Arrange the elements with the following electronic configurations in the increasing order of electron affinity

(i) $1 s^{2} 2 s^{2} 2 p^{3}$ (ii) $1 s^{2} 2 s^{2} 2 p^{4}$ (iii) $1 s^{2} 2 s^{2} 2 p^{5}$ (iv) $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{4}$

i < iv < ii< iii

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iv < iii < ii < i

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i < ii < iv < iii

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ii < iii < i < iv

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Solution

The correct option is C
i < ii < iv < iii

The elements are

(i) N (ii) O (iii) F (iv) S

Now, we can arrange them in the order of electron affinity. Nitrogen has a negative electron affinity - meaning, it would "take" energy to put in one more electron into the already exactly half-filled $2 p^{3}$ orbital. We can consider the electron affinity to be 0 as per the recent convention. Flourine is the most electronegative element and from the given elements, it will have the highest EA value. Sulphur has a higher electron affinity than oxygen. This is an anomaly.

N < O < S < F

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Arrange the following in increasing acidic character
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I II III IV

1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{5}

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

1 s^{2} 2 s^{2} 2 p^{5}

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

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i i . 3 s^{2} 3 p^{5}

i i i . 2 s^{2} 2 p^{4}

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(a) (i) < (ii) < (iii) <(iv) (b) (i) < (iii) < (iv) < (ii)

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