Classification of Elements and Periodicity in Properties
89392
p-orbital in a given shell can accommodate upto-
1 four electrons
2 two electrons with parallel spin
3 six electrons
4 two electrons with opposite spin
Explanation:
Any orbital of any subshell can accommodate a maximum of two electrons with opposite spins. ' $\mathrm{P}$ ' subshell have 3 orbitals (Px, Py, Pz). Each orbital can accommodate two electrons with opposite spin. Therefore, P-orbitals in a given shell can accommodate upto six electrons.
BCECE-2005
Classification of Elements and Periodicity in Properties
89343
Identify the least stable ion amongst the following :
Classification of Elements and Periodicity in Properties
89345
Atomic numbers of elements $X, Y$ and $Z$ are 50 , 78 and 60 respectively. these elements are placed in modern long from of Periodic Table respectively in
1 p-block, d-block and f-block
2 p-block, d- block and s- block
3 s-block, p-block and d-block
4 s-block, d-block and f- block
Explanation:
Atomic number $x=50$ is tin(Sn). It is an P-block element. The electronic configuration of tin (Sn) is $[\mathrm{Kr}] 4 \mathrm{~d}^{10} 5 \mathrm{~s}^2 \mathrm{sp}^2$ Atomic number $\mathrm{y}=78$ is platinum (Pt). It is an d-block element. The electronic configuration of Platinum (Pt) is $[\mathrm{Xe}] 4 \mathrm{f}^{14} 5 \mathrm{~d}^9 6 \mathrm{~s}^1$ Atomic number $\mathrm{z}=60$ is neodymium (Nd). It is an f-block element. The electronic configuration of $\mathrm{Nd}$ is $[\mathrm{Xe}] 4 \mathrm{f}^4 6 \mathrm{~s}^2$. Therefore, In modern periodic table $\mathrm{x}, \mathrm{y}$ and $\mathrm{z}$ are $\mathrm{p}$ block, d-block and f-block element.
CG PET- 2011
Classification of Elements and Periodicity in Properties
89366
An element has $[\mathrm{Ar}] 3 \mathrm{~d}^1$ configuration in its +2 oxidation state. Its position in the periodic table is
1 period-3, group - 3
2 period-3, group - 7
3 period-4, group - 3
4 period-3, group -9
Explanation:
Given that an element with electronic configuration $[\mathrm{Ar}] 3 \mathrm{~d}^1$ in its +2 oxidation state. $\therefore$ The atomic number of element $$ =\mathrm{Ar}+1+2=21 $$ Thus, the element is Scandium. It belong to $4^{\text {th }}$ period and group -3
AP-EAMCET - 2016
Classification of Elements and Periodicity in Properties
89367
Identify the most acidic oxide among the following oxides based on their reaction.
1 $\mathrm{SO}_3$
2 $\mathrm{P}_4 \mathrm{O}_{10}$
3 $\mathrm{Cl}_2 \mathrm{O}_7$
4 $\mathrm{N}_2 \mathrm{O}_5$
Explanation:
On going from left to right in a period, the ionisation energy of element generally increases. Hence, their oxides basic to amphoteric to acidic character. Oxide of more electronegative atom is more acidic in nature. The oxidation state of oxygen is -2 . Let $\mathrm{x}$ be the oxidation state of non metal. For, $\mathrm{SO}_3$ molecule, $\quad \mathrm{x}+(-2) 3=0$ or $\mathrm{x}=+6$ For $\mathrm{P}_4 \mathrm{O}_{10}$ molecule, $4(\mathrm{x})+10(-2)=0, \mathrm{x}=+5$ For, $\mathrm{Cl}_2 \mathrm{O}_7$ molecule, $2(\mathrm{x})+7(-2)=0$ or $\mathrm{x}=+7$ For, $\mathrm{N}_2 \mathrm{O}_5$ molecule, $\quad 2(\mathrm{x})+5(-2)=0$ or $\mathrm{x}=+5$
Classification of Elements and Periodicity in Properties
89392
p-orbital in a given shell can accommodate upto-
1 four electrons
2 two electrons with parallel spin
3 six electrons
4 two electrons with opposite spin
Explanation:
Any orbital of any subshell can accommodate a maximum of two electrons with opposite spins. ' $\mathrm{P}$ ' subshell have 3 orbitals (Px, Py, Pz). Each orbital can accommodate two electrons with opposite spin. Therefore, P-orbitals in a given shell can accommodate upto six electrons.
BCECE-2005
Classification of Elements and Periodicity in Properties
89343
Identify the least stable ion amongst the following :
Classification of Elements and Periodicity in Properties
89345
Atomic numbers of elements $X, Y$ and $Z$ are 50 , 78 and 60 respectively. these elements are placed in modern long from of Periodic Table respectively in
1 p-block, d-block and f-block
2 p-block, d- block and s- block
3 s-block, p-block and d-block
4 s-block, d-block and f- block
Explanation:
Atomic number $x=50$ is tin(Sn). It is an P-block element. The electronic configuration of tin (Sn) is $[\mathrm{Kr}] 4 \mathrm{~d}^{10} 5 \mathrm{~s}^2 \mathrm{sp}^2$ Atomic number $\mathrm{y}=78$ is platinum (Pt). It is an d-block element. The electronic configuration of Platinum (Pt) is $[\mathrm{Xe}] 4 \mathrm{f}^{14} 5 \mathrm{~d}^9 6 \mathrm{~s}^1$ Atomic number $\mathrm{z}=60$ is neodymium (Nd). It is an f-block element. The electronic configuration of $\mathrm{Nd}$ is $[\mathrm{Xe}] 4 \mathrm{f}^4 6 \mathrm{~s}^2$. Therefore, In modern periodic table $\mathrm{x}, \mathrm{y}$ and $\mathrm{z}$ are $\mathrm{p}$ block, d-block and f-block element.
CG PET- 2011
Classification of Elements and Periodicity in Properties
89366
An element has $[\mathrm{Ar}] 3 \mathrm{~d}^1$ configuration in its +2 oxidation state. Its position in the periodic table is
1 period-3, group - 3
2 period-3, group - 7
3 period-4, group - 3
4 period-3, group -9
Explanation:
Given that an element with electronic configuration $[\mathrm{Ar}] 3 \mathrm{~d}^1$ in its +2 oxidation state. $\therefore$ The atomic number of element $$ =\mathrm{Ar}+1+2=21 $$ Thus, the element is Scandium. It belong to $4^{\text {th }}$ period and group -3
AP-EAMCET - 2016
Classification of Elements and Periodicity in Properties
89367
Identify the most acidic oxide among the following oxides based on their reaction.
1 $\mathrm{SO}_3$
2 $\mathrm{P}_4 \mathrm{O}_{10}$
3 $\mathrm{Cl}_2 \mathrm{O}_7$
4 $\mathrm{N}_2 \mathrm{O}_5$
Explanation:
On going from left to right in a period, the ionisation energy of element generally increases. Hence, their oxides basic to amphoteric to acidic character. Oxide of more electronegative atom is more acidic in nature. The oxidation state of oxygen is -2 . Let $\mathrm{x}$ be the oxidation state of non metal. For, $\mathrm{SO}_3$ molecule, $\quad \mathrm{x}+(-2) 3=0$ or $\mathrm{x}=+6$ For $\mathrm{P}_4 \mathrm{O}_{10}$ molecule, $4(\mathrm{x})+10(-2)=0, \mathrm{x}=+5$ For, $\mathrm{Cl}_2 \mathrm{O}_7$ molecule, $2(\mathrm{x})+7(-2)=0$ or $\mathrm{x}=+7$ For, $\mathrm{N}_2 \mathrm{O}_5$ molecule, $\quad 2(\mathrm{x})+5(-2)=0$ or $\mathrm{x}=+5$
Classification of Elements and Periodicity in Properties
89392
p-orbital in a given shell can accommodate upto-
1 four electrons
2 two electrons with parallel spin
3 six electrons
4 two electrons with opposite spin
Explanation:
Any orbital of any subshell can accommodate a maximum of two electrons with opposite spins. ' $\mathrm{P}$ ' subshell have 3 orbitals (Px, Py, Pz). Each orbital can accommodate two electrons with opposite spin. Therefore, P-orbitals in a given shell can accommodate upto six electrons.
BCECE-2005
Classification of Elements and Periodicity in Properties
89343
Identify the least stable ion amongst the following :
Classification of Elements and Periodicity in Properties
89345
Atomic numbers of elements $X, Y$ and $Z$ are 50 , 78 and 60 respectively. these elements are placed in modern long from of Periodic Table respectively in
1 p-block, d-block and f-block
2 p-block, d- block and s- block
3 s-block, p-block and d-block
4 s-block, d-block and f- block
Explanation:
Atomic number $x=50$ is tin(Sn). It is an P-block element. The electronic configuration of tin (Sn) is $[\mathrm{Kr}] 4 \mathrm{~d}^{10} 5 \mathrm{~s}^2 \mathrm{sp}^2$ Atomic number $\mathrm{y}=78$ is platinum (Pt). It is an d-block element. The electronic configuration of Platinum (Pt) is $[\mathrm{Xe}] 4 \mathrm{f}^{14} 5 \mathrm{~d}^9 6 \mathrm{~s}^1$ Atomic number $\mathrm{z}=60$ is neodymium (Nd). It is an f-block element. The electronic configuration of $\mathrm{Nd}$ is $[\mathrm{Xe}] 4 \mathrm{f}^4 6 \mathrm{~s}^2$. Therefore, In modern periodic table $\mathrm{x}, \mathrm{y}$ and $\mathrm{z}$ are $\mathrm{p}$ block, d-block and f-block element.
CG PET- 2011
Classification of Elements and Periodicity in Properties
89366
An element has $[\mathrm{Ar}] 3 \mathrm{~d}^1$ configuration in its +2 oxidation state. Its position in the periodic table is
1 period-3, group - 3
2 period-3, group - 7
3 period-4, group - 3
4 period-3, group -9
Explanation:
Given that an element with electronic configuration $[\mathrm{Ar}] 3 \mathrm{~d}^1$ in its +2 oxidation state. $\therefore$ The atomic number of element $$ =\mathrm{Ar}+1+2=21 $$ Thus, the element is Scandium. It belong to $4^{\text {th }}$ period and group -3
AP-EAMCET - 2016
Classification of Elements and Periodicity in Properties
89367
Identify the most acidic oxide among the following oxides based on their reaction.
1 $\mathrm{SO}_3$
2 $\mathrm{P}_4 \mathrm{O}_{10}$
3 $\mathrm{Cl}_2 \mathrm{O}_7$
4 $\mathrm{N}_2 \mathrm{O}_5$
Explanation:
On going from left to right in a period, the ionisation energy of element generally increases. Hence, their oxides basic to amphoteric to acidic character. Oxide of more electronegative atom is more acidic in nature. The oxidation state of oxygen is -2 . Let $\mathrm{x}$ be the oxidation state of non metal. For, $\mathrm{SO}_3$ molecule, $\quad \mathrm{x}+(-2) 3=0$ or $\mathrm{x}=+6$ For $\mathrm{P}_4 \mathrm{O}_{10}$ molecule, $4(\mathrm{x})+10(-2)=0, \mathrm{x}=+5$ For, $\mathrm{Cl}_2 \mathrm{O}_7$ molecule, $2(\mathrm{x})+7(-2)=0$ or $\mathrm{x}=+7$ For, $\mathrm{N}_2 \mathrm{O}_5$ molecule, $\quad 2(\mathrm{x})+5(-2)=0$ or $\mathrm{x}=+5$
Classification of Elements and Periodicity in Properties
89392
p-orbital in a given shell can accommodate upto-
1 four electrons
2 two electrons with parallel spin
3 six electrons
4 two electrons with opposite spin
Explanation:
Any orbital of any subshell can accommodate a maximum of two electrons with opposite spins. ' $\mathrm{P}$ ' subshell have 3 orbitals (Px, Py, Pz). Each orbital can accommodate two electrons with opposite spin. Therefore, P-orbitals in a given shell can accommodate upto six electrons.
BCECE-2005
Classification of Elements and Periodicity in Properties
89343
Identify the least stable ion amongst the following :
Classification of Elements and Periodicity in Properties
89345
Atomic numbers of elements $X, Y$ and $Z$ are 50 , 78 and 60 respectively. these elements are placed in modern long from of Periodic Table respectively in
1 p-block, d-block and f-block
2 p-block, d- block and s- block
3 s-block, p-block and d-block
4 s-block, d-block and f- block
Explanation:
Atomic number $x=50$ is tin(Sn). It is an P-block element. The electronic configuration of tin (Sn) is $[\mathrm{Kr}] 4 \mathrm{~d}^{10} 5 \mathrm{~s}^2 \mathrm{sp}^2$ Atomic number $\mathrm{y}=78$ is platinum (Pt). It is an d-block element. The electronic configuration of Platinum (Pt) is $[\mathrm{Xe}] 4 \mathrm{f}^{14} 5 \mathrm{~d}^9 6 \mathrm{~s}^1$ Atomic number $\mathrm{z}=60$ is neodymium (Nd). It is an f-block element. The electronic configuration of $\mathrm{Nd}$ is $[\mathrm{Xe}] 4 \mathrm{f}^4 6 \mathrm{~s}^2$. Therefore, In modern periodic table $\mathrm{x}, \mathrm{y}$ and $\mathrm{z}$ are $\mathrm{p}$ block, d-block and f-block element.
CG PET- 2011
Classification of Elements and Periodicity in Properties
89366
An element has $[\mathrm{Ar}] 3 \mathrm{~d}^1$ configuration in its +2 oxidation state. Its position in the periodic table is
1 period-3, group - 3
2 period-3, group - 7
3 period-4, group - 3
4 period-3, group -9
Explanation:
Given that an element with electronic configuration $[\mathrm{Ar}] 3 \mathrm{~d}^1$ in its +2 oxidation state. $\therefore$ The atomic number of element $$ =\mathrm{Ar}+1+2=21 $$ Thus, the element is Scandium. It belong to $4^{\text {th }}$ period and group -3
AP-EAMCET - 2016
Classification of Elements and Periodicity in Properties
89367
Identify the most acidic oxide among the following oxides based on their reaction.
1 $\mathrm{SO}_3$
2 $\mathrm{P}_4 \mathrm{O}_{10}$
3 $\mathrm{Cl}_2 \mathrm{O}_7$
4 $\mathrm{N}_2 \mathrm{O}_5$
Explanation:
On going from left to right in a period, the ionisation energy of element generally increases. Hence, their oxides basic to amphoteric to acidic character. Oxide of more electronegative atom is more acidic in nature. The oxidation state of oxygen is -2 . Let $\mathrm{x}$ be the oxidation state of non metal. For, $\mathrm{SO}_3$ molecule, $\quad \mathrm{x}+(-2) 3=0$ or $\mathrm{x}=+6$ For $\mathrm{P}_4 \mathrm{O}_{10}$ molecule, $4(\mathrm{x})+10(-2)=0, \mathrm{x}=+5$ For, $\mathrm{Cl}_2 \mathrm{O}_7$ molecule, $2(\mathrm{x})+7(-2)=0$ or $\mathrm{x}=+7$ For, $\mathrm{N}_2 \mathrm{O}_5$ molecule, $\quad 2(\mathrm{x})+5(-2)=0$ or $\mathrm{x}=+5$
Classification of Elements and Periodicity in Properties
89392
p-orbital in a given shell can accommodate upto-
1 four electrons
2 two electrons with parallel spin
3 six electrons
4 two electrons with opposite spin
Explanation:
Any orbital of any subshell can accommodate a maximum of two electrons with opposite spins. ' $\mathrm{P}$ ' subshell have 3 orbitals (Px, Py, Pz). Each orbital can accommodate two electrons with opposite spin. Therefore, P-orbitals in a given shell can accommodate upto six electrons.
BCECE-2005
Classification of Elements and Periodicity in Properties
89343
Identify the least stable ion amongst the following :
Classification of Elements and Periodicity in Properties
89345
Atomic numbers of elements $X, Y$ and $Z$ are 50 , 78 and 60 respectively. these elements are placed in modern long from of Periodic Table respectively in
1 p-block, d-block and f-block
2 p-block, d- block and s- block
3 s-block, p-block and d-block
4 s-block, d-block and f- block
Explanation:
Atomic number $x=50$ is tin(Sn). It is an P-block element. The electronic configuration of tin (Sn) is $[\mathrm{Kr}] 4 \mathrm{~d}^{10} 5 \mathrm{~s}^2 \mathrm{sp}^2$ Atomic number $\mathrm{y}=78$ is platinum (Pt). It is an d-block element. The electronic configuration of Platinum (Pt) is $[\mathrm{Xe}] 4 \mathrm{f}^{14} 5 \mathrm{~d}^9 6 \mathrm{~s}^1$ Atomic number $\mathrm{z}=60$ is neodymium (Nd). It is an f-block element. The electronic configuration of $\mathrm{Nd}$ is $[\mathrm{Xe}] 4 \mathrm{f}^4 6 \mathrm{~s}^2$. Therefore, In modern periodic table $\mathrm{x}, \mathrm{y}$ and $\mathrm{z}$ are $\mathrm{p}$ block, d-block and f-block element.
CG PET- 2011
Classification of Elements and Periodicity in Properties
89366
An element has $[\mathrm{Ar}] 3 \mathrm{~d}^1$ configuration in its +2 oxidation state. Its position in the periodic table is
1 period-3, group - 3
2 period-3, group - 7
3 period-4, group - 3
4 period-3, group -9
Explanation:
Given that an element with electronic configuration $[\mathrm{Ar}] 3 \mathrm{~d}^1$ in its +2 oxidation state. $\therefore$ The atomic number of element $$ =\mathrm{Ar}+1+2=21 $$ Thus, the element is Scandium. It belong to $4^{\text {th }}$ period and group -3
AP-EAMCET - 2016
Classification of Elements and Periodicity in Properties
89367
Identify the most acidic oxide among the following oxides based on their reaction.
1 $\mathrm{SO}_3$
2 $\mathrm{P}_4 \mathrm{O}_{10}$
3 $\mathrm{Cl}_2 \mathrm{O}_7$
4 $\mathrm{N}_2 \mathrm{O}_5$
Explanation:
On going from left to right in a period, the ionisation energy of element generally increases. Hence, their oxides basic to amphoteric to acidic character. Oxide of more electronegative atom is more acidic in nature. The oxidation state of oxygen is -2 . Let $\mathrm{x}$ be the oxidation state of non metal. For, $\mathrm{SO}_3$ molecule, $\quad \mathrm{x}+(-2) 3=0$ or $\mathrm{x}=+6$ For $\mathrm{P}_4 \mathrm{O}_{10}$ molecule, $4(\mathrm{x})+10(-2)=0, \mathrm{x}=+5$ For, $\mathrm{Cl}_2 \mathrm{O}_7$ molecule, $2(\mathrm{x})+7(-2)=0$ or $\mathrm{x}=+7$ For, $\mathrm{N}_2 \mathrm{O}_5$ molecule, $\quad 2(\mathrm{x})+5(-2)=0$ or $\mathrm{x}=+5$
