The ratio of the ionization energy of H and Be 3- isA- 1- 16B- 1- 3C- 1- 1D- 1- 9

The correct option is A 1 : 16 Be^3+ is a single electron system. I.E. of Be^3+ = I.E. of H × Z^2 = I.E. of H × 4^2 The ratio, I.E. of H/I.E of Be^3+ = I. E. ...

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Standard XII

Chemistry

Multi-Electron Configurations and Orbital Diagrams

The ratio of ...

Question

The ratio of the ionization energy of H and $B e^{3 +}$ is

1 : 16

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1 : 9

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1 : 3

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1 : 1

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Solution

The correct option is A
1 : 16

$B e^{3 +}$ is a single electron system.
I.E. of $B e^{3 +} =$ I.E. of $H \times Z^{2} =$ I.E. of $H \times 4^{2}$

The ratio, $\frac{I.E. of H}{I . E of B e^{3 +}} = \frac{I. E. of H}{I.E. of H \times 16} = \frac{1}{16}$

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The ratio of the ionization energy of H and Be3+ is

The correct option is A 1 : 16 Be3+ is a single electron system. I.E. of Be3+= I.E. of H×Z2= I.E. of H×42 The ratio, I.E. of HI.E of Be3+=I. E. of HI.E. ofH×16=116

The ratio of the ionization energy of H and $B e^{3 +}$ is

The correct option is A
1 : 16

$B e^{3 +}$ is a single electron system.
I.E. of $B e^{3 +} =$ I.E. of $H \times Z^{2} =$ I.E. of $H \times 4^{2}$

The ratio, $\frac{I.E. of H}{I . E of B e^{3 +}} = \frac{I. E. of H}{I.E. of H \times 16} = \frac{1}{16}$