Classification of Elements and Periodicity in Properties
89634
If the first ionization energy of $\mathrm{H}$ atom is 13.6 $\mathrm{eV}$, then the second ionization energy of $\mathrm{He}$ atom is
1 $27.2 \mathrm{eV}$
2 $40.8 \mathrm{eV}$
3 $54.4 \mathrm{eV}$
4 $108.8 \mathrm{eV}$
Explanation:
We know that, $\mathrm{E}=-13.6 \times \frac{Z^2}{\mathrm{n}^2} \mathrm{eV}$ For second ionisation energy $E=-13.6 \times \frac{2^2}{1^2}$ $\begin{aligned} & =-13.6 \times 4 \\ & =-54.4 \mathrm{eV} \end{aligned}$ Hence, the second ionization energy of $\mathrm{He}$ atom is 54.4 $\mathrm{eV}$.
WB-JEE-2012
Classification of Elements and Periodicity in Properties
89553
The correct order of electron gain enthalpies of Cl, $F, T e$ and $P o$ is:
The correct order of electron gain enthalpies of $\mathrm{Cl}, \mathrm{F}, \mathrm{Te}$ and $\mathrm{Po}$ is - $\mathrm{Po}<\mathrm{Te}<\mathrm{F}<\mathrm{Cl}$ Table of electron gain enthalpies are given below Element Electron Gain Enthalpies $\left(\mathrm{kJmol}^{-1}\right)$ {|ll} | $$ | -174 | |---|---| |$$ | -190 | |$$ | -328 | |$$ | -349 | |
Shift-II
Classification of Elements and Periodicity in Properties
89568
The one electron species having ionization energy of $54.4 \mathrm{eV}$ is:
Classification of Elements and Periodicity in Properties
89554
Elements $X, Y$ and $Z$ have atomic number 19, 37 and 55 respectively. Which of the following statements is true about them ?
1 $\mathrm{Z}$ would have the highest ionization potential
2 $\mathrm{Y}$ would have the highest ionization potential
3 Their ionization potential would increase with increasing atomic number
4 $\mathrm{Y}$ would have an ionization potential between those of $\mathrm{X}$ and $\mathrm{Z}$
Explanation:
Given that, $\mathrm{X}, \mathrm{Y}$ and $\mathrm{Z}$ have atomic number 19,37 and 55 respectively. Then $\mathrm{X}={ }_{19} \mathrm{~K}=$ ionization potential $=4.3407 \mathrm{eV}$. $\mathrm{Y}={ }_{37} \mathrm{Rb}=$ ionization potential $=4.17771 \mathrm{eV}$. $\mathrm{Z}={ }_{55} \mathrm{Cs}=$ ionization potential $=3.8939 \mathrm{eV}$. Therefore, Ionization potential of $\left({ }_{37} \mathrm{Rb}\right)=\frac{\text { I.P.of }(\mathrm{Cs})+\text { I.P.of }(\mathrm{K})}{2}$ By symbolic, $Y=\frac{X+Z}{2}$ So, $\mathrm{Y}$ would have an ionization potential between those of $\mathrm{X}$ and $\mathrm{Z}$.
Shift-II
Classification of Elements and Periodicity in Properties
89572
The first ionisation potenital (in $\mathrm{eV}$ ) of $\mathrm{N}$ and $\mathrm{O}$ atoms are
1 $14.6,13.6$
2 $13.6,14.6$
3 $13.6,13.6$
4 $14.6,14.6$
Explanation:
The electronic configuration of nitrogen and oxygen are- $$ \begin{array}{llll} \mathrm{N}- & 1 \mathrm{~s}^2 & 2 \mathrm{~s}^2 & 2 \mathrm{p}^3 \\ \mathrm{O}- & 1 \mathrm{~s}^2 & 2 \mathrm{~s}^2 & 2 \mathrm{p}^4 \end{array} $$ When oxygen loses one electron it will attain half-filled configuration. So, it can lose an electron easily but nitrogen has already in a half filled configuration and hence losing an electron will require more energy than that of oxygen. Therefore, the first ionization potential of $\mathrm{N}$ and $\mathrm{O}$ atoms are 14.6 and 13.6 respectively.
Classification of Elements and Periodicity in Properties
89634
If the first ionization energy of $\mathrm{H}$ atom is 13.6 $\mathrm{eV}$, then the second ionization energy of $\mathrm{He}$ atom is
1 $27.2 \mathrm{eV}$
2 $40.8 \mathrm{eV}$
3 $54.4 \mathrm{eV}$
4 $108.8 \mathrm{eV}$
Explanation:
We know that, $\mathrm{E}=-13.6 \times \frac{Z^2}{\mathrm{n}^2} \mathrm{eV}$ For second ionisation energy $E=-13.6 \times \frac{2^2}{1^2}$ $\begin{aligned} & =-13.6 \times 4 \\ & =-54.4 \mathrm{eV} \end{aligned}$ Hence, the second ionization energy of $\mathrm{He}$ atom is 54.4 $\mathrm{eV}$.
WB-JEE-2012
Classification of Elements and Periodicity in Properties
89553
The correct order of electron gain enthalpies of Cl, $F, T e$ and $P o$ is:
The correct order of electron gain enthalpies of $\mathrm{Cl}, \mathrm{F}, \mathrm{Te}$ and $\mathrm{Po}$ is - $\mathrm{Po}<\mathrm{Te}<\mathrm{F}<\mathrm{Cl}$ Table of electron gain enthalpies are given below Element Electron Gain Enthalpies $\left(\mathrm{kJmol}^{-1}\right)$ {|ll} | $$ | -174 | |---|---| |$$ | -190 | |$$ | -328 | |$$ | -349 | |
Shift-II
Classification of Elements and Periodicity in Properties
89568
The one electron species having ionization energy of $54.4 \mathrm{eV}$ is:
Classification of Elements and Periodicity in Properties
89554
Elements $X, Y$ and $Z$ have atomic number 19, 37 and 55 respectively. Which of the following statements is true about them ?
1 $\mathrm{Z}$ would have the highest ionization potential
2 $\mathrm{Y}$ would have the highest ionization potential
3 Their ionization potential would increase with increasing atomic number
4 $\mathrm{Y}$ would have an ionization potential between those of $\mathrm{X}$ and $\mathrm{Z}$
Explanation:
Given that, $\mathrm{X}, \mathrm{Y}$ and $\mathrm{Z}$ have atomic number 19,37 and 55 respectively. Then $\mathrm{X}={ }_{19} \mathrm{~K}=$ ionization potential $=4.3407 \mathrm{eV}$. $\mathrm{Y}={ }_{37} \mathrm{Rb}=$ ionization potential $=4.17771 \mathrm{eV}$. $\mathrm{Z}={ }_{55} \mathrm{Cs}=$ ionization potential $=3.8939 \mathrm{eV}$. Therefore, Ionization potential of $\left({ }_{37} \mathrm{Rb}\right)=\frac{\text { I.P.of }(\mathrm{Cs})+\text { I.P.of }(\mathrm{K})}{2}$ By symbolic, $Y=\frac{X+Z}{2}$ So, $\mathrm{Y}$ would have an ionization potential between those of $\mathrm{X}$ and $\mathrm{Z}$.
Shift-II
Classification of Elements and Periodicity in Properties
89572
The first ionisation potenital (in $\mathrm{eV}$ ) of $\mathrm{N}$ and $\mathrm{O}$ atoms are
1 $14.6,13.6$
2 $13.6,14.6$
3 $13.6,13.6$
4 $14.6,14.6$
Explanation:
The electronic configuration of nitrogen and oxygen are- $$ \begin{array}{llll} \mathrm{N}- & 1 \mathrm{~s}^2 & 2 \mathrm{~s}^2 & 2 \mathrm{p}^3 \\ \mathrm{O}- & 1 \mathrm{~s}^2 & 2 \mathrm{~s}^2 & 2 \mathrm{p}^4 \end{array} $$ When oxygen loses one electron it will attain half-filled configuration. So, it can lose an electron easily but nitrogen has already in a half filled configuration and hence losing an electron will require more energy than that of oxygen. Therefore, the first ionization potential of $\mathrm{N}$ and $\mathrm{O}$ atoms are 14.6 and 13.6 respectively.
Classification of Elements and Periodicity in Properties
89634
If the first ionization energy of $\mathrm{H}$ atom is 13.6 $\mathrm{eV}$, then the second ionization energy of $\mathrm{He}$ atom is
1 $27.2 \mathrm{eV}$
2 $40.8 \mathrm{eV}$
3 $54.4 \mathrm{eV}$
4 $108.8 \mathrm{eV}$
Explanation:
We know that, $\mathrm{E}=-13.6 \times \frac{Z^2}{\mathrm{n}^2} \mathrm{eV}$ For second ionisation energy $E=-13.6 \times \frac{2^2}{1^2}$ $\begin{aligned} & =-13.6 \times 4 \\ & =-54.4 \mathrm{eV} \end{aligned}$ Hence, the second ionization energy of $\mathrm{He}$ atom is 54.4 $\mathrm{eV}$.
WB-JEE-2012
Classification of Elements and Periodicity in Properties
89553
The correct order of electron gain enthalpies of Cl, $F, T e$ and $P o$ is:
The correct order of electron gain enthalpies of $\mathrm{Cl}, \mathrm{F}, \mathrm{Te}$ and $\mathrm{Po}$ is - $\mathrm{Po}<\mathrm{Te}<\mathrm{F}<\mathrm{Cl}$ Table of electron gain enthalpies are given below Element Electron Gain Enthalpies $\left(\mathrm{kJmol}^{-1}\right)$ {|ll} | $$ | -174 | |---|---| |$$ | -190 | |$$ | -328 | |$$ | -349 | |
Shift-II
Classification of Elements and Periodicity in Properties
89568
The one electron species having ionization energy of $54.4 \mathrm{eV}$ is:
Classification of Elements and Periodicity in Properties
89554
Elements $X, Y$ and $Z$ have atomic number 19, 37 and 55 respectively. Which of the following statements is true about them ?
1 $\mathrm{Z}$ would have the highest ionization potential
2 $\mathrm{Y}$ would have the highest ionization potential
3 Their ionization potential would increase with increasing atomic number
4 $\mathrm{Y}$ would have an ionization potential between those of $\mathrm{X}$ and $\mathrm{Z}$
Explanation:
Given that, $\mathrm{X}, \mathrm{Y}$ and $\mathrm{Z}$ have atomic number 19,37 and 55 respectively. Then $\mathrm{X}={ }_{19} \mathrm{~K}=$ ionization potential $=4.3407 \mathrm{eV}$. $\mathrm{Y}={ }_{37} \mathrm{Rb}=$ ionization potential $=4.17771 \mathrm{eV}$. $\mathrm{Z}={ }_{55} \mathrm{Cs}=$ ionization potential $=3.8939 \mathrm{eV}$. Therefore, Ionization potential of $\left({ }_{37} \mathrm{Rb}\right)=\frac{\text { I.P.of }(\mathrm{Cs})+\text { I.P.of }(\mathrm{K})}{2}$ By symbolic, $Y=\frac{X+Z}{2}$ So, $\mathrm{Y}$ would have an ionization potential between those of $\mathrm{X}$ and $\mathrm{Z}$.
Shift-II
Classification of Elements and Periodicity in Properties
89572
The first ionisation potenital (in $\mathrm{eV}$ ) of $\mathrm{N}$ and $\mathrm{O}$ atoms are
1 $14.6,13.6$
2 $13.6,14.6$
3 $13.6,13.6$
4 $14.6,14.6$
Explanation:
The electronic configuration of nitrogen and oxygen are- $$ \begin{array}{llll} \mathrm{N}- & 1 \mathrm{~s}^2 & 2 \mathrm{~s}^2 & 2 \mathrm{p}^3 \\ \mathrm{O}- & 1 \mathrm{~s}^2 & 2 \mathrm{~s}^2 & 2 \mathrm{p}^4 \end{array} $$ When oxygen loses one electron it will attain half-filled configuration. So, it can lose an electron easily but nitrogen has already in a half filled configuration and hence losing an electron will require more energy than that of oxygen. Therefore, the first ionization potential of $\mathrm{N}$ and $\mathrm{O}$ atoms are 14.6 and 13.6 respectively.
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Classification of Elements and Periodicity in Properties
89634
If the first ionization energy of $\mathrm{H}$ atom is 13.6 $\mathrm{eV}$, then the second ionization energy of $\mathrm{He}$ atom is
1 $27.2 \mathrm{eV}$
2 $40.8 \mathrm{eV}$
3 $54.4 \mathrm{eV}$
4 $108.8 \mathrm{eV}$
Explanation:
We know that, $\mathrm{E}=-13.6 \times \frac{Z^2}{\mathrm{n}^2} \mathrm{eV}$ For second ionisation energy $E=-13.6 \times \frac{2^2}{1^2}$ $\begin{aligned} & =-13.6 \times 4 \\ & =-54.4 \mathrm{eV} \end{aligned}$ Hence, the second ionization energy of $\mathrm{He}$ atom is 54.4 $\mathrm{eV}$.
WB-JEE-2012
Classification of Elements and Periodicity in Properties
89553
The correct order of electron gain enthalpies of Cl, $F, T e$ and $P o$ is:
The correct order of electron gain enthalpies of $\mathrm{Cl}, \mathrm{F}, \mathrm{Te}$ and $\mathrm{Po}$ is - $\mathrm{Po}<\mathrm{Te}<\mathrm{F}<\mathrm{Cl}$ Table of electron gain enthalpies are given below Element Electron Gain Enthalpies $\left(\mathrm{kJmol}^{-1}\right)$ {|ll} | $$ | -174 | |---|---| |$$ | -190 | |$$ | -328 | |$$ | -349 | |
Shift-II
Classification of Elements and Periodicity in Properties
89568
The one electron species having ionization energy of $54.4 \mathrm{eV}$ is:
Classification of Elements and Periodicity in Properties
89554
Elements $X, Y$ and $Z$ have atomic number 19, 37 and 55 respectively. Which of the following statements is true about them ?
1 $\mathrm{Z}$ would have the highest ionization potential
2 $\mathrm{Y}$ would have the highest ionization potential
3 Their ionization potential would increase with increasing atomic number
4 $\mathrm{Y}$ would have an ionization potential between those of $\mathrm{X}$ and $\mathrm{Z}$
Explanation:
Given that, $\mathrm{X}, \mathrm{Y}$ and $\mathrm{Z}$ have atomic number 19,37 and 55 respectively. Then $\mathrm{X}={ }_{19} \mathrm{~K}=$ ionization potential $=4.3407 \mathrm{eV}$. $\mathrm{Y}={ }_{37} \mathrm{Rb}=$ ionization potential $=4.17771 \mathrm{eV}$. $\mathrm{Z}={ }_{55} \mathrm{Cs}=$ ionization potential $=3.8939 \mathrm{eV}$. Therefore, Ionization potential of $\left({ }_{37} \mathrm{Rb}\right)=\frac{\text { I.P.of }(\mathrm{Cs})+\text { I.P.of }(\mathrm{K})}{2}$ By symbolic, $Y=\frac{X+Z}{2}$ So, $\mathrm{Y}$ would have an ionization potential between those of $\mathrm{X}$ and $\mathrm{Z}$.
Shift-II
Classification of Elements and Periodicity in Properties
89572
The first ionisation potenital (in $\mathrm{eV}$ ) of $\mathrm{N}$ and $\mathrm{O}$ atoms are
1 $14.6,13.6$
2 $13.6,14.6$
3 $13.6,13.6$
4 $14.6,14.6$
Explanation:
The electronic configuration of nitrogen and oxygen are- $$ \begin{array}{llll} \mathrm{N}- & 1 \mathrm{~s}^2 & 2 \mathrm{~s}^2 & 2 \mathrm{p}^3 \\ \mathrm{O}- & 1 \mathrm{~s}^2 & 2 \mathrm{~s}^2 & 2 \mathrm{p}^4 \end{array} $$ When oxygen loses one electron it will attain half-filled configuration. So, it can lose an electron easily but nitrogen has already in a half filled configuration and hence losing an electron will require more energy than that of oxygen. Therefore, the first ionization potential of $\mathrm{N}$ and $\mathrm{O}$ atoms are 14.6 and 13.6 respectively.
Classification of Elements and Periodicity in Properties
89634
If the first ionization energy of $\mathrm{H}$ atom is 13.6 $\mathrm{eV}$, then the second ionization energy of $\mathrm{He}$ atom is
1 $27.2 \mathrm{eV}$
2 $40.8 \mathrm{eV}$
3 $54.4 \mathrm{eV}$
4 $108.8 \mathrm{eV}$
Explanation:
We know that, $\mathrm{E}=-13.6 \times \frac{Z^2}{\mathrm{n}^2} \mathrm{eV}$ For second ionisation energy $E=-13.6 \times \frac{2^2}{1^2}$ $\begin{aligned} & =-13.6 \times 4 \\ & =-54.4 \mathrm{eV} \end{aligned}$ Hence, the second ionization energy of $\mathrm{He}$ atom is 54.4 $\mathrm{eV}$.
WB-JEE-2012
Classification of Elements and Periodicity in Properties
89553
The correct order of electron gain enthalpies of Cl, $F, T e$ and $P o$ is:
The correct order of electron gain enthalpies of $\mathrm{Cl}, \mathrm{F}, \mathrm{Te}$ and $\mathrm{Po}$ is - $\mathrm{Po}<\mathrm{Te}<\mathrm{F}<\mathrm{Cl}$ Table of electron gain enthalpies are given below Element Electron Gain Enthalpies $\left(\mathrm{kJmol}^{-1}\right)$ {|ll} | $$ | -174 | |---|---| |$$ | -190 | |$$ | -328 | |$$ | -349 | |
Shift-II
Classification of Elements and Periodicity in Properties
89568
The one electron species having ionization energy of $54.4 \mathrm{eV}$ is:
Classification of Elements and Periodicity in Properties
89554
Elements $X, Y$ and $Z$ have atomic number 19, 37 and 55 respectively. Which of the following statements is true about them ?
1 $\mathrm{Z}$ would have the highest ionization potential
2 $\mathrm{Y}$ would have the highest ionization potential
3 Their ionization potential would increase with increasing atomic number
4 $\mathrm{Y}$ would have an ionization potential between those of $\mathrm{X}$ and $\mathrm{Z}$
Explanation:
Given that, $\mathrm{X}, \mathrm{Y}$ and $\mathrm{Z}$ have atomic number 19,37 and 55 respectively. Then $\mathrm{X}={ }_{19} \mathrm{~K}=$ ionization potential $=4.3407 \mathrm{eV}$. $\mathrm{Y}={ }_{37} \mathrm{Rb}=$ ionization potential $=4.17771 \mathrm{eV}$. $\mathrm{Z}={ }_{55} \mathrm{Cs}=$ ionization potential $=3.8939 \mathrm{eV}$. Therefore, Ionization potential of $\left({ }_{37} \mathrm{Rb}\right)=\frac{\text { I.P.of }(\mathrm{Cs})+\text { I.P.of }(\mathrm{K})}{2}$ By symbolic, $Y=\frac{X+Z}{2}$ So, $\mathrm{Y}$ would have an ionization potential between those of $\mathrm{X}$ and $\mathrm{Z}$.
Shift-II
Classification of Elements and Periodicity in Properties
89572
The first ionisation potenital (in $\mathrm{eV}$ ) of $\mathrm{N}$ and $\mathrm{O}$ atoms are
1 $14.6,13.6$
2 $13.6,14.6$
3 $13.6,13.6$
4 $14.6,14.6$
Explanation:
The electronic configuration of nitrogen and oxygen are- $$ \begin{array}{llll} \mathrm{N}- & 1 \mathrm{~s}^2 & 2 \mathrm{~s}^2 & 2 \mathrm{p}^3 \\ \mathrm{O}- & 1 \mathrm{~s}^2 & 2 \mathrm{~s}^2 & 2 \mathrm{p}^4 \end{array} $$ When oxygen loses one electron it will attain half-filled configuration. So, it can lose an electron easily but nitrogen has already in a half filled configuration and hence losing an electron will require more energy than that of oxygen. Therefore, the first ionization potential of $\mathrm{N}$ and $\mathrm{O}$ atoms are 14.6 and 13.6 respectively.