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Geometrical Isomerism in Square Planar Geometry

Dipole moment...

Question

Dipole moment will be zero in the complexes

I) $[N i (C N)_{4}]^{2 -}$

II) $C i s - [P t (N H_{3})_{2} C l_{2}]$

III) Trans - $[P t (N H_{3})_{2} C l_{2}]$

I,II

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I,III

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II,III

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I,II and III

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Solution

The correct option is B I,III
$[N i (C N)_{4}]^{2 -}$ have square planar geometry so due to symmetrical geometry it will show zero dipole moment.
Also, Trans - $[P t (N H_{3})_{2} C l_{2}]$ have zero dipole moment due to symmetry.

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Similar questions

(i) [C o (N H_{3})_{4} (H_{2} O) C l]^{2 +} (i i) [P t (N H_{3})_{2} C l_{2}] (i i i) [C o C l_{2} (e n)_{2}]^{+} (i v) [C r (H_{2} O)_{5} C l] C l_{2} (v) [F e (N H_{3})_{2} (C N)_{4}]^{-} (v i) [C r C l_{2} (o x)_{2}]^{3 -} (v i i) K [C o (H_{2} O)_{2} (o x)_{2}] (v i i i) [C o (N H_{3})_{2} C l_{2} (e n)]^{+} (i x) [C o (e n)_{3}] C l_{3}

cis-trans

I I I

2,

I I I I

3,

I I

1,

I I - I I I I + I I I + I I I I I =

In the following molecules:
$C H_{3} B r, C H_{3} F, C H_{3} C l, C H_{3} I I, I I, I I I, I V$
The correct order of dipole moment will be:

[T i (H_{2} O)_{6}]^{3 +}

[P t (N H_{3})_{2} C l_{2}]

C C l_{4}

1, 2

1, 2

C H_{2} C l_{2}

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