238605
The atomic weight of an element is 39. The number of neutrons in its nucleus is one more than the number of protons. The number of protons, neutrons and electrons-respectively in its atoms would be
1 $19,20,19$
2 $19,19,20$
3 $20,19,19$
4 $20,19,20$
Explanation:
Atomic weight of an element $=39$ unit $\therefore$ Atomic weight $=$ no of proton + no of neutron Hence According to question. no. of neutron $=$ no. of proton +1 So, $39=$ no. of neutron + no. of proton $39=$ no. of proton $+1+$ no. of proton $2 \times$ no. of proton $=38$ no. of proton $=19$ Also, no. of electron $=$ no. of proton They, no. of electron $=19$ And no. of neutron $=19+1=20$ Hence, no. of proton $=19$ no. of neutron $=20$ no. of electron $=19$
CG PET - 2004
Structure of Atom
238606
In the following reaction ${ }_3 \mathrm{Li}^6+? \rightarrow{ }_2 \mathrm{He}^4+1 \mathrm{H}^3$ the missing particle is
1 Electron
2 Neutron
3 Proton
4 Deuteron
Explanation:
In the given reaction ${ }_3 \mathrm{Li}^6+? \rightarrow{ }_2 \mathrm{He}^4+1 \mathrm{H}^3$ Mass number and atomic number on LHS and RHS will be equal. Let us assume the missing particle be ${ }_x A^y$ Atomic number $\begin{aligned} & 3+\mathrm{x}=2+1 \\ & \mathrm{x}=0 \end{aligned}$ Mass number, $6+y=4+3$ $y=1$ It represents neutron, ${ }_3 \mathrm{Li}^6+{ }_0 \mathrm{n}^1 \rightarrow{ }_2 \mathrm{He}^4+1 \mathrm{H}^3$
Key idea = Isoelectronic species contains same number of electron. The species with its atomic number and number of electrons are as follows: {|l|c|c|} | {${c} { Species } | |---| | { (ions) }$} | ${c} { Atomic } | | { number (z) }$ | ${c} { Number of } | | { electrons }$ | |$^{3-}$ | 7 | $7+3=10$ | |$^{2-}$ | 8 | $8+2=10$ | |$^{-}$ | 9 | $9+1=10$ | |$^{+}$ | 11 | $11-1=10$ | |$^{+}$ | 3 | $3-1=2$ | |$^{2+}$ | 12 | $12-2=10$ | |Isoelectronic set of ions are $^{3-}, ^{2-}, ^{-}, ^{+}$and $^{2+}$. | |
(JEE Main 2019
Structure of Atom
238615
Which one of the following constitutes a group of the isoelectronic species?
Number of electrons in each species are given below $\begin{array}{ll} \mathrm{N}_2=14 & \mathrm{CN}^{-}=14 \\ \mathrm{O}_2^{-}=17 & \mathrm{C}_2^{2-}=14 \\ \mathrm{NO}^{+}=14 & \mathrm{O}_2^{-}=18 \\ \mathrm{CO}=14 & \mathrm{No}=15 \end{array}$ It is quite evident from the above that $\mathrm{NO}^{+}, \mathrm{C}_2^{2-}, \mathrm{CN}^{-}$, $\mathrm{N}_2$ and $\mathrm{Co}$ are isoelectronic in natures.
238605
The atomic weight of an element is 39. The number of neutrons in its nucleus is one more than the number of protons. The number of protons, neutrons and electrons-respectively in its atoms would be
1 $19,20,19$
2 $19,19,20$
3 $20,19,19$
4 $20,19,20$
Explanation:
Atomic weight of an element $=39$ unit $\therefore$ Atomic weight $=$ no of proton + no of neutron Hence According to question. no. of neutron $=$ no. of proton +1 So, $39=$ no. of neutron + no. of proton $39=$ no. of proton $+1+$ no. of proton $2 \times$ no. of proton $=38$ no. of proton $=19$ Also, no. of electron $=$ no. of proton They, no. of electron $=19$ And no. of neutron $=19+1=20$ Hence, no. of proton $=19$ no. of neutron $=20$ no. of electron $=19$
CG PET - 2004
Structure of Atom
238606
In the following reaction ${ }_3 \mathrm{Li}^6+? \rightarrow{ }_2 \mathrm{He}^4+1 \mathrm{H}^3$ the missing particle is
1 Electron
2 Neutron
3 Proton
4 Deuteron
Explanation:
In the given reaction ${ }_3 \mathrm{Li}^6+? \rightarrow{ }_2 \mathrm{He}^4+1 \mathrm{H}^3$ Mass number and atomic number on LHS and RHS will be equal. Let us assume the missing particle be ${ }_x A^y$ Atomic number $\begin{aligned} & 3+\mathrm{x}=2+1 \\ & \mathrm{x}=0 \end{aligned}$ Mass number, $6+y=4+3$ $y=1$ It represents neutron, ${ }_3 \mathrm{Li}^6+{ }_0 \mathrm{n}^1 \rightarrow{ }_2 \mathrm{He}^4+1 \mathrm{H}^3$
Key idea = Isoelectronic species contains same number of electron. The species with its atomic number and number of electrons are as follows: {|l|c|c|} | {${c} { Species } | |---| | { (ions) }$} | ${c} { Atomic } | | { number (z) }$ | ${c} { Number of } | | { electrons }$ | |$^{3-}$ | 7 | $7+3=10$ | |$^{2-}$ | 8 | $8+2=10$ | |$^{-}$ | 9 | $9+1=10$ | |$^{+}$ | 11 | $11-1=10$ | |$^{+}$ | 3 | $3-1=2$ | |$^{2+}$ | 12 | $12-2=10$ | |Isoelectronic set of ions are $^{3-}, ^{2-}, ^{-}, ^{+}$and $^{2+}$. | |
(JEE Main 2019
Structure of Atom
238615
Which one of the following constitutes a group of the isoelectronic species?
Number of electrons in each species are given below $\begin{array}{ll} \mathrm{N}_2=14 & \mathrm{CN}^{-}=14 \\ \mathrm{O}_2^{-}=17 & \mathrm{C}_2^{2-}=14 \\ \mathrm{NO}^{+}=14 & \mathrm{O}_2^{-}=18 \\ \mathrm{CO}=14 & \mathrm{No}=15 \end{array}$ It is quite evident from the above that $\mathrm{NO}^{+}, \mathrm{C}_2^{2-}, \mathrm{CN}^{-}$, $\mathrm{N}_2$ and $\mathrm{Co}$ are isoelectronic in natures.
238605
The atomic weight of an element is 39. The number of neutrons in its nucleus is one more than the number of protons. The number of protons, neutrons and electrons-respectively in its atoms would be
1 $19,20,19$
2 $19,19,20$
3 $20,19,19$
4 $20,19,20$
Explanation:
Atomic weight of an element $=39$ unit $\therefore$ Atomic weight $=$ no of proton + no of neutron Hence According to question. no. of neutron $=$ no. of proton +1 So, $39=$ no. of neutron + no. of proton $39=$ no. of proton $+1+$ no. of proton $2 \times$ no. of proton $=38$ no. of proton $=19$ Also, no. of electron $=$ no. of proton They, no. of electron $=19$ And no. of neutron $=19+1=20$ Hence, no. of proton $=19$ no. of neutron $=20$ no. of electron $=19$
CG PET - 2004
Structure of Atom
238606
In the following reaction ${ }_3 \mathrm{Li}^6+? \rightarrow{ }_2 \mathrm{He}^4+1 \mathrm{H}^3$ the missing particle is
1 Electron
2 Neutron
3 Proton
4 Deuteron
Explanation:
In the given reaction ${ }_3 \mathrm{Li}^6+? \rightarrow{ }_2 \mathrm{He}^4+1 \mathrm{H}^3$ Mass number and atomic number on LHS and RHS will be equal. Let us assume the missing particle be ${ }_x A^y$ Atomic number $\begin{aligned} & 3+\mathrm{x}=2+1 \\ & \mathrm{x}=0 \end{aligned}$ Mass number, $6+y=4+3$ $y=1$ It represents neutron, ${ }_3 \mathrm{Li}^6+{ }_0 \mathrm{n}^1 \rightarrow{ }_2 \mathrm{He}^4+1 \mathrm{H}^3$
Key idea = Isoelectronic species contains same number of electron. The species with its atomic number and number of electrons are as follows: {|l|c|c|} | {${c} { Species } | |---| | { (ions) }$} | ${c} { Atomic } | | { number (z) }$ | ${c} { Number of } | | { electrons }$ | |$^{3-}$ | 7 | $7+3=10$ | |$^{2-}$ | 8 | $8+2=10$ | |$^{-}$ | 9 | $9+1=10$ | |$^{+}$ | 11 | $11-1=10$ | |$^{+}$ | 3 | $3-1=2$ | |$^{2+}$ | 12 | $12-2=10$ | |Isoelectronic set of ions are $^{3-}, ^{2-}, ^{-}, ^{+}$and $^{2+}$. | |
(JEE Main 2019
Structure of Atom
238615
Which one of the following constitutes a group of the isoelectronic species?
Number of electrons in each species are given below $\begin{array}{ll} \mathrm{N}_2=14 & \mathrm{CN}^{-}=14 \\ \mathrm{O}_2^{-}=17 & \mathrm{C}_2^{2-}=14 \\ \mathrm{NO}^{+}=14 & \mathrm{O}_2^{-}=18 \\ \mathrm{CO}=14 & \mathrm{No}=15 \end{array}$ It is quite evident from the above that $\mathrm{NO}^{+}, \mathrm{C}_2^{2-}, \mathrm{CN}^{-}$, $\mathrm{N}_2$ and $\mathrm{Co}$ are isoelectronic in natures.
238605
The atomic weight of an element is 39. The number of neutrons in its nucleus is one more than the number of protons. The number of protons, neutrons and electrons-respectively in its atoms would be
1 $19,20,19$
2 $19,19,20$
3 $20,19,19$
4 $20,19,20$
Explanation:
Atomic weight of an element $=39$ unit $\therefore$ Atomic weight $=$ no of proton + no of neutron Hence According to question. no. of neutron $=$ no. of proton +1 So, $39=$ no. of neutron + no. of proton $39=$ no. of proton $+1+$ no. of proton $2 \times$ no. of proton $=38$ no. of proton $=19$ Also, no. of electron $=$ no. of proton They, no. of electron $=19$ And no. of neutron $=19+1=20$ Hence, no. of proton $=19$ no. of neutron $=20$ no. of electron $=19$
CG PET - 2004
Structure of Atom
238606
In the following reaction ${ }_3 \mathrm{Li}^6+? \rightarrow{ }_2 \mathrm{He}^4+1 \mathrm{H}^3$ the missing particle is
1 Electron
2 Neutron
3 Proton
4 Deuteron
Explanation:
In the given reaction ${ }_3 \mathrm{Li}^6+? \rightarrow{ }_2 \mathrm{He}^4+1 \mathrm{H}^3$ Mass number and atomic number on LHS and RHS will be equal. Let us assume the missing particle be ${ }_x A^y$ Atomic number $\begin{aligned} & 3+\mathrm{x}=2+1 \\ & \mathrm{x}=0 \end{aligned}$ Mass number, $6+y=4+3$ $y=1$ It represents neutron, ${ }_3 \mathrm{Li}^6+{ }_0 \mathrm{n}^1 \rightarrow{ }_2 \mathrm{He}^4+1 \mathrm{H}^3$
Key idea = Isoelectronic species contains same number of electron. The species with its atomic number and number of electrons are as follows: {|l|c|c|} | {${c} { Species } | |---| | { (ions) }$} | ${c} { Atomic } | | { number (z) }$ | ${c} { Number of } | | { electrons }$ | |$^{3-}$ | 7 | $7+3=10$ | |$^{2-}$ | 8 | $8+2=10$ | |$^{-}$ | 9 | $9+1=10$ | |$^{+}$ | 11 | $11-1=10$ | |$^{+}$ | 3 | $3-1=2$ | |$^{2+}$ | 12 | $12-2=10$ | |Isoelectronic set of ions are $^{3-}, ^{2-}, ^{-}, ^{+}$and $^{2+}$. | |
(JEE Main 2019
Structure of Atom
238615
Which one of the following constitutes a group of the isoelectronic species?
Number of electrons in each species are given below $\begin{array}{ll} \mathrm{N}_2=14 & \mathrm{CN}^{-}=14 \\ \mathrm{O}_2^{-}=17 & \mathrm{C}_2^{2-}=14 \\ \mathrm{NO}^{+}=14 & \mathrm{O}_2^{-}=18 \\ \mathrm{CO}=14 & \mathrm{No}=15 \end{array}$ It is quite evident from the above that $\mathrm{NO}^{+}, \mathrm{C}_2^{2-}, \mathrm{CN}^{-}$, $\mathrm{N}_2$ and $\mathrm{Co}$ are isoelectronic in natures.
