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Standard XII

Chemistry

Electronic Configuration

The correct o...

Question

The correct order for decreasing atomic radius of the five atoms $H, B r, I, H e$ and $S$ is :

I > B r > S > H e > H

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I > S > B r > H > H e

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I > S > B r > H e > H

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B r > I > S > H e > H

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Solution

The correct option is A $I > B r > S > H e > H$
I, Br belong to halogen family of 4 and 5th period so they are larger in atomic size, S belongs to 3rd period and group 16, He is a noble gas with two electrons and smaller in size but H with one electron is of least atomic size.

Hence, $I > B r > S > H e > H$ is the correct order.

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Similar questions

Q. The correct order for decreasing atomic radius of the five atoms H, Br, I, He and S is?

Q. The decreasing order of rate of

S_{N} 2

is :

(I)

B r - C H_{2} - C H_{2} - B r

(II)

N = C - C H_{2} - B r

(III)

C H_{3} - C | B r H - C H_{3}

(IV)

C H_{3} - C H_{2} - B r

Consider the reactions

(i) $(C H_{3})_{2} C H - C H_{2} B r C_{2} H_{5} O H - ---- \to (C H_{3})_{2} C H - C H_{2} O C_{2} H_{5} + H B r$

(ii) $(C H_{3})_{2} C H - C H_{2} B r C_{2} H_{S} O^{-} - ---- \to (C H_{3})_{2} C H C H_{2} O C_{2} H_{5} + B r^{-}$

The mechanism of reactions (i) & (ii) are respectively

Q. Following is the list of four halides. Select correct sequence of decreasing order of reactivity for

S_{N} 1

reaction using the codes given below.

C_{6} H_{5} - C | H C H_{3} - B r

II)

C_{6} H_{5} - C H_{2} - B r

III)

C_{6} H_{5} - C | H C H_{3} - I

IV)

C_{6} H_{5} - C H_{2} - I

Consider the reactions:
$2 S_{2} O_{3}^{2 -} (a q) + I_{2} (s) \to S_{4} O_{6}^{2 -} (a q) + 2 I^{-} (a q)$

$S_{2} O_{3}^{2 -} (a q) + 2 B r_{2} (l) + 5 H_{2} O (l) \to 2 S O_{4}^{2 -} (a q) + 4 B r^{-} (a q) + 10 H^{+} (a q)$
Why does the same reductant, thiosulphate react differently with iodine and bromine?

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The correct order for decreasing atomic radius of the five atoms H, Br, I, He and S is :

The correct order for decreasing atomic radius of the five atoms $H, B r, I, H e$ and $S$ is :