The electron configuration for lead (Pb), which has an atomic number of 82, is normally written as [Xe] 4f² 5d¹⁰ 6s² 6p². When lead is in the +2 oxidation state (Pb²⁺), it loses two electrons, typically from the 6p subshell. Therefore, the electron configuration for Pb²⁺ is [Xe] 4f² 5d¹⁰ 6s².
The electron configuration of copper(II) is [Ar] 3d9 . Copper is [Ar] 3d10 4s1
The shorthand electron configuration for nobelium (No), which has an atomic number of 102, is [Rn] 5f^14 7s^2. This notation indicates that nobelium has the same electron configuration as radon (Rn), plus 14 electrons in the 5f subshell and 2 electrons in the 7s subshell.
The electron configuration for the +2 cation of cobalt (Co) is [Ar] 3d7. This is because cobalt has an atomic number of 27, which means its neutral configuration is 1s2 2s2 2p6 3s2 3p6 4s2 3d7. When it forms a +2 cation, it loses two electrons from its 4s subshell, resulting in the configuration [Ar] 3d7.
It has 39 electrons and ends in a d orbital. Now finish your homework :)
Beryllium electron configuration is [He]2s2.
The electron configuration for neutral Barium is [Xe] 6s2. Barium plus 2 means it has lost 2 electrons, so the electron configuration for Barium plus 2 would be [Xe].
Cobalt electron configuration is [Ar]3d7.4s2.Cobalt(2+) electron configuration is [Ar]3d7.
The electron configuration of copper(II) is [Ar] 3d9 . Copper is [Ar] 3d10 4s1
The electron configuration of lead (Pb) is [Xe] 4f^14 5d^10 6s^2 6p^2, where [Xe] represents the electron configuration of xenon, which is the noble gas before lead in the periodic table. Lead has 82 electrons in total.
The shorthand electron configuration for nobelium (No), which has an atomic number of 102, is [Rn] 5f^14 7s^2. This notation indicates that nobelium has the same electron configuration as radon (Rn), plus 14 electrons in the 5f subshell and 2 electrons in the 7s subshell.
The electron configuration for the +2 cation of cobalt (Co) is [Ar] 3d7. This is because cobalt has an atomic number of 27, which means its neutral configuration is 1s2 2s2 2p6 3s2 3p6 4s2 3d7. When it forms a +2 cation, it loses two electrons from its 4s subshell, resulting in the configuration [Ar] 3d7.
Rhodium (Rh) forms a 3 plus ion that has the electron configuration Kr4d6. Rhodium has oxidation states of 2,3 and 4, so it can loan out 2, 3 or 4 electrons depending on the circumstances of a chemical reaction.
The electron configuration of sulfur (long) is: 1s22s22p63s23p4. The electron configuration of sulfur (short) is: [He]3s23p4.
The electronic configuration of tin is: [Kr]D10.5s2.5p2.The electronic configuration of chromium(2+) is correct.
The longhand electron configuration for sulfur is 1s2.2s2.2p6.3s2.3p4.
There is no element such as 12M. Assuming that you mean 12Mg, the electron configuration is 2, 8, 2.
It has 39 electrons and ends in a d orbital. Now finish your homework :)