2+}<\mathrm{Fe}$
Explanation:
contains more the +ve charge, less the size of metal. The more electrostatic force act in small size of metal. Hence, the correct order of ionic sizes are $-\mathrm{Fe}^{3+}<\mathrm{Fe}^{2+}$ $<\mathrm{Fe}$.
AP-EAMCET (Engg.)-2004
Classification of Elements and Periodicity in Properties
89496
The atomic size of cerium and promethium is quite close, due to:
1 they are in same period in periodic table
2 their electronic configuration is same
3 f-electrons have poor shielding effect
4 nuclear charge is higher on cerium than promethium
Explanation:
For the lanthanide is the contraction of the atomic size with increase in atomic number called as lanthanide contraction while moving along the lanthanide series from $\mathrm{Ce}^{58}$ to $\mathrm{Lu}^{71}$, a regular decrease in the size of the atom/ion and an increase in atomic number. This decrease in size is called as lanthanide contraction. Lanthanide contraction take place due to imperfect shielding of electron which increase by one by one more sub shell so, due to f-electrons have shielding effect.
JCECE - 2004
Classification of Elements and Periodicity in Properties
89500
Which of thr following is thr smallest anion?
1 $\mathrm{O}^{2-}$
2 $\mathrm{S}^{2-}$
3 $\mathrm{Cl}^{-}$
4 $\mathrm{Br}^{-}$
Explanation:
$\mathrm{O}^{2-}$ is belong the second period and $\mathrm{S}^{2-}, \mathrm{Cl}^{-}$ belong the third period and $\mathrm{Br}^{-}$belong to the fourth period on moving from top to bottom in a group atomic radii increase. Hence, the $\mathrm{O}^{2-}$ is the smallest anion.
Classification of Elements and Periodicity in Properties
89492
Identify the correct order in which the ionic radius of the following ions increases: (I) $\mathbf{F}^{-}$(II) $\mathrm{Na}^{+}$ (III) $\mathrm{N}^{3-}$
1 III, I, II
2 I, II, III
3 II, III, I
4 II, I, III
Explanation:
The given species are $\mathrm{F}^{-}$(I), $\mathrm{Na}^{+}$(II) and $\mathrm{N}^{3-}$ (III). These are iso-electronic species. As the-ve charge are increases in the element then the size of anion increases. More the -ve charge more will be size of element. Hence the correct increasing order of ionic sizes are- II $<$ I $<$ III
AP-EAMCET (Engg.)-2005
Classification of Elements and Periodicity in Properties
89493
Which of the following order is correct for the size of $\mathrm{Fe}^{3+}, \mathrm{Fe}$ and $\mathrm{Fe}^{2+}$ ?
1 $\mathrm{Fe}<\mathrm{Fe}^{2+}<\mathrm{Fe}^{3+}$
2 $\mathrm{Fe}^{2+}<\mathrm{Fe}^{3+}<\mathrm{Fe}$
3 $\mathrm{Fe}<\mathrm{Fe}^{3+}<\mathrm{Fe}^{2+}$
4 $\mathrm{Fe}^{3+}<\mathrm{Fe}^{2+}<\mathrm{Fe}$
Explanation:
contains more the +ve charge, less the size of metal. The more electrostatic force act in small size of metal. Hence, the correct order of ionic sizes are $-\mathrm{Fe}^{3+}<\mathrm{Fe}^{2+}$ $<\mathrm{Fe}$.
AP-EAMCET (Engg.)-2004
Classification of Elements and Periodicity in Properties
89496
The atomic size of cerium and promethium is quite close, due to:
1 they are in same period in periodic table
2 their electronic configuration is same
3 f-electrons have poor shielding effect
4 nuclear charge is higher on cerium than promethium
Explanation:
For the lanthanide is the contraction of the atomic size with increase in atomic number called as lanthanide contraction while moving along the lanthanide series from $\mathrm{Ce}^{58}$ to $\mathrm{Lu}^{71}$, a regular decrease in the size of the atom/ion and an increase in atomic number. This decrease in size is called as lanthanide contraction. Lanthanide contraction take place due to imperfect shielding of electron which increase by one by one more sub shell so, due to f-electrons have shielding effect.
JCECE - 2004
Classification of Elements and Periodicity in Properties
89500
Which of thr following is thr smallest anion?
1 $\mathrm{O}^{2-}$
2 $\mathrm{S}^{2-}$
3 $\mathrm{Cl}^{-}$
4 $\mathrm{Br}^{-}$
Explanation:
$\mathrm{O}^{2-}$ is belong the second period and $\mathrm{S}^{2-}, \mathrm{Cl}^{-}$ belong the third period and $\mathrm{Br}^{-}$belong to the fourth period on moving from top to bottom in a group atomic radii increase. Hence, the $\mathrm{O}^{2-}$ is the smallest anion.
Classification of Elements and Periodicity in Properties
89492
Identify the correct order in which the ionic radius of the following ions increases: (I) $\mathbf{F}^{-}$(II) $\mathrm{Na}^{+}$ (III) $\mathrm{N}^{3-}$
1 III, I, II
2 I, II, III
3 II, III, I
4 II, I, III
Explanation:
The given species are $\mathrm{F}^{-}$(I), $\mathrm{Na}^{+}$(II) and $\mathrm{N}^{3-}$ (III). These are iso-electronic species. As the-ve charge are increases in the element then the size of anion increases. More the -ve charge more will be size of element. Hence the correct increasing order of ionic sizes are- II $<$ I $<$ III
AP-EAMCET (Engg.)-2005
Classification of Elements and Periodicity in Properties
89493
Which of the following order is correct for the size of $\mathrm{Fe}^{3+}, \mathrm{Fe}$ and $\mathrm{Fe}^{2+}$ ?
1 $\mathrm{Fe}<\mathrm{Fe}^{2+}<\mathrm{Fe}^{3+}$
2 $\mathrm{Fe}^{2+}<\mathrm{Fe}^{3+}<\mathrm{Fe}$
3 $\mathrm{Fe}<\mathrm{Fe}^{3+}<\mathrm{Fe}^{2+}$
4 $\mathrm{Fe}^{3+}<\mathrm{Fe}^{2+}<\mathrm{Fe}$
Explanation:
contains more the +ve charge, less the size of metal. The more electrostatic force act in small size of metal. Hence, the correct order of ionic sizes are $-\mathrm{Fe}^{3+}<\mathrm{Fe}^{2+}$ $<\mathrm{Fe}$.
AP-EAMCET (Engg.)-2004
Classification of Elements and Periodicity in Properties
89496
The atomic size of cerium and promethium is quite close, due to:
1 they are in same period in periodic table
2 their electronic configuration is same
3 f-electrons have poor shielding effect
4 nuclear charge is higher on cerium than promethium
Explanation:
For the lanthanide is the contraction of the atomic size with increase in atomic number called as lanthanide contraction while moving along the lanthanide series from $\mathrm{Ce}^{58}$ to $\mathrm{Lu}^{71}$, a regular decrease in the size of the atom/ion and an increase in atomic number. This decrease in size is called as lanthanide contraction. Lanthanide contraction take place due to imperfect shielding of electron which increase by one by one more sub shell so, due to f-electrons have shielding effect.
JCECE - 2004
Classification of Elements and Periodicity in Properties
89500
Which of thr following is thr smallest anion?
1 $\mathrm{O}^{2-}$
2 $\mathrm{S}^{2-}$
3 $\mathrm{Cl}^{-}$
4 $\mathrm{Br}^{-}$
Explanation:
$\mathrm{O}^{2-}$ is belong the second period and $\mathrm{S}^{2-}, \mathrm{Cl}^{-}$ belong the third period and $\mathrm{Br}^{-}$belong to the fourth period on moving from top to bottom in a group atomic radii increase. Hence, the $\mathrm{O}^{2-}$ is the smallest anion.
