[Kr] 5s1 4d5
Mo is an exception to the usual rules of electron configuration. You would think that the configuration would be
[Kr] 5s2 4d4
But Mo is more stable with and extra electron in it's d orbital, rather than it's s orbital.
Mo: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d4
Mo 3+: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s0 4d3
Mo is 42 on the Periodic Table, since the question asks for Mo3+, you have to subtract 3 electrons. That's why Mo3+ has 39 electrons instead of 42 in its ground state electron configuration. You first have to subtract the electrons from the highest energy level, aka 5s2, then from the next highest, aka 4d4. This gives you the correct configuration.
[Kr] 5s1 4d5
[Kr]4d55s1
Pallidium
yes, the number of electrons in the outermost shell of the carbon atom is filled with electrons, leaving no electrons unpaired and therefore making it diamagnetic.
Thai mo Thai mo thai mo
This formula is Mo(ClO3)4.
ambot ninyo nganu mangutana man mo bogo deay mo wahahhahaha
Mo^3+ = [Kr] 4d^3
Molybdenium: (K,L,M,N,O) = 2,8,18,13,1or:Mo, complete electron configuration: [1s2, 2s2 2p6, 3s2 3p6 3d10, 4s2 4p6] 4d5, 5s1 shorthand: (Kr),5s1,4d5(It is a 4d-block 'transition' element, atom no.42)
It's neither. It's a noble gas, and it's very inert.
Molybdenum's electon configuration is [Kr] 5s1 4d5 this is because all elements want to be half-full or full. so the 2nd electron in the 5s2 moves to make the 4d4 complete so it turns to 5s1 4d5....making Mo half-full and stable.
If it is an element, it could be either chromium (Cr) or molybdenum (Mo). You need to know more to know which. The electronic configuration for Cr is:[Ar]3d54s1and for Mo it is[Kr]4d55s1So both elements have 5 d-electrons and 1 s-electron. I'm not sure if that what you mean however by "s1d5"...
Pallidium
Bromine has 7 electrons in its fourth energy level.
parot mo dude
Molybdenium: (K,L,M,N,O) = 2,8,18,13,1 or: Mo complete electron configuration: [1s2, 2s2 2p6, 3s2 3p6 3d10, 4s2 4p6] 4d5, 5s1 shorthand: (Kr),5s1,4d5 (It is a 4d-block 'transition' element, atom no.42)
No, it is not correct to say that the bond energy always decreases when a diatomic molecule loses an electron. F2 and O2 are counterexamples to this point. When a molecule loses an electron, it will come from the highest occupied molecular orbital. In both O2 and F2, this MO is an antibonding MO. Removing an electron from an antibonding MO *increases* the bond energy.
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