NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Classification of Elements and Periodicity in Properties
89559
The correct order in which the first ionization potential increases is
1 $\mathrm{Na}, \mathrm{K}, \mathrm{Be}$
2 $\mathrm{K}, \mathrm{Na}, \mathrm{Be}$
3 $\mathrm{K}, \mathrm{Be}, \mathrm{Na}$
4 $\mathrm{Be}, \mathrm{Na}, \mathrm{K}$
Explanation:
The first ionization potential generally increases in a period from left to right and decreases in a group from up to down. Thus, the correct order of first ionization potential is $$ \mathrm{K}<\mathrm{Na}<\mathrm{Be} $$
WB JEE-2015
Classification of Elements and Periodicity in Properties
89564
For the second period elements the correct increasing order of first ionization enthalpy is
1 Li $<$ Be $<$ B $<$ C $<$ O $<$ N $<$ F $<\mathrm{Ne}$
2 Li $<$ Be $<$ B $<$ C $<$ N $<$ O $<$ F $<\mathrm{Ne}$
3 Li $<$ B $<\mathrm{Be}<$ C $<$ O $<\mathrm{N}<$ F $<\mathrm{Ne}$
4 Li $<$ B $<$ Be $<$ C $<$ N $<$ O $<$ F $<\mathrm{Ne}$
Explanation:
When we move from left to right in a period ionisaiton energy increase. Be has full filled $2 \mathrm{~s}$ subshell so the ionisation energy is maximum in comparison to $\mathrm{B}, \mathrm{N}$ is half filled $2 \mathrm{p}$ orbital so ionisation energy is maximum in comparison to oxygen atom. Thus, order of first ionisation enthalpy is- $$ \mathrm{Li}<\mathrm{B}<\mathrm{Be}<\mathrm{C}<\mathrm{O}<\mathrm{N}<\mathrm{F}<\mathrm{Ne} $$
JEE Main-2013
Classification of Elements and Periodicity in Properties
89576
The increasing order of the first ionisation enthalpies of the elements $B, P, S$ and $F$ (lowest first) is
1 F $<$ S $<$ P $<$ B
2 P $<$ S $<$ B $<$ F
3 B $<$ P $<$ S $<$ F
4 B $<$ S $<$ P $<$ F
Explanation:
On moving from left to right in a period ionisaiton energy increase but on moving from top to bottom in group ionisation energy decreases. Phosphorous (P) has half-filled configuration, it has higher $\mathrm{IE}_1$ than of S-atom. Hence, the correct order of first ionisation energy is $\underset{800}{\mathrm{~B}}<\underset{999.4}{\mathrm{~S}}<\underset{1012}{\mathrm{P}}<\underset{1680.8 \mathrm{~kJ} / \text { mole }}{\mathrm{F}}$
UPTU/UPSEE-2017
Classification of Elements and Periodicity in Properties
89579
Which of the order for ionization energy is correct?
1 Be $<$ B $<$ C $<\mathrm{N}<\mathrm{O}$
2 B $<\mathrm{Be}<\mathrm{C}<\mathrm{O}<\mathrm{N}$
3 Be $>$ B $>\mathrm{C}>\mathrm{N}>\mathrm{O}$
4 B $<\mathrm{Be}<\mathrm{N}<\mathrm{C}<\mathrm{O}$
Explanation:
The energy required to remove the most loosely bound electron from an isolated gaseous atom is called the ionization energy. The ionization potential decreases as the size of the atom decrease. Atom will fully or partly filled orbitals have high ionization potential. Hence, the order- $$ \mathrm{B}<\mathrm{Be}<\mathrm{C}<\mathrm{O}<\mathrm{N} $$
Classification of Elements and Periodicity in Properties
89559
The correct order in which the first ionization potential increases is
1 $\mathrm{Na}, \mathrm{K}, \mathrm{Be}$
2 $\mathrm{K}, \mathrm{Na}, \mathrm{Be}$
3 $\mathrm{K}, \mathrm{Be}, \mathrm{Na}$
4 $\mathrm{Be}, \mathrm{Na}, \mathrm{K}$
Explanation:
The first ionization potential generally increases in a period from left to right and decreases in a group from up to down. Thus, the correct order of first ionization potential is $$ \mathrm{K}<\mathrm{Na}<\mathrm{Be} $$
WB JEE-2015
Classification of Elements and Periodicity in Properties
89564
For the second period elements the correct increasing order of first ionization enthalpy is
1 Li $<$ Be $<$ B $<$ C $<$ O $<$ N $<$ F $<\mathrm{Ne}$
2 Li $<$ Be $<$ B $<$ C $<$ N $<$ O $<$ F $<\mathrm{Ne}$
3 Li $<$ B $<\mathrm{Be}<$ C $<$ O $<\mathrm{N}<$ F $<\mathrm{Ne}$
4 Li $<$ B $<$ Be $<$ C $<$ N $<$ O $<$ F $<\mathrm{Ne}$
Explanation:
When we move from left to right in a period ionisaiton energy increase. Be has full filled $2 \mathrm{~s}$ subshell so the ionisation energy is maximum in comparison to $\mathrm{B}, \mathrm{N}$ is half filled $2 \mathrm{p}$ orbital so ionisation energy is maximum in comparison to oxygen atom. Thus, order of first ionisation enthalpy is- $$ \mathrm{Li}<\mathrm{B}<\mathrm{Be}<\mathrm{C}<\mathrm{O}<\mathrm{N}<\mathrm{F}<\mathrm{Ne} $$
JEE Main-2013
Classification of Elements and Periodicity in Properties
89576
The increasing order of the first ionisation enthalpies of the elements $B, P, S$ and $F$ (lowest first) is
1 F $<$ S $<$ P $<$ B
2 P $<$ S $<$ B $<$ F
3 B $<$ P $<$ S $<$ F
4 B $<$ S $<$ P $<$ F
Explanation:
On moving from left to right in a period ionisaiton energy increase but on moving from top to bottom in group ionisation energy decreases. Phosphorous (P) has half-filled configuration, it has higher $\mathrm{IE}_1$ than of S-atom. Hence, the correct order of first ionisation energy is $\underset{800}{\mathrm{~B}}<\underset{999.4}{\mathrm{~S}}<\underset{1012}{\mathrm{P}}<\underset{1680.8 \mathrm{~kJ} / \text { mole }}{\mathrm{F}}$
UPTU/UPSEE-2017
Classification of Elements and Periodicity in Properties
89579
Which of the order for ionization energy is correct?
1 Be $<$ B $<$ C $<\mathrm{N}<\mathrm{O}$
2 B $<\mathrm{Be}<\mathrm{C}<\mathrm{O}<\mathrm{N}$
3 Be $>$ B $>\mathrm{C}>\mathrm{N}>\mathrm{O}$
4 B $<\mathrm{Be}<\mathrm{N}<\mathrm{C}<\mathrm{O}$
Explanation:
The energy required to remove the most loosely bound electron from an isolated gaseous atom is called the ionization energy. The ionization potential decreases as the size of the atom decrease. Atom will fully or partly filled orbitals have high ionization potential. Hence, the order- $$ \mathrm{B}<\mathrm{Be}<\mathrm{C}<\mathrm{O}<\mathrm{N} $$
Classification of Elements and Periodicity in Properties
89559
The correct order in which the first ionization potential increases is
1 $\mathrm{Na}, \mathrm{K}, \mathrm{Be}$
2 $\mathrm{K}, \mathrm{Na}, \mathrm{Be}$
3 $\mathrm{K}, \mathrm{Be}, \mathrm{Na}$
4 $\mathrm{Be}, \mathrm{Na}, \mathrm{K}$
Explanation:
The first ionization potential generally increases in a period from left to right and decreases in a group from up to down. Thus, the correct order of first ionization potential is $$ \mathrm{K}<\mathrm{Na}<\mathrm{Be} $$
WB JEE-2015
Classification of Elements and Periodicity in Properties
89564
For the second period elements the correct increasing order of first ionization enthalpy is
1 Li $<$ Be $<$ B $<$ C $<$ O $<$ N $<$ F $<\mathrm{Ne}$
2 Li $<$ Be $<$ B $<$ C $<$ N $<$ O $<$ F $<\mathrm{Ne}$
3 Li $<$ B $<\mathrm{Be}<$ C $<$ O $<\mathrm{N}<$ F $<\mathrm{Ne}$
4 Li $<$ B $<$ Be $<$ C $<$ N $<$ O $<$ F $<\mathrm{Ne}$
Explanation:
When we move from left to right in a period ionisaiton energy increase. Be has full filled $2 \mathrm{~s}$ subshell so the ionisation energy is maximum in comparison to $\mathrm{B}, \mathrm{N}$ is half filled $2 \mathrm{p}$ orbital so ionisation energy is maximum in comparison to oxygen atom. Thus, order of first ionisation enthalpy is- $$ \mathrm{Li}<\mathrm{B}<\mathrm{Be}<\mathrm{C}<\mathrm{O}<\mathrm{N}<\mathrm{F}<\mathrm{Ne} $$
JEE Main-2013
Classification of Elements and Periodicity in Properties
89576
The increasing order of the first ionisation enthalpies of the elements $B, P, S$ and $F$ (lowest first) is
1 F $<$ S $<$ P $<$ B
2 P $<$ S $<$ B $<$ F
3 B $<$ P $<$ S $<$ F
4 B $<$ S $<$ P $<$ F
Explanation:
On moving from left to right in a period ionisaiton energy increase but on moving from top to bottom in group ionisation energy decreases. Phosphorous (P) has half-filled configuration, it has higher $\mathrm{IE}_1$ than of S-atom. Hence, the correct order of first ionisation energy is $\underset{800}{\mathrm{~B}}<\underset{999.4}{\mathrm{~S}}<\underset{1012}{\mathrm{P}}<\underset{1680.8 \mathrm{~kJ} / \text { mole }}{\mathrm{F}}$
UPTU/UPSEE-2017
Classification of Elements and Periodicity in Properties
89579
Which of the order for ionization energy is correct?
1 Be $<$ B $<$ C $<\mathrm{N}<\mathrm{O}$
2 B $<\mathrm{Be}<\mathrm{C}<\mathrm{O}<\mathrm{N}$
3 Be $>$ B $>\mathrm{C}>\mathrm{N}>\mathrm{O}$
4 B $<\mathrm{Be}<\mathrm{N}<\mathrm{C}<\mathrm{O}$
Explanation:
The energy required to remove the most loosely bound electron from an isolated gaseous atom is called the ionization energy. The ionization potential decreases as the size of the atom decrease. Atom will fully or partly filled orbitals have high ionization potential. Hence, the order- $$ \mathrm{B}<\mathrm{Be}<\mathrm{C}<\mathrm{O}<\mathrm{N} $$
Classification of Elements and Periodicity in Properties
89559
The correct order in which the first ionization potential increases is
1 $\mathrm{Na}, \mathrm{K}, \mathrm{Be}$
2 $\mathrm{K}, \mathrm{Na}, \mathrm{Be}$
3 $\mathrm{K}, \mathrm{Be}, \mathrm{Na}$
4 $\mathrm{Be}, \mathrm{Na}, \mathrm{K}$
Explanation:
The first ionization potential generally increases in a period from left to right and decreases in a group from up to down. Thus, the correct order of first ionization potential is $$ \mathrm{K}<\mathrm{Na}<\mathrm{Be} $$
WB JEE-2015
Classification of Elements and Periodicity in Properties
89564
For the second period elements the correct increasing order of first ionization enthalpy is
1 Li $<$ Be $<$ B $<$ C $<$ O $<$ N $<$ F $<\mathrm{Ne}$
2 Li $<$ Be $<$ B $<$ C $<$ N $<$ O $<$ F $<\mathrm{Ne}$
3 Li $<$ B $<\mathrm{Be}<$ C $<$ O $<\mathrm{N}<$ F $<\mathrm{Ne}$
4 Li $<$ B $<$ Be $<$ C $<$ N $<$ O $<$ F $<\mathrm{Ne}$
Explanation:
When we move from left to right in a period ionisaiton energy increase. Be has full filled $2 \mathrm{~s}$ subshell so the ionisation energy is maximum in comparison to $\mathrm{B}, \mathrm{N}$ is half filled $2 \mathrm{p}$ orbital so ionisation energy is maximum in comparison to oxygen atom. Thus, order of first ionisation enthalpy is- $$ \mathrm{Li}<\mathrm{B}<\mathrm{Be}<\mathrm{C}<\mathrm{O}<\mathrm{N}<\mathrm{F}<\mathrm{Ne} $$
JEE Main-2013
Classification of Elements and Periodicity in Properties
89576
The increasing order of the first ionisation enthalpies of the elements $B, P, S$ and $F$ (lowest first) is
1 F $<$ S $<$ P $<$ B
2 P $<$ S $<$ B $<$ F
3 B $<$ P $<$ S $<$ F
4 B $<$ S $<$ P $<$ F
Explanation:
On moving from left to right in a period ionisaiton energy increase but on moving from top to bottom in group ionisation energy decreases. Phosphorous (P) has half-filled configuration, it has higher $\mathrm{IE}_1$ than of S-atom. Hence, the correct order of first ionisation energy is $\underset{800}{\mathrm{~B}}<\underset{999.4}{\mathrm{~S}}<\underset{1012}{\mathrm{P}}<\underset{1680.8 \mathrm{~kJ} / \text { mole }}{\mathrm{F}}$
UPTU/UPSEE-2017
Classification of Elements and Periodicity in Properties
89579
Which of the order for ionization energy is correct?
1 Be $<$ B $<$ C $<\mathrm{N}<\mathrm{O}$
2 B $<\mathrm{Be}<\mathrm{C}<\mathrm{O}<\mathrm{N}$
3 Be $>$ B $>\mathrm{C}>\mathrm{N}>\mathrm{O}$
4 B $<\mathrm{Be}<\mathrm{N}<\mathrm{C}<\mathrm{O}$
Explanation:
The energy required to remove the most loosely bound electron from an isolated gaseous atom is called the ionization energy. The ionization potential decreases as the size of the atom decrease. Atom will fully or partly filled orbitals have high ionization potential. Hence, the order- $$ \mathrm{B}<\mathrm{Be}<\mathrm{C}<\mathrm{O}<\mathrm{N} $$
