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Byju's Answer

Standard XII

Chemistry

Shape and Geometry Using VSEPR

Iso-electroni...

Question

Iso-electronic structures among the following structures are:
(I) $C H_{3}^{+}$
(II) $H_{3} O^{+}$
(III) $N H_{3}$
(IV) $C H_{3}^{-}$

I

and

I I I

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I I I

and

I V

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I

and

I I

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I I, I I I

and

I V

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Solution

The correct option is D $I I, I I I$ and $I V$
species no of electron
(I) $C H_{3}^{+}$ $6 + 3 - 1 = 8$
(II) $H_{3} O^{+}$ $8 + 3 - 1 = 10$
(III) $N H_{3}$ $7 + 3 = 10$
(IV) $C H_{3}^{-}$ $6 + 3 + 1 = 10$

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Similar questions

Q. Pick out the isoelectronic structures from the following.
(I)

C H_{3}^{+}

(II)

H_{3} O^{+}

(III)

N H_{3}

(IV)

C H_{3}^{-}

Arrange the following: $C H_{3} C H_{2} C H_{2} C l (I), C H_{3} C H_{2} C H C l C H_{3} (I I) {(C H_{3})}_{2} C H C H_{2} C l (I I I) a n d {(C H_{3})}_{3} C - C l (I V)$ in order of decreasing tendency towards $S N^{2}$ reaction

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(I),

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(II),

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(III) and

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Q. Arrange the following compounds in increasing order of their boiling points.
I.

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

II.

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III.

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IV.

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Iso-electronic structures among the following structures are:(I)CH+3(II)H3O+(III)NH3(IV)CH−3

The correct option is D II,III and IVspecies no of electron(I)CH+3 6+3−1=8 (II)H3O+ 8+3−1=10(III)NH3 7+3=10(IV)CH−3 6+3+1=10

Iso-electronic structures among the following structures are:
(I) $C H_{3}^{+}$
(II) $H_{3} O^{+}$
(III) $N H_{3}$
(IV) $C H_{3}^{-}$

The correct option is D $I I, I I I$ and $I V$
species no of electron
(I) $C H_{3}^{+}$ $6 + 3 - 1 = 8$
(II) $H_{3} O^{+}$ $8 + 3 - 1 = 10$
(III) $N H_{3}$ $7 + 3 = 10$
(IV) $C H_{3}^{-}$ $6 + 3 + 1 = 10$