Which of the following elements is most likely to form an oxide with the formula\(M{O_3}\): Zr, Nb, or Mo?

Short Answer

Expert verified

\(Mo\), is most likely to form an oxide with the formula \(M{O_3}\).

Step by step solution

01

Oxidation agent:

The oxidation state, or oxidation number, is the hypothetical charge of an atom if all of its bonds to different atoms were fully ionic.

02

Compairing the given oxidation state:

We need to look at the oxidation state of this three elements. In order to form molecule with the formula \(M{O_3}\)our element needs to have oxidation state 6+, since oxidation state of oxygen is 2-

The most highest oxidation state of Zr is 4+ , of Nb is 5+ and of Mo is 6+. Which means Mo can form oxide: \(Mo{O_3}\)

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Most popular questions from this chapter

The standard reduction potential for the reaction \({(Co{({H_2}O)_6})^{3 + }}(aq) + {e^ - } \to {(Co{({H_2}O)_6})^{2 + }}(aq)\)is about 1.8 V. The reduction potential for the reaction \({(Co{(N{H_3})_6})^{3 + }}(aq) + {e^ - } \to {(Co{(N{H_3})_6})^{2 + }}(aq)\) is +0.1 V. Calculate the cell potentials to show whether the complex ions, \({(Co{({H_2}O)_6})^{2 + }}\) and/or\({(Co{(N{H_3})_6})^{2 + }}\), can be oxidized to the corresponding cobalt (III) complex by oxygen.

Specify whether the following complexes have isomers.

\(\begin{aligned}{}(a)tetrahedral\left( {Ni{{(CO)}_2}{{(Cl)}_2}} \right)\\(b)trigonalbipyramidal\left( {Mn{{(CO)}_4}NO} \right)\\(c)\left( {Pt{{(en)}_2}C{l_2}} \right)C{l_2}\end{aligned}\)

Trimethylphosphine, \(P{\left( {C{H_3}} \right)_3}\) can act as a ligand by donating the lone pair of electrons on the phosphorus atom. If trimethylphosphine is added to a solution of nickel \(\left( {II} \right)\) chloride in acetone, a blue compound that has a molecular mass of approximately \(270 g\) and contains \(21.5\% Ni,26.0\% Cl,\)and \(52.5\% P{\left( {C{H_3}} \right)_3}\) can be isolated. This blue compound does not have any isomeric forms. What are the geometry and molecular formula of the blue compound?

Indicate the coordination number for the central metal atom in each of the following coordination compounds:

\(\begin{aligned}{}(a)\left( {Pt{{\left( {{H_2}O} \right)}_2}B{r_2}} \right)\\(b)\left( {Pt\left( {N{H_3}} \right)(py)(Cl)(Br)} \right)\left( {py = } \right.pyridine,\left. {{C_5}{H_5}\;N} \right)\\(c)\left( {Zn{{\left( {N{H_3}} \right)}_2}C{l_2}} \right)\\(d)\left( {Zn\left( {N{H_3}} \right)(py)(Cl)(Br)} \right)\\(e)\left( {Ni{{\left( {{H_2}O} \right)}_4}C{l_2}} \right)\\(f){\left( {Fe{{(en)}_2}{{(CN)}_2}} \right)^ + }\left( {en = } \right.ethylenediamine,\left. {{C_2}{H_8}\;{N_2}} \right)\end{aligned}\)

Give the oxidation state of the metal, number of \(d\)electrons, and the number of unpaired electrons predicted for \(\left( {Co{{\left( {N{H_3}} \right)}_6}} \right)C{l_3}. \)

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