An electron in an atom has the following set of four quantum numbers to which orbital it belong to:
$n$ $l$ $m_{l}$ $m_{s}$
2 0 0 $+ \frac{1}{2}$

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Solution

The correct option is D 2s
Now for n = 1, $l = 0$ we have 1s orbital.
for n = 2, $l = 0$ we have 2s orbital.
l = 1 we have 2p
Thus $n = 2, l = 0, m_{l} = 0, m_{s} = + \frac{1}{2}$
is the set of quantum number for 2s orbitals.

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Similar questions

n = 2, l = 0

Which one of the following represents an acceptable possible set of quantum numbers (in the order n, l, m_land m_sfor an electron in an atom?

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 = 4,

m_{l}

m_{s}

n = 4, | m_{l} | = 1

m_{s} = - \frac{1}{2}

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