What is the assigned oxidation state of molybdenum in indium molybdate?
955612
In2(MoO4)3\noxidation number of Mo is +6
1088014
你很漂亮. ni3 hen3 piao4 liang
279021
There's nothing surprising about the normal Group oxidation state of +4.\n\nAll of the elements in the group have the outer electronic structure ns2npx1npy1, where n varies from 2 (for carbon) to 6 (for lead). The oxidation state of +4 is where all these outer electrons are directly involved in the bonding.\n\nAs you get closer to the bottom of the Group, there is an increasing tendency for the s2 pair not to be used in the bonding. This is often known as the inert pair effect - and is dominant in lead chemistry.\n\nHowever, just giving it a name like "inert pair effect" explains nothing. You need to look at two different explanations depending on whether you are talking about the formation of ionic or covalent bonds.
92319
Molybdenum is a silvery-white, hard, transition metal. Scheele discovered it in 1778. It was often confused with graphite and lead ore. Molybdenum is used in alloys, electrodes and catalysts. The World War 2 German artillery piece called "Big Bertha" contains molybdenum as an essential component of its steel.
677341
The reaction is\n\nH2O2(aq) + 3I-(aq) + 2 H+(aq) I3-(aq) + 2 H2O(aq)\n\nThe reaction above is often very slow, so a catalyst is usually added. Ammonium molybdate works; so do other transition metal ions. The catalytic action probably involves a free radical mechanism. Hydrogen peroxide can reduce MnO4-, but molybdate isn't a strong enough oxidizing agent to be reduced.
819193
If you look at how the -CN is bonded [C≡N], you can see that the oxidation number of N is -3.  From that it will follow, as you note, that C is +2.
791522
Rules!!\n\n1)the number of MOs formed is always equal to the number of AOs used to construct them.\n2)The more stable the bonding MO, the less stable the corresponding anti-bonding MO.\n3)Like AOs, each MO can accommodate a maximum of two electrons provided both electrons have opposing spins (Pauli’s Exclusion Principle.)\n4)When electrons are added to MOs that are degenerate (ie. Have the same energy) the most stable arrangement is the one in which the electrons enter each degenerate compondent unpaired and with spins parallel (Hund’s Rules)\n5)The neutral molecules, the number of electrons in all the MOs must equal the sum of all electrons in the constituent atoms. For charged molecules, the total number of electrons in the MOs must equal the sum of all electrons from the constituent atoms minus charge.\n6)Qualitative MO theory predicts that in a stable molecule the number of electrons in the bonding MOs is always greater than the number of electrons in anti-bonding MOs. It should be noted that this is an assumption and so is not always accurate.\n\n\ncheers,
174141
"Redox" metals have more than one oxidation state.  e.g. Fe2+/Fe3+ or Cu1+/Cu2+.  Zn and Ca only occur in biological systems in the 2+ state (therefore they cannot redox cycle).
745323
Sb shows -3 +3 and +5 oxidation states, it belongs to N family. Having more metallic character, it exhibits cations as in Sb2O3
7062
it's outer electrons are in the d orbitals.  the s and p orbital elements (the first 2 columns and the last 6) are the more reactive ones that really want to form octets. none of those d-block transition metals are all that reactive, especially the ones in the middle there where gold is.
382554
It's not polar; It's ionic.\n\nBaO consists of Barium in the +2 oxidation state and Oxygen in the -2 oxidation state.
984431
The oxidation state of Vanadium in the NH4VO3 complex is +5 because oxygen has a -2 charge and NH4 has a +1 charge. 3 oxygens give you a -6 charge from oxygen. -6+1=-5. Since the complex is a neutral complex, vanadium would have to be +5 to match.
737570
Whether it's uranium-235 or 238 makes no difference to the chemical formula. What determines the formula of the oxide is the most stable oxidation state of the metal, and with uranium these are +4 and+6, leading to formulae of UO2 and UO3.
477021
Not an element; yes its still the king of hardness\nThere are experiments ongoing with low-speed nuclear bombardment of iridium with smaller nuclei which are producing very thin layers of great density that approaches the hardness of diamond.
87280
MBrO3-9H2O where M can be Na,K,Li, other +1 cations\n\nIf M is other than these you have to adjust subscripts
781203
2H3PO4(aq) + 3Mn(OH)2(s)  --> Mn3(PO4)2(s) + 6H2O(l)
697898
pull out your periodic table and this becomes very easy.  elements have an oxidation # of 0.