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Ionization Enthalpy

Which of foll...

Question

Which of following orbital has maximum energy?

n = 4, l = 0, m = 0

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n = 3, l = 2, m = 0

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n = 3, l = 0, m = 1

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n = 3, l = 1, m = 0

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Solution

The correct option is C $n = 3, l = 2, m = 0$
According to Aufbau principle, or $n + l$ rule, $n = 3, l = 2$ has maximum energy because $3 + 2 = 5$ which is maximum

Suggest Corrections

Similar questions

n = 4, l = 0, m_{l} = 0, m_{s} = + \frac{1}{2}

n = 3, l = 1, m_{l} = 1, m_{s} = - \frac{1}{2}

n = 3, l = 2, m_{l} = 0, m_{s} = + \frac{1}{2}

n = 3, l = 0, m_{l} = 0, m_{s} = - \frac{1}{2}

n = 1, l = 0, m = 0

n = 2, l = 0, m = 0

n = 3, l = 1, m = 1

n = 3, l = 2, m = 1

n = 3, l = 2, m = 0

n = 4, l = 0, m = 0, s = + \frac{1}{2}

n = 3, l = 1, m = 1, s = - \frac{1}{2}

n = 3, l = 2, m = 0, s = + \frac{1}{2}

n = 3, l = 0, m = 0, s = - \frac{1}{2}

n = 4, l = 0, m = 0, s = + \frac{1}{2}

n = 3, l = 1, m = 1, s = - \frac{1}{2}

n = 3, l = 2, m = 0, s = + \frac{1}{2}

n = 3, l = 0, m = 0, s = - \frac{1}{2}

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