When the isoelectronic species the number of electron is increases the effective nuclear charge decrease and size of ion increase. size of ion $\propto \frac{1}{\text { nuclear charge }}$ Hence, the order are $\mathrm{N}^{3-}>\mathrm{O}^{2-}>\mathrm{F}^{-}>\mathrm{Na}^{+}$
2007
Classification of Elements and Periodicity in Properties
89420
$\mathrm{Ce}^{3+}, \mathrm{La}^{3+}, \mathrm{Pm}^{3+}$ and $\mathrm{Yb}^{3+}$ have ionic radii in the increasing order as
Ionic radii decrease on moving along a lanthanide series due to lanthanide contraction. As all ion are in +3 oxidation state. Thus, the ionic radii follow the trend are- $$ \mathrm{Yb}^{3+}<\mathrm{Pm}^{3+}<\mathrm{Ce}^{3+}<\mathrm{La}^{3+} $$
[BITSAT-2005]
Classification of Elements and Periodicity in Properties
89425
Effective nuclear charge is Maximum in case of
1 Lithium
2 Beryllium
3 Oxygen
4 Fluorine
Explanation:
On going from left to right in a period effective nuclear charge is maximum. So, the fluorine (F) has maximum effective nuclear charge. The effective nuclear charge is the net positive charge experienced by valence electrons. $$ \mathrm{Z}_{\text {eff. }}=\mathrm{Z}-\mathrm{S} $$ Where $\mathrm{Z}=$ Atomic number $\mathrm{S}=$ Number of shielding electrons.
AP EAPCET 24.08.2021 Shift-II
Classification of Elements and Periodicity in Properties
89430
Which among the following compound shows the highest lattice energy?
1 $\mathrm{LiF}$
2 $\mathrm{CsF}$
3 $\mathrm{NaF}$
4 $\mathrm{KF}$
Explanation:
According to Born-Lande equation- $\mathrm{U}=\frac{\mathrm{N}_{\mathrm{A}} \mathrm{Mz}^{+} \mathrm{z}^{-} \mathrm{e}^2}{4 \pi \varepsilon_0 \mathrm{r}}\left(1-\frac{1}{\mathrm{n}}\right)$ Lattice energy is inversely proportional to radius of ion. So, smaller the size of ions, larger the magnitude of charges, and having the highest lattice energy. Hence, $\mathrm{LiF}$ has maximum lattice energy.
When the isoelectronic species the number of electron is increases the effective nuclear charge decrease and size of ion increase. size of ion $\propto \frac{1}{\text { nuclear charge }}$ Hence, the order are $\mathrm{N}^{3-}>\mathrm{O}^{2-}>\mathrm{F}^{-}>\mathrm{Na}^{+}$
2007
Classification of Elements and Periodicity in Properties
89420
$\mathrm{Ce}^{3+}, \mathrm{La}^{3+}, \mathrm{Pm}^{3+}$ and $\mathrm{Yb}^{3+}$ have ionic radii in the increasing order as
Ionic radii decrease on moving along a lanthanide series due to lanthanide contraction. As all ion are in +3 oxidation state. Thus, the ionic radii follow the trend are- $$ \mathrm{Yb}^{3+}<\mathrm{Pm}^{3+}<\mathrm{Ce}^{3+}<\mathrm{La}^{3+} $$
[BITSAT-2005]
Classification of Elements and Periodicity in Properties
89425
Effective nuclear charge is Maximum in case of
1 Lithium
2 Beryllium
3 Oxygen
4 Fluorine
Explanation:
On going from left to right in a period effective nuclear charge is maximum. So, the fluorine (F) has maximum effective nuclear charge. The effective nuclear charge is the net positive charge experienced by valence electrons. $$ \mathrm{Z}_{\text {eff. }}=\mathrm{Z}-\mathrm{S} $$ Where $\mathrm{Z}=$ Atomic number $\mathrm{S}=$ Number of shielding electrons.
AP EAPCET 24.08.2021 Shift-II
Classification of Elements and Periodicity in Properties
89430
Which among the following compound shows the highest lattice energy?
1 $\mathrm{LiF}$
2 $\mathrm{CsF}$
3 $\mathrm{NaF}$
4 $\mathrm{KF}$
Explanation:
According to Born-Lande equation- $\mathrm{U}=\frac{\mathrm{N}_{\mathrm{A}} \mathrm{Mz}^{+} \mathrm{z}^{-} \mathrm{e}^2}{4 \pi \varepsilon_0 \mathrm{r}}\left(1-\frac{1}{\mathrm{n}}\right)$ Lattice energy is inversely proportional to radius of ion. So, smaller the size of ions, larger the magnitude of charges, and having the highest lattice energy. Hence, $\mathrm{LiF}$ has maximum lattice energy.
When the isoelectronic species the number of electron is increases the effective nuclear charge decrease and size of ion increase. size of ion $\propto \frac{1}{\text { nuclear charge }}$ Hence, the order are $\mathrm{N}^{3-}>\mathrm{O}^{2-}>\mathrm{F}^{-}>\mathrm{Na}^{+}$
2007
Classification of Elements and Periodicity in Properties
89420
$\mathrm{Ce}^{3+}, \mathrm{La}^{3+}, \mathrm{Pm}^{3+}$ and $\mathrm{Yb}^{3+}$ have ionic radii in the increasing order as
Ionic radii decrease on moving along a lanthanide series due to lanthanide contraction. As all ion are in +3 oxidation state. Thus, the ionic radii follow the trend are- $$ \mathrm{Yb}^{3+}<\mathrm{Pm}^{3+}<\mathrm{Ce}^{3+}<\mathrm{La}^{3+} $$
[BITSAT-2005]
Classification of Elements and Periodicity in Properties
89425
Effective nuclear charge is Maximum in case of
1 Lithium
2 Beryllium
3 Oxygen
4 Fluorine
Explanation:
On going from left to right in a period effective nuclear charge is maximum. So, the fluorine (F) has maximum effective nuclear charge. The effective nuclear charge is the net positive charge experienced by valence electrons. $$ \mathrm{Z}_{\text {eff. }}=\mathrm{Z}-\mathrm{S} $$ Where $\mathrm{Z}=$ Atomic number $\mathrm{S}=$ Number of shielding electrons.
AP EAPCET 24.08.2021 Shift-II
Classification of Elements and Periodicity in Properties
89430
Which among the following compound shows the highest lattice energy?
1 $\mathrm{LiF}$
2 $\mathrm{CsF}$
3 $\mathrm{NaF}$
4 $\mathrm{KF}$
Explanation:
According to Born-Lande equation- $\mathrm{U}=\frac{\mathrm{N}_{\mathrm{A}} \mathrm{Mz}^{+} \mathrm{z}^{-} \mathrm{e}^2}{4 \pi \varepsilon_0 \mathrm{r}}\left(1-\frac{1}{\mathrm{n}}\right)$ Lattice energy is inversely proportional to radius of ion. So, smaller the size of ions, larger the magnitude of charges, and having the highest lattice energy. Hence, $\mathrm{LiF}$ has maximum lattice energy.
When the isoelectronic species the number of electron is increases the effective nuclear charge decrease and size of ion increase. size of ion $\propto \frac{1}{\text { nuclear charge }}$ Hence, the order are $\mathrm{N}^{3-}>\mathrm{O}^{2-}>\mathrm{F}^{-}>\mathrm{Na}^{+}$
2007
Classification of Elements and Periodicity in Properties
89420
$\mathrm{Ce}^{3+}, \mathrm{La}^{3+}, \mathrm{Pm}^{3+}$ and $\mathrm{Yb}^{3+}$ have ionic radii in the increasing order as
Ionic radii decrease on moving along a lanthanide series due to lanthanide contraction. As all ion are in +3 oxidation state. Thus, the ionic radii follow the trend are- $$ \mathrm{Yb}^{3+}<\mathrm{Pm}^{3+}<\mathrm{Ce}^{3+}<\mathrm{La}^{3+} $$
[BITSAT-2005]
Classification of Elements and Periodicity in Properties
89425
Effective nuclear charge is Maximum in case of
1 Lithium
2 Beryllium
3 Oxygen
4 Fluorine
Explanation:
On going from left to right in a period effective nuclear charge is maximum. So, the fluorine (F) has maximum effective nuclear charge. The effective nuclear charge is the net positive charge experienced by valence electrons. $$ \mathrm{Z}_{\text {eff. }}=\mathrm{Z}-\mathrm{S} $$ Where $\mathrm{Z}=$ Atomic number $\mathrm{S}=$ Number of shielding electrons.
AP EAPCET 24.08.2021 Shift-II
Classification of Elements and Periodicity in Properties
89430
Which among the following compound shows the highest lattice energy?
1 $\mathrm{LiF}$
2 $\mathrm{CsF}$
3 $\mathrm{NaF}$
4 $\mathrm{KF}$
Explanation:
According to Born-Lande equation- $\mathrm{U}=\frac{\mathrm{N}_{\mathrm{A}} \mathrm{Mz}^{+} \mathrm{z}^{-} \mathrm{e}^2}{4 \pi \varepsilon_0 \mathrm{r}}\left(1-\frac{1}{\mathrm{n}}\right)$ Lattice energy is inversely proportional to radius of ion. So, smaller the size of ions, larger the magnitude of charges, and having the highest lattice energy. Hence, $\mathrm{LiF}$ has maximum lattice energy.
