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Alecsey [184]
3 years ago
9

Draw the structures of all possible isomers for the following complexes. Indicate which isomers are enantiomer pairs. (a) Bromoc

hloro(ethylenediamine)platinum(II) (square-planar) (b) Tetra-amminedichloroiron(III) ion (octahedral) (c) Amminechlorobis(ethylenediamine)iron(III) ion (octahedral)
Chemistry
1 answer:
andriy [413]3 years ago
7 0

Answer:

structures B and C are: enantiomer pair.

Explanation:

Attached below are the structures of all possible Isomers

structures B and C  under solution 3 are enantiomer pair. <em>this is because they posses the same connectivity but exhibit opposite three dimensional shapes .</em>

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Might be solar but I may be wrong
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Answer:

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Explanation:

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olga55 [171]

Answer:

_{93}^{232}\text{Np} + _{-1}^{0}\text{e} \longrightarrow _{92}^{232}\text{U}

Explanation:

The unbalanced nuclear equation is

_{93}^{232}\text{Np} + _{1}^{0}\text{e} \longrightarrow ?

It is convenient to replace the question by an atomic symbol, _{x}^{y}\text{Z}, where <em>x </em>= the atomic number, <em>y</em> = the mass number, and Z = the symbol of the element.

_{93}^{232}\text{Np} + _{1}^{0}\text{e} \longrightarrow _{x}^{y}\text{Z}

Then your equation becomes

_{93}^{232}\text{Np} + _{1}^{0}\text{e} \longrightarrow _{x}^{y}\text{Z}

The main point to remember in balancing nuclear equations is that the <em>sums of the superscripts and of the subscripts</em> must be the same on each side of the equation.  

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_{93}^{232}\text{Np} + _{-1}^{0}\text{e} \longrightarrow _{92}^{232}\text{U}

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