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The valence electrons fill in 4d orbital The electron configuration of yttrium is [Kr]4d15s2.
The element is Yttrium, with the symbol Y
The d sublevel always contains 5 orbitals. Therefore the d sublevel can accommodate 10 electrons just the same as 3d and 4d orbitals. Each of the 5 separate d orbitals can only contain two electrons.
there is only 1 lone pair present in BrF5 there are 7 electrons in Br valence shell from which 5 electrons went to make bond with F, while the remaining two makes a lone pair. and thus anly 1 lone pair exist in BrF5
since it is in period 5 it contains 4d and 3d levels which both hold 10 electrons each. so therefore I (53) has 20 d electrons
The valence electrons fill in 4d orbital The electron configuration of yttrium is [Kr]4d15s2.
The element is Yttrium, with the symbol Y
Niobium (Nb) Because three 4d electrons = 3d^3
In the fifth period of the periodic table, the atoms of the elements in the first two groups are adding 1 and 2 electrons, respectively, to their highest energy 5s sublevel. Starting in group 3/IIIB and going through group 12/IIB, the atoms of those elements are adding electrons to their highest energy 4d sublevel. Since the d sublevel can contain a maxium of 10 electrons, there are 10 elements whose atoms are filling the 4d sublevel. Once the 4d sublevel is filled, the next higher energy sublevel is the 5p sublevel. Starting with the group 13/IIIA elements, the 5p sublevel is being filled. Since a p sublevel can contain a maximum of 6 electrons, there are six elements whose atoms are filling the 5p sublevel. 5s sublevel filling: 2 elements 4d sublevel filling: 10 elements 5p sublevel filling: 6 elements --------------------------------------- Total: 18 elements For a printable periodic table that includes electron configurations, go to the following link: http://www.nist.gov/pml/data/periodic.cfm
7 Orbitals
I am pretty sure its Pd (Palladium)
The d sublevel always contains 5 orbitals. Therefore the d sublevel can accommodate 10 electrons just the same as 3d and 4d orbitals. Each of the 5 separate d orbitals can only contain two electrons.
Ruthenium has a electron configuration of *[Kr]4d^7 5s^1. It has only one valence electron. Ruthenium is a rare transition metal that belongs with the platinum group on the periodic table.
Indium has 3 valence electrons. The electronic configuration for indium (In) which has 49 electrons will end up as:[Kr] (5s^2) (4d^6) (5p^1)The 5s and 5p are the largest valence electron n{n=5} by summing the Powers (2) + (1) = 3 Valence electron.
It goes 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p... it goes according to the principal quantum number, l. the subshells (or energy levels) have corresponding letters... 0th sublevel: s 1st sublevel: p 2nd sublevel: d 3rd sublevel: f 4th: g 5th: h 6th: i etc. usually, we dont use energy levels above the f subshell.
there is only 1 lone pair present in BrF5 there are 7 electrons in Br valence shell from which 5 electrons went to make bond with F, while the remaining two makes a lone pair. and thus anly 1 lone pair exist in BrF5
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