Classification of Elements and Periodicity in Properties
89644
The incorrect order of second ionization energies in the following is
1 $\mathrm{Rb}>\mathrm{K}$
2 $\mathrm{Na}>\mathrm{Mg}$
3 $\mathrm{Cr}>\mathrm{Mn}$
4 $\mathrm{S}>\mathrm{P}$
Explanation:
As we know that ionisation energies decreases down the group due to increase of size of metal and increases left to right in periodic table. $\mathrm{Rb}$ and $\mathrm{K}$ are belongs to the $1^{\text {st }}$ group of the periodic table. The correct order of second ionisation energy will be $\mathrm{K}>\mathrm{Rb}$. The correct order is given as follows - $\mathrm{Na}>\mathrm{Mg}$ (Third period metal) $\mathrm{Cr}>\mathrm{Mn}$ (Fourth period metals) $\mathrm{S}>$ P (Third Period metal)
AP- EAMCET(Medical) -2010
Classification of Elements and Periodicity in Properties
89648
Which one of the following order is correct for the first ionisation energies of the elements?
1 B $<$ Be $<\mathrm{N}<\mathrm{O}$
2 $\mathrm{Be}<\mathrm{B}<\mathrm{N}<\mathrm{O}$
3 $\mathrm{B}<\mathrm{Be}<\mathrm{O}<\mathrm{N}$
4 $\mathrm{B}<\mathrm{O}<\mathrm{Be}<\mathrm{N}$
Explanation:
Ionisation energy of B than that of Be because $\mathrm{B}\left(1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}^2\right)$ electron is to removed from $2 \mathrm{p}$ orbitals while $\mathrm{Be}\left(1 \mathrm{~s}^2, 2 \mathrm{~s}^2\right)$ electron is to be removed from $2 \mathrm{~s}$ is difficult due to filled s-orbitals. The ionization energy of oxygen is lower than $\mathrm{N}$ due to half filled $2 \mathrm{p}$ orbitals in $\mathrm{N}\left(1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}^3\right)$ Therefore, the increasing order are- $$ \mathrm{B}<\mathrm{Be}<\mathrm{O}<\mathrm{N} $$
AP EAMCET (Engg.)-2009
Classification of Elements and Periodicity in Properties
89654
The comparatively high b.pt. of $\mathrm{HF}$ is due to
1 high reactivity of fluorine
2 small size of hydrogen atom
3 formation of hydrogen bonds and consequent association
4 high IE of fluorine
Explanation:
Intermolecular hydrogen bonding is found in $(\mathrm{HF})_{\mathrm{n}}$ due to higher electronegativity of fluorine atoms $\mathrm{H}-\mathrm{F}---\mathrm{H}-\mathrm{F}---\mathrm{H}$ Hydrogen bonding is associated of HF molecule. So, the high boiling point of $\mathrm{HF}$ is due to formation of hydrogen bonds and consequent association.
MHT CET-2007
Classification of Elements and Periodicity in Properties
89656
Ionisation energy decreases down the group due to
1 increase in charge
2 increase in atomic size
3 decrease in atomic size
4 decrease in shilding effect
Explanation:
Ionisation energy is defined as the energy required to remove an electron from the outermost orbit of an isolated gaseous atom in its ground state. $\mathrm{A} \rightarrow \mathrm{A}^{+}+\mathrm{e}^{-} \quad \text { [first IE }$ Ionisation energy decrease in a group as the atomic number increases. It is based on the fact that we move down a group, the size of atom increases, and the outer electrons became further away from the nucleus thus reducing the force of attraction and hence ionization energy decreases with increase in the atomic size.
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Classification of Elements and Periodicity in Properties
89644
The incorrect order of second ionization energies in the following is
1 $\mathrm{Rb}>\mathrm{K}$
2 $\mathrm{Na}>\mathrm{Mg}$
3 $\mathrm{Cr}>\mathrm{Mn}$
4 $\mathrm{S}>\mathrm{P}$
Explanation:
As we know that ionisation energies decreases down the group due to increase of size of metal and increases left to right in periodic table. $\mathrm{Rb}$ and $\mathrm{K}$ are belongs to the $1^{\text {st }}$ group of the periodic table. The correct order of second ionisation energy will be $\mathrm{K}>\mathrm{Rb}$. The correct order is given as follows - $\mathrm{Na}>\mathrm{Mg}$ (Third period metal) $\mathrm{Cr}>\mathrm{Mn}$ (Fourth period metals) $\mathrm{S}>$ P (Third Period metal)
AP- EAMCET(Medical) -2010
Classification of Elements and Periodicity in Properties
89648
Which one of the following order is correct for the first ionisation energies of the elements?
1 B $<$ Be $<\mathrm{N}<\mathrm{O}$
2 $\mathrm{Be}<\mathrm{B}<\mathrm{N}<\mathrm{O}$
3 $\mathrm{B}<\mathrm{Be}<\mathrm{O}<\mathrm{N}$
4 $\mathrm{B}<\mathrm{O}<\mathrm{Be}<\mathrm{N}$
Explanation:
Ionisation energy of B than that of Be because $\mathrm{B}\left(1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}^2\right)$ electron is to removed from $2 \mathrm{p}$ orbitals while $\mathrm{Be}\left(1 \mathrm{~s}^2, 2 \mathrm{~s}^2\right)$ electron is to be removed from $2 \mathrm{~s}$ is difficult due to filled s-orbitals. The ionization energy of oxygen is lower than $\mathrm{N}$ due to half filled $2 \mathrm{p}$ orbitals in $\mathrm{N}\left(1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}^3\right)$ Therefore, the increasing order are- $$ \mathrm{B}<\mathrm{Be}<\mathrm{O}<\mathrm{N} $$
AP EAMCET (Engg.)-2009
Classification of Elements and Periodicity in Properties
89654
The comparatively high b.pt. of $\mathrm{HF}$ is due to
1 high reactivity of fluorine
2 small size of hydrogen atom
3 formation of hydrogen bonds and consequent association
4 high IE of fluorine
Explanation:
Intermolecular hydrogen bonding is found in $(\mathrm{HF})_{\mathrm{n}}$ due to higher electronegativity of fluorine atoms $\mathrm{H}-\mathrm{F}---\mathrm{H}-\mathrm{F}---\mathrm{H}$ Hydrogen bonding is associated of HF molecule. So, the high boiling point of $\mathrm{HF}$ is due to formation of hydrogen bonds and consequent association.
MHT CET-2007
Classification of Elements and Periodicity in Properties
89656
Ionisation energy decreases down the group due to
1 increase in charge
2 increase in atomic size
3 decrease in atomic size
4 decrease in shilding effect
Explanation:
Ionisation energy is defined as the energy required to remove an electron from the outermost orbit of an isolated gaseous atom in its ground state. $\mathrm{A} \rightarrow \mathrm{A}^{+}+\mathrm{e}^{-} \quad \text { [first IE }$ Ionisation energy decrease in a group as the atomic number increases. It is based on the fact that we move down a group, the size of atom increases, and the outer electrons became further away from the nucleus thus reducing the force of attraction and hence ionization energy decreases with increase in the atomic size.
Classification of Elements and Periodicity in Properties
89644
The incorrect order of second ionization energies in the following is
1 $\mathrm{Rb}>\mathrm{K}$
2 $\mathrm{Na}>\mathrm{Mg}$
3 $\mathrm{Cr}>\mathrm{Mn}$
4 $\mathrm{S}>\mathrm{P}$
Explanation:
As we know that ionisation energies decreases down the group due to increase of size of metal and increases left to right in periodic table. $\mathrm{Rb}$ and $\mathrm{K}$ are belongs to the $1^{\text {st }}$ group of the periodic table. The correct order of second ionisation energy will be $\mathrm{K}>\mathrm{Rb}$. The correct order is given as follows - $\mathrm{Na}>\mathrm{Mg}$ (Third period metal) $\mathrm{Cr}>\mathrm{Mn}$ (Fourth period metals) $\mathrm{S}>$ P (Third Period metal)
AP- EAMCET(Medical) -2010
Classification of Elements and Periodicity in Properties
89648
Which one of the following order is correct for the first ionisation energies of the elements?
1 B $<$ Be $<\mathrm{N}<\mathrm{O}$
2 $\mathrm{Be}<\mathrm{B}<\mathrm{N}<\mathrm{O}$
3 $\mathrm{B}<\mathrm{Be}<\mathrm{O}<\mathrm{N}$
4 $\mathrm{B}<\mathrm{O}<\mathrm{Be}<\mathrm{N}$
Explanation:
Ionisation energy of B than that of Be because $\mathrm{B}\left(1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}^2\right)$ electron is to removed from $2 \mathrm{p}$ orbitals while $\mathrm{Be}\left(1 \mathrm{~s}^2, 2 \mathrm{~s}^2\right)$ electron is to be removed from $2 \mathrm{~s}$ is difficult due to filled s-orbitals. The ionization energy of oxygen is lower than $\mathrm{N}$ due to half filled $2 \mathrm{p}$ orbitals in $\mathrm{N}\left(1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}^3\right)$ Therefore, the increasing order are- $$ \mathrm{B}<\mathrm{Be}<\mathrm{O}<\mathrm{N} $$
AP EAMCET (Engg.)-2009
Classification of Elements and Periodicity in Properties
89654
The comparatively high b.pt. of $\mathrm{HF}$ is due to
1 high reactivity of fluorine
2 small size of hydrogen atom
3 formation of hydrogen bonds and consequent association
4 high IE of fluorine
Explanation:
Intermolecular hydrogen bonding is found in $(\mathrm{HF})_{\mathrm{n}}$ due to higher electronegativity of fluorine atoms $\mathrm{H}-\mathrm{F}---\mathrm{H}-\mathrm{F}---\mathrm{H}$ Hydrogen bonding is associated of HF molecule. So, the high boiling point of $\mathrm{HF}$ is due to formation of hydrogen bonds and consequent association.
MHT CET-2007
Classification of Elements and Periodicity in Properties
89656
Ionisation energy decreases down the group due to
1 increase in charge
2 increase in atomic size
3 decrease in atomic size
4 decrease in shilding effect
Explanation:
Ionisation energy is defined as the energy required to remove an electron from the outermost orbit of an isolated gaseous atom in its ground state. $\mathrm{A} \rightarrow \mathrm{A}^{+}+\mathrm{e}^{-} \quad \text { [first IE }$ Ionisation energy decrease in a group as the atomic number increases. It is based on the fact that we move down a group, the size of atom increases, and the outer electrons became further away from the nucleus thus reducing the force of attraction and hence ionization energy decreases with increase in the atomic size.
Classification of Elements and Periodicity in Properties
89644
The incorrect order of second ionization energies in the following is
1 $\mathrm{Rb}>\mathrm{K}$
2 $\mathrm{Na}>\mathrm{Mg}$
3 $\mathrm{Cr}>\mathrm{Mn}$
4 $\mathrm{S}>\mathrm{P}$
Explanation:
As we know that ionisation energies decreases down the group due to increase of size of metal and increases left to right in periodic table. $\mathrm{Rb}$ and $\mathrm{K}$ are belongs to the $1^{\text {st }}$ group of the periodic table. The correct order of second ionisation energy will be $\mathrm{K}>\mathrm{Rb}$. The correct order is given as follows - $\mathrm{Na}>\mathrm{Mg}$ (Third period metal) $\mathrm{Cr}>\mathrm{Mn}$ (Fourth period metals) $\mathrm{S}>$ P (Third Period metal)
AP- EAMCET(Medical) -2010
Classification of Elements and Periodicity in Properties
89648
Which one of the following order is correct for the first ionisation energies of the elements?
1 B $<$ Be $<\mathrm{N}<\mathrm{O}$
2 $\mathrm{Be}<\mathrm{B}<\mathrm{N}<\mathrm{O}$
3 $\mathrm{B}<\mathrm{Be}<\mathrm{O}<\mathrm{N}$
4 $\mathrm{B}<\mathrm{O}<\mathrm{Be}<\mathrm{N}$
Explanation:
Ionisation energy of B than that of Be because $\mathrm{B}\left(1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}^2\right)$ electron is to removed from $2 \mathrm{p}$ orbitals while $\mathrm{Be}\left(1 \mathrm{~s}^2, 2 \mathrm{~s}^2\right)$ electron is to be removed from $2 \mathrm{~s}$ is difficult due to filled s-orbitals. The ionization energy of oxygen is lower than $\mathrm{N}$ due to half filled $2 \mathrm{p}$ orbitals in $\mathrm{N}\left(1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}^3\right)$ Therefore, the increasing order are- $$ \mathrm{B}<\mathrm{Be}<\mathrm{O}<\mathrm{N} $$
AP EAMCET (Engg.)-2009
Classification of Elements and Periodicity in Properties
89654
The comparatively high b.pt. of $\mathrm{HF}$ is due to
1 high reactivity of fluorine
2 small size of hydrogen atom
3 formation of hydrogen bonds and consequent association
4 high IE of fluorine
Explanation:
Intermolecular hydrogen bonding is found in $(\mathrm{HF})_{\mathrm{n}}$ due to higher electronegativity of fluorine atoms $\mathrm{H}-\mathrm{F}---\mathrm{H}-\mathrm{F}---\mathrm{H}$ Hydrogen bonding is associated of HF molecule. So, the high boiling point of $\mathrm{HF}$ is due to formation of hydrogen bonds and consequent association.
MHT CET-2007
Classification of Elements and Periodicity in Properties
89656
Ionisation energy decreases down the group due to
1 increase in charge
2 increase in atomic size
3 decrease in atomic size
4 decrease in shilding effect
Explanation:
Ionisation energy is defined as the energy required to remove an electron from the outermost orbit of an isolated gaseous atom in its ground state. $\mathrm{A} \rightarrow \mathrm{A}^{+}+\mathrm{e}^{-} \quad \text { [first IE }$ Ionisation energy decrease in a group as the atomic number increases. It is based on the fact that we move down a group, the size of atom increases, and the outer electrons became further away from the nucleus thus reducing the force of attraction and hence ionization energy decreases with increase in the atomic size.
