Question

calcubte the effective atomic number or copper in (cu(NH3)4)2+ (z of cu =29)explain linkage is amorism in coordination compound with suitable exam


​

Answer 1

Answer:

$+ 79 {x56. \binom{\pi\pi}{?} \div \times 6 + 15 {52123 \cos( \tan( \cot( \beta \sec(ee \gamma ) ) ) ) }^{2} }^{?} \sec(\pi \infty)$

Explanation:

$\cos( \tan( \cot(y {y222 \sq\pi \gamma \beta \gamma e \beta \beta \alpha \beta e log( \beta \gamma 014 {?}^{2} ) r{ | > ln( log( \beta \gamma ) ) | } \times \frac{?}{?} }^{2} ) ) )$

Answer 2

Answer:

In [Cu(NH

3

)

4

]

+2

, its Cu

+2

ion

We know that Cu has atomic number=29

∴ No. of electrons in Cu

+2

=29−2=27

Number of monodentate ligands=4

∴ EAN=Number of electrons in the central metal ion+2(No. of monodentate ligands)

=27+2(4)

=27+8

∴ EAN=35

Ionisation isomers are the isomers with different possible ions formation in the solution but the ions must balance the change of the comples.

For example:

[Co(H

2

O)

5

Br]Cl→[Co(H

2

O)

5

Br]

+

+Cl

−

[Co(H

2

O)Cl]Br→[Co(H

2

O)

5

Cl]

+

+Br

−

.