If the azimuthal quantum number could have the value of $n$ also then:

S c (Z = 21)

would have configuration as

1 s^{2} 1 p^{6} 2 s^{2} 2 p^{6} 2 d^{3} 3 s^{2}

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F e (Z = 26)

would have two unpaired electrons

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F e^{2 +}

and

F e^{3 +}

would be paramagnetic and coloured

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C u^{+}

would be coloured and paramagnetic

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Solution

The correct options are
A $S c (Z = 21)$ would have configuration as $1 s^{2} 1 p^{6} 2 s^{2} 2 p^{6} 2 d^{3} 3 s^{2}$
B $F e (Z = 26)$ would have two unpaired electrons
C $F e^{2 +}$ and $F e^{3 +}$ would be paramagnetic and coloured
If $n = n$ then $l = 0, 1, 2, 3, \dots \dots n$

Energy order, $1 s < 1 p < 2 s < 2 p < 3 s < 2 d$

$a) (Z = 21), 1 s^{2} 1 p^{6} 2 s^{2} 2 p^{6} 3 s^{2} 2 d^{3} : Correct$

$b) (F e = 26), 1 s^{2} 1 p^{6} 2 s^{2} 2 p^{6} 3 s^{2} 2 d^{8}$

Two unpaired electrons : correct

$c) F e^{2 +} = 1 s^{2} 1 p^{6} 2 s^{2} 2 p^{6} 2 d^{8}$
(Valence shell $3 s$ electrons are taken out)
$F e^{3 +} = 1 s^{2} 1 p^{6} 2 s^{2} 2 p^{6} 2 d^{7}$
Both have unpaired electrons in the $d -$ orbital
Thus, coloured: Correct

d) $C u = 1 s^{2} 1 p^{6} 2 s^{2} 2 p^{6} 3 s^{2} 2 d^{10} 3 p^{1}$
$C u^{+} = 1 s^{2} 1 p^{6} 2 s^{2} 2 p^{6} 3 s^{2} 2 d^{10}$
Colourless: Incorrect

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Similar questions

(n + l)

If the value of azimuthal quantum number is 3, the possible values of magnetic quantum numbers would be:

F e^{+ 3} (z = 26)

F e^{+ 3}, M n^{+}

C r

Electronic configuration of four neutral atoms are given below.

Which of the following elements would not be paramagnetic

m_{s}

- \frac{1}{2}

0

+ \frac{1}{2}

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