long hand: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p2
Short hand: [Kr] 5s2 4d10 5p6
*Remember the numbers after the letter is written as an exponent*
The noble gas configuration for tin (Sn) is [Kr] 5s^2 4d^10 5p^2, where [Kr] represents the electron configuration of krypton.
In the electron configuration of tin (Sn), the 3d electrons are not present. Tin has a configuration of [Kr] 5s2 4d10 5p2, which means it has 2 electrons in the 5s orbital, 10 electrons in the 4d orbital, and 2 electrons in the 5p orbital.
The element with this electron configuration is tin (Sn), which has an atomic number of 50. The electron configuration provided corresponds to the electron distribution in its various electron orbitals.
The noble gas electron configuration for W (tungsten) is [Xe] 6s2 4f14 5d4. This configuration represents the arrangement of electrons in tungsten, with the outermost electrons following the pattern of the noble gas xenon.
(Xe)6s24f145d4
Tin. {http://en.wikipedia.org/wiki/Tin}.
The electronic configuration of tin is: [Kr]D10.5s2.5p2.The electronic configuration of chromium(2+) is correct.
5s25p2 The 2's are exponents on the s and the p
The electron configuration of the element with the symbol W (tungsten) is Xe 4f14 5d4 6s2.
The noble gas configuration for tin (Sn) is [Kr] 5s^2 4d^10 5p^2, where [Kr] represents the electron configuration of krypton.
In the electron configuration of tin (Sn), the 3d electrons are not present. Tin has a configuration of [Kr] 5s2 4d10 5p2, which means it has 2 electrons in the 5s orbital, 10 electrons in the 4d orbital, and 2 electrons in the 5p orbital.
Tin has 4 valence electrons. Because of this, Tin needs to lose the 4 electrons to make it stable. Thus the answer is SN4+
The element with this electron configuration is tin (Sn), which has an atomic number of 50. The electron configuration provided corresponds to the electron distribution in its various electron orbitals.
Tin typically has a configuration of [Kr] 5s^2 4d^10 5p^2 when it is a neutral atom. If tin forms an ion by losing two electrons, it would have a configuration of [Kr] 5s^2 4d^10, as it would lose its two outermost electrons from the 5p orbital.
It could be either Sn2+ ion (stannous ion) or Sn4+ ion (stannic ion)
The element with the electron configuration Kr 5s2 4d10 5p5 is iodine, which has 53 electrons in total. The electron configuration indicates that iodine has 7 valence electrons in its outermost shell, which is in the 5p subshell.
The noble gas electron configuration for W (tungsten) is [Xe] 6s2 4f14 5d4. This configuration represents the arrangement of electrons in tungsten, with the outermost electrons following the pattern of the noble gas xenon.