The quantum numbers of differentiating electron of copper atom in 4s-orbital are; n = 4, l = 0, m = 0 and s = +1/2.

True

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False

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Solution

The correct option is A True
Electronic configuration of Cu is : $[A r] 3 d^{10} 4 s^{1}$
Therefore the quantum no's of the differentiating electron in 4s-orbital are; n = 4, l = 0, m = 0 and s = +1/2

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Similar questions

The quantum numbers for an electron in the outermost orbital of an element are given as $n = 2, l = 0, m = 0, s = - \frac{1}{2}$ . The atom could be:

n

l

(i) n = 4, m_{s} = - \frac{1}{2}

(i i) n = 3, l = 0

(i i i) n = 2, l = 1

(i v) n = 2, l = 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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